Optimal. Leaf size=367 \[ -\frac{2 d (c+d x) \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \sqrt{a^2-b^2}}+\frac{2 d (c+d x) \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \sqrt{a^2-b^2}}-\frac{2 i d^2 \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^3 \sqrt{a^2-b^2}}+\frac{2 i d^2 \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^3 \sqrt{a^2-b^2}}-\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}}+\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f \sqrt{a^2-b^2}} \]
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Rubi [A] time = 0.821562, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac{2 d (c+d x) \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \sqrt{a^2-b^2}}+\frac{2 d (c+d x) \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \sqrt{a^2-b^2}}-\frac{2 i d^2 \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^3 \sqrt{a^2-b^2}}+\frac{2 i d^2 \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^3 \sqrt{a^2-b^2}}-\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}}+\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f \sqrt{a^2-b^2}} \]
Antiderivative was successfully verified.
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Rule 3323
Rule 2264
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{(c+d x)^2}{a+b \sin (e+f x)} \, dx &=2 \int \frac{e^{i (e+f x)} (c+d x)^2}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx\\ &=-\frac{(2 i b) \int \frac{e^{i (e+f x)} (c+d x)^2}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\sqrt{a^2-b^2}}+\frac{(2 i b) \int \frac{e^{i (e+f x)} (c+d x)^2}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}+\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}+\frac{(2 i d) \int (c+d x) \log \left (1-\frac{2 i b e^{i (e+f x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} f}-\frac{(2 i d) \int (c+d x) \log \left (1-\frac{2 i b e^{i (e+f x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} f}\\ &=-\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}+\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}-\frac{2 d (c+d x) \text{Li}_2\left (\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^2}+\frac{2 d (c+d x) \text{Li}_2\left (\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^2}+\frac{\left (2 d^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (e+f x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} f^2}-\frac{\left (2 d^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (e+f x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} f^2}\\ &=-\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}+\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}-\frac{2 d (c+d x) \text{Li}_2\left (\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^2}+\frac{2 d (c+d x) \text{Li}_2\left (\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^2}-\frac{\left (2 i d^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 (e+f x)}\right )}{\sqrt{a^2-b^2} f^3}+\frac{\left (2 i d^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 (e+f x)}\right )}{\sqrt{a^2-b^2} f^3}\\ &=-\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}+\frac{i (c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f}-\frac{2 d (c+d x) \text{Li}_2\left (\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^2}+\frac{2 d (c+d x) \text{Li}_2\left (\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^2}-\frac{2 i d^2 \text{Li}_3\left (\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^3}+\frac{2 i d^2 \text{Li}_3\left (\frac{i b e^{i (e+f x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} f^3}\\ \end{align*}
Mathematica [A] time = 0.191191, size = 296, normalized size = 0.81 \[ -\frac{i \left (\frac{2 d \left (d \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )-i f (c+d x) \text{PolyLog}\left (2,-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )\right )}{f^2}+\frac{2 i d \left (f (c+d x) \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )+i d \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )\right )}{f^2}+(c+d x)^2 \log \left (1+\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )-(c+d x)^2 \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )\right )}{f \sqrt{a^2-b^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.283, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{2}}{a+b\sin \left ( fx+e \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 = 3.14299, size = 3729, normalized size = 10.16 \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{\left (c + d x\right )^{2}}{a + b \sin{\left (e + f x \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{{\left (d x + c\right )}^{2}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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