Optimal. Leaf size=115 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}+\frac{d x}{4 b}+\frac{i (c+d x)^2}{2 d} \]
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Rubi [A] time = 0.127852, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4407, 4404, 2635, 8, 3719, 2190, 2279, 2391} \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}+\frac{d x}{4 b}+\frac{i (c+d x)^2}{2 d} \]
Antiderivative was successfully verified.
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Rule 4407
Rule 4404
Rule 2635
Rule 8
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int (c+d x) \sin ^2(a+b x) \tan (a+b x) \, dx &=-\int (c+d x) \cos (a+b x) \sin (a+b x) \, dx+\int (c+d x) \tan (a+b x) \, dx\\ &=\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}-2 i \int \frac{e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}} \, dx+\frac{d \int \sin ^2(a+b x) \, dx}{2 b}\\ &=\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}+\frac{d \int 1 \, dx}{4 b}+\frac{d \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{d x}{4 b}+\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^2}\\ &=\frac{d x}{4 b}+\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i d \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{d \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{(c+d x) \sin ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 0.303667, size = 134, normalized size = 1.17 \[ \frac{d \left (\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )+\frac{1}{2} i (a+b x)^2-(a+b x) \log \left (1+e^{2 i (a+b x)}\right )\right )}{b^2}-\frac{d \sin (2 (a+b x))}{8 b^2}+\frac{a d \log (\cos (a+b x))}{b^2}-\frac{c \left (\log (\cos (a+b x))-\frac{1}{2} \cos ^2(a+b x)\right )}{b}+\frac{d x \cos (2 (a+b x))}{4 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.355, size = 179, normalized size = 1.6 \begin{align*}{\frac{i}{2}}d{x}^{2}-icx+{\frac{ \left ( 2\,dxb+id+2\,bc \right ){{\rm e}^{2\,i \left ( bx+a \right ) }}}{16\,{b}^{2}}}+{\frac{ \left ( 2\,dxb-id+2\,bc \right ){{\rm e}^{-2\,i \left ( bx+a \right ) }}}{16\,{b}^{2}}}+2\,{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{2\,idax}{b}}+{\frac{id{a}^{2}}{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-2\,{\frac{ad\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.77019, size = 196, normalized size = 1.7 \begin{align*} -\frac{-4 i \, b^{2} d x^{2} - 8 i \, b^{2} c x +{\left (8 i \, b d x + 8 i \, b c\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \,{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 i \, d{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 4 \,{\left (b d x + b c\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.61366, size = 965, normalized size = 8.39 \begin{align*} -\frac{b d x - 2 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 2 i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \,{\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 2 \,{\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 2 \,{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 2 \,{\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{4 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 )} \sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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