Integrand size = 20, antiderivative size = 157 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d}+\frac {\log (c+d x)}{2 a d}-\frac {\text {Chi}\left (\frac {2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{2 a d}-\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d}+\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d} \]
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Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3807, 3384, 3379, 3382} \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=-\frac {\text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{2 a d}+\frac {\text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )}{2 a d}+\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{2 a d}-\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{2 a d}+\frac {\log (c+d x)}{2 a d} \]
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Rule 3379
Rule 3382
Rule 3384
Rule 3807
Rubi steps \begin{align*} \text {integral}& = \frac {\log (c+d x)}{2 a d}+\frac {\int \frac {\cosh (2 e+2 f x)}{c+d x} \, dx}{2 a}-\frac {\int \frac {\sinh (2 e+2 f x)}{c+d x} \, dx}{2 a} \\ & = \frac {\log (c+d x)}{2 a d}+\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\cosh \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a}-\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\sinh \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a}-\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\cosh \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a}+\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\sinh \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a} \\ & = \frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d}+\frac {\log (c+d x)}{2 a d}-\frac {\text {Chi}\left (\frac {2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{2 a d}-\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d}+\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d} \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=\frac {\text {sech}(e+f x) (\cosh (f x)+\sinh (f x)) \left (\log (f (c+d x)) (\cosh (e)+\sinh (e))+\text {Chi}\left (\frac {2 f (c+d x)}{d}\right ) \left (\cosh \left (e-\frac {2 c f}{d}\right )-\sinh \left (e-\frac {2 c f}{d}\right )\right )+\left (-\cosh \left (e-\frac {2 c f}{d}\right )+\sinh \left (e-\frac {2 c f}{d}\right )\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )\right )}{2 a d (1+\tanh (e+f x))} \]
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Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.39
method | result | size |
risch | \(\frac {\ln \left (d x +c \right )}{2 a d}-\frac {{\mathrm e}^{\frac {2 c f -2 d e}{d}} \operatorname {Ei}_{1}\left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{2 a d}\) | \(61\) |
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Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=\frac {{\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + \log \left (d x + c\right )}{2 \, a d} \]
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\[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=\frac {\int \frac {1}{c \tanh {\left (e + f x \right )} + c + d x \tanh {\left (e + f x \right )} + d x}\, dx}{a} \]
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Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.31 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=-\frac {e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{2 \, a d} + \frac {\log \left (d x + c\right )}{2 \, a d} \]
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Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.30 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=\frac {{\left ({\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {2 \, c f}{d}\right )} + e^{\left (2 \, e\right )} \log \left (d x + c\right )\right )} e^{\left (-2 \, e\right )}}{2 \, a d} \]
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Timed out. \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))} \, dx=\int \frac {1}{\left (a+a\,\mathrm {tanh}\left (e+f\,x\right )\right )\,\left (c+d\,x\right )} \,d x \]
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