Integrand size = 18, antiderivative size = 133 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {3 d x}{16 a^2 f}-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {d}{16 f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {3 d}{16 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )} \]
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Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3560, 8, 3811} \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=-\frac {c+d x}{4 f \left (a^2 \tanh (e+f x)+a^2\right )}+\frac {x (c+d x)}{4 a^2}-\frac {3 d}{16 f^2 \left (a^2 \tanh (e+f x)+a^2\right )}+\frac {3 d x}{16 a^2 f}-\frac {d x^2}{8 a^2}-\frac {c+d x}{4 f (a \tanh (e+f x)+a)^2}-\frac {d}{16 f^2 (a \tanh (e+f x)+a)^2} \]
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Rule 8
Rule 3560
Rule 3811
Rubi steps \begin{align*} \text {integral}& = \frac {x (c+d x)}{4 a^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}-d \int \left (\frac {x}{4 a^2}-\frac {1}{4 f (a+a \tanh (e+f x))^2}-\frac {1}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}\right ) \, dx \\ & = -\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}+\frac {d \int \frac {1}{(a+a \tanh (e+f x))^2} \, dx}{4 f}+\frac {d \int \frac {1}{a^2+a^2 \tanh (e+f x)} \, dx}{4 f} \\ & = -\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {d}{16 f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {d}{8 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}+\frac {d \int 1 \, dx}{8 a^2 f}+\frac {d \int \frac {1}{a+a \tanh (e+f x)} \, dx}{8 a f} \\ & = \frac {d x}{8 a^2 f}-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {d}{16 f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {3 d}{16 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}+\frac {d \int 1 \, dx}{16 a^2 f} \\ & = \frac {3 d x}{16 a^2 f}-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {d}{16 f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {3 d}{16 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )} \\ \end{align*}
Time = 0.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {\text {sech}^2(e+f x) \left (-8 (d+2 c f+2 d f x)+\left (4 c f (-1+4 f x)+d \left (-1-4 f x+8 f^2 x^2\right )\right ) \cosh (2 (e+f x))+\left (4 c f (1+4 f x)+d \left (1+4 f x+8 f^2 x^2\right )\right ) \sinh (2 (e+f x))\right )}{64 a^2 f^2 (1+\tanh (e+f x))^2} \]
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Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.56
method | result | size |
risch | \(\frac {d \,x^{2}}{8 a^{2}}+\frac {c x}{4 a^{2}}-\frac {\left (2 d x f +2 c f +d \right ) {\mathrm e}^{-2 f x -2 e}}{8 a^{2} f^{2}}-\frac {\left (4 d x f +4 c f +d \right ) {\mathrm e}^{-4 f x -4 e}}{64 a^{2} f^{2}}\) | \(74\) |
parallelrisch | \(\frac {-3 d \tanh \left (f x +e \right )-4 d +4 c x \,f^{2}+2 d \,x^{2} f^{2}-4 c \tanh \left (f x +e \right ) f +2 d \tanh \left (f x +e \right ) x f -8 c f -5 d x f +4 d \tanh \left (f x +e \right ) x^{2} f^{2}+8 x \tanh \left (f x +e \right ) c \,f^{2}+2 d \tanh \left (f x +e \right )^{2} x^{2} f^{2}+4 x \tanh \left (f x +e \right )^{2} c \,f^{2}+3 x \tanh \left (f x +e \right )^{2} d f}{16 f^{2} a^{2} \left (1+\tanh \left (f x +e \right )\right )^{2}}\) | \(151\) |
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Time = 0.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.44 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=-\frac {16 \, d f x - {\left (8 \, d f^{2} x^{2} - 4 \, c f + 4 \, {\left (4 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right )^{2} - 2 \, {\left (8 \, d f^{2} x^{2} + 4 \, c f + 4 \, {\left (4 \, c f^{2} + d f\right )} x + d\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) - {\left (8 \, d f^{2} x^{2} - 4 \, c f + 4 \, {\left (4 \, c f^{2} - d f\right )} x - d\right )} \sinh \left (f x + e\right )^{2} + 16 \, c f + 8 \, d}{64 \, {\left (a^{2} f^{2} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{2} \sinh \left (f x + e\right )^{2}\right )}} \]
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\[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {\int \frac {c}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
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Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.80 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {1}{16} \, c {\left (\frac {4 \, {\left (f x + e\right )}}{a^{2} f} - \frac {4 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac {{\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} - 8 \, {\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - {\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d e^{\left (-4 \, e\right )}}{64 \, a^{2} f^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {{\left (8 \, d f^{2} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 16 \, c f^{2} x e^{\left (4 \, f x + 4 \, e\right )} - 16 \, d f x e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d f x - 16 \, c f e^{\left (2 \, f x + 2 \, e\right )} - 4 \, c f - 8 \, d e^{\left (2 \, f x + 2 \, e\right )} - d\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{64 \, a^{2} f^{2}} \]
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Time = 1.79 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.67 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {d\,x^2}{8\,a^2}-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {d+4\,c\,f}{64\,a^2\,f^2}+\frac {d\,x}{16\,a^2\,f}\right )-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {d+2\,c\,f}{8\,a^2\,f^2}+\frac {d\,x}{4\,a^2\,f}\right )+\frac {c\,x}{4\,a^2} \]
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