3.24 \(\int x^3 \sin ^3(a+b x^2) \, dx\)

Optimal. Leaf size=79 \[ \frac {\sin ^3\left (a+b x^2\right )}{18 b^2}+\frac {\sin \left (a+b x^2\right )}{3 b^2}-\frac {x^2 \cos \left (a+b x^2\right )}{3 b}-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{6 b} \]

[Out]

-1/3*x^2*cos(b*x^2+a)/b+1/3*sin(b*x^2+a)/b^2-1/6*x^2*cos(b*x^2+a)*sin(b*x^2+a)^2/b+1/18*sin(b*x^2+a)^3/b^2

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Rubi [A]  time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3379, 3310, 3296, 2637} \[ \frac {\sin ^3\left (a+b x^2\right )}{18 b^2}+\frac {\sin \left (a+b x^2\right )}{3 b^2}-\frac {x^2 \cos \left (a+b x^2\right )}{3 b}-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sin[a + b*x^2]^3,x]

[Out]

-(x^2*Cos[a + b*x^2])/(3*b) + Sin[a + b*x^2]/(3*b^2) - (x^2*Cos[a + b*x^2]*Sin[a + b*x^2]^2)/(6*b) + Sin[a + b
*x^2]^3/(18*b^2)

Rule 2637

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

Rule 3296

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 3310

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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

\begin {align*} \int x^3 \sin ^3\left (a+b x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x \sin ^3(a+b x) \, dx,x,x^2\right )\\ &=-\frac {x^2 \cos \left (a+b x^2\right ) \sin ^2\left (a+b x^2\right )}{6 b}+\frac {\sin ^3\left (a+b x^2\right )}{18 b^2}+\frac {1}{3} \operatorname {Subst}\left (\int x \sin (a+b x) \, dx,x,x^2\right )\\ &=-\frac {x^2 \cos \left (a+b x^2\right )}{3 b}-\frac {x^2 \cos \left (a+b x^2\right ) \sin ^2\left (a+b x^2\right )}{6 b}+\frac {\sin ^3\left (a+b x^2\right )}{18 b^2}+\frac {\operatorname {Subst}\left (\int \cos (a+b x) \, dx,x,x^2\right )}{3 b}\\ &=-\frac {x^2 \cos \left (a+b x^2\right )}{3 b}+\frac {\sin \left (a+b x^2\right )}{3 b^2}-\frac {x^2 \cos \left (a+b x^2\right ) \sin ^2\left (a+b x^2\right )}{6 b}+\frac {\sin ^3\left (a+b x^2\right )}{18 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 58, normalized size = 0.73 \[ -\frac {-27 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )+27 b x^2 \cos \left (a+b x^2\right )-3 b x^2 \cos \left (3 \left (a+b x^2\right )\right )}{72 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sin[a + b*x^2]^3,x]

[Out]

-1/72*(27*b*x^2*Cos[a + b*x^2] - 3*b*x^2*Cos[3*(a + b*x^2)] - 27*Sin[a + b*x^2] + Sin[3*(a + b*x^2)])/b^2

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fricas [A]  time = 0.61, size = 58, normalized size = 0.73 \[ \frac {3 \, b x^{2} \cos \left (b x^{2} + a\right )^{3} - 9 \, b x^{2} \cos \left (b x^{2} + a\right ) - {\left (\cos \left (b x^{2} + a\right )^{2} - 7\right )} \sin \left (b x^{2} + a\right )}{18 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/18*(3*b*x^2*cos(b*x^2 + a)^3 - 9*b*x^2*cos(b*x^2 + a) - (cos(b*x^2 + a)^2 - 7)*sin(b*x^2 + a))/b^2

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giac [A]  time = 0.51, size = 60, normalized size = 0.76 \[ \frac {3 \, b x^{2} \cos \left (3 \, b x^{2} + 3 \, a\right ) - 27 \, b x^{2} \cos \left (b x^{2} + a\right ) - \sin \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, \sin \left (b x^{2} + a\right )}{72 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/72*(3*b*x^2*cos(3*b*x^2 + 3*a) - 27*b*x^2*cos(b*x^2 + a) - sin(3*b*x^2 + 3*a) + 27*sin(b*x^2 + a))/b^2

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maple [A]  time = 0.02, size = 66, normalized size = 0.84 \[ -\frac {3 x^{2} \cos \left (b \,x^{2}+a \right )}{8 b}+\frac {3 \sin \left (b \,x^{2}+a \right )}{8 b^{2}}+\frac {x^{2} \cos \left (3 b \,x^{2}+3 a \right )}{24 b}-\frac {\sin \left (3 b \,x^{2}+3 a \right )}{72 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sin(b*x^2+a)^3,x)

[Out]

-3/8*x^2*cos(b*x^2+a)/b+3/8*sin(b*x^2+a)/b^2+1/24/b*x^2*cos(3*b*x^2+3*a)-1/72/b^2*sin(3*b*x^2+3*a)

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maxima [A]  time = 0.34, size = 60, normalized size = 0.76 \[ \frac {3 \, b x^{2} \cos \left (3 \, b x^{2} + 3 \, a\right ) - 27 \, b x^{2} \cos \left (b x^{2} + a\right ) - \sin \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, \sin \left (b x^{2} + a\right )}{72 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/72*(3*b*x^2*cos(3*b*x^2 + 3*a) - 27*b*x^2*cos(b*x^2 + a) - sin(3*b*x^2 + 3*a) + 27*sin(b*x^2 + a))/b^2

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mupad [B]  time = 4.75, size = 66, normalized size = 0.84 \[ \frac {\frac {7\,\sin \left (b\,x^2+a\right )}{18}-\frac {{\cos \left (b\,x^2+a\right )}^2\,\sin \left (b\,x^2+a\right )}{18}+b\,\left (\frac {x^2\,{\cos \left (b\,x^2+a\right )}^3}{6}-\frac {x^2\,\cos \left (b\,x^2+a\right )}{2}\right )}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sin(a + b*x^2)^3,x)

[Out]

((7*sin(a + b*x^2))/18 - (cos(a + b*x^2)^2*sin(a + b*x^2))/18 + b*((x^2*cos(a + b*x^2)^3)/6 - (x^2*cos(a + b*x
^2))/2))/b^2

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sympy [A]  time = 3.53, size = 92, normalized size = 1.16 \[ \begin {cases} - \frac {x^{2} \sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{2 b} - \frac {x^{2} \cos ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {7 \sin ^{3}{\left (a + b x^{2} \right )}}{18 b^{2}} + \frac {\sin {\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sin ^{3}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sin(b*x**2+a)**3,x)

[Out]

Piecewise((-x**2*sin(a + b*x**2)**2*cos(a + b*x**2)/(2*b) - x**2*cos(a + b*x**2)**3/(3*b) + 7*sin(a + b*x**2)*
*3/(18*b**2) + sin(a + b*x**2)*cos(a + b*x**2)**2/(3*b**2), Ne(b, 0)), (x**4*sin(a)**3/4, True))

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