Integrand size = 15, antiderivative size = 55 \[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x^2}{2}-x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5658, 5656, 470, 371} \[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x^2}{2}-x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right ) \]
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Rule 371
Rule 470
Rule 5656
Rule 5658
Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int x^{-1+\frac {2}{n}} \tanh (d (a+b \log (x))) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {2}{n}} \left (-1+e^{2 a d} x^{2 b d}\right )}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{n} \\ & = \frac {x^2}{2}-\frac {\left (2 x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {2}{n}}}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{n} \\ & = \frac {x^2}{2}-x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(122\) vs. \(2(55)=110\).
Time = 7.93 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.22 \[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x^2 \left (e^{2 d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{b d n},2+\frac {1}{b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )-(1+b d n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{2+2 b d n} \]
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\[\int x \tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )d x\]
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\[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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\[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x \tanh {\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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\[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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Timed out. \[ \int x \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x\,\mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right ) \,d x \]
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