Integrand size = 18, antiderivative size = 74 \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {d x}{4 a f}+\frac {(c+d x)^2}{4 a d}-\frac {d}{4 f^2 (a+a \tanh (e+f x))}-\frac {c+d x}{2 f (a+a \tanh (e+f x))} \]
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Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3804, 3560, 8} \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=-\frac {c+d x}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^2}{4 a d}-\frac {d}{4 f^2 (a \tanh (e+f x)+a)}+\frac {d x}{4 a f} \]
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Rule 8
Rule 3560
Rule 3804
Rubi steps \begin{align*} \text {integral}& = \frac {(c+d x)^2}{4 a d}-\frac {c+d x}{2 f (a+a \tanh (e+f x))}+\frac {d \int \frac {1}{a+a \tanh (e+f x)} \, dx}{2 f} \\ & = \frac {(c+d x)^2}{4 a d}-\frac {d}{4 f^2 (a+a \tanh (e+f x))}-\frac {c+d x}{2 f (a+a \tanh (e+f x))}+\frac {d \int 1 \, dx}{4 a f} \\ & = \frac {d x}{4 a f}+\frac {(c+d x)^2}{4 a d}-\frac {d}{4 f^2 (a+a \tanh (e+f x))}-\frac {c+d x}{2 f (a+a \tanh (e+f x))} \\ \end{align*}
Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {2 c f (-1+2 f x)+d \left (-1-2 f x+2 f^2 x^2\right )+\left (2 c f (1+2 f x)+d \left (1+2 f x+2 f^2 x^2\right )\right ) \tanh (e+f x)}{8 a f^2 (1+\tanh (e+f x))} \]
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Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.62
method | result | size |
risch | \(\frac {d \,x^{2}}{4 a}+\frac {c x}{2 a}-\frac {\left (2 d x f +2 c f +d \right ) {\mathrm e}^{-2 f x -2 e}}{8 a \,f^{2}}\) | \(46\) |
parallelrisch | \(\frac {d \tanh \left (f x +e \right ) x^{2} f^{2}+d \,x^{2} f^{2}+2 x \tanh \left (f x +e \right ) c \,f^{2}+d \tanh \left (f x +e \right ) x f +2 c x \,f^{2}-d x f -2 c f -d}{4 f^{2} a \left (1+\tanh \left (f x +e \right )\right )}\) | \(84\) |
default | \(\frac {-\frac {d \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right )^{2}}{2}-\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{4}-\frac {f x}{4}-\frac {e}{4}\right )}{f}+\frac {d \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {\left (f x +e \right )^{2}}{4}-\frac {\cosh \left (f x +e \right )^{2}}{4}\right )}{f}+\frac {d e \cosh \left (f x +e \right )^{2}}{2 f}-\frac {d e \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {c \cosh \left (f x +e \right )^{2}}{2}+c \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{a f}\) | \(175\) |
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Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36 \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {{\left (2 \, d f^{2} x^{2} - 2 \, c f + 2 \, {\left (2 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right ) + {\left (2 \, d f^{2} x^{2} + 2 \, c f + 2 \, {\left (2 \, c f^{2} + d f\right )} x + d\right )} \sinh \left (f x + e\right )}{8 \, {\left (a f^{2} \cosh \left (f x + e\right ) + a f^{2} \sinh \left (f x + e\right )\right )}} \]
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\[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {\int \frac {c}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d x}{\tanh {\left (e + f x \right )} + 1}\, dx}{a} \]
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Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {1}{4} \, c {\left (\frac {2 \, {\left (f x + e\right )}}{a f} - \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{a f}\right )} + \frac {{\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} - {\left (2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} d e^{\left (-2 \, e\right )}}{8 \, a f^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {{\left (2 \, d f^{2} x^{2} e^{\left (2 \, f x + 2 \, e\right )} + 4 \, c f^{2} x e^{\left (2 \, f x + 2 \, e\right )} - 2 \, d f x - 2 \, c f - d\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{8 \, a f^{2}} \]
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Time = 1.77 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx=\frac {\frac {d\,x^2}{4}+\left (\frac {c}{2}+\frac {d}{4\,f}\right )\,x}{a}-\frac {\frac {\frac {d}{4}+\frac {c\,f}{2}}{f^2}-x\,\left (\frac {c}{2}-\frac {d}{4\,f}\right )+x\,\left (\frac {c}{2}+\frac {d}{4\,f}\right )}{a+a\,\mathrm {tanh}\left (e+f\,x\right )} \]
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