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

Optimal. Leaf size=117 \[ \frac{\sqrt{\frac{\pi }{2}} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt{b}}+\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0586227, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6720, 3353, 3352, 3351} \[ \frac{\sqrt{\frac{\pi }{2}} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt{b}}+\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

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

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 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[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 \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 \sin \left (a+b x^2\right ) \, dx\\ &=\left (\cos (a) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \sin \left (b x^2\right ) \, dx+\left (\csc \left (a+b x^2\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \cos \left (b x^2\right ) \, dx\\ &=\frac{\sqrt{\frac{\pi }{2}} \cos (a) \csc \left (a+b x^2\right ) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt{b}}+\frac{\sqrt{\frac{\pi }{2}} \csc \left (a+b x^2\right ) C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{\sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.12134, size = 80, normalized size = 0.68 \[ \frac{\sqrt{\frac{\pi }{2}} \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (\sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )+\cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.083, size = 157, normalized size = 1.3 \begin{align*}{\frac{\sqrt{\pi }{{\rm e}^{i \left ( b{x}^{2}+2\,a \right ) }}}{4\,{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-4}\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}}}}-{\frac{{{\rm e}^{ib{x}^{2}}}\sqrt{\pi }}{4\,{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-4}\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((c*sin(b*x^2+a)^3)^(1/3),x)

[Out]

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

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Maxima [C]  time = 1.71268, size = 444, normalized size = 3.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(pi)*((((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*arctan2(0, b)))*cos(a
) - ((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)) + (sq
rt(3) + I)*sin(1/4*pi + 1/2*arctan2(0, b)) + (sqrt(3) - I)*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*erf(sqrt(
I*b)*x) + (((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*arctan2(0, b)))*cos(a) -
((-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)))*sin(a))*erf(sqrt(-I*b
)*x))*c^(1/3)/sqrt(abs(b))

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Fricas [A]  time = 1.81931, size = 356, normalized size = 3.04 \begin{align*} -\frac{4^{\frac{1}{3}}{\left (4^{\frac{2}{3}} \sqrt{2} \pi \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\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{C}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (b x^{2} + a\right ) \sin \left (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}}}{8 \,{\left (b \cos \left (b x^{2} + a\right )^{2} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*4^(1/3)*(4^(2/3)*sqrt(2)*pi*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a) + 4^(2/3)*
sqrt(2)*pi*sqrt(b/pi)*fresnel_cos(sqrt(2)*x*sqrt(b/pi))*sin(b*x^2 + a)*sin(a))*(-(c*cos(b*x^2 + a)^2 - c)*sin(
b*x^2 + a))^(1/3)/(b*cos(b*x^2 + a)^2 - b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((c*sin(b*x**2+a)**3)**(1/3),x)

[Out]

Integral((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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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