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

Optimal. Leaf size=155 \[ \frac {\sqrt {\frac {\pi }{2}} \cos (a) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^{3/2}}-\frac {x \cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \]

[Out]

-1/2*x*cot(b*x^2+a)*(c*sin(b*x^2+a)^3)^(1/3)/b+1/4*cos(a)*csc(b*x^2+a)*FresnelC(x*b^(1/2)*2^(1/2)/Pi^(1/2))*(c
*sin(b*x^2+a)^3)^(1/3)*2^(1/2)*Pi^(1/2)/b^(3/2)-1/4*csc(b*x^2+a)*FresnelS(x*b^(1/2)*2^(1/2)/Pi^(1/2))*sin(a)*(
c*sin(b*x^2+a)^3)^(1/3)*2^(1/2)*Pi^(1/2)/b^(3/2)

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Rubi [A]  time = 0.21, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6720, 3385, 3354, 3352, 3351} \[ \frac {\sqrt {\frac {\pi }{2}} \cos (a) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^{3/2}}-\frac {x \cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

-(x*Cot[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(1/3))/(2*b) + (Sqrt[Pi/2]*Cos[a]*Csc[a + b*x^2]*FresnelC[Sqrt[b]*Sqrt
[2/Pi]*x]*(c*Sin[a + b*x^2]^3)^(1/3))/(2*b^(3/2)) - (Sqrt[Pi/2]*Csc[a + b*x^2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*
Sin[a]*(c*Sin[a + b*x^2]^3)^(1/3))/(2*b^(3/2))

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

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

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Mathematica [A]  time = 0.27, size = 105, normalized size = 0.68 \[ -\frac {\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (-\sqrt {2 \pi } \cos (a) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )+\sqrt {2 \pi } \sin (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )+2 \sqrt {b} x \cos \left (a+b x^2\right )\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Csc[a + b*x^2]*(2*Sqrt[b]*x*Cos[a + b*x^2] - Sqrt[2*Pi]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x] + Sqrt[2*P
i]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a])*(c*Sin[a + b*x^2]^3)^(1/3))/b^(3/2)

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fricas [A]  time = 0.60, size = 156, normalized size = 1.01 \[ -\frac {4^{\frac {1}{3}} {\left (4^{\frac {2}{3}} \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) - 4^{\frac {2}{3}} \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) \sin \relax (a) - 2 \cdot 4^{\frac {2}{3}} b x \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {1}{3}}}{16 \, {\left (b^{2} \cos \left (b x^{2} + a\right )^{2} - b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*4^(1/3)*(4^(2/3)*sqrt(2)*pi*sqrt(b/pi)*cos(a)*fresnel_cos(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a) - 4^(2/3)
*sqrt(2)*pi*sqrt(b/pi)*fresnel_sin(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a)*sin(a) - 2*4^(2/3)*b*x*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*cos(b*x^2 + a)^2 - b^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {1}{3}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.28, size = 240, normalized size = 1.55 \[ \frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} \left (-\frac {i x \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{2 b}+\frac {i \sqrt {\pi }\, \erf \left (\sqrt {-i b}\, x \right ) {\mathrm e}^{i \left (b \,x^{2}+2 a \right )}}{4 b \sqrt {-i b}}\right )}{2 \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-2}-\frac {i x \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}}}{4 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i b \,x^{2}} \sqrt {\pi }\, \erf \left (\sqrt {i b}\, x \right )}{8 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right ) b \sqrt {i b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(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)*(-1/2*I/b*x*exp(2*I*(b*x^2
+a))+1/4*I/b*Pi^(1/2)/(-I*b)^(1/2)*erf((-I*b)^(1/2)*x)*exp(I*(b*x^2+2*a)))-1/4*I*x/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)+1/8*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)*exp(I*b*x^2)/b*Pi^(1/2)/(I*b)^(1/2)*erf((I*b)^(1/2)*x)

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maxima [C]  time = 0.96, size = 73, normalized size = 0.47 \[ \frac {8 \, b^{2} c^{\frac {1}{3}} x \cos \left (b x^{2} + a\right ) + \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \relax (a) + \left (i + 1\right ) \, \sin \relax (a)\right )} \operatorname {erf}\left (\sqrt {i \, b} x\right ) + {\left (-\left (i + 1\right ) \, \cos \relax (a) - \left (i - 1\right ) \, \sin \relax (a)\right )} \operatorname {erf}\left (\sqrt {-i \, b} x\right )\right )} b^{\frac {3}{2}} c^{\frac {1}{3}}}{32 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*b^2*c^(1/3)*x*cos(b*x^2 + a) + sqrt(2)*sqrt(pi)*(((I - 1)*cos(a) + (I + 1)*sin(a))*erf(sqrt(I*b)*x) +
(-(I + 1)*cos(a) - (I - 1)*sin(a))*erf(sqrt(-I*b)*x))*b^(3/2)*c^(1/3))/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*sin(a + b*x^2)^3)^(1/3),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt [3]{c \sin ^{3}{\left (a + b x^{2} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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