3.4.19 \(\int x^3 \sqrt [3]{c \sin ^3(a+b x^2)} \, dx\) [319]

Optimal. Leaf size=58 \[ \frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^2}-\frac {x^2 \cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6852, 3460, 3377, 2717} \begin {gather*} \frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^2}-\frac {x^2 \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^3*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

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

Rule 2717

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

Rule 3377

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 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]))

Rule 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m)^FracPart[p]/v^(m*FracPart[p])), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int x^3 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx &=\left (\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int x^3 \sin \left (a+b x^2\right ) \, dx\\ &=\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 x \sin (a+b x) \, dx,x,x^2\right )\\ &=-\frac {x^2 \cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b}+\frac {\left (\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \text {Subst}\left (\int \cos (a+b x) \, dx,x,x^2\right )}{2 b}\\ &=\frac {\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^2}-\frac {x^2 \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 = 0.06, size = 38, normalized size = 0.66 \begin {gather*} -\frac {\left (-1+b x^2 \cot \left (a+b x^2\right )\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.16, size = 135, normalized size = 2.33

method	result	size
risch	\(-\frac {i \left (b \,x^{2}+i\right ) \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{4 b^{2} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} \left (b \,x^{2}-i\right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right ) b^{2}}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I/b^2*(b*x^2+I)/(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*(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*x^2-I)/b^2

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Maxima [A]
time = 0.53, size = 32, normalized size = 0.55 \begin {gather*} \frac {{\left (b x^{2} \cos \left (b x^{2} + a\right ) - \sin \left (b x^{2} + a\right )\right )} c^{\frac {1}{3}}}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2*cos(b*x^2 + a) - sin(b*x^2 + a))*c^(1/3)/b^2

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Fricas [A]
time = 0.35, size = 67, normalized size = 1.16 \begin {gather*} -\frac {{\left (b x^{2} \cos \left (b x^{2} + a\right ) - \sin \left (b x^{2} + a\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {1}{3}}}{2 \, b^{2} \sin \left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 2.37, size = 85, normalized size = 1.47 \begin {gather*} \begin {cases} \frac {x^{4} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{4} & \text {for}\: b = 0 \\0 & \text {for}\: a = - b x^{2} \vee a = - b x^{2} + \pi \\- \frac {x^{2} \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 )}} + \frac {\sqrt [3]{c \sin ^{3}{\left (a + b x^{2} \right )}}}{2 b^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*(c*sin(a)**3)**(1/3)/4, Eq(b, 0)), (0, Eq(a, -b*x**2) | Eq(a, -b*x**2 + pi)), (-x**2*(c*sin(a
+ b*x**2)**3)**(1/3)*cos(a + b*x**2)/(2*b*sin(a + b*x**2)) + (c*sin(a + b*x**2)**3)**(1/3)/(2*b**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^3*(c*sin(b*x^2+a)^3)^(1/3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.19, size = 71, normalized size = 1.22 \begin {gather*} \frac {\left (\frac {{\sin \left (b\,x^2+a\right )}^2}{4}-\frac {b\,x^2\,\sin \left (2\,b\,x^2+2\,a\right )}{8}\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}}{b^2\,{\sin \left (b\,x^2+a\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*sin(a + b*x^2)^3)^(1/3),x)

[Out]

((sin(a + b*x^2)^2/4 - (b*x^2*sin(2*a + 2*b*x^2))/8)*(-2*c*(sin(3*a + 3*b*x^2) - 3*sin(a + b*x^2)))^(1/3))/(b^
2*sin(a + b*x^2)^2)

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