Optimal. Leaf size=495 \[ -\frac {6 i d^2 (c+d x) \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}-\frac {3 d (c+d x)^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {3 d (c+d x)^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}}+\frac {6 d^3 \text {Li}_4\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}}-\frac {6 d^3 \text {Li}_4\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}} \]
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Rubi [A] time = 0.97, antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3323, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac {6 i d^2 (c+d x) \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {6 i d^2 (c+d x) \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \sqrt {a^2-b^2}}-\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {6 d^3 \text {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}}-\frac {6 d^3 \text {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^4 \sqrt {a^2-b^2}}-\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3323
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx &=2 \int \frac {e^{i (e+f x)} (c+d x)^3}{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)^3}{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)^3}{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)^3 \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)^3 \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 {(3 i d) \int (c+d x)^2 \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 {(3 i d) \int (c+d x)^2 \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)^3 \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)^3 \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 {3 d (c+d x)^2 \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 {3 d (c+d x)^2 \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 (6 d^2\right ) \int (c+d x) \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 (6 d^2\right ) \int (c+d x) \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)^3 \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)^3 \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 {3 d (c+d x)^2 \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 {3 d (c+d x)^2 \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 {6 i d^2 (c+d x) \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 {6 i d^2 (c+d x) \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 {\left (6 i d^3\right ) \int \text {Li}_3\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^3}-\frac {\left (6 i d^3\right ) \int \text {Li}_3\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^3}\\ &=-\frac {i (c+d x)^3 \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)^3 \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 {3 d (c+d x)^2 \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 {3 d (c+d x)^2 \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 {6 i d^2 (c+d x) \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 {6 i d^2 (c+d x) \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 {\left (6 d^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 (e+f x)}\right )}{\sqrt {a^2-b^2} f^4}-\frac {\left (6 d^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 (e+f x)}\right )}{\sqrt {a^2-b^2} f^4}\\ &=-\frac {i (c+d x)^3 \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)^3 \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 {3 d (c+d x)^2 \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 {3 d (c+d x)^2 \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 {6 i d^2 (c+d x) \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 {6 i d^2 (c+d x) \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 {6 d^3 \text {Li}_4\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4}-\frac {6 d^3 \text {Li}_4\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 401, normalized size = 0.81 \[ -\frac {i \left (\frac {3 i d \left (f^2 (c+d x)^2 \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+2 i d f (c+d x) \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )-2 d^2 \text {Li}_4\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{f^3}+\frac {3 d \left (2 d \left (f (c+d x) \text {Li}_3\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )+i d \text {Li}_4\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )\right )-i f^2 (c+d x)^2 \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )\right )}{f^3}+(c+d x)^3 \log \left (1+\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )-(c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )\right )}{f \sqrt {a^2-b^2}} \]
Antiderivative was successfully verified.
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fricas [C] time = 1.09, size = 2197, normalized size = 4.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{3}}{b \sin \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.76, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{3}}{a +b \sin \left (f x +e \right )}\, dx \]
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] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^3}{a+b\,\sin \left (e+f\,x\right )} \,d x \]
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 \[ \int \frac {\left (c + d x\right )^{3}}{a + b \sin {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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