Optimal. Leaf size=362 \[ -\frac {i \text {Li}_3\left (\frac {i b e^{i \left (d x^2+c\right )}}{a-\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}+\frac {i \text {Li}_3\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}-\frac {x^2 \text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}+\frac {x^2 \text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}-\frac {i x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d \sqrt {a^2-b^2}}+\frac {i x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d \sqrt {a^2-b^2}} \]
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Rubi [A] time = 0.88, antiderivative size = 362, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3379, 3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac {x^2 \text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}+\frac {x^2 \text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{d^2 \sqrt {a^2-b^2}}-\frac {i \text {PolyLog}\left (3,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}+\frac {i \text {PolyLog}\left (3,\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{d^3 \sqrt {a^2-b^2}}-\frac {i x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d \sqrt {a^2-b^2}}+\frac {i x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d \sqrt {a^2-b^2}} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3323
Rule 3379
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{a+b \sin (c+d x)} \, dx,x,x^2\right )\\ &=\operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{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}{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}{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^4 \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^4 \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 x \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 )}{\sqrt {a^2-b^2} d}-\frac {i \operatorname {Subst}\left (\int x \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 )}{\sqrt {a^2-b^2} d}\\ &=-\frac {i x^4 \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^4 \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 {x^2 \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {x^2 \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {\operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx,x,x^2\right )}{\sqrt {a^2-b^2} d^2}-\frac {\operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx,x,x^2\right )}{\sqrt {a^2-b^2} d^2}\\ &=-\frac {i x^4 \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^4 \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 {x^2 \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {x^2 \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}-\frac {i \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 \left (c+d x^2\right )}\right )}{\sqrt {a^2-b^2} d^3}+\frac {i \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 \left (c+d x^2\right )}\right )}{\sqrt {a^2-b^2} d^3}\\ &=-\frac {i x^4 \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^4 \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 {x^2 \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {x^2 \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}-\frac {i \text {Li}_3\left (\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3}+\frac {i \text {Li}_3\left (\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 289, normalized size = 0.80 \[ \frac {-i \left (d^2 x^4 \log \left (1+\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}-a}\right )-d^2 x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )+2 i d x^2 \text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )+2 \text {Li}_3\left (\frac {i b e^{i \left (d x^2+c\right )}}{a-\sqrt {a^2-b^2}}\right )-2 \text {Li}_3\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )\right )-2 d x^2 \text {Li}_2\left (-\frac {i b e^{i \left (d x^2+c\right )}}{\sqrt {a^2-b^2}-a}\right )}{2 d^3 \sqrt {a^2-b^2}} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.85, size = 1453, normalized size = 4.01 \[ \text {result too large to display} \]
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 \[ \int \frac {x^{5}}{b \sin \left (d x^{2} + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{a +b \sin \left (d \,x^{2}+c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{b \sin \left (d x^{2} + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5}{a+b\,\sin \left (d\,x^2+c\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{a + b \sin {\left (c + d x^{2} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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