Integrand size = 18, antiderivative size = 183 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {11 d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}-\frac {d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {11 d}{96 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )} \]
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Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3560, 8, 3811} \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=-\frac {c+d x}{8 f \left (a^3 \tanh (e+f x)+a^3\right )}+\frac {x (c+d x)}{8 a^3}-\frac {11 d}{96 f^2 \left (a^3 \tanh (e+f x)+a^3\right )}+\frac {11 d x}{96 a^3 f}-\frac {d x^2}{16 a^3}-\frac {c+d x}{8 a f (a \tanh (e+f x)+a)^2}-\frac {c+d x}{6 f (a \tanh (e+f x)+a)^3}-\frac {5 d}{96 a f^2 (a \tanh (e+f x)+a)^2}-\frac {d}{36 f^2 (a \tanh (e+f x)+a)^3} \]
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Rule 8
Rule 3560
Rule 3811
Rubi steps \begin{align*} \text {integral}& = \frac {x (c+d x)}{8 a^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}-d \int \left (\frac {x}{8 a^3}-\frac {1}{6 f (a+a \tanh (e+f x))^3}-\frac {1}{8 a f (a+a \tanh (e+f x))^2}-\frac {1}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}\right ) \, dx \\ & = -\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac {d \int \frac {1}{a^3+a^3 \tanh (e+f x)} \, dx}{8 f}+\frac {d \int \frac {1}{(a+a \tanh (e+f x))^3} \, dx}{6 f}+\frac {d \int \frac {1}{(a+a \tanh (e+f x))^2} \, dx}{8 a f} \\ & = -\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}-\frac {d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {d}{32 a f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {d}{16 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac {d \int 1 \, dx}{16 a^3 f}+\frac {d \int \frac {1}{a+a \tanh (e+f x)} \, dx}{16 a^2 f}+\frac {d \int \frac {1}{(a+a \tanh (e+f x))^2} \, dx}{12 a f} \\ & = \frac {d x}{16 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}-\frac {d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {3 d}{32 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac {d \int 1 \, dx}{32 a^3 f}+\frac {d \int \frac {1}{a+a \tanh (e+f x)} \, dx}{24 a^2 f} \\ & = \frac {3 d x}{32 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}-\frac {d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {11 d}{96 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac {d \int 1 \, dx}{48 a^3 f} \\ & = \frac {11 d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}-\frac {d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac {c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac {5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac {11 d}{96 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac {c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )} \\ \end{align*}
Time = 1.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {\text {sech}^3(e+f x) \left (-27 (12 c f+d (5+12 f x)) \cosh (e+f x)+4 \left (6 c f (-1+6 f x)+d \left (-1-6 f x+18 f^2 x^2\right )\right ) \cosh (3 (e+f x))-81 d \sinh (e+f x)-108 c f \sinh (e+f x)-108 d f x \sinh (e+f x)+4 d \sinh (3 (e+f x))+24 c f \sinh (3 (e+f x))+24 d f x \sinh (3 (e+f x))+144 c f^2 x \sinh (3 (e+f x))+72 d f^2 x^2 \sinh (3 (e+f x))\right )}{1152 a^3 f^2 (1+\tanh (e+f x))^3} \]
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Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.56
method | result | size |
risch | \(\frac {d \,x^{2}}{16 a^{3}}+\frac {c x}{8 a^{3}}-\frac {3 \left (2 d x f +2 c f +d \right ) {\mathrm e}^{-2 f x -2 e}}{32 a^{3} f^{2}}-\frac {3 \left (4 d x f +4 c f +d \right ) {\mathrm e}^{-4 f x -4 e}}{128 a^{3} f^{2}}-\frac {\left (6 d x f +6 c f +d \right ) {\mathrm e}^{-6 f x -6 e}}{288 a^{3} f^{2}}\) | \(102\) |
parallelrisch | \(\frac {-81 d \tanh \left (f x +e \right )-56 d -33 \tanh \left (f x +e \right )^{2} d +36 x \tanh \left (f x +e \right )^{3} c \,f^{2}+33 x \tanh \left (f x +e \right )^{3} d f +18 d \tanh \left (f x +e \right )^{3} x^{2} f^{2}+36 c x \,f^{2}+18 d \,x^{2} f^{2}-108 c \tanh \left (f x +e \right ) f -9 d \tanh \left (f x +e \right ) x f -36 \tanh \left (f x +e \right )^{2} c f -120 c f -87 d x f +54 d \tanh \left (f x +e \right ) x^{2} f^{2}+108 x \tanh \left (f x +e \right ) c \,f^{2}+54 d \tanh \left (f x +e \right )^{2} x^{2} f^{2}+108 x \tanh \left (f x +e \right )^{2} c \,f^{2}+63 x \tanh \left (f x +e \right )^{2} d f}{288 f^{2} a^{3} \left (1+\tanh \left (f x +e \right )\right )^{3}}\) | \(219\) |
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Time = 0.27 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.56 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {4 \, {\left (18 \, d f^{2} x^{2} - 6 \, c f + 6 \, {\left (6 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right )^{3} + 12 \, {\left (18 \, d f^{2} x^{2} - 6 \, c f + 6 \, {\left (6 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{2} + 4 \, {\left (18 \, d f^{2} x^{2} + 6 \, c f + 6 \, {\left (6 \, c f^{2} + d f\right )} x + d\right )} \sinh \left (f x + e\right )^{3} - 27 \, {\left (12 \, d f x + 12 \, c f + 5 \, d\right )} \cosh \left (f x + e\right ) - 3 \, {\left (36 \, d f x - 4 \, {\left (18 \, d f^{2} x^{2} + 6 \, c f + 6 \, {\left (6 \, c f^{2} + d f\right )} x + d\right )} \cosh \left (f x + e\right )^{2} + 36 \, c f + 27 \, d\right )} \sinh \left (f x + e\right )}{1152 \, {\left (a^{3} f^{2} \cosh \left (f x + e\right )^{3} + 3 \, a^{3} f^{2} \cosh \left (f x + e\right )^{2} \sinh \left (f x + e\right ) + 3 \, a^{3} f^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{2} + a^{3} f^{2} \sinh \left (f x + e\right )^{3}\right )}} \]
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\[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {\int \frac {c}{\tanh ^{3}{\left (e + f x \right )} + 3 \tanh ^{2}{\left (e + f x \right )} + 3 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d x}{\tanh ^{3}{\left (e + f x \right )} + 3 \tanh ^{2}{\left (e + f x \right )} + 3 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.73 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {1}{96} \, c {\left (\frac {12 \, {\left (f x + e\right )}}{a^{3} f} - \frac {18 \, e^{\left (-2 \, f x - 2 \, e\right )} + 9 \, e^{\left (-4 \, f x - 4 \, e\right )} + 2 \, e^{\left (-6 \, f x - 6 \, e\right )}}{a^{3} f}\right )} + \frac {{\left (72 \, f^{2} x^{2} e^{\left (6 \, e\right )} - 108 \, {\left (2 \, f x e^{\left (4 \, e\right )} + e^{\left (4 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 27 \, {\left (4 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-4 \, f x\right )} - 4 \, {\left (6 \, f x + 1\right )} e^{\left (-6 \, f x\right )}\right )} d e^{\left (-6 \, e\right )}}{1152 \, a^{3} f^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {{\left (72 \, d f^{2} x^{2} e^{\left (6 \, f x + 6 \, e\right )} + 144 \, c f^{2} x e^{\left (6 \, f x + 6 \, e\right )} - 216 \, d f x e^{\left (4 \, f x + 4 \, e\right )} - 108 \, d f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, d f x - 216 \, c f e^{\left (4 \, f x + 4 \, e\right )} - 108 \, c f e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c f - 108 \, d e^{\left (4 \, f x + 4 \, e\right )} - 27 \, d e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d\right )} e^{\left (-6 \, f x - 6 \, e\right )}}{1152 \, a^{3} f^{2}} \]
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Time = 1.83 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^3} \, dx=\frac {d\,x^2}{16\,a^3}-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {3\,d+6\,c\,f}{32\,a^3\,f^2}+\frac {3\,d\,x}{16\,a^3\,f}\right )-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {3\,d+12\,c\,f}{128\,a^3\,f^2}+\frac {3\,d\,x}{32\,a^3\,f}\right )-{\mathrm {e}}^{-6\,e-6\,f\,x}\,\left (\frac {d+6\,c\,f}{288\,a^3\,f^2}+\frac {d\,x}{48\,a^3\,f}\right )+\frac {c\,x}{8\,a^3} \]
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