3.1.32 \(\int (d+e x)^2 \cos ^2(a+b x+c x^2) \, dx\) [32]

3.1.32.1 Optimal result
3.1.32.2 Mathematica [A] (verified)
3.1.32.3 Rubi [A] (verified)
3.1.32.4 Maple [A] (verified)
3.1.32.5 Fricas [A] (verification not implemented)
3.1.32.6 Sympy [F]
3.1.32.7 Maxima [C] (verification not implemented)
3.1.32.8 Giac [C] (verification not implemented)
3.1.32.9 Mupad [F(-1)]

3.1.32.1 Optimal result

Integrand size = 21, antiderivative size = 291 \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {(d+e x)^3}{6 e}+\frac {(2 c d-b e)^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {e^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {e^2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}-\frac {(2 c d-b e)^2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{5/2}}+\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c} \]

output
1/6*(e*x+d)^3/e+1/16*e*(-b*e+2*c*d)*sin(2*c*x^2+2*b*x+2*a)/c^2+1/8*e*(e*x+ 
d)*sin(2*c*x^2+2*b*x+2*a)/c+1/16*(-b*e+2*c*d)^2*cos(2*a-1/2*b^2/c)*Fresnel 
C((2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(1/2)/c^(5/2)-1/16*e^2*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*e^2*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*e+ 
2*c*d)^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)
 
3.1.32.2 Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.74 \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left ((-2 c d+b e)^2 \cos \left (2 a-\frac {b^2}{2 c}\right )-c e^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right )-3 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left (c e^2 \cos \left (2 a-\frac {b^2}{2 c}\right )+(-2 c d+b e)^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right )+\sqrt {c} \left (8 c^2 x \left (3 d^2+3 d e x+e^2 x^2\right )+3 e (4 c d-b e+2 c e x) \sin (2 (a+x (b+c x)))\right )}{48 c^{5/2}} \]

input
Integrate[(d + e*x)^2*Cos[a + b*x + c*x^2]^2,x]
 
output
(3*Sqrt[Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*((-2*c*d + b*e)^2*Cos 
[2*a - b^2/(2*c)] - c*e^2*Sin[2*a - b^2/(2*c)]) - 3*Sqrt[Pi]*FresnelS[(b + 
 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*(c*e^2*Cos[2*a - b^2/(2*c)] + (-2*c*d + b*e)^2 
*Sin[2*a - b^2/(2*c)]) + Sqrt[c]*(8*c^2*x*(3*d^2 + 3*d*e*x + e^2*x^2) + 3* 
e*(4*c*d - b*e + 2*c*e*x)*Sin[2*(a + x*(b + c*x))]))/(48*c^(5/2))
 
3.1.32.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

\(\displaystyle \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx\)

\(\Big \downarrow \) 3949

\(\displaystyle \int \left (\frac {1}{2} (d+e x)^2 \cos \left (2 a+2 b x+2 c x^2\right )+\frac {1}{2} (d+e x)^2\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } e^2 \sin \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {\sqrt {\pi } e^2 \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}+\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {(d+e x)^3}{6 e}\)

input
Int[(d + e*x)^2*Cos[a + b*x + c*x^2]^2,x]
 
output
(d + e*x)^3/(6*e) + ((2*c*d - b*e)^2*Sqrt[Pi]*Cos[2*a - b^2/(2*c)]*Fresnel 
C[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(5/2)) - (e^2*Sqrt[Pi]*Cos[2*a - 
b^2/(2*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(3/2)) - (e^2*S 
qrt[Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*Sin[2*a - b^2/(2*c)])/(16 
*c^(3/2)) - ((2*c*d - b*e)^2*Sqrt[Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[P 
i])]*Sin[2*a - b^2/(2*c)])/(16*c^(5/2)) + (e*(2*c*d - b*e)*Sin[2*a + 2*b*x 
 + 2*c*x^2])/(16*c^2) + (e*(d + e*x)*Sin[2*a + 2*b*x + 2*c*x^2])/(8*c)
 

3.1.32.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3949
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]
 
3.1.32.4 Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.30

method result size
default \(\frac {e^{2} x \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}-\frac {e^{2} b \left (\frac {\sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{4 c}-\frac {e^{2} \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )-\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{16 c^{\frac {3}{2}}}+\frac {d e \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {d e b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}+\frac {\sqrt {\pi }\, d^{2} \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}}+\frac {d e \,x^{2}}{2}+\frac {d^{2} x}{2}+\frac {e^{2} x^{3}}{6}\) \(378\)
risch \(\frac {d e \,x^{2}}{2}+\frac {d^{2} x}{2}+\frac {e^{2} x^{3}}{6}+\frac {\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, d^{2} {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {i c}}+\frac {e^{2} \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, b^{2} {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{64 \sqrt {i c}\, c^{2}}-\frac {i e^{2} \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{64 \sqrt {i c}\, c}-\frac {d e \,\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, b \,{\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {i c}\, c}-\frac {\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, d^{2} {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}}-\frac {e^{2} \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, b^{2} {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{32 \sqrt {-2 i c}\, c^{2}}-\frac {i e^{2} \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{32 \sqrt {-2 i c}\, c}+\frac {d e \,\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, b \,{\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}\, c}+2 i \left (\frac {e^{2} \left (-\frac {i x}{4 c}+\frac {i b}{8 c^{2}}\right )}{4}-\frac {i d e}{8 c}\right ) \sin \left (2 c \,x^{2}+2 b x +2 a \right )\) \(544\)

input
int((e*x+d)^2*cos(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
 
output
1/8*e^2/c*x*sin(2*c*x^2+2*b*x+2*a)-1/4*e^2*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*e^2/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)))+1/4*d*e/c*sin(2*c*x^2+2*b*x+2*a)-1/4*d*e*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)))+1/4*Pi^(1/2)/c^(1/2)*d^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)))+1/2*d*e*x^2+1/2*d^2*x+1/6*e^2*x^ 
3
 
3.1.32.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.88 \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {8 \, c^{3} e^{2} x^{3} + 24 \, c^{3} d e x^{2} + 24 \, c^{3} d^{2} x + 6 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \cos \left (c x^{2} + b x + a\right ) \sin \left (c x^{2} + b x + a\right ) - 3 \, {\left (\pi c e^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) - \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) - 3 \, {\left (\pi c e^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right )}{48 \, c^{3}} \]

input
integrate((e*x+d)^2*cos(c*x^2+b*x+a)^2,x, algorithm="fricas")
 
output
1/48*(8*c^3*e^2*x^3 + 24*c^3*d*e*x^2 + 24*c^3*d^2*x + 6*(2*c^2*e^2*x + 4*c 
^2*d*e - b*c*e^2)*cos(c*x^2 + b*x + a)*sin(c*x^2 + b*x + a) - 3*(pi*c*e^2* 
sin(-1/2*(b^2 - 4*a*c)/c) - pi*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*cos(-1/2* 
(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_cos((2*c*x + b)*sqrt(c/pi)/c) - 3*(pi 
*c*e^2*cos(-1/2*(b^2 - 4*a*c)/c) + pi*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*si 
n(-1/2*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_sin((2*c*x + b)*sqrt(c/pi)/c)) 
/c^3
 
3.1.32.6 Sympy [F]

\[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\int \left (d + e x\right )^{2} \cos ^{2}{\left (a + b x + c x^{2} \right )}\, dx \]

input
integrate((e*x+d)**2*cos(c*x**2+b*x+a)**2,x)
 
output
Integral((d + e*x)**2*cos(a + b*x + c*x**2)**2, x)
 
3.1.32.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.38 (sec) , antiderivative size = 2360, normalized size of antiderivative = 8.11 \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\text {Too large to display} \]

input
integrate((e*x+d)^2*cos(c*x^2+b*x+a)^2,x, algorithm="maxima")
 
output
-1/32*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I + 
 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*c*x + I*b)/sqrt(2*I*c)) + ((I + 1) 
*cos(-1/2*(b^2 - 4*a*c)/c) + (I - 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*c 
*x + I*b)/sqrt(-2*I*c)))*c^(3/2) - 16*c^2*x)*d^2/c^2 + 1/32*sqrt(2)*(((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*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*sin(-1/2*(b^2 - 4*a*c)/c) - 2*((-(I - 1)*sqrt(2)*sq 
rt(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*c*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)*sq 
rt(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*c*sin(-1/2*(b^2 - 4*a*c)/c))*x + 2*sqrt(2)*(4*c^2*x^2 - c*(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) + 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))*sin(-1/2*( 
b^2 - 4*a*c)/c))*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/c))*d*e/(c^2*sqrt((4*...
 
3.1.32.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00 \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {1}{6} \, e^{2} x^{3} + \frac {1}{2} \, d e x^{2} + \frac {1}{2} \, d^{2} x - \frac {i \, {\left (c e^{2} {\left (2 \, x + \frac {b}{c}\right )} + 4 \, c d e - 2 \, b e^{2}\right )} e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )} + \frac {\sqrt {\pi } {\left (-4 i \, c^{2} d^{2} + 4 i \, b c d e - i \, b^{2} e^{2} + c e^{2}\right )} \operatorname {erf}\left (-\frac {1}{2} i \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} - \frac {-i \, {\left (c e^{2} {\left (2 \, x + \frac {b}{c}\right )} + 4 \, c d e - 2 \, b e^{2}\right )} e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )} + \frac {\sqrt {\pi } {\left (4 i \, c^{2} d^{2} - 4 i \, b c d e + i \, b^{2} e^{2} + c e^{2}\right )} \operatorname {erf}\left (\frac {1}{2} i \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} \]

input
integrate((e*x+d)^2*cos(c*x^2+b*x+a)^2,x, algorithm="giac")
 
output
1/6*e^2*x^3 + 1/2*d*e*x^2 + 1/2*d^2*x - 1/32*(I*(c*e^2*(2*x + b/c) + 4*c*d 
*e - 2*b*e^2)*e^(2*I*c*x^2 + 2*I*b*x + 2*I*a) + sqrt(pi)*(-4*I*c^2*d^2 + 4 
*I*b*c*d*e - I*b^2*e^2 + c*e^2)*erf(-1/2*I*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 - 1/32 
*(-I*(c*e^2*(2*x + b/c) + 4*c*d*e - 2*b*e^2)*e^(-2*I*c*x^2 - 2*I*b*x - 2*I 
*a) + sqrt(pi)*(4*I*c^2*d^2 - 4*I*b*c*d*e + I*b^2*e^2 + c*e^2)*erf(1/2*I*s 
qrt(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
 
3.1.32.9 Mupad [F(-1)]

Timed out. \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\int {\cos \left (c\,x^2+b\,x+a\right )}^2\,{\left (d+e\,x\right )}^2 \,d x \]

input
int(cos(a + b*x + c*x^2)^2*(d + e*x)^2,x)
 
output
int(cos(a + b*x + c*x^2)^2*(d + e*x)^2, x)