Optimal. Leaf size=84 \[ \frac {d (c+d x) \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}+\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^3}{3 d}-\frac {d^2 \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3} \]
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Rubi [A] time = 0.16, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3718, 2190, 2531, 2282, 6589} \[ \frac {d (c+d x) \text {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^3}{3 d} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 6589
Rubi steps
\begin {align*} \int (c+d x)^2 \tanh (e+f x) \, dx &=-\frac {(c+d x)^3}{3 d}+2 \int \frac {e^{2 (e+f x)} (c+d x)^2}{1+e^{2 (e+f x)}} \, dx\\ &=-\frac {(c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac {(2 d) \int (c+d x) \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f}\\ &=-\frac {(c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {d (c+d x) \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}-\frac {d^2 \int \text {Li}_2\left (-e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=-\frac {(c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {d (c+d x) \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{2 f^3}\\ &=-\frac {(c+d x)^3}{3 d}+\frac {(c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {d (c+d x) \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}-\frac {d^2 \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}\\ \end {align*}
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Mathematica [A] time = 1.46, size = 143, normalized size = 1.70 \[ \frac {1}{3} x \tanh (e) \left (3 c^2+3 c d x+d^2 x^2\right )+\frac {e^{2 e} \left (-\frac {3 d \left (e^{-2 e}+1\right ) \left (2 f (c+d x) \text {Li}_2\left (-e^{-2 (e+f x)}\right )+d \text {Li}_3\left (-e^{-2 (e+f x)}\right )\right )}{f^3}+\frac {6 \left (e^{-2 e}+1\right ) (c+d x)^2 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac {4 e^{-2 e} (c+d x)^3}{d}\right )}{6 \left (e^{2 e}+1\right )} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.50, size = 329, normalized size = 3.92 \[ -\frac {d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x + 6 \, d^{2} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 6 \, d^{2} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 6 \, {\left (d^{2} f x + c d f\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 6 \, {\left (d^{2} f x + c d f\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 3 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 3 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) - 3 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) - 3 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right )}{3 \, f^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \tanh \left (f x + e\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.19, size = 234, normalized size = 2.79 \[ -\frac {d^{2} x^{3}}{3}-c d \,x^{2}+c^{2} x +\frac {c^{2} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right )}{f}-\frac {2 c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}-\frac {2 d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 d^{2} e^{2} x}{f^{2}}+\frac {4 d^{2} e^{3}}{3 f^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x^{2}}{f}+\frac {d^{2} \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {4 c d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {4 d c e x}{f}-\frac {2 d c \,e^{2}}{f^{2}}+\frac {2 c d \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x}{f}+\frac {c d \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 176, normalized size = 2.10 \[ \frac {1}{3} \, d^{2} x^{3} + c d x^{2} + \frac {c^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{2 \, f} + \frac {c^{2} \log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{2 \, f} + \frac {{\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} c d}{f^{2}} + \frac {{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} d^{2}}{2 \, f^{3}} - \frac {2 \, {\left (d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2}\right )}}{3 \, f^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {tanh}\left (e+f\,x\right )\,{\left (c+d\,x\right )}^2 \,d x \]
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 \[ \int \left (c + d x\right )^{2} \tanh {\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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