3.4.21 \(\int x \sqrt [3]{c \sin ^3(a+b x^2)} \, dx\) [321]

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]

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

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Rubi [A]
time = 0.06, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 = {6847, 3286, 2718} \begin {gather*} -\frac {\cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

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

Rule 2718

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

Rule 3286

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 6847

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]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 31, normalized size = 1.00 \begin {gather*} -\frac {\cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.22, size = 119, normalized size = 3.84

method	result	size
risch	\(-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{4 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}}}{4 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 16, normalized size = 0.52 \begin {gather*} \frac {c^{\frac {1}{3}} \cos \left (b x^{2} + a\right )}{4 \, b} \end {gather*}

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 = 0.38, size = 51, normalized size = 1.65 \begin {gather*} -\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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (27) = 54\).
time = 0.81, size = 63, normalized size = 2.03 \begin {gather*} \begin {cases} \frac {x^{2} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{2} & \text {for}\: b = 0 \\0 & \text {for}\: a = - b x^{2} \vee a = - b x^{2} + \pi \\- \frac {\sqrt [3]{c \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 {gather*}

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((x**2*(c*sin(a)**3)**(1/3)/2, Eq(b, 0)), (0, Eq(a, -b*x**2) | Eq(a, -b*x**2 + pi)), (-(c*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 4.82, size = 53, normalized size = 1.71 \begin {gather*} -\frac {\sin \left (2\,b\,x^2+2\,a\right )\,{\left (-2\,c\,\left (\sin \left (3\,b\,x^2+3\,a\right )-3\,\sin \left (b\,x^2+a\right )\right )\right )}^{1/3}}{8\,b\,{\sin \left (b\,x^2+a\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*sin(a + b*x^2)^3)^(1/3),x)

[Out]

-(sin(2*a + 2*b*x^2)*(-2*c*(sin(3*a + 3*b*x^2) - 3*sin(a + b*x^2)))^(1/3))/(8*b*sin(a + b*x^2)^2)

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