Optimal. Leaf size=329 \[ -\frac {2 i \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}+\frac {2 i \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}-\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}+\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \sqrt {a^2-b^2}} \]
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Rubi [A] time = 0.66, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3321, 2264, 2190, 2531, 2282, 6589} \[ -\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}+\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}-\frac {2 i \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}+\frac {2 i \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^3 \sqrt {a^2-b^2}}-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \sqrt {a^2-b^2}} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3321
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^2}{a+b \cos (c+d x)} \, dx &=2 \int \frac {e^{i (c+d x)} x^2}{b+2 a e^{i (c+d x)}+b e^{2 i (c+d x)}} \, dx\\ &=\frac {(2 b) \int \frac {e^{i (c+d x)} x^2}{2 a-2 \sqrt {a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt {a^2-b^2}}-\frac {(2 b) \int \frac {e^{i (c+d x)} x^2}{2 a+2 \sqrt {a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt {a^2-b^2}}\\ &=-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {(2 i) \int x \log \left (1+\frac {2 b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} d}-\frac {(2 i) \int x \log \left (1+\frac {2 b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} d}\\ &=-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}-\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {2 \int \text {Li}_2\left (-\frac {2 b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} d^2}-\frac {2 \int \text {Li}_2\left (-\frac {2 b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} d^2}\\ &=-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}-\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b x}{-a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt {a^2-b^2} d^3}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt {a^2-b^2} d^3}\\ &=-\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x^2 \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}-\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {2 x \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}-\frac {2 i \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3}+\frac {2 i \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3}\\ \end {align*}
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Mathematica [A] time = 0.72, size = 379, normalized size = 1.15 \[ \frac {e^{i c} \left (-i \left (d^2 x^2 \log \left (1+\frac {b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {e^{2 i c} \left (a^2-b^2\right )}}\right )-d^2 x^2 \log \left (1+\frac {b e^{i (2 c+d x)}}{\sqrt {e^{2 i c} \left (a^2-b^2\right )}+a e^{i c}}\right )+2 i d x \text {Li}_2\left (-\frac {b e^{i (2 c+d x)}}{e^{i c} a+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )+2 \text {Li}_3\left (-\frac {b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )-2 \text {Li}_3\left (-\frac {b e^{i (2 c+d x)}}{e^{i c} a+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )\right )-2 d x \text {Li}_2\left (-\frac {b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )\right )}{d^3 \sqrt {e^{2 i c} \left (a^2-b^2\right )}} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.79, size = 1267, normalized size = 3.85 \[ \text {result too large to display} \]
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 \[ \int \frac {x^{2}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{a +b \cos \left (d x +c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int \frac {x^2}{a+b\,\cos \left (c+d\,x\right )} \,d x \]
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 \[ \int \frac {x^{2}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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