Optimal. Leaf size=107 \[ \frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{8 b^4}-\frac {3 \sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{4 b^3 x}-\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{4 b^2 x^2}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^3}+\frac {3}{8 b^2 x^2}-\frac {1}{8 x^4} \]
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Rubi [A] time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3379, 3311, 30, 3310} \[ -\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{4 b^2 x^2}+\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{8 b^4}-\frac {3 \sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{4 b^3 x}+\frac {\sin \left (a+\frac {b}{x}\right ) \cos \left (a+\frac {b}{x}\right )}{2 b x^3}+\frac {3}{8 b^2 x^2}-\frac {1}{8 x^4} \]
Antiderivative was successfully verified.
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Rule 30
Rule 3310
Rule 3311
Rule 3379
Rubi steps
\begin {align*} \int \frac {\sin ^2\left (a+\frac {b}{x}\right )}{x^5} \, dx &=-\operatorname {Subst}\left (\int x^3 \sin ^2(a+b x) \, dx,x,\frac {1}{x}\right )\\ &=\frac {\cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{2 b x^3}-\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{4 b^2 x^2}-\frac {1}{2} \operatorname {Subst}\left (\int x^3 \, dx,x,\frac {1}{x}\right )+\frac {3 \operatorname {Subst}\left (\int x \sin ^2(a+b x) \, dx,x,\frac {1}{x}\right )}{2 b^2}\\ &=-\frac {1}{8 x^4}+\frac {\cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{2 b x^3}-\frac {3 \cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{4 b^3 x}+\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{8 b^4}-\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{4 b^2 x^2}+\frac {3 \operatorname {Subst}\left (\int x \, dx,x,\frac {1}{x}\right )}{4 b^2}\\ &=-\frac {1}{8 x^4}+\frac {3}{8 b^2 x^2}+\frac {\cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{2 b x^3}-\frac {3 \cos \left (a+\frac {b}{x}\right ) \sin \left (a+\frac {b}{x}\right )}{4 b^3 x}+\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{8 b^4}-\frac {3 \sin ^2\left (a+\frac {b}{x}\right )}{4 b^2 x^2}\\ \end {align*}
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Mathematica [A] time = 0.20, size = 65, normalized size = 0.61 \[ -\frac {3 \left (x^4-2 b^2 x^2\right ) \cos \left (2 \left (a+\frac {b}{x}\right )\right )+2 b \left (\left (3 x^3-2 b^2 x\right ) \sin \left (2 \left (a+\frac {b}{x}\right )\right )+b^3\right )}{16 b^4 x^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 90, normalized size = 0.84 \[ -\frac {2 \, b^{4} + 6 \, b^{2} x^{2} - 3 \, x^{4} - 6 \, {\left (2 \, b^{2} x^{2} - x^{4}\right )} \cos \left (\frac {a x + b}{x}\right )^{2} - 4 \, {\left (2 \, b^{3} x - 3 \, b x^{3}\right )} \cos \left (\frac {a x + b}{x}\right ) \sin \left (\frac {a x + b}{x}\right )}{16 \, b^{4} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.69, size = 255, normalized size = 2.38 \[ -\frac {4 \, a^{3} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {8 \, {\left (a x + b\right )} a^{3}}{x} - 6 \, a^{2} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) - \frac {12 \, {\left (a x + b\right )} a^{2} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {12 \, {\left (a x + b\right )}^{2} a^{2}}{x^{2}} + \frac {12 \, {\left (a x + b\right )} a \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} - 6 \, a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right ) + \frac {12 \, {\left (a x + b\right )}^{2} a \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {8 \, {\left (a x + b\right )}^{3} a}{x^{3}} - \frac {6 \, {\left (a x + b\right )}^{2} \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x^{3}} + \frac {6 \, {\left (a x + b\right )} \sin \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{x} + \frac {2 \, {\left (a x + b\right )}^{4}}{x^{4}} + 3 \, \cos \left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}{16 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 334, normalized size = 3.12 \[ -\frac {\left (a +\frac {b}{x}\right )^{3} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {3 \left (a +\frac {b}{x}\right )^{2} \left (\cos ^{2}\left (a +\frac {b}{x}\right )\right )}{4}+\frac {3 \left (a +\frac {b}{x}\right ) \left (\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {b}{2 x}+\frac {a}{2}\right )}{2}-\frac {3 \left (a +\frac {b}{x}\right )^{2}}{8}-\frac {3 \left (\sin ^{2}\left (a +\frac {b}{x}\right )\right )}{8}-\frac {3 \left (a +\frac {b}{x}\right )^{4}}{8}-3 a \left (\left (a +\frac {b}{x}\right )^{2} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right ) \left (\cos ^{2}\left (a +\frac {b}{x}\right )\right )}{2}+\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{4}+\frac {b}{4 x}+\frac {a}{4}-\frac {\left (a +\frac {b}{x}\right )^{3}}{3}\right )+3 a^{2} \left (\left (a +\frac {b}{x}\right ) \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )-\frac {\left (a +\frac {b}{x}\right )^{2}}{4}+\frac {\left (\sin ^{2}\left (a +\frac {b}{x}\right )\right )}{4}\right )-a^{3} \left (-\frac {\cos \left (a +\frac {b}{x}\right ) \sin \left (a +\frac {b}{x}\right )}{2}+\frac {a}{2}+\frac {b}{2 x}\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.37, size = 68, normalized size = 0.64 \[ -\frac {{\left ({\left (\Gamma \left (4, \frac {2 i \, b}{x}\right ) + \Gamma \left (4, -\frac {2 i \, b}{x}\right )\right )} \cos \left (2 \, a\right ) - {\left (i \, \Gamma \left (4, \frac {2 i \, b}{x}\right ) - i \, \Gamma \left (4, -\frac {2 i \, b}{x}\right )\right )} \sin \left (2 \, a\right )\right )} x^{4} + 8 \, b^{4}}{64 \, b^{4} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.72, size = 84, normalized size = 0.79 \[ -\frac {3\,\cos \left (2\,a+\frac {2\,b}{x}\right )}{16\,b^4}-\frac {\frac {b^4}{8}-\frac {3\,b^2\,x^2\,\cos \left (2\,a+\frac {2\,b}{x}\right )}{8}+\frac {3\,b\,x^3\,\sin \left (2\,a+\frac {2\,b}{x}\right )}{8}-\frac {b^3\,x\,\sin \left (2\,a+\frac {2\,b}{x}\right )}{4}}{b^4\,x^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 8.21, size = 726, normalized size = 6.79 \[ \begin {cases} - \frac {b^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {2 b^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {b^{4}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {8 b^{3} x \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {8 b^{3} x \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {3 b^{2} x^{2} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {18 b^{2} x^{2} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {3 b^{2} x^{2}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {12 b x^{3} \tan ^{3}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} - \frac {12 b x^{3} \tan {\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} + \frac {12 x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )}}{8 b^{4} x^{4} \tan ^{4}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 16 b^{4} x^{4} \tan ^{2}{\left (\frac {a}{2} + \frac {b}{2 x} \right )} + 8 b^{4} x^{4}} & \text {for}\: b \neq 0 \\- \frac {\sin ^{2}{\relax (a )}}{4 x^{4}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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