3.4.12 \(\int x^2 \sqrt [3]{c \sin ^3(a+b x)} \, dx\) [312]

Optimal. Leaf size=74 \[ \frac {2 x \sqrt [3]{c \sin ^3(a+b x)}}{b^2}+\frac {2 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac {x^2 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6852, 3377, 2718} \begin {gather*} \frac {2 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}+\frac {2 x \sqrt [3]{c \sin ^3(a+b x)}}{b^2}-\frac {x^2 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(c*Sin[a + b*x]^3)^(1/3),x]

[Out]

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

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Mathematica [A]
time = 0.16, size = 40, normalized size = 0.54 \begin {gather*} \frac {\left (2 b x+\left (2-b^2 x^2\right ) \cot (a+b x)\right ) \sqrt [3]{c \sin ^3(a+b x)}}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.13, size = 133, normalized size = 1.80

method	result	size
risch	\(-\frac {i \left (x^{2} b^{2}+2 i b x -2\right ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 b^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {1}{3}} \left (x^{2} b^{2}-2 i b x -2\right )}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right ) b^{3}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/b^3*(x^2*b^2+2*I*b*x-2)/(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)*(x^2*b^2-2*I*b*x-2
)/b^3

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Maxima [A]
time = 0.54, size = 99, normalized size = 1.34 \begin {gather*} -\frac {2 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a c^{\frac {1}{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 )} c^{\frac {1}{3}} + \frac {4 \, a^{2} c^{\frac {1}{3}}}{\frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1}}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 64, normalized size = 0.86 \begin {gather*} \frac {{\left (2 \, b x \sin \left (b x + a\right ) - {\left (b^{2} x^{2} - 2\right )} \cos \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^{3} \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.31, size = 107, normalized size = 1.45 \begin {gather*} \begin {cases} \frac {x^{3} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{3} & \text {for}\: b = 0 \\0 & \text {for}\: a = - b x \vee a = - b x + \pi \\- \frac {x^{2} \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} + \frac {2 x \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}}}{b^{2}} + \frac {2 \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b^{3} \sin {\left (a + b x \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 5.49, size = 88, normalized size = 1.19 \begin {gather*} -\frac {{\left (2\,c\,\left (3\,\sin \left (a+b\,x\right )-\sin \left (3\,a+3\,b\,x\right )\right )\right )}^{1/3}\,\left (\sin \left (2\,a+2\,b\,x\right )+b\,x-\frac {b^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{2}-b\,x\,\cos \left (2\,a+2\,b\,x\right )\right )}{b^3\,\left (\cos \left (2\,a+2\,b\,x\right )-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*sin(a + b*x)^3)^(1/3),x)

[Out]

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

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