Integrand size = 21, antiderivative size = 169 \[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(1+m+b d n) (e x)^{1+m}}{b d e (1+m) n}+\frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n} \]
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Time = 0.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5658, 5656, 516, 470, 371} \[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2 b d n},\frac {m+1}{2 b d n}+1,-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n}+\frac {(e x)^{m+1} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}+\frac {(e x)^{m+1} (b d n+m+1)}{b d e (m+1) n} \]
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Rule 371
Rule 470
Rule 516
Rule 5656
Rule 5658
Rubi steps \begin{align*} \text {integral}& = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \tanh ^2(d (a+b \log (x))) \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {1+m}{n}} \left (-1+e^{2 a d} x^{2 b d}\right )^2}{\left (1+e^{2 a d} x^{2 b d}\right )^2} \, dx,x,c x^n\right )}{e n} \\ & = \frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {\left (e^{-2 a d} (e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {1+m}{n}} \left (\frac {2 e^{2 a d} (1+m-b d n)}{n}-\frac {2 e^{4 a d} (1+m+b d n) x^{2 b d}}{n}\right )}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{2 b d e n} \\ & = \frac {(1+m+b d n) (e x)^{1+m}}{b d e (1+m) n}+\frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {\left (2 (1+m) (e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {1+m}{n}}}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{b d e n^2} \\ & = \frac {(1+m+b d n) (e x)^{1+m}}{b d e (1+m) n}+\frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(516\) vs. \(2(169)=338\).
Time = 16.97 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.05 \[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m}{1+m}-\frac {x (e x)^m \text {sech}\left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \text {sech}\left (b d n \log (x)+d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \sinh (b d n \log (x))}{b d n}+\frac {(1+m) x^{-m} (e x)^m \text {sech}\left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (\frac {x^{1+m} \text {sech}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \sinh (b d n \log (x))}{1+m}-\frac {e^{-\frac {(1+2 m) \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{b n}} \cosh \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 b d n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )-e^{\frac {a (1+2 m+2 b d n)}{b n}+(1+m+2 b d n) \log (x)+\frac {(1+2 m+2 b d n) \left (-n \log (x)+\log \left (c x^n\right )\right )}{n}} (1+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+2 b d n}{2 b d n},\frac {1+m+4 b d n}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 b d n) \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )}{(1+m) (1+m+2 b d n)}\right )}{b d n} \]
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\[\int \left (e x \right )^{m} {\tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]
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\[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
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\[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \tanh ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
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\[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
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Timed out. \[ \int (e x)^m \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2\,{\left (e\,x\right )}^m \,d x \]
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