Optimal. Leaf size=115 \[ \frac{i b d (c+d x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b (c+d x)^3}{3 d} \]
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Rubi [A] time = 0.208222, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3722, 3719, 2190, 2531, 2282, 6589} \[ \frac{a (c+d x)^3}{3 d}+\frac{i b d (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b (c+d x)^3}{3 d}-\frac{b d^2 \text{Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3} \]
Antiderivative was successfully verified.
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Rule 3722
Rule 3719
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int (c+d x)^2 (a+b \tan (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \tan (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^3}{3 d}+b \int (c+d x)^2 \tan (e+f x) \, dx\\ &=\frac{a (c+d x)^3}{3 d}+\frac{i b (c+d x)^3}{3 d}-(2 i b) \int \frac{e^{2 i (e+f x)} (c+d x)^2}{1+e^{2 i (e+f x)}} \, dx\\ &=\frac{a (c+d x)^3}{3 d}+\frac{i b (c+d x)^3}{3 d}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{(2 b d) \int (c+d x) \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^3}{3 d}+\frac{i b (c+d x)^3}{3 d}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b d (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac{\left (i b d^2\right ) \int \text{Li}_2\left (-e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac{a (c+d x)^3}{3 d}+\frac{i b (c+d x)^3}{3 d}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b d (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^3}\\ &=\frac{a (c+d x)^3}{3 d}+\frac{i b (c+d x)^3}{3 d}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b d (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac{b d^2 \text{Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}\\ \end{align*}
Mathematica [A] time = 0.115045, size = 191, normalized size = 1.66 \[ \frac{i b c d \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}+\frac{i b d^2 x \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+a c^2 x+a c d x^2+\frac{1}{3} a d^2 x^3-\frac{b c^2 \log (\cos (e+f x))}{f}-\frac{2 b c d x \log \left (1+e^{2 i (e+f x)}\right )}{f}+i b c d x^2-\frac{b d^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{1}{3} i b d^2 x^3 \]
Antiderivative was successfully verified.
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Maple [B] time = 0.074, size = 295, normalized size = 2.6 \begin{align*}{\frac{4\,ibcdex}{f}}-{\frac{{\frac{4\,i}{3}}b{d}^{2}{e}^{3}}{{f}^{3}}}+{\frac{a{d}^{2}{x}^{3}}{3}}+ibcd{x}^{2}+acd{x}^{2}+a{c}^{2}x-{\frac{b{c}^{2}\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ) }{f}}+2\,{\frac{b{c}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}+2\,{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}-ib{c}^{2}x+{\frac{ibcd{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-2\,{\frac{bcd\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ) x}{f}}+{\frac{2\,ibcd{e}^{2}}{{f}^{2}}}-{\frac{b{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ){x}^{2}}{f}}+{\frac{ib{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}-{\frac{b{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }{2\,{f}^{3}}}-4\,{\frac{bcde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ib{d}^{2}{e}^{2}x}{{f}^{2}}}+{\frac{i}{3}}b{d}^{2}{x}^{3} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.6821, size = 498, normalized size = 4.33 \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 )\right ) + \frac{6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac{12 \, b c d e \log \left (\sec \left (f x + e\right )\right )}{f} - \frac{-2 i \,{\left (f x + e\right )}^{3} b d^{2} + 3 \, b d^{2}{\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )}) +{\left (6 i \, b d^{2} e - 6 i \, b c d f\right )}{\left (f x + e\right )}^{2} +{\left (6 i \,{\left (f x + e\right )}^{2} b d^{2} +{\left (-12 i \, b d^{2} e + 12 i \, b c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) +{\left (-6 i \,{\left (f x + e\right )} b d^{2} + 6 i \, b d^{2} e - 6 i \, b c d f\right )}{\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 3 \,{\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 (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}{f^{2}}}{6 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.68288, size = 814, normalized size = 7.08 \begin{align*} \frac{4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x - 3 \, b d^{2}{\rm polylog}\left (3, \frac{\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b d^{2}{\rm polylog}\left (3, \frac{\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) +{\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )}{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) +{\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )}{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, f^{3}} \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 \tan{\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 \tan \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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