Optimal. Leaf size=293 \[ \frac{8 i x \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{8 i x \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{16 \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt{a \sin (c+d x)+a}}+\frac{16 \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt{a \sin (c+d x)+a}}-\frac{4 x^2 \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt{a \sin (c+d x)+a}} \]
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Rubi [A] time = 0.183318, antiderivative size = 293, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {3319, 4183, 2531, 2282, 6589} \[ \frac{8 i x \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{8 i x \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{16 \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt{a \sin (c+d x)+a}}+\frac{16 \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt{a \sin (c+d x)+a}}-\frac{4 x^2 \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt{a \sin (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 3319
Rule 4183
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{x^2}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \int x^2 \csc \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx}{\sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d \sqrt{a+a \sin (c+d x)}}-\frac{\left (4 \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \int x \log \left (1-e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right ) \, dx}{d \sqrt{a+a \sin (c+d x)}}+\frac{\left (4 \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \int x \log \left (1+e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right ) \, dx}{d \sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d \sqrt{a+a \sin (c+d x)}}+\frac{8 i x \text{Li}_2\left (-e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{8 i x \text{Li}_2\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{\left (8 i \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \int \text{Li}_2\left (-e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right ) \, dx}{d^2 \sqrt{a+a \sin (c+d x)}}+\frac{\left (8 i \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \int \text{Li}_2\left (e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right ) \, dx}{d^2 \sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d \sqrt{a+a \sin (c+d x)}}+\frac{8 i x \text{Li}_2\left (-e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{8 i x \text{Li}_2\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{\left (16 \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right )}{d^3 \sqrt{a+a \sin (c+d x)}}+\frac{\left (16 \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right )}{d^3 \sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d \sqrt{a+a \sin (c+d x)}}+\frac{8 i x \text{Li}_2\left (-e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{8 i x \text{Li}_2\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{16 \text{Li}_3\left (-e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^3 \sqrt{a+a \sin (c+d x)}}+\frac{16 \text{Li}_3\left (e^{\frac{1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{d^3 \sqrt{a+a \sin (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.594169, size = 245, normalized size = 0.84 \[ \frac{\sqrt [4]{-1} \sqrt{2} e^{-\frac{1}{2} i (c+d x)} \left (e^{i (c+d x)}+i\right ) \left (4 d x \text{PolyLog}\left (2,-\sqrt [4]{-1} e^{\frac{1}{2} i (c+d x)}\right )-i \left (-4 i d x \text{PolyLog}\left (2,\sqrt [4]{-1} e^{\frac{1}{2} i (c+d x)}\right )-8 \text{PolyLog}\left (3,-\sqrt [4]{-1} e^{\frac{1}{2} i (c+d x)}\right )+8 \text{PolyLog}\left (3,\sqrt [4]{-1} e^{\frac{1}{2} i (c+d x)}\right )+d^2 x^2 \log \left (1-\sqrt [4]{-1} e^{\frac{1}{2} i (c+d x)}\right )-d^2 x^2 \log \left (1+\sqrt [4]{-1} e^{\frac{1}{2} i (c+d x)}\right )\right )\right )}{d^3 \sqrt{-i a e^{-i (c+d x)} \left (e^{i (c+d x)}+i\right )^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.064, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{a \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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{\sqrt{a \sin \left (d x + c\right ) + a}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \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{x^{2}}{\sqrt{a \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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