Optimal. Leaf size=557 \[ -\frac {2 i f^2 \sqrt {a^2-b^2} \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^3}+\frac {2 i f^2 \sqrt {a^2-b^2} \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^3}-\frac {2 f \sqrt {a^2-b^2} (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^2}+\frac {2 f \sqrt {a^2-b^2} (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^2}-\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d}+\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b d}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(e+f x)^3}{3 b f} \]
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Rubi [A] time = 1.19, antiderivative size = 557, normalized size of antiderivative = 1.00, 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 32
Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 2638
Rule 3296
Rule 3323
Rule 4183
Rule 4408
Rule 4525
Rule 4543
Rule 6589
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*}
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Mathematica [A] time = 1.67, size = 607, normalized size = 1.09 \[ \frac {i \left (a^2-b^2\right ) \left (-i \left (d^2 \left (2 e^2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+f x \sqrt {a^2-b^2} (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}+i a}\right )\right )\right )+2 f^2 \sqrt {a^2-b^2} \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-2 f^2 \sqrt {a^2-b^2} \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )-2 d f \sqrt {a^2-b^2} (e+f x) \text {Li}_2\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )+2 d f \sqrt {a^2-b^2} (e+f x) \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )}{a b d^3 \sqrt {-\left (a^2-b^2\right )^2}}+\frac {\frac {2 i f \left (d (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )+i f \text {Li}_3\left (-e^{i (c+d x)}\right )\right )}{d^2}+\frac {2 f \left (f \text {Li}_3\left (e^{i (c+d x)}\right )-i d (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )\right )}{d^2}+(e+f x)^2 \log \left (1-e^{i (c+d x)}\right )-(e+f x)^2 \log \left (1+e^{i (c+d x)}\right )}{a d}-\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.70, size = 2123, normalized size = 3.81 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.79, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \cos \left (d x +c \right ) \cot \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx \]
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(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
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 \[ \int \frac {\left (e + f x\right )^{2} \cos {\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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