3.1.1 \(\int x^2 \cos (a+b x+c x^2) \, dx\) [1]

Optimal. Leaf size=249 \[ \frac {b^2 \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}-\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )}{2 c^{3/2}}-\frac {b^2 \sqrt {\frac {\pi }{2}} S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )}{4 c^{5/2}}-\frac {b \sin \left (a+b x+c x^2\right )}{4 c^2}+\frac {x \sin \left (a+b x+c x^2\right )}{2 c} \]

[Out]

-1/4*b*sin(c*x^2+b*x+a)/c^2+1/2*x*sin(c*x^2+b*x+a)/c+1/8*b^2*cos(a-1/4*b^2/c)*FresnelC(1/2*(2*c*x+b)/c^(1/2)*2
^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/c^(5/2)-1/4*cos(a-1/4*b^2/c)*FresnelS(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2)
)*2^(1/2)*Pi^(1/2)/c^(3/2)-1/4*FresnelC(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a-1/4*b^2/c)*2^(1/2)*Pi^(1
/2)/c^(3/2)-1/8*b^2*FresnelS(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a-1/4*b^2/c)*2^(1/2)*Pi^(1/2)/c^(5/2)

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Rubi [A]
time = 0.16, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3545, 3543, 3529, 3433, 3432, 3528} \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} \sin \left (a-\frac {b^2}{4 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} b^2 \cos \left (a-\frac {b^2}{4 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{4 c^{5/2}}-\frac {\sqrt {\frac {\pi }{2}} b^2 \sin \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}-\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {b \sin \left (a+b x+c x^2\right )}{4 c^2}+\frac {x \sin \left (a+b x+c x^2\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x + c*x^2],x]

[Out]

(b^2*Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(4*c^(5/2)) - (Sqrt[Pi/2]*Cos[a
- b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(2*c^(3/2)) - (Sqrt[Pi/2]*FresnelC[(b + 2*c*x)/(Sqrt
[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(2*c^(3/2)) - (b^2*Sqrt[Pi/2]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*
Sin[a - b^2/(4*c)])/(4*c^(5/2)) - (b*Sin[a + b*x + c*x^2])/(4*c^2) + (x*Sin[a + b*x + c*x^2])/(2*c)

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 3528

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

Rule 3529

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

Rule 3543

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] + Dist[(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 3545

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

Rubi steps

\begin {align*} \int x^2 \cos \left (a+b x+c x^2\right ) \, dx &=\frac {x \sin \left (a+b x+c x^2\right )}{2 c}-\frac {\int \sin \left (a+b x+c x^2\right ) \, dx}{2 c}-\frac {b \int x \cos \left (a+b x+c x^2\right ) \, dx}{2 c}\\ &=-\frac {b \sin \left (a+b x+c x^2\right )}{4 c^2}+\frac {x \sin \left (a+b x+c x^2\right )}{2 c}+\frac {b^2 \int \cos \left (a+b x+c x^2\right ) \, dx}{4 c^2}-\frac {\cos \left (a-\frac {b^2}{4 c}\right ) \int \sin \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{2 c}-\frac {\sin \left (a-\frac {b^2}{4 c}\right ) \int \cos \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{2 c}\\ &=-\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )}{2 c^{3/2}}-\frac {b \sin \left (a+b x+c x^2\right )}{4 c^2}+\frac {x \sin \left (a+b x+c x^2\right )}{2 c}+\frac {\left (b^2 \cos \left (a-\frac {b^2}{4 c}\right )\right ) \int \cos \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{4 c^2}-\frac {\left (b^2 \sin \left (a-\frac {b^2}{4 c}\right )\right ) \int \sin \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{4 c^2}\\ &=\frac {b^2 \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}-\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )}{2 c^{3/2}}-\frac {b^2 \sqrt {\frac {\pi }{2}} S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )}{4 c^{5/2}}-\frac {b \sin \left (a+b x+c x^2\right )}{4 c^2}+\frac {x \sin \left (a+b x+c x^2\right )}{2 c}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 160, normalized size = 0.64 \begin {gather*} \frac {-\sqrt {2 \pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (2 c \cos \left (a-\frac {b^2}{4 c}\right )+b^2 \sin \left (a-\frac {b^2}{4 c}\right )\right )+\sqrt {2 \pi } \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (b^2 \cos \left (a-\frac {b^2}{4 c}\right )-2 c \sin \left (a-\frac {b^2}{4 c}\right )\right )+2 \sqrt {c} (-b+2 c x) \sin (a+x (b+c x))}{8 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x + c*x^2],x]

[Out]

(-(Sqrt[2*Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*(2*c*Cos[a - b^2/(4*c)] + b^2*Sin[a - b^2/(4*c)])) +
Sqrt[2*Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*(b^2*Cos[a - b^2/(4*c)] - 2*c*Sin[a - b^2/(4*c)]) + 2*Sq
rt[c]*(-b + 2*c*x)*Sin[a + x*(b + c*x)])/(8*c^(5/2))

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Maple [A]
time = 0.15, size = 204, normalized size = 0.82

method	result	size
default	\(\frac {x \sin \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {b \left (\frac {\sin \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-a c}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}-a c}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{2 c}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-a c}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-a c}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\)	\(204\)
risch	\(\frac {b^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{4 c}} \erf \left (\sqrt {i c}\, x +\frac {i b}{2 \sqrt {i c}}\right )}{16 c^{2} \sqrt {i c}}-\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{4 c}} \erf \left (\sqrt {i c}\, x +\frac {i b}{2 \sqrt {i c}}\right )}{8 c \sqrt {i c}}-\frac {b^{2} \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{4 c}} \erf \left (-\sqrt {-i c}\, x +\frac {i b}{2 \sqrt {-i c}}\right )}{16 c^{2} \sqrt {-i c}}-\frac {i \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{4 c}} \erf \left (-\sqrt {-i c}\, x +\frac {i b}{2 \sqrt {-i c}}\right )}{8 c \sqrt {-i c}}-2 i \left (\frac {i x}{4 c}-\frac {i b}{8 c^{2}}\right ) \sin \left (c \,x^{2}+b x +a \right )\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 1.16, size = 1570, normalized size = 6.31 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*(((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (I + 1)*sqrt(2
)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + 4*((I + 1)*sqrt(2)*gamma(3/2,
1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I - 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c
))*c^4)*cos(-1/4*(b^2 - 4*a*c)/c) + ((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2
)/c)) - 1) + (I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + 4*(
-(I - 1)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^
2*x^2 + 4*I*b*c*x + I*b^2)/c))*c^4)*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + 12*(((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*
sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b
*c*x + I*b^2)/c)) - 1))*b^3*c^2 + 4*((I + 1)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I
- 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*cos(-1/4*(b^2 - 4*a*c)/c) + ((-(I +
1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (I - 1)*sqrt(2)*sqrt(pi)*(erf(1
/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + 4*(-(I - 1)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*sin(-1
/4*(b^2 - 4*a*c)/c))*x^2 + 8*(b*c^2*(I*e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - I*e^(-1/4*(4*I*c^2*x^2 +
4*I*b*c*x + I*b^2)/c))*cos(-1/4*(b^2 - 4*a*c)/c) - b*c^2*(e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + e^(-1/
4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*sin(-1/4*(b^2 - 4*a*c)/c))*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2) + 6*(
((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (I + 1)*sqrt(2)*sqrt(pi
)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + 4*((I + 1)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^
2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I - 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)
*cos(-1/4*(b^2 - 4*a*c)/c) + ((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) -
1) + (I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + 4*(-(I - 1)*
sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4
*I*b*c*x + I*b^2)/c))*b^2*c^2)*sin(-1/4*(b^2 - 4*a*c)/c))*x - (((I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^
2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2
)/c)) - 1))*b^5 - 4*((I + 1)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I - 1)*sqrt(2)*gam
ma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*cos(-1/4*(b^2 - 4*a*c)/c) - (((I + 1)*sqrt(2)*sqrt(p
i)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2
*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5 - 4*(-(I - 1)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)
/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*sin(-1/4*(b^2 - 4*a*c)/c))/
(c^4*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2))

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Fricas [A]
time = 0.51, size = 173, normalized size = 0.69 \begin {gather*} \frac {\sqrt {2} {\left (\pi b^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - 2 \, \pi c \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} {\left (\pi b^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + 2 \, \pi c \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + 2 \, {\left (2 \, c^{2} x - b c\right )} \sin \left (c x^{2} + b x + a\right )}{8 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \cos {\left (a + b x + c x^{2} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cos(c*x**2+b*x+a),x)

[Out]

Integral(x**2*cos(a + b*x + c*x**2), x)

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Giac [C] Result contains complex when optimal does not.
time = 0.45, size = 225, normalized size = 0.90 \begin {gather*} -\frac {\frac {\sqrt {2} \sqrt {\pi } {\left (b^{2} + 2 i \, c\right )} \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 \, {\left (c {\left (-2 i \, x - \frac {i \, b}{c}\right )} + 2 i \, b\right )} e^{\left (i \, c x^{2} + i \, b x + i \, a\right )}}{16 \, c^{2}} - \frac {\frac {\sqrt {2} \sqrt {\pi } {\left (b^{2} - 2 i \, c\right )} \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 \, {\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (-i \, c x^{2} - i \, b x - i \, a\right )}}{16 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(2)*sqrt(pi)*(b^2 + 2*I*c)*erf(-1/4*sqrt(2)*(2*x + b/c)*(-I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(I*
b^2 - 4*I*a*c)/c)/((-I*c/abs(c) + 1)*sqrt(abs(c))) - 2*(c*(-2*I*x - I*b/c) + 2*I*b)*e^(I*c*x^2 + I*b*x + I*a))
/c^2 - 1/16*(sqrt(2)*sqrt(pi)*(b^2 - 2*I*c)*erf(-1/4*sqrt(2)*(2*x + b/c)*(I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/
4*(-I*b^2 + 4*I*a*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) - 2*(c*(2*I*x + I*b/c) - 2*I*b)*e^(-I*c*x^2 - I*b*x -
I*a))/c^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\cos \left (c\,x^2+b\,x+a\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a + b*x + c*x^2),x)

[Out]

int(x^2*cos(a + b*x + c*x^2), x)

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