3.1.16 \(\int x^2 \cos ^2(a+b x+c x^2) \, dx\) [16]

Optimal. Leaf size=248 \[ \frac {x^3}{6}+\frac {b^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}-\frac {b^2 \sqrt {\pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{5/2}}-\frac {b \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c} \]

[Out]

1/6*x^3-1/16*b*sin(2*c*x^2+2*b*x+2*a)/c^2+1/8*x*sin(2*c*x^2+2*b*x+2*a)/c+1/16*b^2*cos(2*a-1/2*b^2/c)*FresnelC(
(2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(1/2)/c^(5/2)-1/16*cos(2*a-1/2*b^2/c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(
1/2)/c^(3/2)-1/16*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a-1/2*b^2/c)*Pi^(1/2)/c^(3/2)-1/16*b^2*FresnelS((
2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a-1/2*b^2/c)*Pi^(1/2)/c^(5/2)

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Rubi [A]
time = 0.18, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3549, 3545, 3543, 3529, 3433, 3432, 3528} \begin {gather*} -\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } b^2 \cos \left (2 a-\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } b^2 \sin \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {b \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {x^3}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x + c*x^2]^2,x]

[Out]

x^3/6 + (b^2*Sqrt[Pi]*Cos[2*a - b^2/(2*c)]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(5/2)) - (Sqrt[Pi]*
Cos[2*a - b^2/(2*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(3/2)) - (Sqrt[Pi]*FresnelC[(b + 2*c*x)/(
Sqrt[c]*Sqrt[Pi])]*Sin[2*a - b^2/(2*c)])/(16*c^(3/2)) - (b^2*Sqrt[Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]
*Sin[2*a - b^2/(2*c)])/(16*c^(5/2)) - (b*Sin[2*a + 2*b*x + 2*c*x^2])/(16*c^2) + (x*Sin[2*a + 2*b*x + 2*c*x^2])
/(8*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]

Rule 3549

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]

Rubi steps

\begin {align*} \int x^2 \cos ^2\left (a+b x+c x^2\right ) \, dx &=\int \left (\frac {x^2}{2}+\frac {1}{2} x^2 \cos \left (2 a+2 b x+2 c x^2\right )\right ) \, dx\\ &=\frac {x^3}{6}+\frac {1}{2} \int x^2 \cos \left (2 a+2 b x+2 c x^2\right ) \, dx\\ &=\frac {x^3}{6}+\frac {x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}-\frac {\int \sin \left (2 a+2 b x+2 c x^2\right ) \, dx}{8 c}-\frac {b \int x \cos \left (2 a+2 b x+2 c x^2\right ) \, dx}{4 c}\\ &=\frac {x^3}{6}-\frac {b \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {b^2 \int \cos \left (2 a+2 b x+2 c x^2\right ) \, dx}{8 c^2}-\frac {\cos \left (2 a-\frac {b^2}{2 c}\right ) \int \sin \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c}-\frac {\sin \left (2 a-\frac {b^2}{2 c}\right ) \int \cos \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c}\\ &=\frac {x^3}{6}-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {\sqrt {\pi } C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}-\frac {b \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {\left (b^2 \cos \left (2 a-\frac {b^2}{2 c}\right )\right ) \int \cos \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c^2}-\frac {\left (b^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right ) \int \sin \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c^2}\\ &=\frac {x^3}{6}+\frac {b^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {\sqrt {\pi } C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}-\frac {b^2 \sqrt {\pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{5/2}}-\frac {b \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}\\ \end {align*}

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Mathematica [A]
time = 0.65, size = 170, normalized size = 0.69 \begin {gather*} \frac {-3 \sqrt {\pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left (c \cos \left (2 a-\frac {b^2}{2 c}\right )+b^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right )+3 \sqrt {\pi } \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left (b^2 \cos \left (2 a-\frac {b^2}{2 c}\right )-c \sin \left (2 a-\frac {b^2}{2 c}\right )\right )+\sqrt {c} \left (8 c^2 x^3-3 (b-2 c x) \sin (2 (a+x (b+c x)))\right )}{48 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x + c*x^2]^2,x]

[Out]

(-3*Sqrt[Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*(c*Cos[2*a - b^2/(2*c)] + b^2*Sin[2*a - b^2/(2*c)]) + 3*
Sqrt[Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*(b^2*Cos[2*a - b^2/(2*c)] - c*Sin[2*a - b^2/(2*c)]) + Sqrt[c
]*(8*c^2*x^3 - 3*(b - 2*c*x)*Sin[2*(a + x*(b + c*x))]))/(48*c^(5/2))

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Maple [A]
time = 0.21, size = 191, normalized size = 0.77

method	result	size
default	\(\frac {x^{3}}{6}+\frac {x \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}-\frac {b \left (\frac {\sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{4 c}-\frac {\sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )-\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{16 c^{\frac {3}{2}}}\)	\(191\)
risch	\(\frac {x^{3}}{6}+\frac {b^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}} \sqrt {2}\, \erf \left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right )}{64 c^{2} \sqrt {i c}}-\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}} \sqrt {2}\, \erf \left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right )}{64 c \sqrt {i c}}-\frac {b^{2} \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}} \erf \left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right )}{32 c^{2} \sqrt {-2 i c}}-\frac {i \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}} \erf \left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right )}{32 c \sqrt {-2 i c}}+2 i \left (-\frac {i x}{16 c}+\frac {i b}{32 c^{2}}\right ) \sin \left (2 c \,x^{2}+2 b x +2 a \right )\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3+1/8*x*sin(2*c*x^2+2*b*x+2*a)/c-1/4*b/c*(1/4*sin(2*c*x^2+2*b*x+2*a)/c-1/4*b/c^(3/2)*Pi^(1/2)*(cos(1/2*(
-4*a*c+b^2)/c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))+sin(1/2*(-4*a*c+b^2)/c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2
))))-1/16/c^(3/2)*Pi^(1/2)*(cos(1/2*(-4*a*c+b^2)/c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))-sin(1/2*(-4*a*c+b^2)/
c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2)))

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Maxima [C] Result contains complex when optimal does not.
time = 1.13, size = 1617, normalized size = 6.52 \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)^2,x, algorithm="maxima")

[Out]

-1/384*sqrt(2)*(24*((((I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + 2*(-(I +
1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I - 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c))*c^4)*cos(-1/2*(b^2 - 4*a*c)/c) + (((I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*
x + I*b^2)/c)) - 1))*b^2*c^3 + 2*((I - 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I + 1
)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*c^4)*sin(-1/2*(b^2 - 4*a*c)/c))*x^3 + 36*((((I
- 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (I + 1)*sqrt(2)*sqrt(p
i)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + 2*(-(I + 1)*sqrt(2)*gamma(3/2, 1
/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I - 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)
)*b*c^3)*cos(-1/2*(b^2 - 4*a*c)/c) + (((I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b
^3*c^2 + 2*((I - 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I + 1)*sqrt(2)*gamma(3/2, -
1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*sin(-1/2*(b^2 - 4*a*c)/c))*x^2 - 4*sqrt(2)*(8*c^4*x^3 - 3*b*c
^2*(-I*e^(1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + I*e^(-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*cos(-1/2*
(b^2 - 4*a*c)/c) - 3*b*c^2*(e^(1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + e^(-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I
*b^2)/c))*sin(-1/2*(b^2 - 4*a*c)/c))*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2) + 18*((((I - 1)*sqrt(2)*sqrt(pi)*(e
rf(sqrt(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(sqrt(1/2)*sqrt(-(
4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + 2*(-(I + 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c) + (I - 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*cos(-1/2*(b^2 -
4*a*c)/c) + (((I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + 2*((I - 1)*sqrt(2)
*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I + 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*
x + I*b^2)/c))*b^2*c^2)*sin(-1/2*(b^2 - 4*a*c)/c))*x + 3*(((I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c)) - 1))*b^5 + 2*(-(I + 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I - 1)*sqr
t(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*cos(-1/2*(b^2 - 4*a*c)/c) + 3*(((I + 1)*sqrt
(2)*sqrt(pi)*(erf(sqrt(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(sq
rt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5 + 2*((I - 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c) - (I + 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*sin(-1
/2*(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 = 0.43, size = 178, normalized size = 0.72 \begin {gather*} \frac {8 \, c^{3} x^{3} + 6 \, {\left (2 \, c^{2} x - b c\right )} \cos \left (c x^{2} + b x + a\right ) \sin \left (c x^{2} + b x + a\right ) + 3 \, {\left (\pi b^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) - \pi c \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) - 3 \, {\left (\pi b^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \pi c \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right )}{48 \, 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)^2,x, algorithm="fricas")

[Out]

1/48*(8*c^3*x^3 + 6*(2*c^2*x - b*c)*cos(c*x^2 + b*x + a)*sin(c*x^2 + b*x + a) + 3*(pi*b^2*cos(-1/2*(b^2 - 4*a*
c)/c) - pi*c*sin(-1/2*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_cos((2*c*x + b)*sqrt(c/pi)/c) - 3*(pi*b^2*sin(-1/2*
(b^2 - 4*a*c)/c) + pi*c*cos(-1/2*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_sin((2*c*x + b)*sqrt(c/pi)/c))/c^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \cos ^{2}{\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)**2,x)

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.47, size = 212, normalized size = 0.85 \begin {gather*} \frac {1}{6} \, x^{3} - \frac {{\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )} + \frac {\sqrt {\pi } {\left (b^{2} + i \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} - \frac {{\left (c {\left (-2 i \, x - \frac {i \, b}{c}\right )} + 2 i \, b\right )} e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )} + \frac {\sqrt {\pi } {\left (b^{2} - i \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, 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)^2,x, algorithm="giac")

[Out]

1/6*x^3 - 1/32*((c*(2*I*x + I*b/c) - 2*I*b)*e^(2*I*c*x^2 + 2*I*b*x + 2*I*a) + sqrt(pi)*(b^2 + I*c)*erf(-1/2*sq
rt(c)*(2*x + b/c)*(-I*c/abs(c) + 1))*e^(-1/2*(I*b^2 - 4*I*a*c)/c)/(sqrt(c)*(-I*c/abs(c) + 1)))/c^2 - 1/32*((c*
(-2*I*x - I*b/c) + 2*I*b)*e^(-2*I*c*x^2 - 2*I*b*x - 2*I*a) + sqrt(pi)*(b^2 - I*c)*erf(-1/2*sqrt(c)*(2*x + b/c)
*(I*c/abs(c) + 1))*e^(-1/2*(-I*b^2 + 4*I*a*c)/c)/(sqrt(c)*(I*c/abs(c) + 1)))/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 )}^2 \,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)^2,x)

[Out]

int(x^2*cos(a + b*x + c*x^2)^2, x)

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