Optimal. Leaf size=825 \[ \frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{(e+f x)^3}{6 b f}-\frac{2 \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a d}+\frac{\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^2}{a b^2 d}+\frac{\cos (c+d x) (e+f x)^2}{a d}+\frac{i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac{i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac{\cos (c+d x) \sin (c+d x) (e+f x)^2}{2 b d}-\frac{f \cos ^2(c+d x) (e+f x)}{2 b d^2}+\frac{2 i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right ) (e+f x)}{a d^2}-\frac{2 i f \text{PolyLog}\left (2,e^{i (c+d x)}\right ) (e+f x)}{a d^2}+\frac{2 \left (a^2-b^2\right )^{3/2} f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac{2 \left (a^2-b^2\right )^{3/2} f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac{2 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)}{a b^2 d^2}-\frac{2 f \sin (c+d x) (e+f x)}{a d^2}+\frac{f^2 x}{4 b d^2}-\frac{2 \left (a^2-b^2\right ) f^2 \cos (c+d x)}{a b^2 d^3}-\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{2 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{2 i \left (a^2-b^2\right )^{3/2} f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d^3}-\frac{2 i \left (a^2-b^2\right )^{3/2} f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 d^3}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3} \]
[Out]
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Rubi [A] time = 1.6327, antiderivative size = 825, normalized size of antiderivative = 1., number of steps used = 41, number of rules used = 18, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {4543, 4408, 4405, 3310, 3296, 2638, 4183, 2531, 2282, 6589, 4525, 3311, 32, 2635, 8, 3323, 2264, 2190} \[ \frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{(e+f x)^3}{6 b f}-\frac{2 \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a d}+\frac{\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^2}{a b^2 d}+\frac{\cos (c+d x) (e+f x)^2}{a d}+\frac{i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac{i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac{\cos (c+d x) \sin (c+d x) (e+f x)^2}{2 b d}-\frac{f \cos ^2(c+d x) (e+f x)}{2 b d^2}+\frac{2 i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right ) (e+f x)}{a d^2}-\frac{2 i f \text{PolyLog}\left (2,e^{i (c+d x)}\right ) (e+f x)}{a d^2}+\frac{2 \left (a^2-b^2\right )^{3/2} f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac{2 \left (a^2-b^2\right )^{3/2} f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac{2 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)}{a b^2 d^2}-\frac{2 f \sin (c+d x) (e+f x)}{a d^2}+\frac{f^2 x}{4 b d^2}-\frac{2 \left (a^2-b^2\right ) f^2 \cos (c+d x)}{a b^2 d^3}-\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{2 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{2 i \left (a^2-b^2\right )^{3/2} f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d^3}-\frac{2 i \left (a^2-b^2\right )^{3/2} f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 d^3}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4543
Rule 4408
Rule 4405
Rule 3310
Rule 3296
Rule 2638
Rule 4183
Rule 2531
Rule 2282
Rule 6589
Rule 4525
Rule 3311
Rule 32
Rule 2635
Rule 8
Rule 3323
Rule 2264
Rule 2190
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \cos ^3(c+d x) \cot (c+d x) \, dx}{a}-\frac{b \int \frac{(e+f x)^2 \cos ^4(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac{\int (e+f x)^2 \cos (c+d x) \cot (c+d x) \, dx}{a}-\frac{\int (e+f x)^2 \cos ^2(c+d x) \, dx}{b}+\left (\frac{a}{b}-\frac{b}{a}\right ) \int \frac{(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx\\ &=-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int (e+f x)^2 \csc (c+d x) \, dx}{a}-\frac{\int (e+f x)^2 \sin (c+d x) \, dx}{a}+\left (\frac{1}{a}-\frac{a}{b^2}\right ) \int (e+f x)^2 \sin (c+d x) \, dx-\frac{\int (e+f x)^2 \, dx}{2 b}+\frac{\left (a^2-b^2\right ) \int (e+f x)^2 \, dx}{b^3}-\frac{\left (a^2-b^2\right )^2 \int \frac{(e+f x)^2}{a+b \sin (c+d x)} \, dx}{a b^3}+\frac{f^2 \int \cos ^2(c+d x) \, dx}{2 b d^2}\\ &=-\frac{(e+f x)^3}{6 b f}+\frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\left (\frac{1}{a}-\frac{a}{b^2}\right ) (e+f x)^2 \cos (c+d x)}{d}-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (2 \left (a^2-b^2\right )^2\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}{a b^3}-\frac{(2 f) \int (e+f x) \cos (c+d x) \, dx}{a d}-\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{\left (2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f\right ) \int (e+f x) \cos (c+d x) \, dx}{d}+\frac{f^2 \int 1 \, dx}{4 b d^2}\\ &=\frac{f^2 x}{4 b d^2}-\frac{(e+f x)^3}{6 b f}+\frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\left (\frac{1}{a}-\frac{a}{b^2}\right ) (e+f x)^2 \cos (c+d x)}{d}-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}+\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 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f (e+f x) \sin (c+d x)}{d^2}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\left (2 i \left (a^2-b^2\right )^{3/2}\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 b^2}-\frac{\left (2 i \left (a^2-b^2\right )^{3/2}\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 b^2}-\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{\left (2 f^2\right ) \int \sin (c+d x) \, dx}{a d^2}-\frac{\left (2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f^2\right ) \int \sin (c+d x) \, dx}{d^2}\\ &=\frac{f^2 x}{4 b d^2}-\frac{(e+f x)^3}{6 b f}+\frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{2 f^2 \cos (c+d x)}{a d^3}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f^2 \cos (c+d x)}{d^3}+\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\left (\frac{1}{a}-\frac{a}{b^2}\right ) (e+f x)^2 \cos (c+d x)}{d}-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}+\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d}-\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 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 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f (e+f x) \sin (c+d x)}{d^2}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (2 i \left (a^2-b^2\right )^{3/2} 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^3 d}+\frac{\left (2 i \left (a^2-b^2\right )^{3/2} 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^3 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{f^2 x}{4 b d^2}-\frac{(e+f x)^3}{6 b f}+\frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{2 f^2 \cos (c+d x)}{a d^3}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f^2 \cos (c+d x)}{d^3}+\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\left (\frac{1}{a}-\frac{a}{b^2}\right ) (e+f x)^2 \cos (c+d x)}{d}-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}+\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d}-\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 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 \left (a^2-b^2\right )^{3/2} 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^3 d^2}-\frac{2 \left (a^2-b^2\right )^{3/2} 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^3 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 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f (e+f x) \sin (c+d x)}{d^2}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (2 \left (a^2-b^2\right )^{3/2} 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^3 d^2}+\frac{\left (2 \left (a^2-b^2\right )^{3/2} 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^3 d^2}\\ &=\frac{f^2 x}{4 b d^2}-\frac{(e+f x)^3}{6 b f}+\frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{2 f^2 \cos (c+d x)}{a d^3}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f^2 \cos (c+d x)}{d^3}+\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\left (\frac{1}{a}-\frac{a}{b^2}\right ) (e+f x)^2 \cos (c+d x)}{d}-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}+\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d}-\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 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 \left (a^2-b^2\right )^{3/2} 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^3 d^2}-\frac{2 \left (a^2-b^2\right )^{3/2} 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^3 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 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f (e+f x) \sin (c+d x)}{d^2}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\left (2 i \left (a^2-b^2\right )^{3/2} 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^3 d^3}-\frac{\left (2 i \left (a^2-b^2\right )^{3/2} 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^3 d^3}\\ &=\frac{f^2 x}{4 b d^2}-\frac{(e+f x)^3}{6 b f}+\frac{\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{2 f^2 \cos (c+d x)}{a d^3}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f^2 \cos (c+d x)}{d^3}+\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\left (\frac{1}{a}-\frac{a}{b^2}\right ) (e+f x)^2 \cos (c+d x)}{d}-\frac{f (e+f x) \cos ^2(c+d x)}{2 b d^2}+\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d}-\frac{i \left (a^2-b^2\right )^{3/2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 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 \left (a^2-b^2\right )^{3/2} 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^3 d^2}-\frac{2 \left (a^2-b^2\right )^{3/2} 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^3 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 \left (a^2-b^2\right )^{3/2} f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b^3 d^3}-\frac{2 i \left (a^2-b^2\right )^{3/2} f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a b^3 d^3}-\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 \left (\frac{1}{a}-\frac{a}{b^2}\right ) f (e+f x) \sin (c+d x)}{d^2}+\frac{f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac{(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}
Mathematica [A] time = 4.98437, size = 1254, normalized size = 1.52 \[ -\frac{-24 d^2 e^2 \log \left (1-e^{i (c+d x)}\right ) b^3-24 d^2 f^2 x^2 \log \left (1-e^{i (c+d x)}\right ) b^3-48 d^2 e f x \log \left (1-e^{i (c+d x)}\right ) b^3+24 d^2 e^2 \log \left (1+e^{i (c+d x)}\right ) b^3+24 d^2 f^2 x^2 \log \left (1+e^{i (c+d x)}\right ) b^3+48 d^2 e f x \log \left (1+e^{i (c+d x)}\right ) b^3-48 i d e f \text{PolyLog}\left (2,-e^{i (c+d x)}\right ) b^3-48 i d f^2 x \text{PolyLog}\left (2,-e^{i (c+d x)}\right ) b^3+48 i d e f \text{PolyLog}\left (2,e^{i (c+d x)}\right ) b^3+48 i d f^2 x \text{PolyLog}\left (2,e^{i (c+d x)}\right ) b^3+48 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right ) b^3-48 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right ) b^3+12 a d^3 f^2 x^3 b^2+36 a d^3 e f x^2 b^2+36 a d^3 e^2 x b^2+6 a d e f \cos (2 (c+d x)) b^2+6 a d f^2 x \cos (2 (c+d x)) b^2+6 a d^2 e^2 \sin (2 (c+d x)) b^2-3 a f^2 \sin (2 (c+d x)) b^2+6 a d^2 f^2 x^2 \sin (2 (c+d x)) b^2+12 a d^2 e f x \sin (2 (c+d x)) b^2-24 a^2 d^2 e^2 \cos (c+d x) b+48 a^2 f^2 \cos (c+d x) b-24 a^2 d^2 f^2 x^2 \cos (c+d x) b-48 a^2 d^2 e f x \cos (c+d x) b+48 a^2 d e f \sin (c+d x) b+48 a^2 d f^2 x \sin (c+d x) b-8 a^3 d^3 f^2 x^3-24 a^3 d^3 e f x^2-24 a^3 d^3 e^2 x+48 \left (a^2-b^2\right )^{3/2} d^2 e^2 \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right )-24 i \left (a^2-b^2\right )^{3/2} d^2 f^2 x^2 \log \left (\frac{i e^{i (c+d x)} b}{\sqrt{a^2-b^2}-a}+1\right )-48 i \left (a^2-b^2\right )^{3/2} d^2 e f x \log \left (\frac{i e^{i (c+d x)} b}{\sqrt{a^2-b^2}-a}+1\right )+24 i \left (a^2-b^2\right )^{3/2} d^2 f^2 x^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )+48 i \left (a^2-b^2\right )^{3/2} d^2 e f x \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )-48 \left (a^2-b^2\right )^{3/2} d e f \text{PolyLog}\left (2,-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right )-48 \left (a^2-b^2\right )^{3/2} d f^2 x \text{PolyLog}\left (2,-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right )+48 \left (a^2-b^2\right )^{3/2} d e f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )+48 \left (a^2-b^2\right )^{3/2} d f^2 x \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )-48 i \left (a^2-b^2\right )^{3/2} f^2 \text{PolyLog}\left (3,-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right )+48 i \left (a^2-b^2\right )^{3/2} f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{24 a b^3 d^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 3.208, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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 = 5.14962, size = 6433, normalized size = 7.8 \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(-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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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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