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]

-(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)

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Rubi [A]  time = 0.073866, antiderivative size = 79, normalized size of antiderivative = 1., 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 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]))

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 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 2637

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

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*}

Mathematica [A]  time = 0.153363, 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]

-(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)])/(72*b^2)

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Maple [A]  time = 0.011, size = 66, normalized size = 0.8 \begin{align*} -{\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}}} \end{align*}

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 = 0.984789, size = 81, normalized size = 1.03 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 2.20764, size = 138, normalized size = 1.75 \begin{align*} \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}} \end{align*}

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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Sympy [A]  time = 4.04069, size = 92, normalized size = 1.16 \begin{align*} \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}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.12334, size = 81, normalized size = 1.03 \begin{align*} \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}} \end{align*}

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