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

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

[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))

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Rubi [A]  time = 0.2145, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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])

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 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 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 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]

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*}

Mathematica [A]  time = 0.271106, size = 105, normalized size = 0.68 \[ -\frac{\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (-\sqrt{2 \pi } \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )+\sqrt{2 \pi } \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )+2 \sqrt{b} x \cos \left (a+b x^2\right )\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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*Pi]*F
resnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a])*(c*Sin[a + b*x^2]^3)^(1/3))/(4*b^(3/2))

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Maple [C]  time = 0.123, size = 240, normalized size = 1.6 \begin{align*}{\frac{1}{2\,{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-2}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}} \left ({\frac{-{\frac{i}{2}}x{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}}{b}}+{\frac{{\frac{i}{4}}\sqrt{\pi }{{\rm e}^{i \left ( b{x}^{2}+2\,a \right ) }}}{b}{\it Erf} \left ( \sqrt{-ib}x \right ){\frac{1}{\sqrt{-ib}}}} \right ) }-{\frac{{\frac{i}{4}}x}{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) b}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}}}+{\frac{{\frac{i}{8}}{{\rm e}^{ib{x}^{2}}}\sqrt{\pi }}{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) b}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}}{\it Erf} \left ( \sqrt{ib}x \right ){\frac{1}{\sqrt{ib}}}} \end{align*}

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*(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)*(-1/2*I*x/b*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*(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+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 = 1.71994, size = 481, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

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*x*abs(b)*cos(b*x^2 + a) - sqrt(pi)*((((-I*sqrt(3) + 1)*cos(1/4*pi + 1/2*arctan2(0, b)) + (I*sqrt(3) +
1)*cos(-1/4*pi + 1/2*arctan2(0, b)) - (sqrt(3) + I)*sin(1/4*pi + 1/2*arctan2(0, b)) - (sqrt(3) - I)*sin(-1/4*p
i + 1/2*arctan2(0, b)))*cos(a) - ((sqrt(3) + I)*cos(1/4*pi + 1/2*arctan2(0, b)) - (sqrt(3) - I)*cos(-1/4*pi +
1/2*arctan2(0, b)) - (I*sqrt(3) - 1)*sin(1/4*pi + 1/2*arctan2(0, b)) - (I*sqrt(3) + 1)*sin(-1/4*pi + 1/2*arcta
n2(0, b)))*sin(a))*erf(sqrt(I*b)*x) + (((I*sqrt(3) + 1)*cos(1/4*pi + 1/2*arctan2(0, b)) + (-I*sqrt(3) + 1)*cos
(-1/4*pi + 1/2*arctan2(0, b)) - (sqrt(3) - I)*sin(1/4*pi + 1/2*arctan2(0, b)) - (sqrt(3) + I)*sin(-1/4*pi + 1/
2*arctan2(0, b)))*cos(a) - ((sqrt(3) - I)*cos(1/4*pi + 1/2*arctan2(0, b)) - (sqrt(3) + I)*cos(-1/4*pi + 1/2*ar
ctan2(0, b)) - (-I*sqrt(3) - 1)*sin(1/4*pi + 1/2*arctan2(0, b)) - (-I*sqrt(3) + 1)*sin(-1/4*pi + 1/2*arctan2(0
, b)))*sin(a))*erf(sqrt(-I*b)*x))*sqrt(abs(b)))*c^(1/3)/(b*abs(b))

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Fricas [A]  time = 1.7084, size = 425, normalized size = 2.74 \begin{align*} -\frac{4^{\frac{1}{3}}{\left (4^{\frac{2}{3}} \sqrt{2} \pi \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \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 \left (a\right ) - 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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt [3]{c \sin ^{3}{\left (a + b x^{2} \right )}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac{1}{3}} x^{2}\,{d x} \end{align*}

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)