Optimal. Leaf size=97 \[ \frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {2}{3} b^3 \text {Ci}\left (\frac {2 b}{x}\right ) \sin (2 a)+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {2}{3} b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right ) \]
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Rubi [A]
time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3506, 3461,
3378, 3384, 3380, 3383} \begin {gather*} \frac {2}{3} b^3 \sin (2 a) \text {CosIntegral}\left (\frac {2 b}{x}\right )+\frac {2}{3} b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {x^3}{6} \end {gather*}
Antiderivative was successfully verified.
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Rule 3378
Rule 3380
Rule 3383
Rule 3384
Rule 3461
Rule 3506
Rubi steps
\begin {align*} \int x^2 \sin ^2\left (a+\frac {b}{x}\right ) \, dx &=\int \left (\frac {x^2}{2}-\frac {1}{2} x^2 \cos \left (2 a+\frac {2 b}{x}\right )\right ) \, dx\\ &=\frac {x^3}{6}-\frac {1}{2} \int x^2 \cos \left (2 a+\frac {2 b}{x}\right ) \, dx\\ &=\frac {x^3}{6}+\frac {1}{2} \text {Subst}\left (\int \frac {\cos (2 a+2 b x)}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{3} b \text {Subst}\left (\int \frac {\sin (2 a+2 b x)}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{3} b^2 \text {Subst}\left (\int \frac {\cos (2 a+2 b x)}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{3} \left (2 b^3\right ) \text {Subst}\left (\int \frac {\sin (2 a+2 b x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {1}{3} \left (2 b^3 \cos (2 a)\right ) \text {Subst}\left (\int \frac {\sin (2 b x)}{x} \, dx,x,\frac {1}{x}\right )+\frac {1}{3} \left (2 b^3 \sin (2 a)\right ) \text {Subst}\left (\int \frac {\cos (2 b x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{6}+\frac {1}{3} b^2 x \cos \left (2 \left (a+\frac {b}{x}\right )\right )-\frac {1}{6} x^3 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {2}{3} b^3 \text {Ci}\left (\frac {2 b}{x}\right ) \sin (2 a)+\frac {1}{6} b x^2 \sin \left (2 \left (a+\frac {b}{x}\right )\right )+\frac {2}{3} b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 86, normalized size = 0.89 \begin {gather*} \frac {1}{6} \left (4 b^3 \text {Ci}\left (\frac {2 b}{x}\right ) \sin (2 a)+x \left (x^2+2 b^2 \cos \left (2 \left (a+\frac {b}{x}\right )\right )-x^2 \cos \left (2 \left (a+\frac {b}{x}\right )\right )+b x \sin \left (2 \left (a+\frac {b}{x}\right )\right )\right )+4 b^3 \cos (2 a) \text {Si}\left (\frac {2 b}{x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 96, normalized size = 0.99
method | result | size |
derivativedivides | \(-b^{3} \left (-\frac {x^{3}}{6 b^{3}}+\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x^{3}}{6 b^{3}}-\frac {\sin \left (2 a +\frac {2 b}{x}\right ) x^{2}}{6 b^{2}}-\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x}{3 b}-\frac {2 \sinIntegral \left (\frac {2 b}{x}\right ) \cos \left (2 a \right )}{3}-\frac {2 \cosineIntegral \left (\frac {2 b}{x}\right ) \sin \left (2 a \right )}{3}\right )\) | \(96\) |
default | \(-b^{3} \left (-\frac {x^{3}}{6 b^{3}}+\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x^{3}}{6 b^{3}}-\frac {\sin \left (2 a +\frac {2 b}{x}\right ) x^{2}}{6 b^{2}}-\frac {\cos \left (2 a +\frac {2 b}{x}\right ) x}{3 b}-\frac {2 \sinIntegral \left (\frac {2 b}{x}\right ) \cos \left (2 a \right )}{3}-\frac {2 \cosineIntegral \left (\frac {2 b}{x}\right ) \sin \left (2 a \right )}{3}\right )\) | \(96\) |
risch | \(-\frac {\pi \,\mathrm {csgn}\left (\frac {b}{x}\right ) {\mathrm e}^{-2 i a} b^{3}}{3}+\frac {2 \sinIntegral \left (\frac {2 b}{x}\right ) {\mathrm e}^{-2 i a} b^{3}}{3}-\frac {i \expIntegral \left (1, -\frac {2 i b}{x}\right ) {\mathrm e}^{-2 i a} b^{3}}{3}+\frac {i b^{3} \expIntegral \left (1, -\frac {2 i b}{x}\right ) {\mathrm e}^{2 i a}}{3}+\frac {x^{3}}{6}+\frac {\cos \left (\frac {2 a x +2 b}{x}\right ) b^{2} x}{3}-\frac {\cos \left (\frac {2 a x +2 b}{x}\right ) x^{3}}{6}+\frac {x^{2} b \sin \left (\frac {2 a x +2 b}{x}\right )}{6}\) | \(131\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains complex when optimal does not.
time = 0.36, size = 99, normalized size = 1.02 \begin {gather*} -\frac {1}{3} \, {\left ({\left (i \, {\rm Ei}\left (\frac {2 i \, b}{x}\right ) - i \, {\rm Ei}\left (-\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) - {\left ({\rm Ei}\left (\frac {2 i \, b}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} b^{3} + \frac {1}{6} \, b x^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {1}{6} \, x^{3} + \frac {1}{6} \, {\left (2 \, b^{2} x - x^{3}\right )} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 109, normalized size = 1.12 \begin {gather*} \frac {1}{3} \, b x^{2} \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right ) + \frac {2}{3} \, b^{3} \cos \left (2 \, a\right ) \operatorname {Si}\left (\frac {2 \, b}{x}\right ) - \frac {1}{3} \, b^{2} x + \frac {1}{3} \, x^{3} + \frac {1}{3} \, {\left (2 \, b^{2} x - x^{3}\right )} \cos \left (\frac {a x + b}{x}\right )^{2} + \frac {1}{3} \, {\left (b^{3} \operatorname {Ci}\left (\frac {2 \, b}{x}\right ) + b^{3} \operatorname {Ci}\left (-\frac {2 \, b}{x}\right )\right )} \sin \left (2 \, a\right ) \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 x^{2} \sin ^{2}{\left (a + \frac {b}{x} \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 442 vs.
\(2 (88) = 176\).
time = 4.93, size = 442, normalized size = 4.56 \begin {gather*} \frac {4 \, a^{3} b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right ) - 4 \, a^{3} b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a^{2} b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right )}{x} + \frac {12 \, {\left (a x + b\right )} a^{2} b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - 2 \, a^{2} b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {12 \, {\left (a x + b\right )}^{2} a b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right )}{x^{2}} - \frac {12 \, {\left (a x + b\right )}^{2} a b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} + \frac {4 \, {\left (a x + b\right )} a b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - \frac {4 \, {\left (a x + b\right )}^{3} b^{4} \operatorname {Ci}\left (-2 \, a + \frac {2 \, {\left (a x + b\right )}}{x}\right ) \sin \left (2 \, a\right )}{x^{3}} + a b^{4} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {4 \, {\left (a x + b\right )}^{3} b^{4} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, a - \frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{3}} + b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {2 \, {\left (a x + b\right )}^{2} b^{4} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {{\left (a x + b\right )} b^{4} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - b^{4}}{6 \, {\left (a^{3} - \frac {3 \, {\left (a x + b\right )} a^{2}}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a}{x^{2}} - \frac {{\left (a x + b\right )}^{3}}{x^{3}}\right )} b} \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 = -0.01 \begin {gather*} \int x^2\,{\sin \left (a+\frac {b}{x}\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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