\(\int (d+e x) \cos ^2(a+b x+c x^2) \, dx\) [33]

Optimal result

Integrand size = 19, antiderivative size = 150 \[ \int (d+e x) \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {(d+e x)^2}{4 e}+\frac {(2 c d-b e) \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}-\frac {(2 c d-b e) \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{8 c^{3/2}}+\frac {e \sin \left (2 a+2 b x+2 c x^2\right )}{8 c} \] Output:

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.93 \[ \int (d+e x) \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {(2 c d-b e) \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )-(2 c d-b e) \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )+\sqrt {c} (2 c x (2 d+e x)+e \sin (2 (a+x (b+c x))))}{8 c^{3/2}} \] Input:

Integrate[(d + e*x)*Cos[a + b*x + c*x^2]^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

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

\(\Big \downarrow \) 3949

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}+\frac {e \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {(d+e x)^2}{4 e}\)

Input:

Int[(d + e*x)*Cos[a + b*x + c*x^2]^2,x]
 

Output:

(d + e*x)^2/(4*e) + ((2*c*d - b*e)*Sqrt[Pi]*Cos[2*a - b^2/(2*c)]*FresnelC[ 
(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(8*c^(3/2)) - ((2*c*d - b*e)*Sqrt[Pi]*Fre 
snelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*Sin[2*a - b^2/(2*c)])/(8*c^(3/2)) + 
(e*Sin[2*a + 2*b*x + 2*c*x^2])/(8*c)
 

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]
 

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {e \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}-\frac {e b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelC}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelS}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{8 c^{\frac {3}{2}}}+\frac {\sqrt {\pi }\, d \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelC}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {FresnelS}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}}+\frac {d x}{2}+\frac {e \,x^{2}}{4}\)	\(170\)
risch	\(\frac {e \,x^{2}}{4}+\frac {d x}{2}+\frac {\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, d \,{\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {i c}}-\frac {e \,\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, b \,{\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{32 \sqrt {i c}\, c}-\frac {\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, d \,{\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}}+\frac {e \,\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, b \,{\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {-2 i c}\, c}+\frac {e \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}\)	\(257\)

Input:

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

Output:

1/8*e*sin(2*c*x^2+2*b*x+2*a)/c-1/8*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)*Fresne 
lS((2*c*x+b)/c^(1/2)/Pi^(1/2)))+1/4*Pi^(1/2)/c^(1/2)*d*(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)*Fresnel 
S((2*c*x+b)/c^(1/2)/Pi^(1/2)))+1/2*d*x+1/4*e*x^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int (d+e x) \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {2 \, c^{2} e x^{2} + \pi {\left (2 \, c d - b e\right )} \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) - \pi {\left (2 \, c d - b e\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + 4 \, c^{2} d x + 2 \, c e \cos \left (c x^{2} + b x + a\right ) \sin \left (c x^{2} + b x + a\right )}{8 \, c^{2}} \] Input:

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

Output:

1/8*(2*c^2*e*x^2 + pi*(2*c*d - b*e)*sqrt(c/pi)*cos(-1/2*(b^2 - 4*a*c)/c)*f 
resnel_cos((2*c*x + b)*sqrt(c/pi)/c) - pi*(2*c*d - b*e)*sqrt(c/pi)*fresnel 
_sin((2*c*x + b)*sqrt(c/pi)/c)*sin(-1/2*(b^2 - 4*a*c)/c) + 4*c^2*d*x + 2*c 
*e*cos(c*x^2 + b*x + a)*sin(c*x^2 + b*x + a))/c^2
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 737, normalized size of antiderivative = 4.91 \[ \int (d+e x) \cos ^2\left (a+b x+c x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate((e*x+d)*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/c^2 + 1/64*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)*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*c*cos(-1/2*(b^2 - 4*a*c)/c) + (-(I + 1)*sqrt(2)*sqrt(pi)*(er 
f(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*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))*e/(c^2*sqrt((4*c^2*...
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.30 \[ \int (d+e x) \cos ^2\left (a+b x+c x^2\right ) \, dx=\frac {1}{4} \, e x^{2} + \frac {1}{2} \, d x - \frac {i \, e e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )} - \frac {i \, \sqrt {\pi } {\left (2 i \, c d - i \, b e\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{16 \, c} - \frac {-i \, e e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )} + \frac {i \, \sqrt {\pi } {\left (-2 i \, c d + i \, b e\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{16 \, c} \] Input:

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

Output:

1/4*e*x^2 + 1/2*d*x - 1/16*(I*e*e^(2*I*c*x^2 + 2*I*b*x + 2*I*a) - I*sqrt(p 
i)*(2*I*c*d - I*b*e)*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 - 1/16*(-I*e*e^(-2* 
I*c*x^2 - 2*I*b*x - 2*I*a) + I*sqrt(pi)*(-2*I*c*d + I*b*e)*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
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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