Optimal. Leaf size=382 \[ \frac {a^2 e x}{b^3}+\frac {a^2 f x^2}{2 b^3}+\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2 \sqrt {a^2-b^2}}-\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2 \sqrt {a^2-b^2}}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {a f \sin (c+d x)}{b^2 d^2}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 b d}+\frac {e x}{2 b}+\frac {f x^2}{4 b} \]
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Rubi [A] time = 0.67, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4515, 3310, 3296, 2637, 3323, 2264, 2190, 2279, 2391} \[ \frac {a^3 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2 \sqrt {a^2-b^2}}-\frac {a^3 f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^2 \sqrt {a^2-b^2}}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d \sqrt {a^2-b^2}}+\frac {a^2 e x}{b^3}+\frac {a^2 f x^2}{2 b^3}-\frac {a f \sin (c+d x)}{b^2 d^2}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 b d}+\frac {e x}{2 b}+\frac {f x^2}{4 b} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2279
Rule 2391
Rule 2637
Rule 3296
Rule 3310
Rule 3323
Rule 4515
Rubi steps
\begin {align*} \int \frac {(e+f x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sin ^2(c+d x) \, dx}{b}-\frac {a \int \frac {(e+f x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {a \int (e+f x) \sin (c+d x) \, dx}{b^2}+\frac {a^2 \int \frac {(e+f x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}+\frac {\int (e+f x) \, dx}{2 b}\\ &=\frac {e x}{2 b}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}+\frac {a^2 \int (e+f x) \, dx}{b^3}-\frac {a^3 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{b^3}-\frac {(a f) \int \cos (c+d x) \, dx}{b^2 d}\\ &=\frac {a^2 e x}{b^3}+\frac {e x}{2 b}+\frac {a^2 f x^2}{2 b^3}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {\left (2 a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b^3}\\ &=\frac {a^2 e x}{b^3}+\frac {e x}{2 b}+\frac {a^2 f x^2}{2 b^3}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}+\frac {\left (2 i a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{b^2 \sqrt {a^2-b^2}}-\frac {\left (2 i a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{b^2 \sqrt {a^2-b^2}}\\ &=\frac {a^2 e x}{b^3}+\frac {e x}{2 b}+\frac {a^2 f x^2}{2 b^3}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {\left (i a^3 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{b^3 \sqrt {a^2-b^2} d}+\frac {\left (i a^3 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{b^3 \sqrt {a^2-b^2} d}\\ &=\frac {a^2 e x}{b^3}+\frac {e x}{2 b}+\frac {a^2 f x^2}{2 b^3}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}-\frac {\left (a^3 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^3 \sqrt {a^2-b^2} d^2}+\frac {\left (a^3 f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^3 \sqrt {a^2-b^2} d^2}\\ &=\frac {a^2 e x}{b^3}+\frac {e x}{2 b}+\frac {a^2 f x^2}{2 b^3}+\frac {f x^2}{4 b}+\frac {a (e+f x) \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {a^3 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f \sin ^2(c+d x)}{4 b d^2}\\ \end {align*}
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Mathematica [A] time = 8.74, size = 752, normalized size = 1.97 \[ -\frac {2 \left (2 a^2+b^2\right ) (c+d x) (c f-d (2 e+f x))+\frac {8 a^3 d (e+f x) \left (\frac {2 (d e-c f) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {i f \left (\text {Li}_2\left (\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {b^2-a^2}\right )}\right )+\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {b^2-a^2}-i a+b}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\text {Li}_2\left (\frac {a \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{a-i \left (b+\sqrt {b^2-a^2}\right )}\right )+\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {b^2-a^2}+i a+b}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\text {Li}_2\left (\frac {a \left (\tan \left (\frac {1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt {b^2-a^2}}\right )+\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}-a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {b^2-a^2}+i a-b}\right )\right )}{\sqrt {b^2-a^2}}-\frac {i f \left (\text {Li}_2\left (\frac {i \tan \left (\frac {1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt {b^2-a^2}-b\right )}\right )+\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{-\sqrt {b^2-a^2}+i a+b}\right )\right )}{\sqrt {b^2-a^2}}\right )}{i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )-c f+d e}-8 a b d (e+f x) \cos (c+d x)+8 a b f \sin (c+d x)+2 b^2 d (e+f x) \sin (2 (c+d x))+b^2 f \cos (2 (c+d x))}{8 b^3 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.70, size = 1255, normalized size = 3.29 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (f x + e\right )} \sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.75, size = 686, normalized size = 1.80 \[ \frac {a^{2} f \,x^{2}}{2 b^{3}}+\frac {f \,x^{2}}{4 b}+\frac {a^{2} e x}{b^{3}}+\frac {e x}{2 b}+\frac {a \left (d f x +d e +i f \right ) {\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d^{2}}+\frac {a \left (d f x +d e -i f \right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d^{2}}+\frac {2 i a^{3} f c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{3} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {a^{3} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{b^{3} d \sqrt {-a^{2}+b^{2}}}-\frac {a^{3} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{b^{3} d^{2} \sqrt {-a^{2}+b^{2}}}+\frac {a^{3} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{b^{3} d \sqrt {-a^{2}+b^{2}}}+\frac {a^{3} f \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{b^{3} d^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i a^{3} f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{b^{3} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i a^{3} f \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{b^{3} d^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 i a^{3} e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{3} d \sqrt {-a^{2}+b^{2}}}-\frac {f \cos \left (2 d x +2 c \right )}{8 d^{2} b}-\frac {\left (f x +e \right ) \sin \left (2 d x +2 c \right )}{4 d b} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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