Optimal. Leaf size=557 \[ -\frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d^2}-\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^3}+\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d^3}+\frac{2 i f (e+f x) \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{2 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^3}{3 b f} \]
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Rubi [A] time = 1.1903, antiderivative size = 557, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 13, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.406, Rules used = {4543, 4408, 3296, 2638, 4183, 2531, 2282, 6589, 4525, 32, 3323, 2264, 2190} \[ -\frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d^2}-\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^3}+\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d^3}+\frac{2 i f (e+f x) \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{2 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^3}{3 b f} \]
Antiderivative was successfully verified.
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Rule 4543
Rule 4408
Rule 3296
Rule 2638
Rule 4183
Rule 2531
Rule 2282
Rule 6589
Rule 4525
Rule 32
Rule 3323
Rule 2264
Rule 2190
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \cos (c+d x) \cot (c+d x) \, dx}{a}-\frac{b \int \frac{(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac{\int (e+f x)^2 \csc (c+d x) \, dx}{a}-\frac{\int (e+f x)^2 \, dx}{b}+\left (\frac{a}{b}-\frac{b}{a}\right ) \int \frac{(e+f x)^2}{a+b \sin (c+d x)} \, dx\\ &=-\frac{(e+f x)^3}{3 b f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\left (2 \left (\frac{a}{b}-\frac{b}{a}\right )\right ) \int \frac{e^{i (c+d x)} (e+f x)^2}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx-\frac{(2 f) \int (e+f x) \log \left (1-e^{i (c+d x)}\right ) \, dx}{a d}+\frac{(2 f) \int (e+f x) \log \left (1+e^{i (c+d x)}\right ) \, dx}{a d}\\ &=-\frac{(e+f x)^3}{3 b f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{2 i f (e+f x) \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{\left (2 i \sqrt{a^2-b^2}\right ) \int \frac{e^{i (c+d x)} (e+f x)^2}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a}+\frac{\left (2 i \sqrt{a^2-b^2}\right ) \int \frac{e^{i (c+d x)} (e+f x)^2}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a}-\frac{\left (2 i f^2\right ) \int \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac{\left (2 i f^2\right ) \int \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx}{a d^2}\\ &=-\frac{(e+f x)^3}{3 b f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d}+\frac{2 i f (e+f x) \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{\left (2 i \sqrt{a^2-b^2} f\right ) \int (e+f x) \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{a b d}-\frac{\left (2 i \sqrt{a^2-b^2} f\right ) \int (e+f x) \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{a b d}-\frac{\left (2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^3}+\frac{\left (2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^3}\\ &=-\frac{(e+f x)^3}{3 b f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d}+\frac{2 i f (e+f x) \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{2 \sqrt{a^2-b^2} f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{2 \sqrt{a^2-b^2} f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d^2}-\frac{2 f^2 \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac{\left (2 \sqrt{a^2-b^2} f^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{a b d^2}-\frac{\left (2 \sqrt{a^2-b^2} f^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{a b d^2}\\ &=-\frac{(e+f x)^3}{3 b f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d}+\frac{2 i f (e+f x) \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{2 \sqrt{a^2-b^2} f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{2 \sqrt{a^2-b^2} f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d^2}-\frac{2 f^2 \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac{\left (2 i \sqrt{a^2-b^2} f^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 (c+d x)}\right )}{a b d^3}+\frac{\left (2 i \sqrt{a^2-b^2} f^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 (c+d x)}\right )}{a b d^3}\\ &=-\frac{(e+f x)^3}{3 b f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d}+\frac{2 i f (e+f x) \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{2 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{2 \sqrt{a^2-b^2} f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{2 \sqrt{a^2-b^2} f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d^2}-\frac{2 f^2 \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac{2 i \sqrt{a^2-b^2} f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^3}+\frac{2 i \sqrt{a^2-b^2} f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b d^3}\\ \end{align*}
Mathematica [A] time = 1.75015, size = 607, normalized size = 1.09 \[ \frac{i \left (a^2-b^2\right ) \left (-i \left (2 f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-2 f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )+d^2 \left (2 e^2 \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right )+f x \sqrt{a^2-b^2} (2 e+f x) \left (\log \left (1-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-\log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )\right )\right )-2 d f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )+2 d f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )}{a b d^3 \sqrt{-\left (a^2-b^2\right )^2}}+\frac{\frac{2 i f \left (d (e+f x) \text{PolyLog}\left (2,-e^{i (c+d x)}\right )+i f \text{PolyLog}\left (3,-e^{i (c+d x)}\right )\right )}{d^2}+\frac{2 f \left (f \text{PolyLog}\left (3,e^{i (c+d x)}\right )-i d (e+f x) \text{PolyLog}\left (2,e^{i (c+d x)}\right )\right )}{d^2}+(e+f x)^2 \log \left (1-e^{i (c+d x)}\right )-(e+f x)^2 \log \left (1+e^{i (c+d x)}\right )}{a d}-\frac{x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.974, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2}\cos \left ( dx+c \right ) \cot \left ( dx+c \right ) }{a+b\sin \left ( dx+c \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 = 3.30347, size = 5239, normalized size = 9.41 \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 (e + f x\right )^{2} \cos{\left (c + d x \right )} \cot{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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