Integrand size = 15, antiderivative size = 85 \[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x^3}{6}+\frac {1}{16} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right )-\frac {1}{16} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {\pi }}\right )-\frac {1}{16} \sin \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{8} x \sin \left (\frac {1}{2}+2 x+2 x^2\right ) \] Output:
1/6*x^3+1/16*Pi^(1/2)*FresnelC((1+2*x)/Pi^(1/2))-1/16*Pi^(1/2)*FresnelS((1 +2*x)/Pi^(1/2))-1/16*sin(1/2+2*x+2*x^2)+1/8*x*sin(1/2+2*x+2*x^2)
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{48} \left (8 x^3+3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right )-3 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {\pi }}\right )-3 \sin \left (\frac {1}{2} (1+2 x)^2\right )+6 x \sin \left (\frac {1}{2} (1+2 x)^2\right )\right ) \] Input:
Integrate[x^2*Cos[1/4 + x + x^2]^2,x]
Output:
(8*x^3 + 3*Sqrt[Pi]*FresnelC[(1 + 2*x)/Sqrt[Pi]] - 3*Sqrt[Pi]*FresnelS[(1 + 2*x)/Sqrt[Pi]] - 3*Sin[(1 + 2*x)^2/2] + 6*x*Sin[(1 + 2*x)^2/2])/48
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3949, 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.
| \(\displaystyle \int x^2 \cos ^2\left (x^2+x+\frac {1}{4}\right ) \, dx\) |
| \(\Big \downarrow \) 3949 |
| \(\displaystyle \int \left (\frac {x^2}{2}+\frac {1}{2} x^2 \cos \left (2 x^2+2 x+\frac {1}{2}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{16} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 x+1}{\sqrt {\pi }}\right )-\frac {1}{16} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {\pi }}\right )+\frac {x^3}{6}+\frac {1}{8} x \sin \left (2 x^2+2 x+\frac {1}{2}\right )-\frac {1}{16} \sin \left (2 x^2+2 x+\frac {1}{2}\right )\) |
Input:
Int[x^2*Cos[1/4 + x + x^2]^2,x]
Output:
x^3/6 + (Sqrt[Pi]*FresnelC[(1 + 2*x)/Sqrt[Pi]])/16 - (Sqrt[Pi]*FresnelS[(1 + 2*x)/Sqrt[Pi]])/16 - Sin[1/2 + 2*x + 2*x^2]/16 + (x*Sin[1/2 + 2*x + 2*x ^2])/8
Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]^(n_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(d + e*x)^m, Cos[a + b*x + c*x^2]^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 1]
Time = 2.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {x^{3}}{6}+\frac {\sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right )}{16}-\frac {\sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {\pi }}\right )}{16}-\frac {\sin \left (\frac {1}{2}+2 x +2 x^{2}\right )}{16}+\frac {x \sin \left (\frac {1}{2}+2 x +2 x^{2}\right )}{8}\) | \(64\) |
| risch | \(\frac {x^{3}}{6}-\frac {\sqrt {\pi }\, \sqrt {2}\, \left (-1\right )^{\frac {3}{4}} \operatorname {erf}\left (\sqrt {2}\, \left (-1\right )^{\frac {1}{4}} x +\frac {\sqrt {2}\, \left (-1\right )^{\frac {1}{4}}}{2}\right )}{64}-\frac {\left (-1\right )^{\frac {1}{4}} \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \left (-1\right )^{\frac {1}{4}} x +\frac {\sqrt {2}\, \left (-1\right )^{\frac {1}{4}}}{2}\right )}{64}+\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i}\, x -\frac {i}{\sqrt {-2 i}}\right )}{32 \sqrt {-2 i}}+\frac {i \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i}\, x -\frac {i}{\sqrt {-2 i}}\right )}{32 \sqrt {-2 i}}+2 i \left (-\frac {1}{16} i x +\frac {1}{32} i\right ) \sin \left (\frac {\left (1+2 x \right )^{2}}{2}\right )\) | \(134\) |
Input:
int(x^2*cos(1/4+x+x^2)^2,x,method=_RETURNVERBOSE)
Output:
1/6*x^3+1/16*Pi^(1/2)*FresnelC((1+2*x)/Pi^(1/2))-1/16*Pi^(1/2)*FresnelS((1 +2*x)/Pi^(1/2))-1/16*sin(1/2+2*x+2*x^2)+1/8*x*sin(1/2+2*x+2*x^2)
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{6} \, x^{3} + \frac {1}{8} \, {\left (2 \, x - 1\right )} \cos \left (x^{2} + x + \frac {1}{4}\right ) \sin \left (x^{2} + x + \frac {1}{4}\right ) + \frac {1}{16} \, \sqrt {\pi } \operatorname {C}\left (\frac {2 \, x + 1}{\sqrt {\pi }}\right ) - \frac {1}{16} \, \sqrt {\pi } \operatorname {S}\left (\frac {2 \, x + 1}{\sqrt {\pi }}\right ) \] Input:
integrate(x^2*cos(1/4+x+x^2)^2,x, algorithm="fricas")
Output:
1/6*x^3 + 1/8*(2*x - 1)*cos(x^2 + x + 1/4)*sin(x^2 + x + 1/4) + 1/16*sqrt( pi)*fresnel_cos((2*x + 1)/sqrt(pi)) - 1/16*sqrt(pi)*fresnel_sin((2*x + 1)/ sqrt(pi))
\[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\int x^{2} \cos ^{2}{\left (x^{2} + x + \frac {1}{4} \right )}\, dx \] Input:
integrate(x**2*cos(1/4+x+x**2)**2,x)
Output:
Integral(x**2*cos(x**2 + x + 1/4)**2, x)
Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.01 \[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {128 \, x^{4} + 64 \, x^{3} - 48 \, x {\left (-i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} + i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )}\right )} - 3 \, \sqrt {8 \, x^{2} + 8 \, x + 2} {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {2 i \, x^{2} + 2 i \, x + \frac {1}{2} i}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i}\right ) - 1\right )} - \left (2 i + 2\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, 2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right ) + \left (2 i - 2\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, -2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )\right )} + 24 i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - 24 i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )}}{384 \, {\left (2 \, x + 1\right )}} \] Input:
integrate(x^2*cos(1/4+x+x^2)^2,x, algorithm="maxima")
Output:
1/384*(128*x^4 + 64*x^3 - 48*x*(-I*e^(2*I*x^2 + 2*I*x + 1/2*I) + I*e^(-2*I *x^2 - 2*I*x - 1/2*I)) - 3*sqrt(8*x^2 + 8*x + 2)*((I - 1)*sqrt(2)*sqrt(pi) *(erf(sqrt(2*I*x^2 + 2*I*x + 1/2*I)) - 1) - (I + 1)*sqrt(2)*sqrt(pi)*(erf( sqrt(-2*I*x^2 - 2*I*x - 1/2*I)) - 1) - (2*I + 2)*sqrt(2)*gamma(3/2, 2*I*x^ 2 + 2*I*x + 1/2*I) + (2*I - 2)*sqrt(2)*gamma(3/2, -2*I*x^2 - 2*I*x - 1/2*I )) + 24*I*e^(2*I*x^2 + 2*I*x + 1/2*I) - 24*I*e^(-2*I*x^2 - 2*I*x - 1/2*I)) /(2*x + 1)
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{6} \, x^{3} - \frac {1}{32} \, {\left (2 i \, x - i\right )} e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - \frac {1}{32} \, {\left (-2 i \, x + i\right )} e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )} - \frac {1}{32} i \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, x + \frac {1}{2} i - \frac {1}{2}\right ) + \frac {1}{32} i \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, x - \frac {1}{2} i - \frac {1}{2}\right ) \] Input:
integrate(x^2*cos(1/4+x+x^2)^2,x, algorithm="giac")
Output:
1/6*x^3 - 1/32*(2*I*x - I)*e^(2*I*x^2 + 2*I*x + 1/2*I) - 1/32*(-2*I*x + I) *e^(-2*I*x^2 - 2*I*x - 1/2*I) - 1/32*I*sqrt(pi)*erf((I - 1)*x + 1/2*I - 1/ 2) + 1/32*I*sqrt(pi)*erf(-(I + 1)*x - 1/2*I - 1/2)
Timed out. \[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\int x^2\,{\cos \left (x^2+x+\frac {1}{4}\right )}^2 \,d x \] Input:
int(x^2*cos(x + x^2 + 1/4)^2,x)
Output:
int(x^2*cos(x + x^2 + 1/4)^2, x)
\[ \int x^2 \cos ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {\cos \left (x^{2}+x +\frac {1}{4}\right ) \sin \left (x^{2}+x +\frac {1}{4}\right ) x}{3}-\frac {\cos \left (x^{2}+x +\frac {1}{4}\right ) \sin \left (x^{2}+x +\frac {1}{4}\right )}{6}-\left (\int \frac {\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{2}}{\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{4}+2 \tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{2}+1}d x \right )-\frac {4 \left (\int \frac {x^{2}}{\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{4}+2 \tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{2}+1}d x \right )}{3}-\frac {4 \left (\int \frac {\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )}{\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{4}+2 \tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{2}+1}d x \right )}{3}+\frac {\left (\int \frac {1}{\tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{4}+2 \tan \left (\frac {1}{2} x^{2}+\frac {1}{2} x +\frac {1}{8}\right )^{2}+1}d x \right )}{3}+\frac {\sin \left (x^{2}+x +\frac {1}{4}\right ) x}{3}-\frac {\sin \left (x^{2}+x +\frac {1}{4}\right )}{6}+\frac {x^{3}}{3} \] Input:
int(x^2*cos(1/4+x+x^2)^2,x)
Output:
(2*cos((4*x**2 + 4*x + 1)/4)*sin((4*x**2 + 4*x + 1)/4)*x - cos((4*x**2 + 4 *x + 1)/4)*sin((4*x**2 + 4*x + 1)/4) - 6*int(tan((4*x**2 + 4*x + 1)/8)**2/ (tan((4*x**2 + 4*x + 1)/8)**4 + 2*tan((4*x**2 + 4*x + 1)/8)**2 + 1),x) - 8 *int(x**2/(tan((4*x**2 + 4*x + 1)/8)**4 + 2*tan((4*x**2 + 4*x + 1)/8)**2 + 1),x) - 8*int(tan((4*x**2 + 4*x + 1)/8)/(tan((4*x**2 + 4*x + 1)/8)**4 + 2 *tan((4*x**2 + 4*x + 1)/8)**2 + 1),x) + 2*int(1/(tan((4*x**2 + 4*x + 1)/8) **4 + 2*tan((4*x**2 + 4*x + 1)/8)**2 + 1),x) + 2*sin((4*x**2 + 4*x + 1)/4) *x - sin((4*x**2 + 4*x + 1)/4) + 2*x**3)/6