Optimal. Leaf size=408 \[ \frac{2 a f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2 \sqrt{a^2-b^2}}-\frac{2 a f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^2 \sqrt{a^2-b^2}}+\frac{2 i a f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3 \sqrt{a^2-b^2}}-\frac{2 i a f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^3 \sqrt{a^2-b^2}}+\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d \sqrt{a^2-b^2}}-\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d \sqrt{a^2-b^2}}+\frac{(e+f x)^3}{3 b f} \]
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Rubi [A] time = 0.858607, antiderivative size = 408, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {4515, 32, 3323, 2264, 2190, 2531, 2282, 6589} \[ \frac{2 a f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2 \sqrt{a^2-b^2}}-\frac{2 a f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^2 \sqrt{a^2-b^2}}+\frac{2 i a f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3 \sqrt{a^2-b^2}}-\frac{2 i a f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^3 \sqrt{a^2-b^2}}+\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d \sqrt{a^2-b^2}}-\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d \sqrt{a^2-b^2}}+\frac{(e+f x)^3}{3 b f} \]
Antiderivative was successfully verified.
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Rule 4515
Rule 32
Rule 3323
Rule 2264
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \, dx}{b}-\frac{a \int \frac{(e+f x)^2}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac{(e+f x)^3}{3 b f}-\frac{(2 a) \int \frac{e^{i (c+d x)} (e+f x)^2}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b}\\ &=\frac{(e+f x)^3}{3 b f}+\frac{(2 i a) \int \frac{e^{i (c+d x)} (e+f x)^2}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2}}-\frac{(2 i a) \int \frac{e^{i (c+d x)} (e+f x)^2}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2}}\\ &=\frac{(e+f x)^3}{3 b f}+\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}-\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}-\frac{(2 i a f) \int (e+f x) \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{b \sqrt{a^2-b^2} d}+\frac{(2 i a f) \int (e+f x) \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{b \sqrt{a^2-b^2} d}\\ &=\frac{(e+f x)^3}{3 b f}+\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}-\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}+\frac{2 a f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{\left (2 a f^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{b \sqrt{a^2-b^2} d^2}+\frac{\left (2 a f^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{b \sqrt{a^2-b^2} d^2}\\ &=\frac{(e+f x)^3}{3 b f}+\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}-\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}+\frac{2 a f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}+\frac{\left (2 i a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{i b x}{a-\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \sqrt{a^2-b^2} d^3}-\frac{\left (2 i a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{i b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \sqrt{a^2-b^2} d^3}\\ &=\frac{(e+f x)^3}{3 b f}+\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}-\frac{i a (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d}+\frac{2 a f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}+\frac{2 i a f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^3}-\frac{2 i a f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^3}\\ \end{align*}
Mathematica [A] time = 2.05572, size = 445, normalized size = 1.09 \[ \frac{x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b}-\frac{i a \left (-i \left (2 f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-2 f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )+d^2 \left (2 e^2 \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right )+f x \sqrt{a^2-b^2} (2 e+f x) \left (\log \left (1-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-\log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )\right )\right )-2 d f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )+2 d f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )}{b d^3 \sqrt{-\left (a^2-b^2\right )^2}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.664, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2}\sin \left ( dx+c \right ) }{a+b\sin \left ( dx+c \right ) }}\, dx \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 [C] time = 3.05961, size = 3943, normalized size = 9.66 \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 )}^{2} \sin \left (d x + c\right )}{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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