Optimal. Leaf size=122 \[ -\frac{2 a^2+b^2}{6 x^3}+\frac{2}{3} a b d \cos (c) \text{CosIntegral}\left (d x^3\right )-\frac{2}{3} a b d \sin (c) \text{Si}\left (d x^3\right )-\frac{2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac{1}{3} b^2 d \sin (2 c) \text{CosIntegral}\left (2 d x^3\right )+\frac{1}{3} b^2 d \cos (2 c) \text{Si}\left (2 d x^3\right )+\frac{b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3} \]
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Rubi [A] time = 0.218841, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3403, 6, 3380, 3297, 3303, 3299, 3302, 3379} \[ -\frac{2 a^2+b^2}{6 x^3}+\frac{2}{3} a b d \cos (c) \text{CosIntegral}\left (d x^3\right )-\frac{2}{3} a b d \sin (c) \text{Si}\left (d x^3\right )-\frac{2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac{1}{3} b^2 d \sin (2 c) \text{CosIntegral}\left (2 d x^3\right )+\frac{1}{3} b^2 d \cos (2 c) \text{Si}\left (2 d x^3\right )+\frac{b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3} \]
Antiderivative was successfully verified.
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Rule 3403
Rule 6
Rule 3380
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rule 3379
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^4} \, dx &=\int \left (\frac{a^2}{x^4}+\frac{b^2}{2 x^4}-\frac{b^2 \cos \left (2 c+2 d x^3\right )}{2 x^4}+\frac{2 a b \sin \left (c+d x^3\right )}{x^4}\right ) \, dx\\ &=\int \left (\frac{a^2+\frac{b^2}{2}}{x^4}-\frac{b^2 \cos \left (2 c+2 d x^3\right )}{2 x^4}+\frac{2 a b \sin \left (c+d x^3\right )}{x^4}\right ) \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}+(2 a b) \int \frac{\sin \left (c+d x^3\right )}{x^4} \, dx-\frac{1}{2} b^2 \int \frac{\cos \left (2 c+2 d x^3\right )}{x^4} \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}+\frac{1}{3} (2 a b) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{x^2} \, dx,x,x^3\right )-\frac{1}{6} b^2 \operatorname{Subst}\left (\int \frac{\cos (2 c+2 d x)}{x^2} \, dx,x,x^3\right )\\ &=-\frac{2 a^2+b^2}{6 x^3}+\frac{b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3}-\frac{2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac{1}{3} (2 a b d) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{x} \, dx,x,x^3\right )+\frac{1}{3} \left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{\sin (2 c+2 d x)}{x} \, dx,x,x^3\right )\\ &=-\frac{2 a^2+b^2}{6 x^3}+\frac{b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3}-\frac{2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac{1}{3} (2 a b d \cos (c)) \operatorname{Subst}\left (\int \frac{\cos (d x)}{x} \, dx,x,x^3\right )+\frac{1}{3} \left (b^2 d \cos (2 c)\right ) \operatorname{Subst}\left (\int \frac{\sin (2 d x)}{x} \, dx,x,x^3\right )-\frac{1}{3} (2 a b d \sin (c)) \operatorname{Subst}\left (\int \frac{\sin (d x)}{x} \, dx,x,x^3\right )+\frac{1}{3} \left (b^2 d \sin (2 c)\right ) \operatorname{Subst}\left (\int \frac{\cos (2 d x)}{x} \, dx,x,x^3\right )\\ &=-\frac{2 a^2+b^2}{6 x^3}+\frac{b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3}+\frac{2}{3} a b d \cos (c) \text{Ci}\left (d x^3\right )+\frac{1}{3} b^2 d \text{Ci}\left (2 d x^3\right ) \sin (2 c)-\frac{2 a b \sin \left (c+d x^3\right )}{3 x^3}-\frac{2}{3} a b d \sin (c) \text{Si}\left (d x^3\right )+\frac{1}{3} b^2 d \cos (2 c) \text{Si}\left (2 d x^3\right )\\ \end{align*}
Mathematica [A] time = 0.26715, size = 116, normalized size = 0.95 \[ \frac{-2 a^2+4 a b d x^3 \cos (c) \text{CosIntegral}\left (d x^3\right )-4 a b d x^3 \sin (c) \text{Si}\left (d x^3\right )-4 a b \sin \left (c+d x^3\right )+2 b^2 d x^3 \sin (2 c) \text{CosIntegral}\left (2 d x^3\right )+2 b^2 d x^3 \cos (2 c) \text{Si}\left (2 d x^3\right )+b^2 \cos \left (2 \left (c+d x^3\right )\right )-b^2}{6 x^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.225, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\sin \left ( d{x}^{3}+c \right ) \right ) ^{2}}{{x}^{4}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 1.24392, size = 167, normalized size = 1.37 \begin{align*} \frac{1}{3} \,{\left ({\left (\Gamma \left (-1, i \, d x^{3}\right ) + \Gamma \left (-1, -i \, d x^{3}\right )\right )} \cos \left (c\right ) -{\left (i \, \Gamma \left (-1, i \, d x^{3}\right ) - i \, \Gamma \left (-1, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b d + \frac{{\left ({\left ({\left (i \, \Gamma \left (-1, 2 i \, d x^{3}\right ) - i \, \Gamma \left (-1, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) +{\left (\Gamma \left (-1, 2 i \, d x^{3}\right ) + \Gamma \left (-1, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} - 1\right )} b^{2}}{6 \, x^{3}} - \frac{a^{2}}{3 \, x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.7884, size = 425, normalized size = 3.48 \begin{align*} \frac{2 \, b^{2} d x^{3} \cos \left (2 \, c\right ) \operatorname{Si}\left (2 \, d x^{3}\right ) - 4 \, a b d x^{3} \sin \left (c\right ) \operatorname{Si}\left (d x^{3}\right ) + 2 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 4 \, a b \sin \left (d x^{3} + c\right ) - 2 \, a^{2} - 2 \, b^{2} + 2 \,{\left (a b d x^{3} \operatorname{Ci}\left (d x^{3}\right ) + a b d x^{3} \operatorname{Ci}\left (-d x^{3}\right )\right )} \cos \left (c\right ) +{\left (b^{2} d x^{3} \operatorname{Ci}\left (2 \, d x^{3}\right ) + b^{2} d x^{3} \operatorname{Ci}\left (-2 \, d x^{3}\right )\right )} \sin \left (2 \, c\right )}{6 \, x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sin{\left (c + d x^{3} \right )}\right )^{2}}{x^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.12372, size = 305, normalized size = 2.5 \begin{align*} \frac{4 \,{\left (d x^{3} + c\right )} a b d^{2} \cos \left (c\right ) \operatorname{Ci}\left (d x^{3}\right ) - 4 \, a b c d^{2} \cos \left (c\right ) \operatorname{Ci}\left (d x^{3}\right ) + 2 \,{\left (d x^{3} + c\right )} b^{2} d^{2} \operatorname{Ci}\left (2 \, d x^{3}\right ) \sin \left (2 \, c\right ) - 2 \, b^{2} c d^{2} \operatorname{Ci}\left (2 \, d x^{3}\right ) \sin \left (2 \, c\right ) - 4 \,{\left (d x^{3} + c\right )} a b d^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{3}\right ) + 4 \, a b c d^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{3}\right ) - 2 \,{\left (d x^{3} + c\right )} b^{2} d^{2} \cos \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{3}\right ) + 2 \, b^{2} c d^{2} \cos \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{3}\right ) + b^{2} d^{2} \cos \left (2 \, d x^{3} + 2 \, c\right ) - 4 \, a b d^{2} \sin \left (d x^{3} + c\right ) - 2 \, a^{2} d^{2} - b^{2} d^{2}}{6 \, d^{2} x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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