Integrand size = 10, antiderivative size = 54 \[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} \text {sech}^{-1}\left (a x^5\right )^2-\frac {1}{5} \text {sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text {sech}^{-1}\left (a x^5\right )}\right )-\frac {1}{10} \operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}\left (a x^5\right )}\right ) \]
[Out]
Time = 0.07 (sec) , antiderivative size = 54, 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 = {6416, 5882, 3799, 2221, 2317, 2438} \[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=-\frac {1}{10} \operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}\left (a x^5\right )}\right )+\frac {1}{10} \text {sech}^{-1}\left (a x^5\right )^2-\frac {1}{5} \text {sech}^{-1}\left (a x^5\right ) \log \left (e^{2 \text {sech}^{-1}\left (a x^5\right )}+1\right ) \]
[In]
[Out]
Rule 2221
Rule 2317
Rule 2438
Rule 3799
Rule 5882
Rule 6416
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \text {Subst}\left (\int \frac {\text {sech}^{-1}(a x)}{x} \, dx,x,x^5\right ) \\ & = -\left (\frac {1}{5} \text {Subst}\left (\int \frac {\text {arccosh}\left (\frac {x}{a}\right )}{x} \, dx,x,\frac {1}{x^5}\right )\right ) \\ & = -\left (\frac {1}{5} \text {Subst}\left (\int x \tanh (x) \, dx,x,\text {sech}^{-1}\left (a x^5\right )\right )\right ) \\ & = \frac {1}{10} \text {sech}^{-1}\left (a x^5\right )^2-\frac {2}{5} \text {Subst}\left (\int \frac {e^{2 x} x}{1+e^{2 x}} \, dx,x,\text {sech}^{-1}\left (a x^5\right )\right ) \\ & = \frac {1}{10} \text {sech}^{-1}\left (a x^5\right )^2-\frac {1}{5} \text {sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text {sech}^{-1}\left (a x^5\right )}\right )+\frac {1}{5} \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text {sech}^{-1}\left (a x^5\right )\right ) \\ & = \frac {1}{10} \text {sech}^{-1}\left (a x^5\right )^2-\frac {1}{5} \text {sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text {sech}^{-1}\left (a x^5\right )}\right )+\frac {1}{10} \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \text {sech}^{-1}\left (a x^5\right )}\right ) \\ & = \frac {1}{10} \text {sech}^{-1}\left (a x^5\right )^2-\frac {1}{5} \text {sech}^{-1}\left (a x^5\right ) \log \left (1+e^{2 \text {sech}^{-1}\left (a x^5\right )}\right )-\frac {1}{10} \operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}\left (a x^5\right )}\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\frac {1}{10} \left (-\text {sech}^{-1}\left (a x^5\right ) \left (\text {sech}^{-1}\left (a x^5\right )+2 \log \left (1+e^{-2 \text {sech}^{-1}\left (a x^5\right )}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}\left (a x^5\right )}\right )\right ) \]
[In]
[Out]
\[\int \frac {\operatorname {arcsech}\left (a \,x^{5}\right )}{x}d x\]
[In]
[Out]
\[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arsech}\left (a x^{5}\right )}{x} \,d x } \]
[In]
[Out]
\[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\int \frac {\operatorname {asech}{\left (a x^{5} \right )}}{x}\, dx \]
[In]
[Out]
\[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arsech}\left (a x^{5}\right )}{x} \,d x } \]
[In]
[Out]
\[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\int { \frac {\operatorname {arsech}\left (a x^{5}\right )}{x} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\text {sech}^{-1}\left (a x^5\right )}{x} \, dx=\int \frac {\mathrm {acosh}\left (\frac {1}{a\,x^5}\right )}{x} \,d x \]
[In]
[Out]