Optimal. Leaf size=382 \[ \frac{a^3 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2 \sqrt{a^2-b^2}}-\frac{a^3 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^2 \sqrt{a^2-b^2}}+\frac{i a^3 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d \sqrt{a^2-b^2}}-\frac{i a^3 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d \sqrt{a^2-b^2}}+\frac{a^2 e x}{b^3}+\frac{a^2 f x^2}{2 b^3}-\frac{a f \sin (c+d x)}{b^2 d^2}+\frac{a (e+f x) \cos (c+d x)}{b^2 d}+\frac{f \sin ^2(c+d x)}{4 b d^2}-\frac{(e+f x) \sin (c+d x) \cos (c+d x)}{2 b d}+\frac{e x}{2 b}+\frac{f x^2}{4 b} \]
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Rubi [A] time = 0.665739, antiderivative size = 382, normalized size of antiderivative = 1., 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 4515
Rule 3310
Rule 3296
Rule 2637
Rule 3323
Rule 2264
Rule 2190
Rule 2279
Rule 2391
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*}
Mathematica [A] time = 8.03769, size = 752, normalized size = 1.97 \[ -\frac{\frac{8 a^3 d (e+f x) \left (-\frac{i f \left (\text{PolyLog}\left (2,\frac{a \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a+i \left (\sqrt{b^2-a^2}+b\right )}\right )+\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{\sqrt{b^2-a^2}+a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{b^2-a^2}-i a+b}\right )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\text{PolyLog}\left (2,\frac{a \left (1+i \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a-i \left (\sqrt{b^2-a^2}+b\right )}\right )+\log \left (1+i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{\sqrt{b^2-a^2}+a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{b^2-a^2}+i a+b}\right )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\text{PolyLog}\left (2,\frac{a \left (\tan \left (\frac{1}{2} (c+d x)\right )+i\right )}{\sqrt{b^2-a^2}+i a-b}\right )+\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{\sqrt{b^2-a^2}-a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{b^2-a^2}+i a-b}\right )\right )}{\sqrt{b^2-a^2}}-\frac{i f \left (\text{PolyLog}\left (2,\frac{a+i a \tan \left (\frac{1}{2} (c+d x)\right )}{a+i \left (\sqrt{b^2-a^2}-b\right )}\right )+\log \left (1+i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{-\sqrt{b^2-a^2}+a \tan \left (\frac{1}{2} (c+d x)\right )+b}{-\sqrt{b^2-a^2}+i a+b}\right )\right )}{\sqrt{b^2-a^2}}+\frac{2 (d e-c f) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}\right )}{i f \log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )-i f \log \left (1+i \tan \left (\frac{1}{2} (c+d x)\right )\right )-c f+d e}+2 \left (2 a^2+b^2\right ) (c+d x) (c f-d (2 e+f x))-8 a b d (e+f x) \cos (c+d x)+8 a b f \sin (c+d x)+2 b^2 d (e+f x) \sin (2 (c+d x))+b^2 f \cos (2 (c+d x))}{8 b^3 d^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.43, size = 710, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.59615, size = 2938, normalized size = 7.69 \begin{align*} \text{result too large to display} \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 \frac{{\left (f x + e\right )} \sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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