Optimal. Leaf size=432 \[ \frac{6 f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^2}+\frac{6 i f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^4}+\frac{6 i f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^4}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d}-\frac{i (e+f x)^4}{4 b f} \]
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Rubi [A] time = 0.607751, antiderivative size = 432, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4519, 2190, 2531, 6609, 2282, 6589} \[ \frac{6 f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^2}+\frac{6 i f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^4}+\frac{6 i f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^4}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d}-\frac{i (e+f x)^4}{4 b f} \]
Antiderivative was successfully verified.
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Rule 4519
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac{i (e+f x)^4}{4 b f}+\int \frac{e^{i (c+d x)} (e+f x)^3}{a-\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx+\int \frac{e^{i (c+d x)} (e+f x)^3}{a+\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx\\ &=-\frac{i (e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d}-\frac{(3 f) \int (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b d}-\frac{(3 f) \int (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b d}\\ &=-\frac{i (e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{\left (6 i f^2\right ) \int (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b d^2}+\frac{\left (6 i f^2\right ) \int (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b d^2}\\ &=-\frac{i (e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{\left (6 f^3\right ) \int \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b d^3}-\frac{\left (6 f^3\right ) \int \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b d^3}\\ &=-\frac{i (e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{\left (6 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i b x}{a-\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b d^4}+\frac{\left (6 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b d^4}\\ &=-\frac{i (e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 i f^3 \text{Li}_4\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^4}+\frac{6 i f^3 \text{Li}_4\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b d^4}\\ \end{align*}
Mathematica [A] time = 0.186359, size = 410, normalized size = 0.95 \[ \frac{\frac{12 f \left (2 f \left (d (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )+i f \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )\right )-i d^2 (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )\right )}{d^4}+\frac{12 f \left (2 f \left (d (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )+i f \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )\right )-i d^2 (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )\right )}{d^4}+\frac{4 (e+f x)^3 \log \left (1+\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right )}{d}+\frac{4 (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d}-\frac{i (e+f x)^4}{f}}{4 b} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.921, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}\cos \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 = 2.98501, size = 4316, normalized size = 9.99 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{3} \cos \left (d x + c\right )}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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