Optimal. Leaf size=115 \[ -\frac {2 a^2+b^2}{4 x^2}+a b d \cos (c) \text {Ci}\left (d x^2\right )-a b d \sin (c) \text {Si}\left (d x^2\right )-\frac {a b \sin \left (c+d x^2\right )}{x^2}+\frac {1}{2} b^2 d \sin (2 c) \text {Ci}\left (2 d x^2\right )+\frac {1}{2} b^2 d \cos (2 c) \text {Si}\left (2 d x^2\right )+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2} \]
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Rubi [A] time = 0.22, antiderivative size = 115, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 a^2+b^2}{4 x^2}+a b d \cos (c) \text {CosIntegral}\left (d x^2\right )-a b d \sin (c) \text {Si}\left (d x^2\right )-\frac {a b \sin \left (c+d x^2\right )}{x^2}+\frac {1}{2} b^2 d \sin (2 c) \text {CosIntegral}\left (2 d x^2\right )+\frac {1}{2} b^2 d \cos (2 c) \text {Si}\left (2 d x^2\right )+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2} \]
Antiderivative was successfully verified.
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Rule 6
Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 3379
Rule 3380
Rule 3403
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^3} \, dx &=\int \left (\frac {a^2}{x^3}+\frac {b^2}{2 x^3}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x^3}+\frac {2 a b \sin \left (c+d x^2\right )}{x^3}\right ) \, dx\\ &=\int \left (\frac {a^2+\frac {b^2}{2}}{x^3}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x^3}+\frac {2 a b \sin \left (c+d x^2\right )}{x^3}\right ) \, dx\\ &=-\frac {2 a^2+b^2}{4 x^2}+(2 a b) \int \frac {\sin \left (c+d x^2\right )}{x^3} \, dx-\frac {1}{2} b^2 \int \frac {\cos \left (2 c+2 d x^2\right )}{x^3} \, dx\\ &=-\frac {2 a^2+b^2}{4 x^2}+(a b) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^2} \, dx,x,x^2\right )-\frac {1}{4} b^2 \operatorname {Subst}\left (\int \frac {\cos (2 c+2 d x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{4 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac {a b \sin \left (c+d x^2\right )}{x^2}+(a b d) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\sin (2 c+2 d x)}{x} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{4 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac {a b \sin \left (c+d x^2\right )}{x^2}+(a b d \cos (c)) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 d \cos (2 c)\right ) \operatorname {Subst}\left (\int \frac {\sin (2 d x)}{x} \, dx,x,x^2\right )-(a b d \sin (c)) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 d \sin (2 c)\right ) \operatorname {Subst}\left (\int \frac {\cos (2 d x)}{x} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{4 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2}+a b d \cos (c) \text {Ci}\left (d x^2\right )+\frac {1}{2} b^2 d \text {Ci}\left (2 d x^2\right ) \sin (2 c)-\frac {a b \sin \left (c+d x^2\right )}{x^2}-a b d \sin (c) \text {Si}\left (d x^2\right )+\frac {1}{2} b^2 d \cos (2 c) \text {Si}\left (2 d x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.26, size = 116, normalized size = 1.01 \[ \frac {-2 a^2+4 a b d x^2 \cos (c) \text {Ci}\left (d x^2\right )-4 a b d x^2 \sin (c) \text {Si}\left (d x^2\right )-4 a b \sin \left (c+d x^2\right )+2 b^2 d x^2 \sin (2 c) \text {Ci}\left (2 d x^2\right )+2 b^2 d x^2 \cos (2 c) \text {Si}\left (2 d x^2\right )+b^2 \cos \left (2 \left (c+d x^2\right )\right )-b^2}{4 x^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.80, size = 147, normalized size = 1.28 \[ \frac {2 \, b^{2} d x^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (2 \, d x^{2}\right ) - 4 \, a b d x^{2} \sin \relax (c) \operatorname {Si}\left (d x^{2}\right ) + 2 \, b^{2} \cos \left (d x^{2} + c\right )^{2} - 4 \, a b \sin \left (d x^{2} + c\right ) - 2 \, a^{2} - 2 \, b^{2} + 2 \, {\left (a b d x^{2} \operatorname {Ci}\left (d x^{2}\right ) + a b d x^{2} \operatorname {Ci}\left (-d x^{2}\right )\right )} \cos \relax (c) + {\left (b^{2} d x^{2} \operatorname {Ci}\left (2 \, d x^{2}\right ) + b^{2} d x^{2} \operatorname {Ci}\left (-2 \, d x^{2}\right )\right )} \sin \left (2 \, c\right )}{4 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.49, size = 226, normalized size = 1.97 \[ \frac {4 \, {\left (d x^{2} + c\right )} a b d^{2} \cos \relax (c) \operatorname {Ci}\left (d x^{2}\right ) - 4 \, a b c d^{2} \cos \relax (c) \operatorname {Ci}\left (d x^{2}\right ) + 2 \, {\left (d x^{2} + c\right )} b^{2} d^{2} \operatorname {Ci}\left (2 \, d x^{2}\right ) \sin \left (2 \, c\right ) - 2 \, b^{2} c d^{2} \operatorname {Ci}\left (2 \, d x^{2}\right ) \sin \left (2 \, c\right ) - 4 \, {\left (d x^{2} + c\right )} a b d^{2} \sin \relax (c) \operatorname {Si}\left (d x^{2}\right ) + 4 \, a b c d^{2} \sin \relax (c) \operatorname {Si}\left (d x^{2}\right ) - 2 \, {\left (d x^{2} + c\right )} b^{2} d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) + 2 \, b^{2} c d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) + b^{2} d^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 4 \, a b d^{2} \sin \left (d x^{2} + c\right ) - 2 \, a^{2} d^{2} - b^{2} d^{2}}{4 \, d^{2} x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.62, size = 203, normalized size = 1.77 \[ -\frac {\pi \,\mathrm {csgn}\left (d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d}{4}+\frac {\Si \left (2 d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d}{2}-\frac {i \Ei \left (1, -2 i d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d}{4}+\frac {i b^{2} d \Ei \left (1, -2 i d \,x^{2}\right ) {\mathrm e}^{2 i c}}{4}-\frac {a b d \Ei \left (1, -i d \,x^{2}\right ) {\mathrm e}^{i c}}{2}+\frac {i \pi \,\mathrm {csgn}\left (d \,x^{2}\right ) {\mathrm e}^{-i c} a b d}{2}-i \Si \left (d \,x^{2}\right ) {\mathrm e}^{-i c} a b d -\frac {\Ei \left (1, -i d \,x^{2}\right ) {\mathrm e}^{-i c} a b d}{2}-\frac {a^{2}}{2 x^{2}}-\frac {b^{2}}{4 x^{2}}-\frac {a b \sin \left (d \,x^{2}+c \right )}{x^{2}}+\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{4 x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.49, size = 124, normalized size = 1.08 \[ \frac {1}{2} \, {\left ({\left (\Gamma \left (-1, i \, d x^{2}\right ) + \Gamma \left (-1, -i \, d x^{2}\right )\right )} \cos \relax (c) - {\left (i \, \Gamma \left (-1, i \, d x^{2}\right ) - i \, \Gamma \left (-1, -i \, d x^{2}\right )\right )} \sin \relax (c)\right )} a b d + \frac {{\left ({\left ({\left (i \, \Gamma \left (-1, 2 i \, d x^{2}\right ) - i \, \Gamma \left (-1, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + {\left (\Gamma \left (-1, 2 i \, d x^{2}\right ) + \Gamma \left (-1, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{2} - 1\right )} b^{2}}{4 \, x^{2}} - \frac {a^{2}}{2 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2}{x^3} \,d x \]
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 \[ \int \frac {\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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