\(\int x^3 \sin ^3(x) \, dx\) [484]

Optimal result

Integrand size = 8, antiderivative size = 73 \[ \int x^3 \sin ^3(x) \, dx=\frac {40}{9} x \cos (x)-\frac {2}{3} x^3 \cos (x)-\frac {40 \sin (x)}{9}+2 x^2 \sin (x)+\frac {2}{9} x \cos (x) \sin ^2(x)-\frac {1}{3} x^3 \cos (x) \sin ^2(x)-\frac {2 \sin ^3(x)}{27}+\frac {1}{3} x^2 \sin ^3(x) \]

[Out]

40/9*x*cos(x)-2/3*x^3*cos(x)-40/9*sin(x)+2*x^2*sin(x)+2/9*x*cos(x)*sin(x)^2-1/3*x^3*cos(x)*sin(x)^2-2/27*sin(x
)^3+1/3*x^2*sin(x)^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3392, 3377, 2717, 3391} \[ \int x^3 \sin ^3(x) \, dx=-\frac {2}{3} x^3 \cos (x)-\frac {1}{3} x^3 \sin ^2(x) \cos (x)+\frac {1}{3} x^2 \sin ^3(x)+2 x^2 \sin (x)-\frac {2 \sin ^3(x)}{27}-\frac {40 \sin (x)}{9}+\frac {40}{9} x \cos (x)+\frac {2}{9} x \sin ^2(x) \cos (x) \]

[In]

Int[x^3*Sin[x]^3,x]

[Out]

(40*x*Cos[x])/9 - (2*x^3*Cos[x])/3 - (40*Sin[x])/9 + 2*x^2*Sin[x] + (2*x*Cos[x]*Sin[x]^2)/9 - (x^3*Cos[x]*Sin[
x]^2)/3 - (2*Sin[x]^3)/27 + (x^2*Sin[x]^3)/3

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3} x^3 \cos (x) \sin ^2(x)+\frac {1}{3} x^2 \sin ^3(x)+\frac {2}{3} \int x^3 \sin (x) \, dx-\frac {2}{3} \int x \sin ^3(x) \, dx \\ & = -\frac {2}{3} x^3 \cos (x)+\frac {2}{9} x \cos (x) \sin ^2(x)-\frac {1}{3} x^3 \cos (x) \sin ^2(x)-\frac {2 \sin ^3(x)}{27}+\frac {1}{3} x^2 \sin ^3(x)-\frac {4}{9} \int x \sin (x) \, dx+2 \int x^2 \cos (x) \, dx \\ & = \frac {4}{9} x \cos (x)-\frac {2}{3} x^3 \cos (x)+2 x^2 \sin (x)+\frac {2}{9} x \cos (x) \sin ^2(x)-\frac {1}{3} x^3 \cos (x) \sin ^2(x)-\frac {2 \sin ^3(x)}{27}+\frac {1}{3} x^2 \sin ^3(x)-\frac {4}{9} \int \cos (x) \, dx-4 \int x \sin (x) \, dx \\ & = \frac {40}{9} x \cos (x)-\frac {2}{3} x^3 \cos (x)-\frac {4 \sin (x)}{9}+2 x^2 \sin (x)+\frac {2}{9} x \cos (x) \sin ^2(x)-\frac {1}{3} x^3 \cos (x) \sin ^2(x)-\frac {2 \sin ^3(x)}{27}+\frac {1}{3} x^2 \sin ^3(x)-4 \int \cos (x) \, dx \\ & = \frac {40}{9} x \cos (x)-\frac {2}{3} x^3 \cos (x)-\frac {40 \sin (x)}{9}+2 x^2 \sin (x)+\frac {2}{9} x \cos (x) \sin ^2(x)-\frac {1}{3} x^3 \cos (x) \sin ^2(x)-\frac {2 \sin ^3(x)}{27}+\frac {1}{3} x^2 \sin ^3(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70 \[ \int x^3 \sin ^3(x) \, dx=\frac {1}{108} \left (-81 x \left (-6+x^2\right ) \cos (x)+3 x \left (-2+3 x^2\right ) \cos (3 x)+243 \left (-2+x^2\right ) \sin (x)-\left (-2+9 x^2\right ) \sin (3 x)\right ) \]

[In]

Integrate[x^3*Sin[x]^3,x]

[Out]

(-81*x*(-6 + x^2)*Cos[x] + 3*x*(-2 + 3*x^2)*Cos[3*x] + 243*(-2 + x^2)*Sin[x] - (-2 + 9*x^2)*Sin[3*x])/108

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68

method	result	size
risch	\(\left (-\frac {3}{4} x^{3}+\frac {9}{2} x \right ) \cos \left (x \right )+\frac {9 \left (x^{2}-2\right ) \sin \left (x \right )}{4}+\left (\frac {1}{12} x^{3}-\frac {1}{18} x \right ) \cos \left (3 x \right )-\frac {\left (9 x^{2}-2\right ) \sin \left (3 x \right )}{108}\)	\(50\)
default	\(-\frac {x^{3} \left (2+\sin ^{2}\left (x \right )\right ) \cos \left (x \right )}{3}+2 x^{2} \sin \left (x \right )-\frac {40 \sin \left (x \right )}{9}+4 x \cos \left (x \right )+\frac {x^{2} \left (\sin ^{3}\left (x \right )\right )}{3}+\frac {2 x \left (2+\sin ^{2}\left (x \right )\right ) \cos \left (x \right )}{9}-\frac {2 \left (\sin ^{3}\left (x \right )\right )}{27}\)	\(57\)
norman	\(\frac {\frac {40 x}{9}-\frac {2 x^{3}}{3}-\frac {496 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{27}-\frac {80 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{9}+\frac {16 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{3}-\frac {16 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{3}-\frac {40 x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{9}+4 x^{2} \tan \left (\frac {x}{2}\right )-2 x^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 x^{3} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+\frac {2 x^{3} \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{3}+\frac {32 \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) x^{2}}{3}+4 \left (\tan ^{5}\left (\frac {x}{2}\right )\right ) x^{2}-\frac {80 \tan \left (\frac {x}{2}\right )}{9}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{3}}\)	\(134\)

[In]

int(x^3*sin(x)^3,x,method=_RETURNVERBOSE)

[Out]

(-3/4*x^3+9/2*x)*cos(x)+9/4*(x^2-2)*sin(x)+(1/12*x^3-1/18*x)*cos(3*x)-1/108*(9*x^2-2)*sin(3*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int x^3 \sin ^3(x) \, dx=\frac {1}{9} \, {\left (3 \, x^{3} - 2 \, x\right )} \cos \left (x\right )^{3} - \frac {1}{3} \, {\left (3 \, x^{3} - 14 \, x\right )} \cos \left (x\right ) - \frac {1}{27} \, {\left ({\left (9 \, x^{2} - 2\right )} \cos \left (x\right )^{2} - 63 \, x^{2} + 122\right )} \sin \left (x\right ) \]

[In]

integrate(x^3*sin(x)^3,x, algorithm="fricas")

[Out]

1/9*(3*x^3 - 2*x)*cos(x)^3 - 1/3*(3*x^3 - 14*x)*cos(x) - 1/27*((9*x^2 - 2)*cos(x)^2 - 63*x^2 + 122)*sin(x)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.26 \[ \int x^3 \sin ^3(x) \, dx=- x^{3} \sin ^{2}{\left (x \right )} \cos {\left (x \right )} - \frac {2 x^{3} \cos ^{3}{\left (x \right )}}{3} + \frac {7 x^{2} \sin ^{3}{\left (x \right )}}{3} + 2 x^{2} \sin {\left (x \right )} \cos ^{2}{\left (x \right )} + \frac {14 x \sin ^{2}{\left (x \right )} \cos {\left (x \right )}}{3} + \frac {40 x \cos ^{3}{\left (x \right )}}{9} - \frac {122 \sin ^{3}{\left (x \right )}}{27} - \frac {40 \sin {\left (x \right )} \cos ^{2}{\left (x \right )}}{9} \]

[In]

integrate(x**3*sin(x)**3,x)

[Out]

-x**3*sin(x)**2*cos(x) - 2*x**3*cos(x)**3/3 + 7*x**2*sin(x)**3/3 + 2*x**2*sin(x)*cos(x)**2 + 14*x*sin(x)**2*co
s(x)/3 + 40*x*cos(x)**3/9 - 122*sin(x)**3/27 - 40*sin(x)*cos(x)**2/9

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int x^3 \sin ^3(x) \, dx=\frac {1}{36} \, {\left (3 \, x^{3} - 2 \, x\right )} \cos \left (3 \, x\right ) - \frac {3}{4} \, {\left (x^{3} - 6 \, x\right )} \cos \left (x\right ) - \frac {1}{108} \, {\left (9 \, x^{2} - 2\right )} \sin \left (3 \, x\right ) + \frac {9}{4} \, {\left (x^{2} - 2\right )} \sin \left (x\right ) \]

[In]

integrate(x^3*sin(x)^3,x, algorithm="maxima")

[Out]

1/36*(3*x^3 - 2*x)*cos(3*x) - 3/4*(x^3 - 6*x)*cos(x) - 1/108*(9*x^2 - 2)*sin(3*x) + 9/4*(x^2 - 2)*sin(x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int x^3 \sin ^3(x) \, dx=\frac {1}{36} \, {\left (3 \, x^{3} - 2 \, x\right )} \cos \left (3 \, x\right ) - \frac {3}{4} \, {\left (x^{3} - 6 \, x\right )} \cos \left (x\right ) - \frac {1}{108} \, {\left (9 \, x^{2} - 2\right )} \sin \left (3 \, x\right ) + \frac {9}{4} \, {\left (x^{2} - 2\right )} \sin \left (x\right ) \]

[In]

integrate(x^3*sin(x)^3,x, algorithm="giac")

[Out]

1/36*(3*x^3 - 2*x)*cos(3*x) - 3/4*(x^3 - 6*x)*cos(x) - 1/108*(9*x^2 - 2)*sin(3*x) + 9/4*(x^2 - 2)*sin(x)

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int x^3 \sin ^3(x) \, dx=\frac {7\,x^2\,\sin \left (x\right )}{3}-\frac {2\,x\,{\cos \left (x\right )}^3}{9}-x^3\,\cos \left (x\right )-\frac {122\,\sin \left (x\right )}{27}+\frac {x^3\,{\cos \left (x\right )}^3}{3}+\frac {2\,{\cos \left (x\right )}^2\,\sin \left (x\right )}{27}+\frac {14\,x\,\cos \left (x\right )}{3}-\frac {x^2\,{\cos \left (x\right )}^2\,\sin \left (x\right )}{3} \]

[In]

int(x^3*sin(x)^3,x)

[Out]

(7*x^2*sin(x))/3 - (2*x*cos(x)^3)/9 - x^3*cos(x) - (122*sin(x))/27 + (x^3*cos(x)^3)/3 + (2*cos(x)^2*sin(x))/27
+ (14*x*cos(x))/3 - (x^2*cos(x)^2*sin(x))/3