Optimal. Leaf size=245 \[ -\frac{\text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d^2 \sqrt{a^2-b^2}}+\frac{\text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d^2 \sqrt{a^2-b^2}}-\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d \sqrt{a^2-b^2}}+\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d \sqrt{a^2-b^2}} \]
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Rubi [A] time = 0.514179, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3379, 3323, 2264, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d^2 \sqrt{a^2-b^2}}+\frac{\text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d^2 \sqrt{a^2-b^2}}-\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d \sqrt{a^2-b^2}}+\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d \sqrt{a^2-b^2}} \]
Antiderivative was successfully verified.
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Rule 3379
Rule 3323
Rule 2264
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{x^3}{a+b \sin \left (c+d x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{a+b \sin (c+d x)} \, dx,x,x^2\right )\\ &=\operatorname{Subst}\left (\int \frac{e^{i (c+d x)} x}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx,x,x^2\right )\\ &=-\frac{(i b) \operatorname{Subst}\left (\int \frac{e^{i (c+d x)} x}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt{a^2-b^2}}+\frac{(i b) \operatorname{Subst}\left (\int \frac{e^{i (c+d x)} x}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a+\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d}+\frac{i \operatorname{Subst}\left (\int \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx,x,x^2\right )}{2 \sqrt{a^2-b^2} d}-\frac{i \operatorname{Subst}\left (\int \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx,x,x^2\right )}{2 \sqrt{a^2-b^2} d}\\ &=-\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a+\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d}+\frac{\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 \left (c+d x^2\right )}\right )}{2 \sqrt{a^2-b^2} d^2}-\frac{\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 \left (c+d x^2\right )}\right )}{2 \sqrt{a^2-b^2} d^2}\\ &=-\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a+\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d}-\frac{\text{Li}_2\left (\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d^2}+\frac{\text{Li}_2\left (\frac{i b e^{i \left (c+d x^2\right )}}{a+\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2} d^2}\\ \end{align*}
Mathematica [A] time = 0.0645226, size = 188, normalized size = 0.77 \[ \frac{-\text{PolyLog}\left (2,-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )+\text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )-i d x^2 \left (\log \left (1+\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )-\log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )\right )}{2 d^2 \sqrt{a^2-b^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{a+b\sin \left ( d{x}^{2}+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^{3}}{b \sin \left (d x^{2} + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.44699, size = 2481, normalized size = 10.13 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{a + b \sin{\left (c + d x^{2} \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^{3}}{b \sin \left (d x^{2} + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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