Integrand size = 17, antiderivative size = 71 \[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\frac {i}{x}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{x} \]
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Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4593, 4591, 470, 371} \[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\frac {i}{x}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{x} \]
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Rule 371
Rule 470
Rule 4591
Rule 4593
Rubi steps \begin{align*} \text {integral}& = \frac {\left (c x^n\right )^{\frac {1}{n}} \text {Subst}\left (\int x^{-1-\frac {1}{n}} \tan (d (a+b \log (x))) \, dx,x,c x^n\right )}{n x} \\ & = \frac {\left (c x^n\right )^{\frac {1}{n}} \text {Subst}\left (\int \frac {x^{-1-\frac {1}{n}} \left (i-i e^{2 i a d} x^{2 i b d}\right )}{1+e^{2 i a d} x^{2 i b d}} \, dx,x,c x^n\right )}{n x} \\ & = \frac {i}{x}+\frac {\left (2 i \left (c x^n\right )^{\frac {1}{n}}\right ) \text {Subst}\left (\int \frac {x^{-1-\frac {1}{n}}}{1+e^{2 i a d} x^{2 i b d}} \, dx,x,c x^n\right )}{n x} \\ & = \frac {i}{x}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{x} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(153\) vs. \(2(71)=142\).
Time = 3.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.15 \[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\frac {-e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {i}{2 b d n},2+\frac {i}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(1-2 i b d n) \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )}{(i+2 b d n) x} \]
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\[\int \frac {\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{x^{2}}d x\]
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\[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int { \frac {\tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )}{x^{2}} \,d x } \]
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\[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int \frac {\tan {\left (a d + b d \log {\left (c x^{n} \right )} \right )}}{x^{2}}\, dx \]
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\[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int { \frac {\tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )}{x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int \frac {\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}{x^2} \,d x \]
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