Optimal. Leaf size=280 \[ -\frac {f^2 \cos \left (a+b (c+d x)^3\right )}{3 b d^3}+\frac {i e^{i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (d e-c f) (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{3 d^3 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (d e-c f) (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{3 d^3 \left (i b (c+d x)^3\right )^{2/3}} \]
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Rubi [A]
time = 0.18, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3514, 3436,
2239, 3470, 2250, 3460, 2718} \begin {gather*} \frac {i e^{i a} f (c+d x)^2 (d e-c f) \text {Gamma}\left (\frac {2}{3},-i b (c+d x)^3\right )}{3 d^3 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (c+d x)^2 (d e-c f) \text {Gamma}\left (\frac {2}{3},i b (c+d x)^3\right )}{3 d^3 \left (i b (c+d x)^3\right )^{2/3}}+\frac {i e^{i a} (c+d x) (d e-c f)^2 \text {Gamma}\left (\frac {1}{3},-i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (c+d x) (d e-c f)^2 \text {Gamma}\left (\frac {1}{3},i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{i b (c+d x)^3}}-\frac {f^2 \cos \left (a+b (c+d x)^3\right )}{3 b d^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2239
Rule 2250
Rule 2718
Rule 3436
Rule 3460
Rule 3470
Rule 3514
Rubi steps
\begin {align*} \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx &=\frac {\text {Subst}\left (\int \left (d^2 e^2 \left (1+\frac {c f (-2 d e+c f)}{d^2 e^2}\right ) \sin \left (a+b x^3\right )+2 d e f \left (1-\frac {c f}{d e}\right ) x \sin \left (a+b x^3\right )+f^2 x^2 \sin \left (a+b x^3\right )\right ) \, dx,x,c+d x\right )}{d^3}\\ &=\frac {f^2 \text {Subst}\left (\int x^2 \sin \left (a+b x^3\right ) \, dx,x,c+d x\right )}{d^3}+\frac {(2 f (d e-c f)) \text {Subst}\left (\int x \sin \left (a+b x^3\right ) \, dx,x,c+d x\right )}{d^3}+\frac {(d e-c f)^2 \text {Subst}\left (\int \sin \left (a+b x^3\right ) \, dx,x,c+d x\right )}{d^3}\\ &=\frac {f^2 \text {Subst}\left (\int \sin (a+b x) \, dx,x,(c+d x)^3\right )}{3 d^3}+\frac {(i f (d e-c f)) \text {Subst}\left (\int e^{-i a-i b x^3} x \, dx,x,c+d x\right )}{d^3}-\frac {(i f (d e-c f)) \text {Subst}\left (\int e^{i a+i b x^3} x \, dx,x,c+d x\right )}{d^3}+\frac {\left (i (d e-c f)^2\right ) \text {Subst}\left (\int e^{-i a-i b x^3} \, dx,x,c+d x\right )}{2 d^3}-\frac {\left (i (d e-c f)^2\right ) \text {Subst}\left (\int e^{i a+i b x^3} \, dx,x,c+d x\right )}{2 d^3}\\ &=-\frac {f^2 \cos \left (a+b (c+d x)^3\right )}{3 b d^3}+\frac {i e^{i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {1}{3},-i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{-i b (c+d x)^3}}-\frac {i e^{-i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {1}{3},i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{i b (c+d x)^3}}+\frac {i e^{i a} f (d e-c f) (c+d x)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^3\right )}{3 d^3 \left (-i b (c+d x)^3\right )^{2/3}}-\frac {i e^{-i a} f (d e-c f) (c+d x)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^3\right )}{3 d^3 \left (i b (c+d x)^3\right )^{2/3}}\\ \end {align*}
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Mathematica [F]
time = 42.13, size = 0, normalized size = 0.00 \begin {gather*} \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (f x +e \right )^{2} \sin \left (a +b \left (d x +c \right )^{3}\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.11, size = 321, normalized size = 1.15 \begin {gather*} -\frac {2 \, d^{2} f^{2} \cos \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right ) + \left (i \, b d^{3}\right )^{\frac {2}{3}} {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} e^{\left (-i \, a\right )} \Gamma \left (\frac {1}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac {2}{3}} {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} e^{\left (i \, a\right )} \Gamma \left (\frac {1}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right ) - 2 \, \left (i \, b d^{3}\right )^{\frac {1}{3}} {\left (c d f^{2} - d^{2} f e\right )} e^{\left (-i \, a\right )} \Gamma \left (\frac {2}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) - 2 \, \left (-i \, b d^{3}\right )^{\frac {1}{3}} {\left (c d f^{2} - d^{2} f e\right )} e^{\left (i \, a\right )} \Gamma \left (\frac {2}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right )}{6 \, b d^{5}} \end {gather*}
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 \begin {gather*} \int \left (e + f x\right )^{2} \sin {\left (a + b c^{3} + 3 b c^{2} d x + 3 b c d^{2} x^{2} + b d^{3} x^{3} \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \sin \left (a+b\,{\left (c+d\,x\right )}^3\right )\,{\left (e+f\,x\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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