Integrand size = 19, antiderivative size = 88 \[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m}}{e (1+m)}-\frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{e (1+m)} \]
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Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5658, 5656, 470, 371} \[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{m+1}}{e (m+1)}-\frac {2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2 b d n},\frac {m+1}{2 b d n}+1,-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{e (m+1)} \]
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Rule 371
Rule 470
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 (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 )}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{e n} \\ & = \frac {(e x)^{1+m}}{e (1+m)}-\frac {\left (2 (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 )}{e n} \\ & = \frac {(e x)^{1+m}}{e (1+m)}-\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 )}{e (1+m)} \\ \end{align*}
Time = 13.66 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.82 \[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \left (-\operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+\frac {e^{2 a d} (1+m) \left (c x^n\right )^{2 b d} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+2 b d n}{2 b d n},\frac {1+m+4 b d n}{2 b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{1+m+2 b d n}\right )}{1+m} \]
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\[\int \left (e x \right )^{m} \tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )d x\]
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\[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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\[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \tanh {\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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\[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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Timed out. \[ \int (e x)^m \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )\,{\left (e\,x\right )}^m \,d x \]
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