Optimal. Leaf size=153 \[ -\frac{2^{-m-2} e^{\frac{2 c f}{d}-2 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )}{a^2 f}-\frac{4^{-m-2} e^{\frac{4 c f}{d}-4 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 f (c+d x)}{d}\right )}{a^2 f}+\frac{(c+d x)^{m+1}}{4 a^2 d (m+1)} \]
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Rubi [A] time = 0.178764, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3729, 2181} \[ -\frac{2^{-m-2} e^{\frac{2 c f}{d}-2 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )}{a^2 f}-\frac{4^{-m-2} e^{\frac{4 c f}{d}-4 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 f (c+d x)}{d}\right )}{a^2 f}+\frac{(c+d x)^{m+1}}{4 a^2 d (m+1)} \]
Antiderivative was successfully verified.
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Rule 3729
Rule 2181
Rubi steps
\begin{align*} \int \frac{(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx &=\int \left (\frac{(c+d x)^m}{4 a^2}+\frac{e^{-4 e-4 f x} (c+d x)^m}{4 a^2}+\frac{e^{-2 e-2 f x} (c+d x)^m}{2 a^2}\right ) \, dx\\ &=\frac{(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac{\int e^{-4 e-4 f x} (c+d x)^m \, dx}{4 a^2}+\frac{\int e^{-2 e-2 f x} (c+d x)^m \, dx}{2 a^2}\\ &=\frac{(c+d x)^{1+m}}{4 a^2 d (1+m)}-\frac{2^{-2-m} e^{-2 e+\frac{2 c f}{d}} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{2 f (c+d x)}{d}\right )}{a^2 f}-\frac{4^{-2-m} e^{-4 e+\frac{4 c f}{d}} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{4 f (c+d x)}{d}\right )}{a^2 f}\\ \end{align*}
Mathematica [A] time = 10.4971, size = 195, normalized size = 1.27 \[ \frac{4^{-m-2} (c+d x)^m \text{sech}^2(e+f x) (\sinh (2 f x)+\cosh (2 f x)) \left (-\frac{f (c+d x)}{d}\right )^m \left (-\frac{f^2 (c+d x)^2}{d^2}\right )^{-m} \left (-d (m+1) e^{\frac{4 c f}{d}} (\cosh (2 e)-\sinh (2 e)) \text{Gamma}\left (m+1,\frac{4 f (c+d x)}{d}\right )-d 2^{m+2} (m+1) e^{\frac{2 c f}{d}} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )+d 4^{m+1} (\sinh (e)+\cosh (e))^2 \left (\frac{f (c+d x)}{d}\right )^{m+1}\right )}{a^2 d f (m+1) (\tanh (e+f x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.101, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{m}}{ \left ( a+a\tanh \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{m}}{{\left (a \tanh \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.42764, size = 609, normalized size = 3.98 \begin{align*} -\frac{{\left (d m + d\right )} \cosh \left (\frac{d m \log \left (\frac{4 \, f}{d}\right ) + 4 \, d e - 4 \, c f}{d}\right ) \Gamma \left (m + 1, \frac{4 \,{\left (d f x + c f\right )}}{d}\right ) + 4 \,{\left (d m + d\right )} \cosh \left (\frac{d m \log \left (\frac{2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) \Gamma \left (m + 1, \frac{2 \,{\left (d f x + c f\right )}}{d}\right ) -{\left (d m + d\right )} \Gamma \left (m + 1, \frac{4 \,{\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac{d m \log \left (\frac{4 \, f}{d}\right ) + 4 \, d e - 4 \, c f}{d}\right ) - 4 \,{\left (d m + d\right )} \Gamma \left (m + 1, \frac{2 \,{\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac{d m \log \left (\frac{2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) - 4 \,{\left (d f x + c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 4 \,{\left (d f x + c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{16 \,{\left (a^{2} d f m + a^{2} d f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (c + d x\right )^{m}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{m}}{{\left (a \tanh \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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