Optimal. Leaf size=169 \[ -\frac {2 a^2+b^2}{8 x^4}-\frac {1}{2} a b d^2 \sin (c) \text {Ci}\left (d x^2\right )-\frac {1}{2} a b d^2 \cos (c) \text {Si}\left (d x^2\right )-\frac {a b d \cos \left (c+d x^2\right )}{2 x^2}-\frac {a b \sin \left (c+d x^2\right )}{2 x^4}+\frac {1}{2} b^2 d^2 \cos (2 c) \text {Ci}\left (2 d x^2\right )-\frac {1}{2} b^2 d^2 \sin (2 c) \text {Si}\left (2 d x^2\right )-\frac {b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4} \]
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Rubi [A] time = 0.29, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 15, 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} \[ -\frac {2 a^2+b^2}{8 x^4}-\frac {1}{2} a b d^2 \sin (c) \text {CosIntegral}\left (d x^2\right )-\frac {1}{2} a b d^2 \cos (c) \text {Si}\left (d x^2\right )-\frac {a b \sin \left (c+d x^2\right )}{2 x^4}-\frac {a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac {1}{2} b^2 d^2 \cos (2 c) \text {CosIntegral}\left (2 d x^2\right )-\frac {1}{2} b^2 d^2 \sin (2 c) \text {Si}\left (2 d x^2\right )-\frac {b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4} \]
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^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^2\right )}{2 x^5}+\frac {2 a b \sin \left (c+d x^2\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^2\right )}{2 x^5}+\frac {2 a b \sin \left (c+d x^2\right )}{x^5}\right ) \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}+(2 a b) \int \frac {\sin \left (c+d x^2\right )}{x^5} \, dx-\frac {1}{2} b^2 \int \frac {\cos \left (2 c+2 d x^2\right )}{x^5} \, dx\\ &=-\frac {2 a^2+b^2}{8 x^4}+(a b) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^3} \, dx,x,x^2\right )-\frac {1}{4} b^2 \operatorname {Subst}\left (\int \frac {\cos (2 c+2 d x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{8 x^4}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}-\frac {a b \sin \left (c+d x^2\right )}{2 x^4}+\frac {1}{2} (a b d) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x^2} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\sin (2 c+2 d x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{8 x^4}-\frac {a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}-\frac {a b \sin \left (c+d x^2\right )}{2 x^4}-\frac {b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac {1}{2} \left (a b d^2\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\cos (2 c+2 d x)}{x} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{8 x^4}-\frac {a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}-\frac {a b \sin \left (c+d x^2\right )}{2 x^4}-\frac {b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac {1}{2} \left (a b d^2 \cos (c)\right ) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 d^2 \cos (2 c)\right ) \operatorname {Subst}\left (\int \frac {\cos (2 d x)}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (a b d^2 \sin (c)\right ) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 d^2 \sin (2 c)\right ) \operatorname {Subst}\left (\int \frac {\sin (2 d x)}{x} \, dx,x,x^2\right )\\ &=-\frac {2 a^2+b^2}{8 x^4}-\frac {a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}+\frac {1}{2} b^2 d^2 \cos (2 c) \text {Ci}\left (2 d x^2\right )-\frac {1}{2} a b d^2 \text {Ci}\left (d x^2\right ) \sin (c)-\frac {a b \sin \left (c+d x^2\right )}{2 x^4}-\frac {b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac {1}{2} a b d^2 \cos (c) \text {Si}\left (d x^2\right )-\frac {1}{2} b^2 d^2 \sin (2 c) \text {Si}\left (2 d x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.47, size = 158, normalized size = 0.93 \[ -\frac {2 a^2+4 a b d^2 x^4 \sin (c) \text {Ci}\left (d x^2\right )+4 a b d^2 x^4 \cos (c) \text {Si}\left (d x^2\right )+4 a b \sin \left (c+d x^2\right )+4 a b d x^2 \cos \left (c+d x^2\right )-4 b^2 d^2 x^4 \cos (2 c) \text {Ci}\left (2 d x^2\right )+4 b^2 d^2 x^4 \sin (2 c) \text {Si}\left (2 d x^2\right )+2 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )+b^2}{8 x^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 189, normalized size = 1.12 \[ -\frac {2 \, b^{2} d^{2} x^{4} \sin \left (2 \, c\right ) \operatorname {Si}\left (2 \, d x^{2}\right ) + 2 \, a b d^{2} x^{4} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) + 2 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} + a^{2} + b^{2} - {\left (b^{2} d^{2} x^{4} \operatorname {Ci}\left (2 \, d x^{2}\right ) + b^{2} d^{2} x^{4} \operatorname {Ci}\left (-2 \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + 2 \, {\left (b^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b\right )} \sin \left (d x^{2} + c\right ) + {\left (a b d^{2} x^{4} \operatorname {Ci}\left (d x^{2}\right ) + a b d^{2} x^{4} \operatorname {Ci}\left (-d x^{2}\right )\right )} \sin \relax (c)}{4 \, x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.43, size = 448, normalized size = 2.65 \[ \frac {4 \, {\left (d x^{2} + c\right )}^{2} b^{2} d^{3} \cos \left (2 \, c\right ) \operatorname {Ci}\left (2 \, d x^{2}\right ) - 8 \, {\left (d x^{2} + c\right )} b^{2} c d^{3} \cos \left (2 \, c\right ) \operatorname {Ci}\left (2 \, d x^{2}\right ) + 4 \, b^{2} c^{2} d^{3} \cos \left (2 \, c\right ) \operatorname {Ci}\left (2 \, d x^{2}\right ) - 4 \, {\left (d x^{2} + c\right )}^{2} a b d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \relax (c) + 8 \, {\left (d x^{2} + c\right )} a b c d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \relax (c) - 4 \, a b c^{2} d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \relax (c) - 4 \, {\left (d x^{2} + c\right )}^{2} a b d^{3} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) + 8 \, {\left (d x^{2} + c\right )} a b c d^{3} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) - 4 \, a b c^{2} d^{3} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) + 4 \, {\left (d x^{2} + c\right )}^{2} b^{2} d^{3} \sin \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) - 8 \, {\left (d x^{2} + c\right )} b^{2} c d^{3} \sin \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) + 4 \, b^{2} c^{2} d^{3} \sin \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) - 4 \, {\left (d x^{2} + c\right )} a b d^{3} \cos \left (d x^{2} + c\right ) + 4 \, a b c d^{3} \cos \left (d x^{2} + c\right ) - 2 \, {\left (d x^{2} + c\right )} b^{2} d^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) + 2 \, b^{2} c d^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) + b^{2} d^{3} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 4 \, a b d^{3} \sin \left (d x^{2} + c\right ) - 2 \, a^{2} d^{3} - b^{2} d^{3}}{8 \, {\left ({\left (d x^{2} + c\right )}^{2} - 2 \, {\left (d x^{2} + c\right )} c + c^{2}\right )} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.69, size = 255, normalized size = 1.51 \[ \frac {\pi \,\mathrm {csgn}\left (d \,x^{2}\right ) {\mathrm e}^{-i c} a b \,d^{2}}{4}-\frac {\Si \left (d \,x^{2}\right ) {\mathrm e}^{-i c} a b \,d^{2}}{2}+\frac {i \Ei \left (1, -i d \,x^{2}\right ) {\mathrm e}^{-i c} a b \,d^{2}}{4}-\frac {a^{2}}{4 x^{4}}-\frac {b^{2}}{8 x^{4}}+\frac {i \pi \,\mathrm {csgn}\left (d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d^{2}}{4}-\frac {i \Si \left (2 d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d^{2}}{2}-\frac {\Ei \left (1, -2 i d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d^{2}}{4}-\frac {b^{2} d^{2} \Ei \left (1, -2 i d \,x^{2}\right ) {\mathrm e}^{2 i c}}{4}-\frac {i a b \,d^{2} \Ei \left (1, -i d \,x^{2}\right ) {\mathrm e}^{i c}}{4}-\frac {a b d \cos \left (d \,x^{2}+c \right )}{2 x^{2}}-\frac {a b \sin \left (d \,x^{2}+c \right )}{2 x^{4}}+\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{8 x^{4}}-\frac {b^{2} d \sin \left (2 d \,x^{2}+2 c \right )}{4 x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.54, size = 129, normalized size = 0.76 \[ \frac {1}{2} \, {\left ({\left (i \, \Gamma \left (-2, i \, d x^{2}\right ) - i \, \Gamma \left (-2, -i \, d x^{2}\right )\right )} \cos \relax (c) + {\left (\Gamma \left (-2, i \, d x^{2}\right ) + \Gamma \left (-2, -i \, d x^{2}\right )\right )} \sin \relax (c)\right )} a b d^{2} - \frac {{\left ({\left (4 \, {\left (\Gamma \left (-2, 2 i \, d x^{2}\right ) + \Gamma \left (-2, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) - {\left (4 i \, \Gamma \left (-2, 2 i \, d x^{2}\right ) - 4 i \, \Gamma \left (-2, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d^{2} x^{4} + 1\right )} b^{2}}{8 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \]
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 \[ \int \frac {{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2}{x^5} \,d x \]
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 \[ \int \frac {\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}{x^{5}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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