Integrand size = 10, antiderivative size = 62 \[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^5\right )}\right ) \]
[Out]
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5326, 4722, 3800, 2221, 2317, 2438} \[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right ) \]
[In]
[Out]
Rule 2221
Rule 2317
Rule 2438
Rule 3800
Rule 4722
Rule 5326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \text {Subst}\left (\int \frac {\sec ^{-1}(a x)}{x} \, dx,x,x^5\right ) \\ & = -\left (\frac {1}{5} \text {Subst}\left (\int \frac {\arccos \left (\frac {x}{a}\right )}{x} \, dx,x,\frac {1}{x^5}\right )\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}\left (a x^5\right )\right ) \\ & = \frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {2}{5} i \text {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}\left (a x^5\right )\right ) \\ & = \frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{5} \text {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}\left (a x^5\right )\right ) \\ & = \frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )-\frac {1}{10} i \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}\left (a x^5\right )}\right ) \\ & = \frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^5\right )}\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} i \left (\sec ^{-1}\left (a x^5\right ) \left (\sec ^{-1}\left (a x^5\right )+2 i \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )\right )+\operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^5\right )}\right )\right ) \]
[In]
[Out]
\[\int \frac {\operatorname {arcsec}\left (a \,x^{5}\right )}{x}d x\]
[In]
[Out]
\[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (a x^{5}\right )}{x} \,d x } \]
[In]
[Out]
\[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\int \frac {\operatorname {asec}{\left (a x^{5} \right )}}{x}\, dx \]
[In]
[Out]
\[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (a x^{5}\right )}{x} \,d x } \]
[In]
[Out]
\[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (a x^{5}\right )}{x} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx=\int \frac {\mathrm {acos}\left (\frac {1}{a\,x^5}\right )}{x} \,d x \]
[In]
[Out]