\(\int x^2 \sqrt [3]{c \sin ^3(a+b x^2)} \, dx\) [320]

Optimal result

Integrand size = 20, antiderivative size = 155 \[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=-\frac {x \cot \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) \csc \left (a+b x^2\right ) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \csc \left (a+b x^2\right ) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 b^{3/2}} \] Output:

-1/2*x*cot(b*x^2+a)*(c*sin(b*x^2+a)^3)^(1/3)/b+1/4*2^(1/2)*Pi^(1/2)*cos(a) 
*csc(b*x^2+a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*x)*(c*sin(b*x^2+a)^3)^(1/3 
)/b^(3/2)-1/4*2^(1/2)*Pi^(1/2)*csc(b*x^2+a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1 
/2)*x)*sin(a)*(c*sin(b*x^2+a)^3)^(1/3)/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.68 \[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=-\frac {\csc \left (a+b x^2\right ) \left (2 \sqrt {b} x \cos \left (a+b x^2\right )-\sqrt {2 \pi } \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )+\sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{4 b^{3/2}} \] Input:

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

Output:

-1/4*(Csc[a + b*x^2]*(2*Sqrt[b]*x*Cos[a + b*x^2] - Sqrt[2*Pi]*Cos[a]*Fresn 
elC[Sqrt[b]*Sqrt[2/Pi]*x] + Sqrt[2*Pi]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[ 
a])*(c*Sin[a + b*x^2]^3)^(1/3))/b^(3/2)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {7271, 3866, 3835, 3832, 3833}

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 3866

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

\(\Big \downarrow \) 3835

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

\(\Big \downarrow \) 3832

\(\displaystyle \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (\frac {\cos (a) \int \cos \left (b x^2\right )dx-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{\sqrt {b}}}{2 b}-\frac {x \cos \left (a+b x^2\right )}{2 b}\right )\)

\(\Big \downarrow \) 3833

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3835

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[Cos[c]   In 
t[Cos[d*(e + f*x)^2], x], x] - Simp[Sin[c]   Int[Sin[d*(e + f*x)^2], x], x] 
 /; FreeQ[{c, d, e, f}, x]
 

rule 3866

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] + Simp[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 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 = 1.00 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.55

method	result	size
risch	\(\frac {\left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {1}{3}} \left (-\frac {i x \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{2 b}+\frac {i \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i b}\, x \right ) {\mathrm e}^{i \left (b \,x^{2}+2 a \right )}}{4 b \sqrt {-i b}}\right )}{2 \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-2}-\frac {i x \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {1}{3}}}{4 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}+\frac {i \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {1}{3}} {\mathrm e}^{i b \,x^{2}} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {i b}\, x \right )}{8 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right ) b \sqrt {i b}}\)	\(240\)

Input:

int(x^2*(c*sin(b*x^2+a)^3)^(1/3),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.86 \[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=-\frac {{\left (4 \, b x \cos \left (b x^{2} + a\right ) - \sqrt {2} {\left (\pi e^{\left (i \, a\right )} + \pi e^{\left (-i \, a\right )}\right )} \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) - \sqrt {2} {\left (i \, \pi e^{\left (i \, a\right )} - i \, \pi e^{\left (-i \, a\right )}\right )} \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {1}{3}}}{8 \, b^{2} \sin \left (b x^{2} + a\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\frac {8 \, b^{2} c^{\frac {1}{3}} x \cos \left (b x^{2} + a\right ) + \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \left (a\right ) + \left (i + 1\right ) \, \sin \left (a\right )\right )} \operatorname {erf}\left (\sqrt {i \, b} x\right ) + {\left (-\left (i + 1\right ) \, \cos \left (a\right ) - \left (i - 1\right ) \, \sin \left (a\right )\right )} \operatorname {erf}\left (\sqrt {-i \, b} x\right )\right )} b^{\frac {3}{2}} c^{\frac {1}{3}}}{32 \, b^{3}} \] Input:

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

Output:

1/32*(8*b^2*c^(1/3)*x*cos(b*x^2 + a) + sqrt(2)*sqrt(pi)*(((I - 1)*cos(a) + 
 (I + 1)*sin(a))*erf(sqrt(I*b)*x) + (-(I + 1)*cos(a) - (I - 1)*sin(a))*erf 
(sqrt(-I*b)*x))*b^(3/2)*c^(1/3))/b^3
                                                                                    
                                                                                    
 

Giac [F]

\[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\int { \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {1}{3}} x^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\int x^2\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{1/3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^2 \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=c^{\frac {1}{3}} \left (\int \sin \left (b \,x^{2}+a \right ) x^{2}d x \right ) \] Input:

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

Output:

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