Optimal. Leaf size=382 \[ -\frac {12 f^3 \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {3 f^2 (e+f x) \sin (c+d x) \cos (c+d x)}{4 a d^3}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {(e+f x)^3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f} \]
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Rubi [A] time = 0.62, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {4515, 3311, 32, 3310, 3296, 2637, 3318, 4184, 3717, 2190, 2531, 2282, 6589} \[ \frac {12 i f^2 (e+f x) \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {12 f^3 \text {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {3 f^2 (e+f x) \sin (c+d x) \cos (c+d x)}{4 a d^3}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {(e+f x)^3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f} \]
Antiderivative was successfully verified.
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Rule 32
Rule 2190
Rule 2282
Rule 2531
Rule 2637
Rule 3296
Rule 3310
Rule 3311
Rule 3318
Rule 3717
Rule 4184
Rule 4515
Rule 6589
Rubi steps
\begin {align*} \int \frac {(e+f x)^3 \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x)^3 \sin ^2(c+d x) \, dx}{a}-\int \frac {(e+f x)^3 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}+\frac {\int (e+f x)^3 \, dx}{2 a}-\frac {\int (e+f x)^3 \sin (c+d x) \, dx}{a}-\frac {\left (3 f^2\right ) \int (e+f x) \sin ^2(c+d x) \, dx}{2 a d^2}+\int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=\frac {(e+f x)^4}{8 a f}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}+\frac {\int (e+f x)^3 \, dx}{a}-\frac {(3 f) \int (e+f x)^2 \cos (c+d x) \, dx}{a d}-\frac {\left (3 f^2\right ) \int (e+f x) \, dx}{4 a d^2}-\int \frac {(e+f x)^3}{a+a \sin (c+d x)} \, dx\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {3 (e+f x)^4}{8 a f}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {\int (e+f x)^3 \csc ^2\left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {d x}{2}\right ) \, dx}{2 a}+\frac {\left (6 f^2\right ) \int (e+f x) \sin (c+d x) \, dx}{a d^2}\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {3 (e+f x)^4}{8 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {(3 f) \int (e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}+\frac {\left (6 f^3\right ) \int \cos (c+d x) \, dx}{a d^3}\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {(6 f) \int \frac {e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )} (e+f x)^2}{1-i e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}} \, dx}{a d}\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}+\frac {\left (12 f^2\right ) \int (e+f x) \log \left (1-i e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {\left (12 i f^3\right ) \int \text {Li}_2\left (i e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^3}\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac {\left (12 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{a d^4}\\ &=-\frac {3 e f^2 x}{4 a d^2}-\frac {3 f^3 x^2}{8 a d^2}+\frac {i (e+f x)^3}{a d}+\frac {3 (e+f x)^4}{8 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {12 f^3 \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac {3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}\\ \end {align*}
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Mathematica [A] time = 2.95, size = 538, normalized size = 1.41 \[ \frac {\frac {192 f (\cos (c)+i \sin (c)) \left (\frac {2 f (\cos (c)-i (\sin (c)+1)) (d (e+f x) \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))-i f \text {Li}_3(-i \cos (c+d x)-\sin (c+d x)))}{d^3}-\frac {(\sin (c)+i \cos (c)+1) (e+f x)^2 \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac {(\cos (c)-i \sin (c)) (e+f x)^3}{3 f}\right )}{d (\cos (c)+i (\sin (c)+1))}+\frac {16 \left (d^3 (e+f x)^3-3 i d^2 f (e+f x)^2-6 d f^2 (e+f x)+6 i f^3\right ) (\cos (c+d x)-i \sin (c+d x))}{d^4}+\frac {16 \left (d^3 (e+f x)^3+3 i d^2 f (e+f x)^2-6 d f^2 (e+f x)-6 i f^3\right ) (\cos (c+d x)+i \sin (c+d x))}{d^4}+\frac {\left (-4 i d^3 (e+f x)^3-6 d^2 f (e+f x)^2+6 i d f^2 (e+f x)+3 f^3\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))}{d^4}+\frac {\left (4 i d^3 (e+f x)^3-6 d^2 f (e+f x)^2-6 i d f^2 (e+f x)+3 f^3\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))}{d^4}-\frac {64 \sin \left (\frac {d x}{2}\right ) (e+f x)^3}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+48 e^3 x+72 e^2 f x^2+48 e f^2 x^3+12 f^3 x^4}{32 a} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.56, size = 1563, normalized size = 4.09 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (f x + e\right )}^{3} \sin \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.29, size = 870, normalized size = 2.28 \[ \frac {3 f^{3} x^{4}}{8 a}+\frac {3 e^{3} x}{2 a}+\frac {\left (f^{3} x^{3} d^{3}+3 d^{3} e \,f^{2} x^{2}+3 i d^{2} f^{3} x^{2}+3 d^{3} e^{2} f x +6 i d^{2} e \,f^{2} x +d^{3} e^{3}+3 i d^{2} e^{2} f -6 d \,f^{3} x -6 f^{2} e d -6 i f^{3}\right ) {\mathrm e}^{i \left (d x +c \right )}}{2 a \,d^{4}}+\frac {\left (f^{3} x^{3} d^{3}+3 d^{3} e \,f^{2} x^{2}-3 i d^{2} f^{3} x^{2}+3 d^{3} e^{2} f x -6 i d^{2} e \,f^{2} x +d^{3} e^{3}-3 i d^{2} e^{2} f -6 d \,f^{3} x -6 f^{2} e d +6 i f^{3}\right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 a \,d^{4}}-\frac {12 f^{3} \polylog \left (3, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}-\frac {\left (2 f^{3} x^{3} d^{2}+6 d^{2} e \,f^{2} x^{2}+6 d^{2} e^{2} f x +2 d^{2} e^{3}-3 f^{3} x -3 f^{2} e \right ) \sin \left (2 d x +2 c \right )}{8 d^{3} a}+\frac {12 i f^{2} e c x}{a \,d^{2}}+\frac {3 e \,f^{2} x^{3}}{2 a}+\frac {9 e^{2} f \,x^{2}}{4 a}-\frac {6 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e^{2}}{a \,d^{2}}-\frac {6 f^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{a \,d^{2}}+\frac {6 f^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c^{2}}{a \,d^{4}}+\frac {6 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e^{2}}{a \,d^{2}}+\frac {2 i f^{3} x^{3}}{a d}+\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}-\frac {4 i f^{3} c^{3}}{a \,d^{4}}-\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{4}}+\frac {2 f^{3} x^{3}+6 e \,f^{2} x^{2}+6 e^{2} f x +2 e^{3}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}-\frac {3 f \left (2 f^{2} x^{2} d^{2}+4 d^{2} e f x +2 d^{2} e^{2}-f^{2}\right ) \cos \left (2 d x +2 c \right )}{16 a \,d^{4}}+\frac {12 i f^{3} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{3}}+\frac {6 i f^{2} e \,c^{2}}{a \,d^{3}}-\frac {12 f^{2} e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {12 f^{2} e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}-\frac {6 i f^{3} c^{2} x}{a \,d^{3}}+\frac {12 i f^{2} e \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {12 f^{2} e c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{3}}-\frac {12 f^{2} e c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {6 i f^{2} e \,x^{2}}{a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (c+d\,x\right )}^3\,{\left (e+f\,x\right )}^3}{a+a\,\sin \left (c+d\,x\right )} \,d x \]
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 \[ \frac {\int \frac {e^{3} \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{3} x^{3} \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e f^{2} x^{2} \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e^{2} f x \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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