Integrand size = 20, antiderivative size = 153 \[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=\frac {(c+d x)^{1+m}}{4 a^2 d (1+m)}-\frac {2^{-2-m} e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{a^2 f}-\frac {4^{-2-m} e^{-4 e+\frac {4 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 f (c+d x)}{d}\right )}{a^2 f} \]
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Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3810, 2212} \[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=-\frac {2^{-m-2} e^{\frac {2 c f}{d}-2 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 f (c+d x)}{d}\right )}{a^2 f}-\frac {4^{-m-2} e^{\frac {4 c f}{d}-4 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 f (c+d x)}{d}\right )}{a^2 f}+\frac {(c+d x)^{m+1}}{4 a^2 d (m+1)} \]
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Rule 2212
Rule 3810
Rubi steps \begin{align*} \text {integral}& = \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*}
Time = 1.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=\frac {(c+d x)^m \left (\frac {4 e^{2 e} f (c+d x)}{d (1+m)}-2^{2-m} e^{\frac {2 c f}{d}} \left (f \left (\frac {c}{d}+x\right )\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )-4^{-m} e^{-2 e+\frac {4 c f}{d}} \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 f (c+d x)}{d}\right )\right ) \text {sech}^2(e+f x) (\cosh (f x)+\sinh (f x))^2}{16 a^2 f (1+\tanh (e+f x))^2} \]
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\[\int \frac {\left (d x +c \right )^{m}}{\left (a +a \tanh \left (f x +e \right )\right )^{2}}d x\]
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none
Time = 0.08 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.62 \[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=-\frac {{\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {4 \, f}{d}\right ) + 4 \, d e - 4 \, c f}{d}\right ) \Gamma \left (m + 1, \frac {4 \, {\left (d f x + c f\right )}}{d}\right ) + 4 \, {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - {\left (d m + d\right )} \Gamma \left (m + 1, \frac {4 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {4 \, f}{d}\right ) + 4 \, d e - 4 \, c f}{d}\right ) - 4 \, {\left (d m + d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) - 4 \, {\left (d f x + c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 4 \, {\left (d f x + c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{16 \, {\left (a^{2} d f m + a^{2} d f\right )}} \]
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\[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=\frac {\int \frac {\left (c + d x\right )^{m}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
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\[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{{\left (a \tanh \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{{\left (a \tanh \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^2} \, dx=\int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tanh}\left (e+f\,x\right )\right )}^2} \,d x \]
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