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

Optimal. Leaf size=135 \[ \sqrt{2 \pi } \sqrt{b} \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\sqrt{2 \pi } \sqrt{b} \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x} \]

[Out]

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

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Rubi [A]  time = 0.15641, antiderivative size = 135, 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, 3387, 3354, 3352, 3351} \[ \sqrt{2 \pi } \sqrt{b} \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\sqrt{2 \pi } \sqrt{b} \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3)/x^2,x]

[Out]

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

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 3387

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[((e*x)^(m + 1)*Sin[c + d*x^n])/(e*(m + 1
)), x] - Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[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 \frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x^2} \, dx &=\left (\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \frac{\sin \left (a+b x^2\right )}{x^2} \, dx\\ &=-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\left (2 b \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\\ &=-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\left (2 b \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-\left (2 b \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\\ &=-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{x}+\sqrt{b} \sqrt{2 \pi } \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 )}-\sqrt{b} \sqrt{2 \pi } \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 )}\\ \end{align*}

Mathematica [A]  time = 0.276604, size = 105, normalized size = 0.78 \[ \frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (\sqrt{2 \pi } \sqrt{b} x \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right ) \csc \left (a+b x^2\right )-\sqrt{2 \pi } \sqrt{b} x \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \csc \left (a+b x^2\right )-1\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.092, size = 232, normalized size = 1.7 \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{{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}}{x}}+{ib\sqrt{\pi }{{\rm e}^{i \left ( b{x}^{2}+2\,a \right ) }}{\it Erf} \left ( \sqrt{-ib}x \right ){\frac{1}{\sqrt{-ib}}}} \right ) }+{\frac{1}{ \left ( 2\,{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-2 \right ) x}\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}{2}}{{\rm e}^{ib{x}^{2}}}b\sqrt{\pi }}{{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1}\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^2,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/x*exp(2*I*(b*x^2+a))+I
*b*Pi^(1/2)/(-I*b)^(1/2)*erf((-I*b)^(1/2)*x)*exp(I*(b*x^2+2*a)))+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)/x+1/2*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 = 1.7393, size = 489, normalized size = 3.62 \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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58083, size = 412, normalized size = 3.05 \begin{align*} -\frac{4^{\frac{1}{3}}{\left (4^{\frac{2}{3}} \sqrt{2} \pi x \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 x \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 ) + 4^{\frac{2}{3}} \cos \left (b x^{2} + a\right )^{2} - 4^{\frac{2}{3}}\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}}}{4 \,{\left (x \cos \left (b x^{2} + a\right )^{2} - x\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^2,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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

[Out]

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