3.348 \(\int \frac{(c \sin ^3(a+b x^2))^{2/3}}{x^2} \, dx\)

Optimal. Leaf size=132 \[ \sqrt{\pi } \sqrt{b} \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}+\sqrt{\pi } \sqrt{b} \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x} \]

[Out]

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

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Rubi [A]  time = 0.148321, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6720, 3393, 4573, 3373, 3353, 3352, 3351} \[ \sqrt{\pi } \sqrt{b} \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}+\sqrt{\pi } \sqrt{b} \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(2/3)/x^2,x]

[Out]

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

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 3393

Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> Simp[(x^(m + 1)*Sin[a + b*x^n]^p)/(m + 1), x] -
Dist[(b*n*p)/(m + 1), Int[Sin[a + b*x^n]^(p - 1)*Cos[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] &&
EqQ[m + n, 0] && NeQ[n, 1] && IntegerQ[n]

Rule 4573

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rule 3373

Int[((a_.) + (b_.)*Sin[u_])^(p_.), x_Symbol] :> Int[(a + b*Sin[ExpandToSum[u, x]])^p, x] /; FreeQ[{a, b, p}, x
] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

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 \frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x^2} \, dx &=\left (\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \frac{\sin ^2\left (a+b x^2\right )}{x^2} \, dx\\ &=-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x}+\left (4 b \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \cos \left (a+b x^2\right ) \sin \left (a+b x^2\right ) \, dx\\ &=-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x}+\left (2 b \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \sin \left (2 \left (a+b x^2\right )\right ) \, dx\\ &=-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x}+\left (2 b \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \sin \left (2 a+2 b x^2\right ) \, dx\\ &=-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x}+\left (2 b \cos (2 a) \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \sin \left (2 b x^2\right ) \, dx+\left (2 b \csc ^2\left (a+b x^2\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\right ) \int \cos \left (2 b x^2\right ) \, dx\\ &=-\frac{\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{x}+\sqrt{b} \sqrt{\pi } \cos (2 a) \csc ^2\left (a+b x^2\right ) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}+\sqrt{b} \sqrt{\pi } \csc ^2\left (a+b x^2\right ) C\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}\\ \end{align*}

Mathematica [A]  time = 0.202151, size = 107, normalized size = 0.81 \[ \frac{\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (2 \sqrt{\pi } \sqrt{b} x \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )+2 \sqrt{\pi } \sqrt{b} x \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )+\cos \left (2 \left (a+b x^2\right )\right )-1\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.096, size = 301, normalized size = 2.3 \begin{align*} -{\frac{1}{4\, \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}x} \left ( 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 ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{{\frac{i}{4}}{{\rm e}^{2\,ib{x}^{2}}}b\sqrt{\pi }\sqrt{2}}{ \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}} \left ( 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 ) }} \right ) ^{{\frac{2}{3}}}{\it Erf} \left ( \sqrt{2}\sqrt{ib}x \right ){\frac{1}{\sqrt{ib}}}}+{\frac{1}{4\, \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}} \left ( 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 ) }} \right ) ^{{\frac{2}{3}}} \left ( -{\frac{{{\rm e}^{4\,i \left ( b{x}^{2}+a \right ) }}}{x}}+{2\,ib\sqrt{\pi }{{\rm e}^{2\,i \left ( b{x}^{2}+2\,a \right ) }}{\it Erf} \left ( \sqrt{-2\,ib}x \right ){\frac{1}{\sqrt{-2\,ib}}}} \right ) }+{\frac{{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}}{2\, \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{2}x} \left ( 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 ) }} \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2/x-1/4*I*(I*c*(exp(2*I*(
b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2*exp(2*I*b*x^2)*b*Pi^(1/2)*2^(1/2)/(I*b)^(1/
2)*erf(2^(1/2)*(I*b)^(1/2)*x)+1/4*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(2/3)/(exp(2*I*(b*x^2+a))
-1)^2*(-1/x*exp(4*I*(b*x^2+a))+2*I*b*Pi^(1/2)/(-2*I*b)^(1/2)*erf((-2*I*b)^(1/2)*x)*exp(2*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)))^(2/3)/(exp(2*I*(b*x^2+a))-1)^2/x*exp(2*I*(b*x^2+a))

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Maxima [C]  time = 1.75053, size = 510, normalized size = 3.86 \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)^(2/3)/x^2,x, algorithm="maxima")

[Out]

-1/32*(sqrt(2)*sqrt(x^2*abs(b))*((((I*sqrt(3) + 1)*gamma(-1/2, 2*I*b*x^2) + (-I*sqrt(3) + 1)*gamma(-1/2, -2*I*
b*x^2))*cos(1/4*pi + 1/2*arctan2(0, b)) + ((-I*sqrt(3) + 1)*gamma(-1/2, 2*I*b*x^2) + (I*sqrt(3) + 1)*gamma(-1/
2, -2*I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) - ((sqrt(3) - I)*gamma(-1/2, 2*I*b*x^2) + (sqrt(3) + I)*gamma
(-1/2, -2*I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) - ((sqrt(3) + I)*gamma(-1/2, 2*I*b*x^2) + (sqrt(3) - I)*ga
mma(-1/2, -2*I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(2*a) + (((sqrt(3) - I)*gamma(-1/2, 2*I*b*x^2) + (
sqrt(3) + I)*gamma(-1/2, -2*I*b*x^2))*cos(1/4*pi + 1/2*arctan2(0, b)) - ((sqrt(3) + I)*gamma(-1/2, 2*I*b*x^2)
+ (sqrt(3) - I)*gamma(-1/2, -2*I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + ((I*sqrt(3) + 1)*gamma(-1/2, 2*I*b
*x^2) + (-I*sqrt(3) + 1)*gamma(-1/2, -2*I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) + ((I*sqrt(3) - 1)*gamma(-1/
2, 2*I*b*x^2) + (-I*sqrt(3) - 1)*gamma(-1/2, -2*I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(2*a))*c^(2/3)
- 8*c^(2/3))/x

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Fricas [A]  time = 1.7511, size = 339, normalized size = 2.57 \begin{align*} -\frac{4^{\frac{2}{3}}{\left (4^{\frac{1}{3}} \pi x \sqrt{\frac{b}{\pi }} \cos \left (2 \, a\right ) \operatorname{S}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) + 4^{\frac{1}{3}} \pi x \sqrt{\frac{b}{\pi }} \operatorname{C}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) \sin \left (2 \, a\right ) + 4^{\frac{1}{3}} \cos \left (b x^{2} + a\right )^{2} - 4^{\frac{1}{3}}\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac{2}{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)^(2/3)/x^2,x, algorithm="fricas")

[Out]

-1/4*4^(2/3)*(4^(1/3)*pi*x*sqrt(b/pi)*cos(2*a)*fresnel_sin(2*x*sqrt(b/pi)) + 4^(1/3)*pi*x*sqrt(b/pi)*fresnel_c
os(2*x*sqrt(b/pi))*sin(2*a) + 4^(1/3)*cos(b*x^2 + a)^2 - 4^(1/3))*(-(c*cos(b*x^2 + a)^2 - c)*sin(b*x^2 + a))^(
2/3)/(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{\left (c \sin ^{3}{\left (a + b x^{2} \right )}\right )^{\frac{2}{3}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sin(a + b*x**2)**3)**(2/3)/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{2}{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)^(2/3)/x^2,x, algorithm="giac")

[Out]

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