Optimal. Leaf size=103 \[ \frac {3 d^3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac {3 d^2 (c+d x) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {3 d (c+d x)^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {(c+d x)^3}{b} \]
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Rubi [A] time = 0.21, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4184, 3718, 2190, 2531, 2282, 6589} \[ -\frac {3 d^2 (c+d x) \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}-\frac {3 d (c+d x)^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {(c+d x)^3}{b} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 4184
Rule 6589
Rubi steps
\begin {align*} \int (c+d x)^3 \text {sech}^2(a+b x) \, dx &=\frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {(3 d) \int (c+d x)^2 \tanh (a+b x) \, dx}{b}\\ &=\frac {(c+d x)^3}{b}+\frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {(6 d) \int \frac {e^{2 (a+b x)} (c+d x)^2}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=\frac {(c+d x)^3}{b}-\frac {3 d (c+d x)^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {(c+d x)^3}{b}-\frac {3 d (c+d x)^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {3 d^2 (c+d x) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {\left (3 d^3\right ) \int \text {Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=\frac {(c+d x)^3}{b}-\frac {3 d (c+d x)^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {3 d^2 (c+d x) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=\frac {(c+d x)^3}{b}-\frac {3 d (c+d x)^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {3 d^2 (c+d x) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \tanh (a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 2.07, size = 145, normalized size = 1.41 \[ \frac {2 \text {sech}(a) \sinh (b x) (c+d x)^3 \text {sech}(a+b x)-\frac {e^{2 a} d \left (-\frac {3 \left (e^{-2 a}+1\right ) d \left (2 b (c+d x) \text {Li}_2\left (-e^{-2 (a+b x)}\right )+d \text {Li}_3\left (-e^{-2 (a+b x)}\right )\right )}{b^3}+\frac {6 \left (e^{-2 a}+1\right ) (c+d x)^2 \log \left (e^{-2 (a+b x)}+1\right )}{b}+\frac {4 e^{-2 a} (c+d x)^3}{d}\right )}{e^{2 a}+1}}{2 b} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.76, size = 1332, normalized size = 12.93 \[ \text {result too large to display} \]
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 \[ \int {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.29, size = 298, normalized size = 2.89 \[ -\frac {2 \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {3 d \,c^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {2 d^{3} x^{3}}{b}-\frac {6 d^{3} a^{2} x}{b^{3}}-\frac {4 d^{3} a^{3}}{b^{4}}-\frac {3 d^{3} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x^{2}}{b^{2}}-\frac {3 d^{3} \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right ) x}{b^{3}}+\frac {3 d^{3} \polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}}-\frac {12 d^{2} c a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {6 d^{2} c \,x^{2}}{b}+\frac {12 d^{2} c a x}{b^{2}}+\frac {6 d^{2} c \,a^{2}}{b^{3}}-\frac {6 d^{2} c \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x}{b^{2}}-\frac {3 d^{2} c \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.86, size = 238, normalized size = 2.31 \[ 3 \, c^{2} d {\left (\frac {2 \, x e^{\left (2 \, b x + 2 \, a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {\log \left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, a\right )}\right )}{b^{2}}\right )} - \frac {3 \, {\left (2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )\right )} c d^{2}}{b^{3}} + \frac {2 \, c^{3}}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}} - \frac {2 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2}\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})\right )} d^{3}}{2 \, b^{4}} + \frac {2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2}\right )}}{b^{4}} \]
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 \[ \int \frac {{\left (c+d\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int \left (c + d x\right )^{3} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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