3.1.25 \(\int x \sin ^3(a+b x^2) \, dx\) [25]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 = {3460, 2713} \begin {gather*} \frac {\cos ^3\left (a+b x^2\right )}{6 b}-\frac {\cos \left (a+b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*x^2]^3,x]

[Out]

-1/2*Cos[a + b*x^2]/b + Cos[a + b*x^2]^3/(6*b)

Rule 2713

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]

Rule 3460

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 \sin ^3\left (a+b x^2\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int \sin ^3(a+b x) \, dx,x,x^2\right )\\ &=-\frac {\text {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*}

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Mathematica [A]
time = 0.02, size = 33, normalized size = 1.00 \begin {gather*} -\frac {3 \cos \left (a+b x^2\right )}{8 b}+\frac {\cos \left (3 \left (a+b x^2\right )\right )}{24 b} \end {gather*}

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 = 0.04, size = 26, normalized size = 0.79

method	result	size
derivativedivides	\(-\frac {\left (2+\sin ^{2}\left (b \,x^{2}+a \right )\right ) \cos \left (b \,x^{2}+a \right )}{6 b}\)	\(26\)
default	\(-\frac {\left (2+\sin ^{2}\left (b \,x^{2}+a \right )\right ) \cos \left (b \,x^{2}+a \right )}{6 b}\)	\(26\)
risch	\(-\frac {3 \cos \left (b \,x^{2}+a \right )}{8 b}+\frac {\cos \left (3 b \,x^{2}+3 a \right )}{24 b}\)	\(31\)
norman	\(\frac {-\frac {2 \left (\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{b}-\frac {2}{3 b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )^{3}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 27, normalized size = 0.82 \begin {gather*} \frac {\cos \left (3 \, b x^{2} + 3 \, a\right ) - 9 \, \cos \left (b x^{2} + a\right )}{24 \, b} \end {gather*}

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 = 0.37, size = 26, normalized size = 0.79 \begin {gather*} \frac {\cos \left (b x^{2} + a\right )^{3} - 3 \, \cos \left (b x^{2} + a\right )}{6 \, b} \end {gather*}

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 = 0.19, size = 46, normalized size = 1.39 \begin {gather*} \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 {gather*}

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 = 3.47, size = 26, normalized size = 0.79 \begin {gather*} \frac {\cos \left (b x^{2} + a\right )^{3} - 3 \, \cos \left (b x^{2} + a\right )}{6 \, b} \end {gather*}

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

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Mupad [B]
time = 4.67, size = 28, normalized size = 0.85 \begin {gather*} -\frac {3\,\cos \left (b\,x^2+a\right )-{\cos \left (b\,x^2+a\right )}^3}{6\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*x^2)^3,x)

[Out]

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

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