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

Optimal. Leaf size=96 \[ -\frac {6 \sqrt [3]{c \sin ^3(a+b x)}}{b^4}+\frac {6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}+\frac {3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

[Out]

-6*(c*sin(b*x+a)^3)^(1/3)/b^4+3*x^2*(c*sin(b*x+a)^3)^(1/3)/b^2+6*x*cot(b*x+a)*(c*sin(b*x+a)^3)^(1/3)/b^3-x^3*c
ot(b*x+a)*(c*sin(b*x+a)^3)^(1/3)/b

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Rubi [A]  time = 0.21, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6720, 3296, 2637} \[ \frac {3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac {6 \sqrt [3]{c \sin ^3(a+b x)}}{b^4}+\frac {6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

(-6*(c*Sin[a + b*x]^3)^(1/3))/b^4 + (3*x^2*(c*Sin[a + b*x]^3)^(1/3))/b^2 + (6*x*Cot[a + b*x]*(c*Sin[a + b*x]^3
)^(1/3))/b^3 - (x^3*Cot[a + b*x]*(c*Sin[a + b*x]^3)^(1/3))/b

Rule 2637

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

Rule 3296

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 6720

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(a+b x)} \, dx &=\left (\csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int x^3 \sin (a+b x) \, dx\\ &=-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b}+\frac {\left (3 \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int x^2 \cos (a+b x) \, dx}{b}\\ &=\frac {3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b}-\frac {\left (6 \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int x \sin (a+b x) \, dx}{b^2}\\ &=\frac {3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}+\frac {6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b}-\frac {\left (6 \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int \cos (a+b x) \, dx}{b^3}\\ &=-\frac {6 \sqrt [3]{c \sin ^3(a+b x)}}{b^4}+\frac {3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}+\frac {6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 47, normalized size = 0.49 \[ -\frac {\left (b x \left (b^2 x^2-6\right ) \cot (a+b x)-3 b^2 x^2+6\right ) \sqrt [3]{c \sin ^3(a+b x)}}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((6 - 3*b^2*x^2 + b*x*(-6 + b^2*x^2)*Cot[a + b*x])*(c*Sin[a + b*x]^3)^(1/3))/b^4)

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fricas [A]  time = 0.60, size = 74, normalized size = 0.77 \[ -\frac {{\left ({\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x + a\right ) - 3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x + a\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {1}{3}}}{b^{4} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^3*x^3 - 6*b*x)*cos(b*x + a) - 3*(b^2*x^2 - 2)*sin(b*x + a))*(-(c*cos(b*x + a)^2 - c)*sin(b*x + a))^(1/3)/
(b^4*sin(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )^{3}\right )^{\frac {1}{3}} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.19, size = 151, normalized size = 1.57 \[ -\frac {i \left (b^{3} x^{3}+3 i b^{2} x^{2}-6 b x -6 i\right ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 b^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} \left (b^{3} x^{3}-3 i b^{2} x^{2}-6 b x +6 i\right )}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right ) b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/b^4*(b^3*x^3+3*I*b^2*x^2-6*b*x-6*I)/(exp(2*I*(b*x+a))-1)*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))
^(1/3)*exp(2*I*(b*x+a))-1/2*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*(b^3*x
^3-3*I*b^2*x^2-6*b*x+6*I)/b^4

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maxima [A]  time = 1.15, size = 146, normalized size = 1.52 \[ \frac {3 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a^{2} c^{\frac {1}{3}} - 3 \, {\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} a c^{\frac {1}{3}} + \frac {4 \, a^{3} c^{\frac {1}{3}}}{\frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1} + {\left ({\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \cos \left (b x + a\right ) - 3 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} c^{\frac {1}{3}}}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*((b*x + a)*cos(b*x + a) - sin(b*x + a))*a^2*c^(1/3) - 3*(((b*x + a)^2 - 2)*cos(b*x + a) - 2*(b*x + a)*s
in(b*x + a))*a*c^(1/3) + 4*a^3*c^(1/3)/(sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + 1) + (((b*x + a)^3 - 6*b*x - 6*a
)*cos(b*x + a) - 3*((b*x + a)^2 - 2)*sin(b*x + a))*c^(1/3))/b^4

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mupad [B]  time = 5.71, size = 109, normalized size = 1.14 \[ \frac {2^{1/3}\,{\left (c\,\left (3\,\sin \left (a+b\,x\right )-\sin \left (3\,a+3\,b\,x\right )\right )\right )}^{1/3}\,\left (3\,b^2\,x^2-12\,{\sin \left (a+b\,x\right )}^2+6\,b\,x\,\sin \left (2\,a+2\,b\,x\right )+3\,b^2\,x^2\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )-b^3\,x^3\,\sin \left (2\,a+2\,b\,x\right )\right )}{4\,b^4\,{\sin \left (a+b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*sin(a + b*x)^3)^(1/3),x)

[Out]

(2^(1/3)*(c*(3*sin(a + b*x) - sin(3*a + 3*b*x)))^(1/3)*(3*b^2*x^2 - 12*sin(a + b*x)^2 + 6*b*x*sin(2*a + 2*b*x)
+ 3*b^2*x^2*(2*sin(a + b*x)^2 - 1) - b^3*x^3*sin(2*a + 2*b*x)))/(4*b^4*sin(a + b*x)^2)

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sympy [A]  time = 21.52, size = 143, normalized size = 1.49 \[ \begin {cases} \frac {x^{4} \sqrt [3]{c \sin ^{3}{\relax (a )}}}{4} & \text {for}\: b = 0 \\0 & \text {for}\: a = - b x \vee a = - b x + \pi \\- \frac {\sqrt [3]{c} x^{3} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} + \frac {3 \sqrt [3]{c} x^{2} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}}}{b^{2}} + \frac {6 \sqrt [3]{c} x \sqrt [3]{\sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b^{3} \sin {\left (a + b x \right )}} - \frac {6 \sqrt [3]{c} \sqrt [3]{\sin ^{3}{\left (a + b x \right )}}}{b^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*(c*sin(a)**3)**(1/3)/4, Eq(b, 0)), (0, Eq(a, -b*x) | Eq(a, -b*x + pi)), (-c**(1/3)*x**3*(sin(a
+ b*x)**3)**(1/3)*cos(a + b*x)/(b*sin(a + b*x)) + 3*c**(1/3)*x**2*(sin(a + b*x)**3)**(1/3)/b**2 + 6*c**(1/3)*
x*(sin(a + b*x)**3)**(1/3)*cos(a + b*x)/(b**3*sin(a + b*x)) - 6*c**(1/3)*(sin(a + b*x)**3)**(1/3)/b**4, True))

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