An alternative given by [1] is to compute the interval for the logit AUC:

$ log \left( \frac{AUC}{1-AUC} \right) \pm \phi ^{-1} \left( 1 - \frac{\alpha}{2} \right) \frac{\sqrt{AUC}}{AUC(1 - AUC)} $

so that you get an asymmetric interval.

If you are frequently dealing with high AUCs and small sample sizes, you may want to have a look at [2] that shows there is no single method that can optimally compute confidence interval for all ROC curves.

[1] Pepe MS, The Statistical Evaluation of Medical Tests for Classification and Prediction, OUP 2003, p. 107

[2] Obuchowski NA, Lieber ML, Confidence bounds when the estimated ROC area is 1.0, Acad Radiol. 2002, 9 (5) p. 526-30

An alternative given by [1] is to compute the interval for the logit AUC:

$ log \left( \frac{AUC}{1-AUC} \right) \pm \phi ^{-1} \left( 1 - \frac{\alpha}{2} \right) \frac{\sqrt{AUC}}{AUC(1 - AUC)} $

so that you get an asymmetric interval. In your case, you would get a 95% CI $(0.38, 0.81)$.

If you are frequently dealing with high AUCs and small sample sizes, you may want to have a look at [2] that shows there is no single method that can optimally compute confidence interval for all ROC curves.

[1] Pepe MS, The Statistical Evaluation of Medical Tests for Classification and Prediction, OUP 2003, p. 107

[2] Obuchowski NA, Lieber ML, Confidence bounds when the estimated ROC area is 1.0, Acad Radiol. 2002, 9 (5) p. 526-30