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*(-b*e+2*c*d)^2*Pi^(1/2)*cos(2*a-1/2*b^2/c)*FresnelC(( 2*c*x+b)/c^(1/2)/Pi^(1/2))/c^(5/2)-1/16*e^2*Pi^(1/2)*cos(2*a-1/2*b^2/c)*Fr esnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))/c^(3/2)-1/16*e^2*Pi^(1/2)*FresnelC((2*c *x+b)/c^(1/2)/Pi^(1/2))*sin(2*a-1/2*b^2/c)/c^(3/2)-1/16*(-b*e+2*c*d)^2*Pi^ (1/2)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a-1/2*b^2/c)/c^(5/2)+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
Time = 1.11 (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))
Time = 0.56 (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)
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]
Time = 2.25 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.30
| method | result | size |
| default | \(\frac {e^{2} x \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}-\frac {e^{2} b \left (\frac {\sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelC}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelS}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{4 c}-\frac {e^{2} \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelS}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )-\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelC}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{16 c^{\frac {3}{2}}}+\frac {d e \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {d e b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelC}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelS}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}+\frac {\sqrt {\pi }\, d^{2} \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelC}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelS}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}}+\frac {d e \,x^{2}}{2}+\frac {d^{2} x}{2}+\frac {e^{2} x^{3}}{6}\) | \(378\) |
| risch | \(\frac {d e \,x^{2}}{2}+\frac {d^{2} x}{2}+\frac {e^{2} x^{3}}{6}+\frac {\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, d^{2} {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {i c}}+\frac {e^{2} b^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}} \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right )}{64 c^{2} \sqrt {i c}}-\frac {i e^{2} \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{64 \sqrt {i c}\, c}-\frac {d e b \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}} \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right )}{16 c \sqrt {i c}}-\frac {\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, d^{2} {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}}-\frac {e^{2} b^{2} \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}} \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right )}{32 c^{2} \sqrt {-2 i c}}-\frac {i e^{2} \sqrt {\pi }\, \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{32 c \sqrt {-2 i c}}+\frac {d e \,\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, b \,{\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}\, c}+2 i \left (\frac {e^{2} \left (-\frac {i x}{4 c}+\frac {i b}{8 c^{2}}\right )}{4}-\frac {i d e}{8 c}\right ) \sin \left (2 c \,x^{2}+2 b x +2 a \right )\) | \(544\) |
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
Time = 0.09 (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
\[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\int \left (d + e x\right )^{2} \cos ^{2}{\left (a + b x + c x^{2} \right )}\, dx \] 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)
Result contains complex when optimal does not.
Time = 1.06 (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*...
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.01 \[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {1}{6} \, e^{2} x^{3} + \frac {1}{2} \, d e x^{2} + \frac {1}{2} \, d^{2} x - \frac {i \, {\left (c e^{2} {\left (2 \, x + \frac {b}{c}\right )} + 4 \, c d e - 2 \, b e^{2}\right )} e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )} + \frac {i \, \sqrt {\pi } {\left (-4 i \, c^{2} d^{2} + 4 i \, b c d e - i \, b^{2} e^{2} + c e^{2}\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} - \frac {-i \, {\left (c e^{2} {\left (2 \, x + \frac {b}{c}\right )} + 4 \, c d e - 2 \, b e^{2}\right )} e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )} - \frac {i \, \sqrt {\pi } {\left (4 i \, c^{2} d^{2} - 4 i \, b c d e + i \, b^{2} e^{2} + c e^{2}\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} \] 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) + I*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*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) - I*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*sqrt(c)*(2*x + b/c)*(I*c/abs(c) + 1))*e^(-1/2*(-I*b^2 + 4*I*a*c)/c)/(sqr t(c)*(I*c/abs(c) + 1)))/c^2
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)
\[ \int (d+e x)^2 \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {-\cos \left (c \,x^{2}+b x +a \right ) \sin \left (c \,x^{2}+b x +a \right ) b \,e^{2}+3 \cos \left (c \,x^{2}+b x +a \right ) \sin \left (c \,x^{2}+b x +a \right ) c d e +2 \cos \left (c \,x^{2}+b x +a \right ) \sin \left (c \,x^{2}+b x +a \right ) c \,e^{2} x -6 \left (\int \frac {\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{4}+2 \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) b^{2} e^{2}+24 \left (\int \frac {\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{4}+2 \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) b c d e -24 \left (\int \frac {\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{4}+2 \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) c^{2} d^{2}-8 \left (\int \frac {x^{2}}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{4}+2 \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) c^{2} e^{2}-8 \left (\int \frac {\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{4}+2 \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) c \,e^{2}+2 \left (\int \frac {1}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{4}+2 \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) b^{2} e^{2}-\sin \left (c \,x^{2}+b x +a \right ) b \,e^{2}+2 \sin \left (c \,x^{2}+b x +a \right ) c \,e^{2} x -3 b c d e x +6 c^{2} d^{2} x +3 c^{2} d e \,x^{2}+2 c^{2} e^{2} x^{3}}{6 c^{2}} \] Input:
int((e*x+d)^2*cos(c*x^2+b*x+a)^2,x)
Output:
( - cos(a + b*x + c*x**2)*sin(a + b*x + c*x**2)*b*e**2 + 3*cos(a + b*x + c *x**2)*sin(a + b*x + c*x**2)*c*d*e + 2*cos(a + b*x + c*x**2)*sin(a + b*x + c*x**2)*c*e**2*x - 6*int(tan((a + b*x + c*x**2)/2)**2/(tan((a + b*x + c*x **2)/2)**4 + 2*tan((a + b*x + c*x**2)/2)**2 + 1),x)*b**2*e**2 + 24*int(tan ((a + b*x + c*x**2)/2)**2/(tan((a + b*x + c*x**2)/2)**4 + 2*tan((a + b*x + c*x**2)/2)**2 + 1),x)*b*c*d*e - 24*int(tan((a + b*x + c*x**2)/2)**2/(tan( (a + b*x + c*x**2)/2)**4 + 2*tan((a + b*x + c*x**2)/2)**2 + 1),x)*c**2*d** 2 - 8*int(x**2/(tan((a + b*x + c*x**2)/2)**4 + 2*tan((a + b*x + c*x**2)/2) **2 + 1),x)*c**2*e**2 - 8*int(tan((a + b*x + c*x**2)/2)/(tan((a + b*x + c* x**2)/2)**4 + 2*tan((a + b*x + c*x**2)/2)**2 + 1),x)*c*e**2 + 2*int(1/(tan ((a + b*x + c*x**2)/2)**4 + 2*tan((a + b*x + c*x**2)/2)**2 + 1),x)*b**2*e* *2 - sin(a + b*x + c*x**2)*b*e**2 + 2*sin(a + b*x + c*x**2)*c*e**2*x - 3*b *c*d*e*x + 6*c**2*d**2*x + 3*c**2*d*e*x**2 + 2*c**2*e**2*x**3)/(6*c**2)