\(\int x^2 (c \sin ^3(a+b x^2))^{2/3} \, dx\) [344]

Optimal result

Integrand size = 20, antiderivative size = 195 \[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\frac {1}{6} x^3 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}+\frac {\sqrt {\pi } \cos (2 a) \csc ^2\left (a+b x^2\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{16 b^{3/2}}+\frac {\sqrt {\pi } \csc ^2\left (a+b x^2\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{16 b^{3/2}}-\frac {x \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \sin \left (2 a+2 b x^2\right )}{8 b} \] Output:

1/6*x^3*csc(b*x^2+a)^2*(c*sin(b*x^2+a)^3)^(2/3)+1/16*Pi^(1/2)*cos(2*a)*csc 
(b*x^2+a)^2*FresnelS(2*b^(1/2)*x/Pi^(1/2))*(c*sin(b*x^2+a)^3)^(2/3)/b^(3/2 
)+1/16*Pi^(1/2)*csc(b*x^2+a)^2*FresnelC(2*b^(1/2)*x/Pi^(1/2))*sin(2*a)*(c* 
sin(b*x^2+a)^3)^(2/3)/b^(3/2)-1/8*x*csc(b*x^2+a)^2*(c*sin(b*x^2+a)^3)^(2/3 
)*sin(2*b*x^2+2*a)/b
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.58 \[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\frac {\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (3 \sqrt {\pi } \cos (2 a) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )+3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a)+2 \sqrt {b} x \left (4 b x^2-3 \sin \left (2 \left (a+b x^2\right )\right )\right )\right )}{48 b^{3/2}} \] Input:

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

Output:

(Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3)*(3*Sqrt[Pi]*Cos[2*a]*FresnelS 
[(2*Sqrt[b]*x)/Sqrt[Pi]] + 3*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x)/Sqrt[Pi]]*Sin 
[2*a] + 2*Sqrt[b]*x*(4*b*x^2 - 3*Sin[2*(a + b*x^2)])))/(48*b^(3/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {7271, 3884, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int x^2 \sin ^2\left (b x^2+a\right )dx\)

\(\Big \downarrow \) 3884

\(\displaystyle \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int \left (\frac {x^2}{2}-\frac {1}{2} x^2 \cos \left (2 b x^2+2 a\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\frac {\sqrt {\pi } \sin (2 a) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{16 b^{3/2}}+\frac {\sqrt {\pi } \cos (2 a) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {x \sin \left (2 a+2 b x^2\right )}{8 b}+\frac {x^3}{6}\right )\)

Input:

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

Output:

Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3)*(x^3/6 + (Sqrt[Pi]*Cos[2*a]*Fr 
esnelS[(2*Sqrt[b]*x)/Sqrt[Pi]])/(16*b^(3/2)) + (Sqrt[Pi]*FresnelC[(2*Sqrt[ 
b]*x)/Sqrt[Pi]]*Sin[2*a])/(16*b^(3/2)) - (x*Sin[2*a + 2*b*x^2])/(8*b))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3884

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x 
_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] 
/; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
 

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[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]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.32 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.58

method	result	size
risch	\(\frac {i x \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}}}{16 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {i \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} {\mathrm e}^{2 i b \,x^{2}} \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \sqrt {i b}\, x \right )}{64 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2} b \sqrt {i b}}+\frac {\left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} \left (-\frac {i x \,{\mathrm e}^{4 i \left (b \,x^{2}+a \right )}}{4 b}+\frac {i \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i b}\, x \right ) {\mathrm e}^{2 i \left (b \,x^{2}+2 a \right )}}{8 b \sqrt {-2 i b}}\right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {x^{3} \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{6 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}\)	\(309\)

Input:

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

Output:

1/16*I*x/b/(exp(2*I*(b*x^2+a))-1)^2*(I*c*exp(-3*I*(b*x^2+a))*(exp(2*I*(b*x 
^2+a))-1)^3)^(2/3)-1/64*I*(I*c*exp(-3*I*(b*x^2+a))*(exp(2*I*(b*x^2+a))-1)^ 
3)^(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/(exp(2*I*(b*x^2+a))-1)^2*(I*c*exp(-3* 
I*(b*x^2+a))*(exp(2*I*(b*x^2+a))-1)^3)^(2/3)*(-1/4*I*x/b*exp(4*I*(b*x^2+a) 
)+1/8*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 
)))-1/6*x^3/(exp(2*I*(b*x^2+a))-1)^2*(I*c*exp(-3*I*(b*x^2+a))*(exp(2*I*(b* 
x^2+a))-1)^3)^(2/3)*exp(2*I*(b*x^2+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.76 \[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {{\left (16 \, b^{2} x^{3} - 24 \, b x \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right ) + 3 \, {\left (-i \, \pi e^{\left (2 i \, a\right )} + i \, \pi e^{\left (-2 i \, a\right )}\right )} \sqrt {\frac {b}{\pi }} \operatorname {C}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) + 3 \, {\left (\pi e^{\left (2 i \, a\right )} + \pi e^{\left (-2 i \, a\right )}\right )} \sqrt {\frac {b}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {b}{\pi }}\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {2}{3}}}{96 \, {\left (b^{2} \cos \left (b x^{2} + a\right )^{2} - b^{2}\right )}} \] Input:

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

Output:

-1/96*(16*b^2*x^3 - 24*b*x*cos(b*x^2 + a)*sin(b*x^2 + a) + 3*(-I*pi*e^(2*I 
*a) + I*pi*e^(-2*I*a))*sqrt(b/pi)*fresnel_cos(2*x*sqrt(b/pi)) + 3*(pi*e^(2 
*I*a) + pi*e^(-2*I*a))*sqrt(b/pi)*fresnel_sin(2*x*sqrt(b/pi)))*(-(c*cos(b* 
x^2 + a)^2 - c)*sin(b*x^2 + a))^(2/3)/(b^2*cos(b*x^2 + a)^2 - b^2)
 

Sympy [F]

\[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int x^{2} \left (c \sin ^{3}{\left (a + b x^{2} \right )}\right )^{\frac {2}{3}}\, dx \] Input:

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.51 \[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {3 \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, \cos \left (2 \, a\right ) - \left (i - 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {2 i \, b} x\right ) + {\left (-\left (i - 1\right ) \, \cos \left (2 \, a\right ) + \left (i + 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {-2 i \, b} x\right )\right )} b^{\frac {3}{2}} c^{\frac {2}{3}} + 16 \, {\left (4 \, b^{3} x^{3} - 3 \, b^{2} x \sin \left (2 \, b x^{2} + 2 \, a\right )\right )} c^{\frac {2}{3}}}{768 \, b^{3}} \] Input:

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

Output:

-1/768*(3*4^(1/4)*sqrt(2)*sqrt(pi)*(((I + 1)*cos(2*a) - (I - 1)*sin(2*a))* 
erf(sqrt(2*I*b)*x) + (-(I - 1)*cos(2*a) + (I + 1)*sin(2*a))*erf(sqrt(-2*I* 
b)*x))*b^(3/2)*c^(2/3) + 16*(4*b^3*x^3 - 3*b^2*x*sin(2*b*x^2 + 2*a))*c^(2/ 
3))/b^3
 

Giac [F]

\[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int { \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {2}{3}} x^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int x^2\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{2/3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^2 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=c^{\frac {2}{3}} \left (\int \sin \left (b \,x^{2}+a \right )^{2} x^{2}d x \right ) \] Input:

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

Output:

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