Optimal. Leaf size=285 \[ \frac {-2 a^2-b^2}{8 x^4}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac {3 i a b d^2 e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac {3 i a b d^2 e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac {3 b^2 d^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {3 b^2 d^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x} \]
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Rubi [A]
time = 0.16, antiderivative size = 283, normalized size of antiderivative = 0.99, number of steps
used = 13, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3484, 6, 3469,
3468, 3471, 2250, 3470} \begin {gather*} -\frac {3 i a b e^{i c} d^2 x^2 \text {Gamma}\left (\frac {2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac {3 i a b e^{-i c} d^2 x^2 \text {Gamma}\left (\frac {2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac {3 b^2 e^{2 i c} d^2 x^2 \text {Gamma}\left (\frac {2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {3 b^2 e^{-2 i c} d^2 x^2 \text {Gamma}\left (\frac {2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {2 a^2+b^2}{8 x^4}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 2250
Rule 3468
Rule 3469
Rule 3470
Rule 3471
Rule 3484
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx &=\int \left (\frac {a^2}{x^5}+\frac {b^2}{2 x^5}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^5}+\frac {2 a b \sin \left (c+d x^3\right )}{x^5}\right ) \, dx\\ &=\int \left (\frac {a^2+\frac {b^2}{2}}{x^5}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^5}+\frac {2 a b \sin \left (c+d x^3\right )}{x^5}\right ) \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}+(2 a b) \int \frac {\sin \left (c+d x^3\right )}{x^5} \, dx-\frac {1}{2} b^2 \int \frac {\cos \left (2 c+2 d x^3\right )}{x^5} \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}+\frac {1}{2} (3 a b d) \int \frac {\cos \left (c+d x^3\right )}{x^2} \, dx+\frac {1}{4} \left (3 b^2 d\right ) \int \frac {\sin \left (2 c+2 d x^3\right )}{x^2} \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}-\frac {1}{2} \left (9 a b d^2\right ) \int x \sin \left (c+d x^3\right ) \, dx+\frac {1}{2} \left (9 b^2 d^2\right ) \int x \cos \left (2 c+2 d x^3\right ) \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}-\frac {1}{4} \left (9 i a b d^2\right ) \int e^{-i c-i d x^3} x \, dx+\frac {1}{4} \left (9 i a b d^2\right ) \int e^{i c+i d x^3} x \, dx+\frac {1}{4} \left (9 b^2 d^2\right ) \int e^{-2 i c-2 i d x^3} x \, dx+\frac {1}{4} \left (9 b^2 d^2\right ) \int e^{2 i c+2 i d x^3} x \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac {3 i a b d^2 e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac {3 i a b d^2 e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac {3 b^2 d^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {3 b^2 d^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}\\ \end {align*}
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Mathematica [A]
time = 1.57, size = 292, normalized size = 1.02 \begin {gather*} -\frac {2 a^2+b^2+12 a b d x^3 \cos \left (c+d x^3\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )-3 \sqrt [3]{2} b^2 \left (i d x^3\right )^{4/3} \cos (2 c) \Gamma \left (\frac {2}{3},2 i d x^3\right )+6 i a b \left (i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))+6 i a b \left (i d x^3\right )^{2/3} \sqrt [3]{d^2 x^6} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))-3 \sqrt [3]{2} b^2 \left (-i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))+3 i \sqrt [3]{2} b^2 \left (i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},2 i d x^3\right ) \sin (2 c)+4 a b \sin \left (c+d x^3\right )+6 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )}{8 x^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}}{x^{5}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.35, size = 194, normalized size = 0.68 \begin {gather*} \frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b d}{6 \, x} - \frac {{\left (2 \cdot 2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {4}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {4}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} + 3\right )} b^{2}}{24 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.13, size = 180, normalized size = 0.63 \begin {gather*} \frac {3 i \, b^{2} \left (2 i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (-2 i \, c\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) + 6 \, a b \left (i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (-i \, c\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + 6 \, a b \left (-i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (i \, c\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) - 3 i \, b^{2} \left (-2 i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (2 i \, c\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right ) - 12 \, a b d x^{3} \cos \left (d x^{3} + c\right ) + 2 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2} - 4 \, {\left (3 \, b^{2} d x^{3} \cos \left (d x^{3} + c\right ) + a b\right )} \sin \left (d x^{3} + c\right )}{8 \, x^{4}} \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 \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{5}}\, 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 \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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