Optimal. Leaf size=119 \[ \frac {2 i d^2 \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {2 i d^2 \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {2 i d (c+d x) \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {2 i d (c+d x) \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.08, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4180, 2531, 2282, 6589} \[ -\frac {2 i d (c+d x) \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {2 i d (c+d x) \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}+\frac {2 i d^2 \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac {2 i d^2 \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac {2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4180
Rule 6589
Rubi steps
\begin {align*} \int (c+d x)^2 \text {sech}(a+b x) \, dx &=\frac {2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(2 i d) \int (c+d x) \log \left (1-i e^{a+b x}\right ) \, dx}{b}+\frac {(2 i d) \int (c+d x) \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=\frac {2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {2 i d (c+d x) \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {2 i d (c+d x) \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {\left (2 i d^2\right ) \int \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}-\frac {\left (2 i d^2\right ) \int \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac {2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {2 i d (c+d x) \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {2 i d (c+d x) \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=\frac {2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {2 i d (c+d x) \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {2 i d (c+d x) \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {2 i d^2 \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {2 i d^2 \text {Li}_3\left (i e^{a+b x}\right )}{b^3}\\ \end {align*}
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Mathematica [A] time = 1.52, size = 199, normalized size = 1.67 \[ \frac {i \left (-2 i b^2 c^2 \tan ^{-1}\left (e^{a+b x}\right )+2 b^2 c d x \log \left (1-i e^{a+b x}\right )-2 b^2 c d x \log \left (1+i e^{a+b x}\right )+b^2 d^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 d^2 x^2 \log \left (1+i e^{a+b x}\right )-2 b d (c+d x) \text {Li}_2\left (-i e^{a+b x}\right )+2 b d (c+d x) \text {Li}_2\left (i e^{a+b x}\right )+2 d^2 \text {Li}_3\left (-i e^{a+b x}\right )-2 d^2 \text {Li}_3\left (i e^{a+b x}\right )\right )}{b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.98, size = 303, normalized size = 2.55 \[ \frac {-2 i \, d^{2} {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 2 i \, d^{2} {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + {\left (-i \, b^{2} c^{2} + 2 i \, a b c d - i \, a^{2} d^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + {\left (-i \, b^{2} d^{2} x^{2} - 2 i \, b^{2} c d x - 2 i \, a b c d + i \, a^{2} d^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (i \, b^{2} d^{2} x^{2} + 2 i \, b^{2} c d x + 2 i \, a b c d - i \, a^{2} d^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{3}} \]
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 \[ \int {\left (d x + c\right )}^{2} \operatorname {sech}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.40, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{2} \mathrm {sech}\left (b x +a \right )\, dx \]
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 \[ -\frac {2 \, c^{2} \arctan \left (e^{\left (-b x - a\right )}\right )}{b} + 2 \, \int \frac {{\left (d^{2} x^{2} e^{a} + 2 \, c d x e^{a}\right )} e^{\left (b x\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \]
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 \[ \int \frac {{\left (c+d\,x\right )}^2}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
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 \[ \int \left (c + d x\right )^{2} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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