Optimal. Leaf size=85 \[ -\frac{2 \cos ^3(k x)}{27 k^3}+\frac{14 \cos (k x)}{9 k^3}+\frac{2 x \sin ^3(k x)}{9 k^2}+\frac{4 x \sin (k x)}{3 k^2}-\frac{2 x^2 \cos (k x)}{3 k}-\frac{x^2 \sin ^2(k x) \cos (k x)}{3 k} \]
[Out]
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Rubi [A] time = 0.103559, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4 \[ -\frac{2 \cos ^3(k x)}{27 k^3}+\frac{14 \cos (k x)}{9 k^3}+\frac{2 x \sin ^3(k x)}{9 k^2}+\frac{4 x \sin (k x)}{3 k^2}-\frac{2 x^2 \cos (k x)}{3 k}-\frac{x^2 \sin ^2(k x) \cos (k x)}{3 k} \]
Antiderivative was successfully verified.
[In] Int[x^2*Sin[k*x]^3,x]
[Out]
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Rubi in Sympy [A] time = 4.89368, size = 85, normalized size = 1. \[ - \frac{x^{2} \sin ^{2}{\left (k x \right )} \cos{\left (k x \right )}}{3 k} - \frac{2 x^{2} \cos{\left (k x \right )}}{3 k} + \frac{2 x \sin ^{3}{\left (k x \right )}}{9 k^{2}} + \frac{4 x \sin{\left (k x \right )}}{3 k^{2}} - \frac{2 \cos ^{3}{\left (k x \right )}}{27 k^{3}} + \frac{14 \cos{\left (k x \right )}}{9 k^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate(x**2*sin(k*x)**3,x)
[Out]
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Mathematica [A] time = 0.0852879, size = 55, normalized size = 0.65 \[ \frac{-81 \left (k^2 x^2-2\right ) \cos (k x)+\left (9 k^2 x^2-2\right ) \cos (3 k x)-6 k x (\sin (3 k x)-27 \sin (k x))}{108 k^3} \]
Antiderivative was successfully verified.
[In] Integrate[x^2*Sin[k*x]^3,x]
[Out]
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Maple [A] time = 0.019, size = 64, normalized size = 0.8 \[{\frac{1}{{k}^{3}} \left ( -{\frac{{k}^{2}{x}^{2} \left ( 2+ \left ( \sin \left ( kx \right ) \right ) ^{2} \right ) \cos \left ( kx \right ) }{3}}+{\frac{4\,\cos \left ( kx \right ) }{3}}+{\frac{4\,kx\sin \left ( kx \right ) }{3}}+{\frac{2\,kx \left ( \sin \left ( kx \right ) \right ) ^{3}}{9}}+{\frac{ \left ( 4+2\, \left ( \sin \left ( kx \right ) \right ) ^{2} \right ) \cos \left ( kx \right ) }{27}} \right ) } \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int(x^2*sin(k*x)^3,x)
[Out]
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Maxima [A] time = 1.42707, size = 74, normalized size = 0.87 \[ -\frac{6 \, k x \sin \left (3 \, k x\right ) - 162 \, k x \sin \left (k x\right ) -{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (3 \, k x\right ) + 81 \,{\left (k^{2} x^{2} - 2\right )} \cos \left (k x\right )}{108 \, k^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(x^2*sin(k*x)^3,x, algorithm="maxima")
[Out]
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Fricas [A] time = 0.236567, size = 80, normalized size = 0.94 \[ \frac{{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (k x\right )^{3} - 3 \,{\left (9 \, k^{2} x^{2} - 14\right )} \cos \left (k x\right ) - 6 \,{\left (k x \cos \left (k x\right )^{2} - 7 \, k x\right )} \sin \left (k x\right )}{27 \, k^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(x^2*sin(k*x)^3,x, algorithm="fricas")
[Out]
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Sympy [A] time = 3.49676, size = 100, normalized size = 1.18 \[ \begin{cases} - \frac{x^{2} \sin ^{2}{\left (k x \right )} \cos{\left (k x \right )}}{k} - \frac{2 x^{2} \cos ^{3}{\left (k x \right )}}{3 k} + \frac{14 x \sin ^{3}{\left (k x \right )}}{9 k^{2}} + \frac{4 x \sin{\left (k x \right )} \cos ^{2}{\left (k x \right )}}{3 k^{2}} + \frac{14 \sin ^{2}{\left (k x \right )} \cos{\left (k x \right )}}{9 k^{3}} + \frac{40 \cos ^{3}{\left (k x \right )}}{27 k^{3}} & \text{for}\: k \neq 0 \\0 & \text{otherwise} \end{cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(x**2*sin(k*x)**3,x)
[Out]
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GIAC/XCAS [A] time = 0.198624, size = 81, normalized size = 0.95 \[ -\frac{x \sin \left (3 \, k x\right )}{18 \, k^{2}} + \frac{3 \, x \sin \left (k x\right )}{2 \, k^{2}} + \frac{{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (3 \, k x\right )}{108 \, k^{3}} - \frac{3 \,{\left (k^{2} x^{2} - 2\right )} \cos \left (k x\right )}{4 \, k^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(x^2*sin(k*x)^3,x, algorithm="giac")
[Out]