Optimal. Leaf size=179 \[ -\frac {6 i d^3 \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{a+b x}\right )}{b^4}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.13, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4180, 2531, 6609, 2282, 6589} \[ \frac {6 i d^2 (c+d x) \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac {6 i d^3 \text {PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}+\frac {6 i d^3 \text {PolyLog}\left (4,i e^{a+b x}\right )}{b^4}+\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4180
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int (c+d x)^3 \text {sech}(a+b x) \, dx &=\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(3 i d) \int (c+d x)^2 \log \left (1-i e^{a+b x}\right ) \, dx}{b}+\frac {(3 i d) \int (c+d x)^2 \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}-\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {\left (6 i d^3\right ) \int \text {Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}+\frac {\left (6 i d^3\right ) \int \text {Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {\left (6 i d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {\left (6 i d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {6 i d^3 \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{a+b x}\right )}{b^4}\\ \end {align*}
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Mathematica [A] time = 0.28, size = 343, normalized size = 1.92 \[ \frac {i \left (-2 i b^3 c^3 \tan ^{-1}\left (e^{a+b x}\right )+3 b^3 c^2 d x \log \left (1-i e^{a+b x}\right )-3 b^3 c^2 d x \log \left (1+i e^{a+b x}\right )+3 b^3 c d^2 x^2 \log \left (1-i e^{a+b x}\right )-3 b^3 c d^2 x^2 \log \left (1+i e^{a+b x}\right )+b^3 d^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 d^3 x^3 \log \left (1+i e^{a+b x}\right )-3 b^2 d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )+3 b^2 d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )+6 b c d^2 \text {Li}_3\left (-i e^{a+b x}\right )-6 b c d^2 \text {Li}_3\left (i e^{a+b x}\right )+6 b d^3 x \text {Li}_3\left (-i e^{a+b x}\right )-6 b d^3 x \text {Li}_3\left (i e^{a+b x}\right )-6 d^3 \text {Li}_4\left (-i e^{a+b x}\right )+6 d^3 \text {Li}_4\left (i e^{a+b x}\right )\right )}{b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.45, size = 493, normalized size = 2.75 \[ \frac {6 i \, d^{3} {\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 i \, d^{3} {\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (3 i \, b^{2} d^{3} x^{2} + 6 i \, b^{2} c d^{2} x + 3 i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (-3 i \, b^{2} d^{3} x^{2} - 6 i \, b^{2} c d^{2} x - 3 i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (i \, b^{3} c^{3} - 3 i \, a b^{2} c^{2} d + 3 i \, a^{2} b c d^{2} - i \, a^{3} d^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + {\left (-i \, b^{3} c^{3} + 3 i \, a b^{2} c^{2} d - 3 i \, a^{2} b c d^{2} + i \, a^{3} d^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + {\left (-i \, b^{3} d^{3} x^{3} - 3 i \, b^{3} c d^{2} x^{2} - 3 i \, b^{3} c^{2} d x - 3 i \, a b^{2} c^{2} d + 3 i \, a^{2} b c d^{2} - i \, a^{3} d^{3}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (i \, b^{3} d^{3} x^{3} + 3 i \, b^{3} c d^{2} x^{2} + 3 i \, b^{3} c^{2} d x + 3 i \, a b^{2} c^{2} d - 3 i \, a^{2} b c d^{2} + i \, a^{3} d^{3}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{b^{4}} \]
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 \[ \int {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{3} \mathrm {sech}\left (b x +a \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, c^{3} \arctan \left (e^{\left (-b x - a\right )}\right )}{b} + 2 \, \int \frac {{\left (d^{3} x^{3} e^{a} + 3 \, c d^{2} x^{2} e^{a} + 3 \, c^{2} d x e^{a}\right )} e^{\left (b x\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \]
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 \[ \int \frac {{\left (c+d\,x\right )}^3}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
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 \[ \int \left (c + d x\right )^{3} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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