3.16 \(\int x^2 \cos ^2(a+b x+c x^2) \, dx\)

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

[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)

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Rubi [A]  time = 0.241172, antiderivative size = 248, normalized size of antiderivative = 1., 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 = {3468, 3464, 3447, 3351, 3352, 3462, 3448} \[ -\frac{\sqrt{\pi } \sin \left (2 a-\frac{b^2}{2 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{\pi } \sqrt{c}}\right )}{16 c^{3/2}}+\frac{\sqrt{\pi } b^2 \cos \left (2 a-\frac{b^2}{2 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{\pi } \sqrt{c}}\right )}{16 c^{5/2}}-\frac{\sqrt{\pi } b^2 \sin \left (2 a-\frac{b^2}{2 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right )}{16 c^{5/2}}-\frac{\sqrt{\pi } \cos \left (2 a-\frac{b^2}{2 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right )}{16 c^{3/2}}-\frac{b \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac{x \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac{x^3}{6} \]

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 3468

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]

Rule 3464

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[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Sin[a + b*x + c*x^2], x], x
] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*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]

Rule 3447

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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3462

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 3448

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]

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*}

Mathematica [A]  time = 0.663893, size = 170, normalized size = 0.69 \[ \frac{3 \sqrt{\pi } \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{\pi } \sqrt{c}}\right ) \left (b^2 \cos \left (2 a-\frac{b^2}{2 c}\right )-c \sin \left (2 a-\frac{b^2}{2 c}\right )\right )-3 \sqrt{\pi } S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right ) \left (b^2 \sin \left (2 a-\frac{b^2}{2 c}\right )+c \cos \left (2 a-\frac{b^2}{2 c}\right )\right )+\sqrt{c} \left (8 c^2 x^3-3 (b-2 c x) \sin (2 (a+x (b+c x)))\right )}{48 c^{5/2}} \]

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))

________________________________________________________________________________________

Maple [A]  time = 0.037, size = 191, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{6}}+{\frac{x\sin \left ( 2\,c{x}^{2}+2\,bx+2\,a \right ) }{8\,c}}-{\frac{b}{4\,c} \left ({\frac{\sin \left ( 2\,c{x}^{2}+2\,bx+2\,a \right ) }{4\,c}}-{\frac{b\sqrt{\pi }}{4} \left ( \cos \left ({\frac{-4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelC} \left ({\frac{2\,cx+b}{\sqrt{\pi }}{\frac{1}{\sqrt{c}}}} \right ) +\sin \left ({\frac{-4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelS} \left ({\frac{2\,cx+b}{\sqrt{\pi }}{\frac{1}{\sqrt{c}}}} \right ) \right ){c}^{-{\frac{3}{2}}}} \right ) }-{\frac{\sqrt{\pi }}{16} \left ( \cos \left ({\frac{-4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelS} \left ({\frac{2\,cx+b}{\sqrt{\pi }}{\frac{1}{\sqrt{c}}}} \right ) -\sin \left ({\frac{-4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelC} \left ({\frac{2\,cx+b}{\sqrt{\pi }}{\frac{1}{\sqrt{c}}}} \right ) \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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]  time = 3.49499, size = 4232, normalized size = 17.06 \begin{align*} \text{result too large to display} \end{align*}

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]

1/192*sqrt(2)*(2*sqrt(2)*(8*c^3*x^3*abs(c) + b*c*(3*I*e^(1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 3*I*e^(-1/
2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) - 3*b*c*(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))*abs(c)*sin(-1/2*(b^2 - 4*a*c)/c))*((4*c^2*x^2
+ 4*b*c*x + b^2)/abs(c))^(3/2) - (6*b^3*(gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2
*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) - b^3*(6*I*gamma(3/2, 1/2*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c) - 6*I*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/2*(b^2 - 4*
a*c)/c) + (48*c^3*(gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*
c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) - c^3*(48*I*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)
/c) - 48*I*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/2*(b^2 - 4*a*c)/c))*x^3 + (72*b
*c^2*(gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/
c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) - b*c^2*(72*I*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 72*I*
gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/2*(b^2 - 4*a*c)/c))*x^2 + (36*b^2*c*(gamma
(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*
cos(-1/2*(b^2 - 4*a*c)/c) - b^2*c*(36*I*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 36*I*gamma(3/2,
-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/2*(b^2 - 4*a*c)/c))*x)*cos(3/2*arctan2(2*(4*c^2*x^2 +
4*b*c*x + b^2)/c, 0)) + (3*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 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*cos(-1/2*(b^2 - 4*a*c)/c) + (-3*I*sqrt(
pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 3*I*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*sin(-1/2*(b^2 - 4*a*c)/c) + (24*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^
2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 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*cos(-1/2*(b^2 - 4*a*c)/c) + (-24*I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)
) - 1) + 24*I*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*sin(-1/2*(b^2
- 4*a*c)/c))*x^3 + (36*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(er
f(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2*cos(-1/2*(b^2 - 4*a*c)/c) + (-36*I*sqrt(
pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 36*I*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*sin(-1/2*(b^2 - 4*a*c)/c))*x^2 + (18*(sqrt(pi)*(erf(sqrt(1/2)*sq
rt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 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*cos(-1/2*(b^2 - 4*a*c)/c) + (-18*I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I
*b^2)/c)) - 1) + 18*I*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*sin(-1/2
*(b^2 - 4*a*c)/c))*x)*cos(1/2*arctan2(2*(4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)) + (b^3*(6*I*gamma(3/2, 1/2*(4*I*c^2
*x^2 + 4*I*b*c*x + I*b^2)/c) - 6*I*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2
- 4*a*c)/c) + 6*b^3*(gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*
b*c*x + I*b^2)/c))*abs(c)*sin(-1/2*(b^2 - 4*a*c)/c) + (c^3*(48*I*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b
^2)/c) - 48*I*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) + 48*c^3*
(gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*a
bs(c)*sin(-1/2*(b^2 - 4*a*c)/c))*x^3 + (b*c^2*(72*I*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 72*I
*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) + 72*b*c^2*(gamma(3/2,
1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-
1/2*(b^2 - 4*a*c)/c))*x^2 + (b^2*c*(36*I*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - 36*I*gamma(3/2,
-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*cos(-1/2*(b^2 - 4*a*c)/c) + 36*b^2*c*(gamma(3/2, 1/2*(4*I*c
^2*x^2 + 4*I*b*c*x + I*b^2)/c) + gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*abs(c)*sin(-1/2*(b^2 -
4*a*c)/c))*x)*sin(3/2*arctan2(2*(4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)) + ((-3*I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*
c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 3*I*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*cos(-1/2*(b^2 - 4*a*c)/c) - 3*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))
- 1) + 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*sin(-1/2*(b^2 - 4*a*c)/c)
+ ((-24*I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 24*I*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*cos(-1/2*(b^2 - 4*a*c)/c) - 24*(sqrt(pi)*(erf(sq
rt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 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*sin(-1/2*(b^2 - 4*a*c)/c))*x^3 + ((-36*I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x
^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 36*I*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*cos(-1/2*(b^2 - 4*a*c)/c) - 36*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))
- 1) + 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*sin(-1/2*(b^2 - 4*a*
c)/c))*x^2 + ((-18*I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 18*I*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*cos(-1/2*(b^2 - 4*a*c)/c) - 18*(sqrt(pi)
*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + 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*sin(-1/2*(b^2 - 4*a*c)/c))*x)*sin(1/2*arctan2(2*(4*c^2*x^2 + 4*b*c*x + b^2
)/c, 0)))/(c^3*((4*c^2*x^2 + 4*b*c*x + b^2)/abs(c))^(3/2)*abs(c))

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Fricas [A]  time = 1.59286, size = 440, normalized size = 1.77 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cos ^{2}{\left (a + b x + c x^{2} \right )}\, dx \end{align*}

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]  time = 1.32767, size = 286, normalized size = 1.15 \begin{align*} \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{align*}

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