Optimal. Leaf size=170 \[ -\frac {d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac {d^2 e^{-2 e-2 f x}}{8 a^2 f^3}-\frac {d e^{-4 e-4 f x} (c+d x)}{32 a^2 f^2}-\frac {d e^{-2 e-2 f x} (c+d x)}{4 a^2 f^2}-\frac {e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac {e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d} \]
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Rubi [A]
time = 0.13, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3810, 2207,
2225} \begin {gather*} -\frac {d (c+d x) e^{-4 e-4 f x}}{32 a^2 f^2}-\frac {d (c+d x) e^{-2 e-2 f x}}{4 a^2 f^2}-\frac {(c+d x)^2 e^{-4 e-4 f x}}{16 a^2 f}-\frac {(c+d x)^2 e^{-2 e-2 f x}}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac {d^2 e^{-2 e-2 f x}}{8 a^2 f^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2207
Rule 2225
Rule 3810
Rubi steps
\begin {align*} \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx &=\int \left (\frac {(c+d x)^2}{4 a^2}+\frac {e^{-4 e-4 f x} (c+d x)^2}{4 a^2}+\frac {e^{-2 e-2 f x} (c+d x)^2}{2 a^2}\right ) \, dx\\ &=\frac {(c+d x)^3}{12 a^2 d}+\frac {\int e^{-4 e-4 f x} (c+d x)^2 \, dx}{4 a^2}+\frac {\int e^{-2 e-2 f x} (c+d x)^2 \, dx}{2 a^2}\\ &=-\frac {e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac {e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}+\frac {d \int e^{-4 e-4 f x} (c+d x) \, dx}{8 a^2 f}+\frac {d \int e^{-2 e-2 f x} (c+d x) \, dx}{2 a^2 f}\\ &=-\frac {d e^{-4 e-4 f x} (c+d x)}{32 a^2 f^2}-\frac {d e^{-2 e-2 f x} (c+d x)}{4 a^2 f^2}-\frac {e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac {e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}+\frac {d^2 \int e^{-4 e-4 f x} \, dx}{32 a^2 f^2}+\frac {d^2 \int e^{-2 e-2 f x} \, dx}{4 a^2 f^2}\\ &=-\frac {d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac {d^2 e^{-2 e-2 f x}}{8 a^2 f^3}-\frac {d e^{-4 e-4 f x} (c+d x)}{32 a^2 f^2}-\frac {d e^{-2 e-2 f x} (c+d x)}{4 a^2 f^2}-\frac {e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac {e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}\\ \end {align*}
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Mathematica [A]
time = 0.58, size = 207, normalized size = 1.22 \begin {gather*} \frac {\text {sech}^2(e+f x) \left (-48 \left (2 c^2 f^2+2 c d f (1+2 f x)+d^2 \left (1+2 f x+2 f^2 x^2\right )\right )+\left (24 c^2 f^2 (-1+4 f x)+12 c d f \left (-1-4 f x+8 f^2 x^2\right )+d^2 \left (-3-12 f x-24 f^2 x^2+32 f^3 x^3\right )\right ) \cosh (2 (e+f x))+\left (24 c^2 f^2 (1+4 f x)+12 c d f \left (1+4 f x+8 f^2 x^2\right )+d^2 \left (3+12 f x+24 f^2 x^2+32 f^3 x^3\right )\right ) \sinh (2 (e+f x))\right )}{384 a^2 f^3 (1+\tanh (e+f x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.54, size = 163, normalized size = 0.96
method | result | size |
risch | \(\frac {d^{2} x^{3}}{12 a^{2}}+\frac {d c \,x^{2}}{4 a^{2}}+\frac {c^{2} x}{4 a^{2}}+\frac {c^{3}}{12 a^{2} d}-\frac {\left (2 d^{2} x^{2} f^{2}+4 c d \,f^{2} x +2 c^{2} f^{2}+2 d^{2} f x +2 c d f +d^{2}\right ) {\mathrm e}^{-2 f x -2 e}}{8 a^{2} f^{3}}-\frac {\left (8 d^{2} x^{2} f^{2}+16 c d \,f^{2} x +8 c^{2} f^{2}+4 d^{2} f x +4 c d f +d^{2}\right ) {\mathrm e}^{-4 f x -4 e}}{128 a^{2} f^{3}}\) | \(163\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.44, size = 202, normalized size = 1.19 \begin {gather*} \frac {1}{16} \, c^{2} {\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 )} c d e^{\left (-4 \, e\right )}}{32 \, a^{2} f^{2}} + \frac {{\left (32 \, f^{3} x^{3} e^{\left (4 \, e\right )} - 48 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} + 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 3 \, {\left (8 \, f^{2} x^{2} + 4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d^{2} e^{\left (-4 \, e\right )}}{384 \, a^{2} f^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 385 vs.
\(2 (156) = 312\).
time = 0.41, size = 385, normalized size = 2.26 \begin {gather*} -\frac {96 \, d^{2} f^{2} x^{2} + 96 \, c^{2} f^{2} + 96 \, c d f - {\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \, {\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )^{2} - 2 \, {\left (32 \, d^{2} f^{3} x^{3} + 24 \, c^{2} f^{2} + 12 \, c d f + 24 \, {\left (4 \, c d f^{3} + d^{2} f^{2}\right )} x^{2} + 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} + 4 \, c d f^{2} + d^{2} f\right )} x\right )} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - {\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \, {\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )^{2} + 48 \, d^{2} + 96 \, {\left (2 \, c d f^{2} + d^{2} f\right )} x}{384 \, {\left (a^{2} f^{3} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )^{2} + 2 \, a^{2} f^{3} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + a^{2} f^{3} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )^{2}\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^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 217, normalized size = 1.28 \begin {gather*} \frac {{\left (32 \, d^{2} f^{3} x^{3} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c d f^{3} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c^{2} f^{3} x e^{\left (4 \, f x + 4 \, e\right )} - 96 \, d^{2} f^{2} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 24 \, d^{2} f^{2} x^{2} - 192 \, c d f^{2} x e^{\left (2 \, f x + 2 \, e\right )} - 48 \, c d f^{2} x - 96 \, c^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 96 \, d^{2} f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c^{2} f^{2} - 12 \, d^{2} f x - 96 \, c d f e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d f - 48 \, d^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d^{2}\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{384 \, a^{2} f^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.18, size = 165, normalized size = 0.97 \begin {gather*} \frac {c^2\,x}{4\,a^2}-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {8\,c^2\,f^2+4\,c\,d\,f+d^2}{128\,a^2\,f^3}+\frac {d^2\,x^2}{16\,a^2\,f}+\frac {d\,x\,\left (d+4\,c\,f\right )}{32\,a^2\,f^2}\right )-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {2\,c^2\,f^2+2\,c\,d\,f+d^2}{8\,a^2\,f^3}+\frac {d^2\,x^2}{4\,a^2\,f}+\frac {d\,x\,\left (d+2\,c\,f\right )}{4\,a^2\,f^2}\right )+\frac {d^2\,x^3}{12\,a^2}+\frac {c\,d\,x^2}{4\,a^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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