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

Optimal result

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

[Out]

1/4*cos(1/4+x+x^2)-1/2*x*cos(1/4+x+x^2)+1/4*FresnelC(1/2*(1+2*x)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)+1/8*Fresne
lS(1/2*(1+2*x)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3544, 3542, 3526, 3432, 3527, 3433} \[ \int x^2 \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )+\frac {1}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )-\frac {1}{2} x \cos \left (x^2+x+\frac {1}{4}\right )+\frac {1}{4} \cos \left (x^2+x+\frac {1}{4}\right ) \]

[In]

Int[x^2*Sin[1/4 + x + x^2],x]

[Out]

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

Rule 3432

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 3433

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 3526

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 3527

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 3542

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

Rule 3544

Int[((d_.) + (e_.)*(x_))^(m_)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[(-e)*(d + e*x)^(m - 1)
*(Cos[a + b*x + c*x^2]/(2*c)), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Sin[a + b*x + c*x^2], x]
, x] + Dist[e^2*((m - 1)/(2*c)), Int[(d + e*x)^(m - 2)*Cos[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2} x \cos \left (\frac {1}{4}+x+x^2\right )+\frac {1}{2} \int \cos \left (\frac {1}{4}+x+x^2\right ) \, dx-\frac {1}{2} \int x \sin \left (\frac {1}{4}+x+x^2\right ) \, dx \\ & = \frac {1}{4} \cos \left (\frac {1}{4}+x+x^2\right )-\frac {1}{2} x \cos \left (\frac {1}{4}+x+x^2\right )+\frac {1}{4} \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx+\frac {1}{2} \int \cos \left (\frac {1}{4} (1+2 x)^2\right ) \, dx \\ & = \frac {1}{4} \cos \left (\frac {1}{4}+x+x^2\right )-\frac {1}{2} x \cos \left (\frac {1}{4}+x+x^2\right )+\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )+\frac {1}{4} \int \sin \left (\frac {1}{4} (1+2 x)^2\right ) \, dx \\ & = \frac {1}{4} \cos \left (\frac {1}{4}+x+x^2\right )-\frac {1}{2} x \cos \left (\frac {1}{4}+x+x^2\right )+\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )+\frac {1}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int x^2 \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{8} \left (2 (1-2 x) \cos \left (\frac {1}{4}+x+x^2\right )+2 \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )+\sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right )\right ) \]

[In]

Integrate[x^2*Sin[1/4 + x + x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72

method	result	size
default	\(-\frac {x \cos \left (\frac {1}{4}+x +x^{2}\right )}{2}+\frac {\cos \left (\frac {1}{4}+x +x^{2}\right )}{4}+\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{8}+\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {C}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{4}\)	\(59\)
risch	\(-\frac {i \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i}\, x -\frac {i}{2 \sqrt {-i}}\right )}{16 \sqrt {-i}}+\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i}\, x -\frac {i}{2 \sqrt {-i}}\right )}{8 \sqrt {-i}}+\frac {\left (-1\right )^{\frac {1}{4}} \sqrt {\pi }\, \operatorname {erf}\left (\left (-1\right )^{\frac {1}{4}} x +\frac {\left (-1\right )^{\frac {1}{4}}}{2}\right )}{16}-\frac {\sqrt {\pi }\, \left (-1\right )^{\frac {3}{4}} \operatorname {erf}\left (\left (-1\right )^{\frac {1}{4}} x +\frac {\left (-1\right )^{\frac {1}{4}}}{2}\right )}{8}+2 \left (-\frac {x}{4}+\frac {1}{8}\right ) \cos \left (\frac {\left (1+2 x \right )^{2}}{4}\right )\)	\(108\)
parts	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right ) x^{2}}{2}-\frac {\sqrt {2}\, \pi ^{\frac {3}{2}} \left (\frac {\operatorname {S}\left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \left (\left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )^{2} \sqrt {\pi }-\sqrt {2}\, \left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )\right )}{\sqrt {\pi }}-\frac {-\frac {\left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \cos \left (\frac {\pi \left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )^{2}}{2}\right )}{\sqrt {\pi }}+\frac {\operatorname {C}\left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )}{\sqrt {\pi }}+\frac {\sqrt {2}\, \cos \left (\frac {\pi \left (\frac {x \sqrt {2}}{\sqrt {\pi }}+\frac {\sqrt {2}}{2 \sqrt {\pi }}\right )^{2}}{2}\right )}{\pi }}{\sqrt {\pi }}\right )}{4}\)	\(204\)

[In]

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

[Out]

-1/2*x*cos(1/4+x+x^2)+1/4*cos(1/4+x+x^2)+1/8*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(x+1/2))+1/4*2^(1/2)*P
i^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*(x+1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int x^2 \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=-\frac {1}{4} \, {\left (2 \, x - 1\right )} \cos \left (x^{2} + x + \frac {1}{4}\right ) + \frac {1}{4} \, \sqrt {2} \sqrt {\pi } \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {\pi } \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(x**2*sin(x**2 + x + 1/4), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.91 \[ \int x^2 \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {16 \, x {\left (e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}\right )} - \sqrt {4 \, x^{2} + 4 \, x + 1} {\left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x^{2} + i \, x + \frac {1}{4} i}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x^{2} - i \, x - \frac {1}{4} i}\right ) - 1\right )} - \left (4 i - 4\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, i \, x^{2} + i \, x + \frac {1}{4} i\right ) + \left (4 i + 4\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, -i \, x^{2} - i \, x - \frac {1}{4} i\right )\right )} + 8 \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + 8 \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}}{32 \, {\left (2 \, x + 1\right )}} \]

[In]

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

[Out]

1/32*(16*x*(e^(I*x^2 + I*x + 1/4*I) + e^(-I*x^2 - I*x - 1/4*I)) - sqrt(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)*gamma(3/2, -I*x^2 - I*x - 1/4*I)) + 8*e^(
I*x^2 + I*x + 1/4*I) + 8*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 = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int x^2 \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=-\left (\frac {1}{32} i + \frac {3}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) + \left (\frac {1}{32} i - \frac {3}{32}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) - \frac {1}{8} i \, {\left (-2 i \, x + i\right )} e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} - \frac {1}{8} i \, {\left (-2 i \, x + i\right )} e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int x^2 \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {\cos \left (x^2+x+\frac {1}{4}\right )}{4}-\frac {x\,\cos \left (x^2+x+\frac {1}{4}\right )}{2}+\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\left (2\,x+1\right )}{2\,\sqrt {\pi }}\right )}{4}+\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\left (2\,x+1\right )}{2\,\sqrt {\pi }}\right )}{8} \]

[In]

int(x^2*sin(x + x^2 + 1/4),x)

[Out]

cos(x + x^2 + 1/4)/4 - (x*cos(x + x^2 + 1/4))/2 + (2^(1/2)*pi^(1/2)*fresnelc((2^(1/2)*(2*x + 1))/(2*pi^(1/2)))
)/4 + (2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*(2*x + 1))/(2*pi^(1/2))))/8