Optimal. Leaf size=157 \[ \frac{2 i b d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i b d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{2 b d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{2 b d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac{a (c+d x)^3}{3 d}-\frac{2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]
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Rubi [A] time = 0.140572, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4190, 4181, 2531, 2282, 6589} \[ \frac{2 i b d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i b d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{2 b d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{2 b d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac{a (c+d x)^3}{3 d}-\frac{2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 4190
Rule 4181
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int (c+d x)^2 (a+b \sec (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \sec (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^3}{3 d}+b \int (c+d x)^2 \sec (e+f x) \, dx\\ &=\frac{a (c+d x)^3}{3 d}-\frac{2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac{(2 b d) \int (c+d x) \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac{(2 b d) \int (c+d x) \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{2 i b d (c+d x) \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i b d (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{\left (2 i b d^2\right ) \int \text{Li}_2\left (-i e^{i (e+f x)}\right ) \, dx}{f^2}+\frac{\left (2 i b d^2\right ) \int \text{Li}_2\left (i e^{i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{2 i b d (c+d x) \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i b d (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3}+\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{2 i b d (c+d x) \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i b d (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac{2 b d^2 \text{Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac{2 b d^2 \text{Li}_3\left (i e^{i (e+f x)}\right )}{f^3}\\ \end{align*}
Mathematica [A] time = 0.249977, size = 203, normalized size = 1.29 \[ \frac{2 i b d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i b d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{2 b d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{2 b d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+a c^2 x+a c d x^2+\frac{1}{3} a d^2 x^3+\frac{b c^2 \tanh ^{-1}(\sin (e+f x))}{f}-\frac{4 i b c d x \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac{2 i b d^2 x^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.144, size = 431, normalized size = 2.8 \begin{align*}{\frac{a{d}^{2}{x}^{3}}{3}}+acd{x}^{2}+a{c}^{2}x-2\,{\frac{b{d}^{2}{\it polylog} \left ( 3,-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+{\frac{b{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{f}}+2\,{\frac{b{d}^{2}{\it polylog} \left ( 3,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}-2\,{\frac{bcd\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{f}}+{\frac{2\,ibcd{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ib{c}^{2}\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}+{\frac{2\,ib{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}-{\frac{2\,ib{d}^{2}{\it polylog} \left ( 2,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}+2\,{\frac{bcd\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{{f}^{2}}}-2\,{\frac{bcd\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{{f}^{2}}}-{\frac{2\,ib{d}^{2}{e}^{2}\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}-{\frac{b{d}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{f}}+{\frac{b{d}^{2}{e}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+{\frac{4\,ibcde\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ibcd{\it polylog} \left ( 2,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}+2\,{\frac{bcd\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{f}}-{\frac{b{d}^{2}{e}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.23438, size = 689, normalized size = 4.39 \begin{align*} \frac{6 \,{\left (f x + e\right )} a c^{2} + \frac{2 \,{\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac{6 \,{\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac{6 \,{\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac{6 \,{\left (f x + e\right )}^{2} a c d}{f} - \frac{12 \,{\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + \frac{6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f^{2}} - \frac{12 \, b c d e \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f} + \frac{3 \,{\left (4 \, b d^{2}{\rm Li}_{3}(i \, e^{\left (i \, f x + i \, e\right )}) - 4 \, b d^{2}{\rm Li}_{3}(-i \, e^{\left (i \, f x + i \, e\right )}) +{\left (-2 i \,{\left (f x + e\right )}^{2} b d^{2} +{\left (4 i \, b d^{2} e - 4 i \, b c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) +{\left (-2 i \,{\left (f x + e\right )}^{2} b d^{2} +{\left (4 i \, b d^{2} e - 4 i \, b c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) +{\left (-4 i \,{\left (f x + e\right )} b d^{2} + 4 i \, b d^{2} e - 4 i \, b c d f\right )}{\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) +{\left (4 i \,{\left (f x + e\right )} b d^{2} - 4 i \, b d^{2} e + 4 i \, b c d f\right )}{\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) +{\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \,{\left (b d^{2} e - b c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) -{\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \,{\left (b d^{2} e - b c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{f^{2}}}{6 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.17941, size = 1732, normalized size = 11.03 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (e + f x \right )}\right ) \left (c + d x\right )^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2}{\left (b \sec \left (f x + e\right ) + a\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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