3.321 \(\int x \sqrt [3]{c \sin ^3(a+b x^2)} \, dx\)

Optimal. Leaf size=31 \[ -\frac{\cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \]

[Out]

-(Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(1/3))/(2*b)

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Rubi [A]  time = 0.104194, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6715, 3207, 2638} \[ -\frac{\cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

-(Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(1/3))/(2*b)

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2638

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

Rubi steps

\begin{align*} \int x \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \sqrt [3]{c \sin ^3(a+b x)} \, dx,x,x^2\right )\\ &=\frac{1}{2} \left (\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \operatorname{Subst}\left (\int \sin (a+b x) \, dx,x,x^2\right )\\ &=-\frac{\cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0517917, size = 31, normalized size = 1. \[ -\frac{\cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

-(Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(1/3))/(2*b)

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Maple [C]  time = 0.071, size = 119, normalized size = 3.8 \begin{align*}{\frac{-{\frac{i}{4}}{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}}{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) b}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}}}-{\frac{{\frac{i}{4}}}{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) b}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*sin(b*x^2+a)^3)^(1/3),x)

[Out]

-1/4*I*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(1/3)/(exp(2*I*(b*x^2+a))-1)/b*exp(2*I*(b*x^2+a))-1/
4*I*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(1/3)/(exp(2*I*(b*x^2+a))-1)/b

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Maxima [A]  time = 1.53615, size = 22, normalized size = 0.71 \begin{align*} \frac{c^{\frac{1}{3}} \cos \left (b x^{2} + a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*sin(b*x^2+a)^3)^(1/3),x, algorithm="maxima")

[Out]

1/4*c^(1/3)*cos(b*x^2 + a)/b

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Fricas [A]  time = 1.60993, size = 120, normalized size = 3.87 \begin{align*} -\frac{\left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac{1}{3}} \cos \left (b x^{2} + a\right )}{2 \, b \sin \left (b x^{2} + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*sin(b*x^2+a)^3)^(1/3),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 5.09722, size = 66, normalized size = 2.13 \begin{align*} \begin{cases} 0 & \text{for}\: a = - b x^{2} \vee a = - b x^{2} + \pi \\\frac{x^{2} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{2} & \text{for}\: b = 0 \\- \frac{\sqrt [3]{c} \sqrt [3]{\sin ^{3}{\left (a + b x^{2} \right )}} \cos{\left (a + b x^{2} \right )}}{2 b \sin{\left (a + b x^{2} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*sin(b*x**2+a)**3)**(1/3),x)

[Out]

Piecewise((0, Eq(a, -b*x**2) | Eq(a, -b*x**2 + pi)), (x**2*(c*sin(a)**3)**(1/3)/2, Eq(b, 0)), (-c**(1/3)*(sin(
a + b*x**2)**3)**(1/3)*cos(a + b*x**2)/(2*b*sin(a + b*x**2)), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac{1}{3}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*sin(b*x^2+a)^3)^(1/3),x, algorithm="giac")

[Out]

integrate((c*sin(b*x^2 + a)^3)^(1/3)*x, x)