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

Optimal. Leaf size=33 \[ \frac{\cos ^3\left (a+b x^2\right )}{6 b}-\frac{\cos \left (a+b x^2\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0307496, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3379, 2633} \[ \frac{\cos ^3\left (a+b x^2\right )}{6 b}-\frac{\cos \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*x^2]^3,x]

[Out]

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

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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0281061, size = 33, normalized size = 1. \[ \frac{\cos \left (3 \left (a+b x^2\right )\right )}{24 b}-\frac{3 \cos \left (a+b x^2\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b*x^2]^3,x]

[Out]

(-3*Cos[a + b*x^2])/(8*b) + Cos[3*(a + b*x^2)]/(24*b)

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Maple [A]  time = 0.006, size = 26, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 2+ \left ( \sin \left ( b{x}^{2}+a \right ) \right ) ^{2} \right ) \cos \left ( b{x}^{2}+a \right ) }{6\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.963564, size = 36, normalized size = 1.09 \begin{align*} \frac{\cos \left (3 \, b x^{2} + 3 \, a\right ) - 9 \, \cos \left (b x^{2} + a\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(cos(3*b*x^2 + 3*a) - 9*cos(b*x^2 + a))/b

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Fricas [A]  time = 2.16631, size = 61, normalized size = 1.85 \begin{align*} \frac{\cos \left (b x^{2} + a\right )^{3} - 3 \, \cos \left (b x^{2} + a\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.03725, size = 46, normalized size = 1.39 \begin{align*} \begin{cases} - \frac{\sin ^{2}{\left (a + b x^{2} \right )} \cos{\left (a + b x^{2} \right )}}{2 b} - \frac{\cos ^{3}{\left (a + b x^{2} \right )}}{3 b} & \text{for}\: b \neq 0 \\\frac{x^{2} \sin ^{3}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10891, size = 35, normalized size = 1.06 \begin{align*} \frac{\cos \left (b x^{2} + a\right )^{3} - 3 \, \cos \left (b x^{2} + a\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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