Optimal. Leaf size=115 \[ \frac {i d \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac {(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {d x}{4 b}+\frac {i (c+d x)^2}{2 d} \]
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Rubi [A] time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 8
Rule 2190
Rule 2279
Rule 2391
Rule 2635
Rule 3719
Rule 4404
Rule 4407
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*}
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Mathematica [A] time = 0.30, size = 134, normalized size = 1.17 \[ \frac {d \left (\frac {1}{2} i \text {Li}_2\left (-e^{2 i (a+b x)}\right )+\frac {1}{2} i (a+b x)^2-(a+b x) \log \left (1+e^{2 i (a+b x)}\right )\right )}{b^2}-\frac {d \sin (2 (a+b x))}{8 b^2}+\frac {a d \log (\cos (a+b x))}{b^2}-\frac {c \left (\log (\cos (a+b x))-\frac {1}{2} \cos ^2(a+b x)\right )}{b}+\frac {d x \cos (2 (a+b x))}{4 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.49, size = 346, normalized size = 3.01 \[ -\frac {b d x - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 2 \, {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{4 \, b^{2}} \]
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 \[ \int {\left (d x + c\right )} \sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.34, size = 179, normalized size = 1.56 \[ \frac {i d \,x^{2}}{2}-i c x +\frac {\left (2 b d x +2 c b +i d \right ) {\mathrm e}^{2 i \left (b x +a \right )}}{16 b^{2}}+\frac {\left (2 b d x +2 c b -i d \right ) {\mathrm e}^{-2 i \left (b x +a \right )}}{16 b^{2}}-\frac {c \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {2 c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {2 i d a x}{b}+\frac {i d \,a^{2}}{b^{2}}-\frac {d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b}+\frac {i d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {2 d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.52, size = 145, normalized size = 1.26 \[ -\frac {-4 i \, b^{2} d x^{2} - 8 i \, b^{2} c x + {\left (8 i \, b d x + 8 i \, b c\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 i \, d {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 4 \, {\left (b d x + b c\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]
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 \[ \int \frac {{\sin \left (a+b\,x\right )}^3\,\left (c+d\,x\right )}{\cos \left (a+b\,x\right )} \,d x \]
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 \[ \int \left (c + d x\right ) \sin ^{3}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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