Optimal. Leaf size=74 \[ \frac {d x}{4 a f}+\frac {(c+d x)^2}{4 a d}-\frac {d}{4 f^2 (a+a \tanh (e+f x))}-\frac {c+d x}{2 f (a+a \tanh (e+f x))} \]
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Rubi [A]
time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3804, 3560, 8}
\begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3560
Rule 3804
Rubi steps
\begin {align*} \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx &=\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*}
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Mathematica [A]
time = 0.17, size = 81, normalized size = 1.09 \begin {gather*} \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))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.59, size = 46, normalized size = 0.62
method | result | size |
risch | \(\frac {d \,x^{2}}{4 a}+\frac {c x}{2 a}-\frac {\left (2 d x f +2 c f +d \right ) {\mathrm e}^{-2 f x -2 e}}{8 a \,f^{2}}\) | \(46\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 78, normalized size = 1.05 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 113, normalized size = 1.53 \begin {gather*} \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 + \cosh \left (1\right ) + \sinh \left (1\right )\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 + \cosh \left (1\right ) + \sinh \left (1\right )\right )}{8 \, {\left (a f^{2} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + a f^{2} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 64, normalized size = 0.86 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.13, size = 76, normalized size = 1.03 \begin {gather*} \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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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