\(\int x^2 \cos (\frac {1}{4}+x+x^2) \, dx\) [11]

Optimal result

Integrand size = 13, antiderivative size = 82 \[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )-\frac {1}{4} \sin \left (\frac {1}{4}+x+x^2\right )+\frac {1}{2} x \sin \left (\frac {1}{4}+x+x^2\right ) \] Output:

1/8*2^(1/2)*Pi^(1/2)*FresnelC(1/2*(1+2*x)*2^(1/2)/Pi^(1/2))-1/4*2^(1/2)*Pi 
^(1/2)*FresnelS(1/2*(1+2*x)*2^(1/2)/Pi^(1/2))-1/4*sin(1/4+x+x^2)+1/2*x*sin 
(1/4+x+x^2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{8} \left (\sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )-2 \left (\sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )+(1-2 x) \sin \left (\frac {1}{4}+x+x^2\right )\right )\right ) \] Input:

Integrate[x^2*Cos[1/4 + x + x^2],x]
 

Output:

(Sqrt[2*Pi]*FresnelC[(1 + 2*x)/Sqrt[2*Pi]] - 2*(Sqrt[2*Pi]*FresnelS[(1 + 2 
*x)/Sqrt[2*Pi]] + (1 - 2*x)*Sin[1/4 + x + x^2]))/8
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3945, 3926, 3832, 3943, 3927, 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.

\(\displaystyle \int x^2 \cos \left (x^2+x+\frac {1}{4}\right ) \, dx\)

\(\Big \downarrow \) 3945

\(\displaystyle -\frac {1}{2} \int \sin \left (x^2+x+\frac {1}{4}\right )dx-\frac {1}{2} \int x \cos \left (x^2+x+\frac {1}{4}\right )dx+\frac {1}{2} x \sin \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 3926

\(\displaystyle -\frac {1}{2} \int x \cos \left (x^2+x+\frac {1}{4}\right )dx-\frac {1}{2} \int \sin \left (\frac {1}{4} (2 x+1)^2\right )dx+\frac {1}{2} x \sin \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 3832

\(\displaystyle -\frac {1}{2} \int x \cos \left (x^2+x+\frac {1}{4}\right )dx-\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )+\frac {1}{2} x \sin \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 3943

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \cos \left (x^2+x+\frac {1}{4}\right )dx-\frac {1}{2} \sin \left (x^2+x+\frac {1}{4}\right )\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )+\frac {1}{2} x \sin \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 3927

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \cos \left (\frac {1}{4} (2 x+1)^2\right )dx-\frac {1}{2} \sin \left (x^2+x+\frac {1}{4}\right )\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )+\frac {1}{2} x \sin \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )-\frac {1}{2} \sin \left (x^2+x+\frac {1}{4}\right )\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )+\frac {1}{2} x \sin \left (x^2+x+\frac {1}{4}\right )\)

Input:

Int[x^2*Cos[1/4 + x + x^2],x]
 

Output:

-1/2*(Sqrt[Pi/2]*FresnelS[(1 + 2*x)/Sqrt[2*Pi]]) + ((Sqrt[Pi/2]*FresnelC[( 
1 + 2*x)/Sqrt[2*Pi]])/2 - Sin[1/4 + x + x^2]/2)/2 + (x*Sin[1/4 + x + x^2]) 
/2
 

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 3926

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Int[Sin[(b + 2*c*x 
)^2/(4*c)], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]
 

rule 3927

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Int[Cos[(b + 2*c*x 
)^2/(4*c)], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]
 

rule 3943

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_)), x_Symbol] 
:> Simp[e*(Sin[a + b*x + c*x^2]/(2*c)), x] + Simp[(2*c*d - b*e)/(2*c)   Int 
[Cos[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b 
*e, 0]
 

rule 3945

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_))^(m_), x_Sym 
bol] :> Simp[e*(d + e*x)^(m - 1)*(Sin[a + b*x + c*x^2]/(2*c)), x] + (-Simp[ 
(b*e - 2*c*d)/(2*c)   Int[(d + e*x)^(m - 1)*Cos[a + b*x + c*x^2], x], x] - 
Simp[e^2*((m - 1)/(2*c))   Int[(d + e*x)^(m - 2)*Sin[a + b*x + c*x^2], x], 
x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]
 

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {x \sin \left (\frac {1}{4}+x +x^{2}\right )}{2}-\frac {\sin \left (\frac {1}{4}+x +x^{2}\right )}{4}+\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{8}-\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{4}\)	\(59\)
risch	\(-\frac {\sqrt {\pi }\, \left (-1\right )^{\frac {3}{4}} \operatorname {erf}\left (\left (-1\right )^{\frac {1}{4}} x +\frac {\left (-1\right )^{\frac {1}{4}}}{2}\right )}{16}-\frac {\left (-1\right )^{\frac {1}{4}} \sqrt {\pi }\, \operatorname {erf}\left (\left (-1\right )^{\frac {1}{4}} x +\frac {\left (-1\right )^{\frac {1}{4}}}{2}\right )}{8}+\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i}\, x -\frac {i}{2 \sqrt {-i}}\right )}{16 \sqrt {-i}}+\frac {i \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i}\, x -\frac {i}{2 \sqrt {-i}}\right )}{8 \sqrt {-i}}+2 i \left (-\frac {1}{4} i x +\frac {1}{8} i\right ) \sin \left (\frac {\left (1+2 x \right )^{2}}{4}\right )\)	\(111\)
parts	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right ) x^{2}}{2}-\frac {\sqrt {2}\, \pi ^{\frac {3}{2}} \left (-\frac {\operatorname {FresnelC}\left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \left (-\left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )^{2} \sqrt {\pi }+\sqrt {2}\, \left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )\right )}{\sqrt {\pi }}+\frac {-\frac {\left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \sin \left (\frac {\pi \left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )^{2}}{2}\right )}{\sqrt {\pi }}+\frac {\operatorname {FresnelS}\left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )}{\sqrt {\pi }}+\frac {\sqrt {2}\, \sin \left (\frac {\pi \left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )^{2}}{2}\right )}{\pi }}{\sqrt {\pi }}\right )}{4}\)	\(204\)

Input:

int(x^2*cos(1/4+x+x^2),x,method=_RETURNVERBOSE)
 

Output:

1/2*x*sin(1/4+x+x^2)-1/4*sin(1/4+x+x^2)+1/8*2^(1/2)*Pi^(1/2)*FresnelC(2^(1 
/2)/Pi^(1/2)*(x+1/2))-1/4*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(x+1/ 
2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{8} \, \sqrt {2} \sqrt {\pi } \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) - \frac {1}{4} \, \sqrt {2} \sqrt {\pi } \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) + \frac {1}{4} \, {\left (2 \, x - 1\right )} \sin \left (x^{2} + x + \frac {1}{4}\right ) \] Input:

integrate(x^2*cos(1/4+x+x^2),x, algorithm="fricas")
 

Output:

1/8*sqrt(2)*sqrt(pi)*fresnel_cos(1/2*sqrt(2)*(2*x + 1)/sqrt(pi)) - 1/4*sqr 
t(2)*sqrt(pi)*fresnel_sin(1/2*sqrt(2)*(2*x + 1)/sqrt(pi)) + 1/4*(2*x - 1)* 
sin(x^2 + x + 1/4)
 

Sympy [F]

\[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=\int x^{2} \cos {\left (x^{2} + x + \frac {1}{4} \right )}\, dx \] Input:

integrate(x**2*cos(1/4+x+x**2),x)
 

Output:

Integral(x**2*cos(x**2 + x + 1/4), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.95 \[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=-\frac {16 \, x {\left (-i \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + i \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}\right )} + \sqrt {4 \, x^{2} + 4 \, x + 1} {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x^{2} + i \, x + \frac {1}{4} i}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x^{2} - i \, x - \frac {1}{4} i}\right ) - 1\right )} - \left (4 i + 4\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, i \, x^{2} + i \, x + \frac {1}{4} i\right ) + \left (4 i - 4\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, -i \, x^{2} - i \, x - \frac {1}{4} i\right )\right )} - 8 i \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + 8 i \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}}{32 \, {\left (2 \, x + 1\right )}} \] Input:

integrate(x^2*cos(1/4+x+x^2),x, algorithm="maxima")
 

Output:

-1/32*(16*x*(-I*e^(I*x^2 + I*x + 1/4*I) + I*e^(-I*x^2 - I*x - 1/4*I)) + sq 
rt(4*x^2 + 4*x + 1)*((I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*x^2 + I*x + 1/4* 
I)) - 1) - (I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*x^2 - I*x - 1/4*I)) - 1) 
- (4*I + 4)*sqrt(2)*gamma(3/2, I*x^2 + I*x + 1/4*I) + (4*I - 4)*sqrt(2)*ga 
mma(3/2, -I*x^2 - I*x - 1/4*I)) - 8*I*e^(I*x^2 + I*x + 1/4*I) + 8*I*e^(-I* 
x^2 - I*x - 1/4*I))/(2*x + 1)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=-\left (\frac {3}{32} i - \frac {1}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) + \left (\frac {3}{32} i + \frac {1}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) + \frac {1}{8} \, {\left (-2 i \, x + i\right )} e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + \frac {1}{8} \, {\left (2 i \, x - i\right )} e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )} \] Input:

integrate(x^2*cos(1/4+x+x^2),x, algorithm="giac")
 

Output:

-(3/32*I - 1/32)*sqrt(2)*sqrt(pi)*erf((1/4*I - 1/4)*sqrt(2)*(2*x + 1)) + ( 
3/32*I + 1/32)*sqrt(2)*sqrt(pi)*erf(-(1/4*I + 1/4)*sqrt(2)*(2*x + 1)) + 1/ 
8*(-2*I*x + I)*e^(I*x^2 + I*x + 1/4*I) + 1/8*(2*I*x - I)*e^(-I*x^2 - I*x - 
 1/4*I)
 

Mupad [B] (verification not implemented)

Time = 41.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x\,\sin \left (x^2+x+\frac {1}{4}\right )}{2}-\frac {\sin \left (x^2+x+\frac {1}{4}\right )}{4}+\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\left (2\,x+1\right )}{2\,\sqrt {\pi }}\right )}{8}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\left (2\,x+1\right )}{2\,\sqrt {\pi }}\right )}{4} \] Input:

int(x^2*cos(x + x^2 + 1/4),x)
 

Output:

(x*sin(x + x^2 + 1/4))/2 - sin(x + x^2 + 1/4)/4 + (2^(1/2)*pi^(1/2)*fresne 
lc((2^(1/2)*(2*x + 1))/(2*pi^(1/2))))/8 - (2^(1/2)*pi^(1/2)*fresnels((2^(1 
/2)*(2*x + 1))/(2*pi^(1/2))))/4
 

Reduce [F]

\[ \int x^2 \cos \left (\frac {1}{4}+x+x^2\right ) \, dx=2 \left (\int \frac {x^{2}}{\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{2}+1}d x \right )-\frac {x^{3}}{3} \] Input:

int(x^2*cos(1/4+x+x^2),x)
 

Output:

(6*int(x**2/(tan((4*x**2 + 4*x + 1)/8)**2 + 1),x) - x**3)/3