Optimal. Leaf size=169 \[ -\frac {3 d^2 (c+d x)}{4 f^3 (a \coth (e+f x)+a)}-\frac {3 d (c+d x)^2}{4 f^2 (a \coth (e+f x)+a)}-\frac {(c+d x)^3}{2 f (a \coth (e+f x)+a)}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a \coth (e+f x)+a)}+\frac {3 d^3 x}{8 a f^3} \]
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Rubi [A] time = 0.19, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3723, 3479, 8} \[ -\frac {3 d^2 (c+d x)}{4 f^3 (a \coth (e+f x)+a)}-\frac {3 d (c+d x)^2}{4 f^2 (a \coth (e+f x)+a)}-\frac {(c+d x)^3}{2 f (a \coth (e+f x)+a)}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a \coth (e+f x)+a)}+\frac {3 d^3 x}{8 a f^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3479
Rule 3723
Rubi steps
\begin {align*} \int \frac {(c+d x)^3}{a+a \coth (e+f x)} \, dx &=\frac {(c+d x)^4}{8 a d}-\frac {(c+d x)^3}{2 f (a+a \coth (e+f x))}+\frac {(3 d) \int \frac {(c+d x)^2}{a+a \coth (e+f x)} \, dx}{2 f}\\ &=\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \coth (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \coth (e+f x))}+\frac {\left (3 d^2\right ) \int \frac {c+d x}{a+a \coth (e+f x)} \, dx}{2 f^2}\\ &=\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \coth (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \coth (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \coth (e+f x))}+\frac {\left (3 d^3\right ) \int \frac {1}{a+a \coth (e+f x)} \, dx}{4 f^3}\\ &=\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+a \coth (e+f x))}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \coth (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \coth (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \coth (e+f x))}+\frac {\left (3 d^3\right ) \int 1 \, dx}{8 a f^3}\\ &=\frac {3 d^3 x}{8 a f^3}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+a \coth (e+f x))}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \coth (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \coth (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \coth (e+f x))}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 244, normalized size = 1.44 \[ \frac {\text {csch}(e+f x) (\sinh (f x)+\cosh (f x)) \left (2 f^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (\sinh (e)+\cosh (e))+(\cosh (e)-\sinh (e)) \cosh (2 f x) \left (4 c^3 f^3+6 c^2 d f^2 (2 f x+1)+6 c d^2 f \left (2 f^2 x^2+2 f x+1\right )+d^3 \left (4 f^3 x^3+6 f^2 x^2+6 f x+3\right )\right )+(\sinh (e)-\cosh (e)) \sinh (2 f x) \left (4 c^3 f^3+6 c^2 d f^2 (2 f x+1)+6 c d^2 f \left (2 f^2 x^2+2 f x+1\right )+d^3 \left (4 f^3 x^3+6 f^2 x^2+6 f x+3\right )\right )\right )}{16 a f^4 (\coth (e+f x)+1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 304, normalized size = 1.80 \[ \frac {{\left (2 \, d^{3} f^{4} x^{4} + 4 \, c^{3} f^{3} + 6 \, c^{2} d f^{2} + 6 \, c d^{2} f + 4 \, {\left (2 \, c d^{2} f^{4} + d^{3} f^{3}\right )} x^{3} + 3 \, d^{3} + 6 \, {\left (2 \, c^{2} d f^{4} + 2 \, c d^{2} f^{3} + d^{3} f^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} f^{4} + 6 \, c^{2} d f^{3} + 6 \, c d^{2} f^{2} + 3 \, d^{3} f\right )} x\right )} \cosh \left (f x + e\right ) + {\left (2 \, d^{3} f^{4} x^{4} - 4 \, c^{3} f^{3} - 6 \, c^{2} d f^{2} - 6 \, c d^{2} f + 4 \, {\left (2 \, c d^{2} f^{4} - d^{3} f^{3}\right )} x^{3} - 3 \, d^{3} + 6 \, {\left (2 \, c^{2} d f^{4} - 2 \, c d^{2} f^{3} - d^{3} f^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} f^{4} - 6 \, c^{2} d f^{3} - 6 \, c d^{2} f^{2} - 3 \, d^{3} f\right )} x\right )} \sinh \left (f x + e\right )}{16 \, {\left (a f^{4} \cosh \left (f x + e\right ) + a f^{4} \sinh \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 193, normalized size = 1.14 \[ \frac {{\left (2 \, d^{3} f^{4} x^{4} e^{\left (2 \, f x + 2 \, e\right )} + 8 \, c d^{2} f^{4} x^{3} e^{\left (2 \, f x + 2 \, e\right )} + 12 \, c^{2} d f^{4} x^{2} e^{\left (2 \, f x + 2 \, e\right )} + 4 \, d^{3} f^{3} x^{3} + 8 \, c^{3} f^{4} x e^{\left (2 \, f x + 2 \, e\right )} + 12 \, c d^{2} f^{3} x^{2} + 12 \, c^{2} d f^{3} x + 6 \, d^{3} f^{2} x^{2} + 4 \, c^{3} f^{3} + 12 \, c d^{2} f^{2} x + 6 \, c^{2} d f^{2} + 6 \, d^{3} f x + 6 \, c d^{2} f + 3 \, d^{3}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, a f^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.43, size = 959, normalized size = 5.67 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 183, normalized size = 1.08 \[ \frac {1}{4} \, c^{3} {\left (\frac {2 \, {\left (f x + e\right )}}{a f} + \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{a f}\right )} + \frac {3 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} + {\left (2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} c^{2} d e^{\left (-2 \, e\right )}}{8 \, a f^{2}} + \frac {{\left (4 \, f^{3} x^{3} e^{\left (2 \, e\right )} + 3 \, {\left (2 \, f^{2} x^{2} + 2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} c d^{2} e^{\left (-2 \, e\right )}}{8 \, a f^{3}} + \frac {{\left (2 \, f^{4} x^{4} e^{\left (2 \, e\right )} + {\left (4 \, f^{3} x^{3} + 6 \, f^{2} x^{2} + 6 \, f x + 3\right )} e^{\left (-2 \, f x\right )}\right )} d^{3} e^{\left (-2 \, e\right )}}{16 \, a f^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.45, size = 223, normalized size = 1.32 \[ \frac {4\,c^3\,f^4\,x+6\,c^2\,d\,f^4\,x^2+6\,c^2\,d\,f^3\,x+4\,c\,d^2\,f^4\,x^3+6\,c\,d^2\,f^3\,x^2+6\,c\,d^2\,f^2\,x+d^3\,f^4\,x^4+2\,d^3\,f^3\,x^3+3\,d^3\,f^2\,x^2+3\,d^3\,f\,x}{8\,a\,f^4}-\frac {4\,c^3\,f^3+12\,c^2\,d\,f^3\,x+6\,c^2\,d\,f^2+12\,c\,d^2\,f^3\,x^2+12\,c\,d^2\,f^2\,x+6\,c\,d^2\,f+4\,d^3\,f^3\,x^3+6\,d^3\,f^2\,x^2+6\,d^3\,f\,x+3\,d^3}{8\,a\,f^4\,\left (\mathrm {coth}\left (e+f\,x\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.74, size = 864, normalized size = 5.11 \[ \begin {cases} \frac {4 c^{3} f^{4} x \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {4 c^{3} f^{4} x}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {4 c^{3} f^{3}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c^{2} d f^{4} x^{2} \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c^{2} d f^{4} x^{2}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} - \frac {6 c^{2} d f^{3} x \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c^{2} d f^{3} x}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c^{2} d f^{2}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {4 c d^{2} f^{4} x^{3} \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {4 c d^{2} f^{4} x^{3}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} - \frac {6 c d^{2} f^{3} x^{2} \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c d^{2} f^{3} x^{2}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} - \frac {6 c d^{2} f^{2} x \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c d^{2} f^{2} x}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {6 c d^{2} f}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {d^{3} f^{4} x^{4} \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {d^{3} f^{4} x^{4}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} - \frac {2 d^{3} f^{3} x^{3} \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {2 d^{3} f^{3} x^{3}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} - \frac {3 d^{3} f^{2} x^{2} \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {3 d^{3} f^{2} x^{2}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} - \frac {3 d^{3} f x \tanh {\left (e + f x \right )}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {3 d^{3} f x}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} + \frac {3 d^{3}}{8 a f^{4} \tanh {\left (e + f x \right )} + 8 a f^{4}} & \text {for}\: f \neq 0 \\\frac {c^{3} x + \frac {3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac {d^{3} x^{4}}{4}}{a \coth {\relax (e )} + a} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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