\(\int x \cos (a+b x-c x^2) \, dx\) [7]

Optimal result

Integrand size = 14, antiderivative size = 124 \[ \int x \cos \left (a+b x-c x^2\right ) \, dx=-\frac {b \sqrt {\frac {\pi }{2}} \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {b \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right )}{2 c^{3/2}}-\frac {\sin \left (a+b x-c x^2\right )}{2 c} \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94 \[ \int x \cos \left (a+b x-c x^2\right ) \, dx=\frac {b \sqrt {2 \pi } \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+b \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right )-2 \sqrt {c} \sin (a+x (b-c x))}{4 c^{3/2}} \] Input:

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

Output:

(b*Sqrt[2*Pi]*Cos[a + b^2/(4*c)]*FresnelC[(-b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi] 
)] + b*Sqrt[2*Pi]*FresnelS[(-b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a + b^2/ 
(4*c)] - 2*Sqrt[c]*Sin[a + x*(b - c*x)])/(4*c^(3/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3943, 3929, 25, 3832, 3833}

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.

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

\(\Big \downarrow \) 3943

\(\displaystyle \frac {b \int \cos \left (-c x^2+b x+a\right )dx}{2 c}-\frac {\sin \left (a+b x-c x^2\right )}{2 c}\)

\(\Big \downarrow \) 3929

\(\displaystyle \frac {b \left (\cos \left (a+\frac {b^2}{4 c}\right ) \int \cos \left (\frac {(b-2 c x)^2}{4 c}\right )dx-\sin \left (a+\frac {b^2}{4 c}\right ) \int -\sin \left (\frac {(b-2 c x)^2}{4 c}\right )dx\right )}{2 c}-\frac {\sin \left (a+b x-c x^2\right )}{2 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b \left (\sin \left (a+\frac {b^2}{4 c}\right ) \int \sin \left (\frac {(b-2 c x)^2}{4 c}\right )dx+\cos \left (a+\frac {b^2}{4 c}\right ) \int \cos \left (\frac {(b-2 c x)^2}{4 c}\right )dx\right )}{2 c}-\frac {\sin \left (a+b x-c x^2\right )}{2 c}\)

\(\Big \downarrow \) 3832

\(\displaystyle \frac {b \left (\cos \left (a+\frac {b^2}{4 c}\right ) \int \cos \left (\frac {(b-2 c x)^2}{4 c}\right )dx-\frac {\sqrt {\frac {\pi }{2}} \sin \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{\sqrt {c}}\right )}{2 c}-\frac {\sin \left (a+b x-c x^2\right )}{2 c}\)

\(\Big \downarrow \) 3833

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

Input:

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

Output:

(b*(-((Sqrt[Pi/2]*Cos[a + b^2/(4*c)]*FresnelC[(b - 2*c*x)/(Sqrt[c]*Sqrt[2* 
Pi])])/Sqrt[c]) - (Sqrt[Pi/2]*FresnelS[(b - 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*S 
in[a + b^2/(4*c)])/Sqrt[c]))/(2*c) - Sin[a + b*x - c*x^2]/(2*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3832

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 3833

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 3929

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[Cos[(b^2 - 4* 
a*c)/(4*c)]   Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] + Simp[Sin[(b^2 - 4*a*c) 
/(4*c)]   Int[Sin[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] && Ne 
Q[b^2 - 4*a*c, 0]
 

rule 3943

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] + Simp[(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]
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\sin \left (-c \,x^{2}+b x +a \right )}{2 c}+\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )\right )}{4 c \sqrt {-c}}\)	\(110\)
risch	\(\frac {b \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c +b^{2}\right )}{4 c}} \operatorname {erf}\left (\sqrt {-i c}\, x +\frac {i b}{2 \sqrt {-i c}}\right )}{8 c \sqrt {-i c}}-\frac {b \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c +b^{2}\right )}{4 c}} \operatorname {erf}\left (-\sqrt {i c}\, x +\frac {i b}{2 \sqrt {i c}}\right )}{8 c \sqrt {i c}}-\frac {\sin \left (-c \,x^{2}+b x +a \right )}{2 c}\)	\(120\)
parts	\(\frac {\sqrt {2}\, \sqrt {\pi }\, x \cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )}{2 \sqrt {-c}}-\frac {\sqrt {2}\, \sqrt {\pi }\, x \sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )}{2 \sqrt {-c}}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (-\frac {\cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-c}\, \left (\operatorname {FresnelC}\left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right ) \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )-\frac {\sin \left (\frac {\pi \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )^{2}}{2}\right )}{\pi }\right )}{2 c}+\frac {\sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-c}\, \left (\operatorname {FresnelS}\left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right ) \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )+\frac {\cos \left (\frac {\pi \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )^{2}}{2}\right )}{\pi }\right )}{2 c}\right )}{2 \sqrt {-c}}\)	\(381\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.01 \[ \int x \cos \left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + \sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + 2 \, c \sin \left (c x^{2} - b x - a\right )}{4 \, c^{2}} \] Input:

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

Output:

1/4*(sqrt(2)*pi*b*sqrt(c/pi)*cos(1/4*(b^2 + 4*a*c)/c)*fresnel_cos(1/2*sqrt 
(2)*(2*c*x - b)*sqrt(c/pi)/c) + sqrt(2)*pi*b*sqrt(c/pi)*fresnel_sin(1/2*sq 
rt(2)*(2*c*x - b)*sqrt(c/pi)/c)*sin(1/4*(b^2 + 4*a*c)/c) + 2*c*sin(c*x^2 - 
 b*x - a))/c^2
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 580, normalized size of antiderivative = 4.68 \[ \int x \cos \left (a+b x-c x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate(x*cos(-c*x^2+b*x+a),x, algorithm="maxima")
 

Output:

-1/16*((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(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(1/2*sqrt(-(4*I*c^2*x^2 - 
4*I*b*c*x + I*b^2)/c)) - 1))*b^2*cos(1/4*(b^2 + 4*a*c)/c) + ((I + 1)*sqrt( 
2)*sqrt(pi)*(erf(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(1/2*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) 
 - 1))*b^2*sin(1/4*(b^2 + 4*a*c)/c) - 2*((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(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(1/2*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*cos(1/4 
*(b^2 + 4*a*c)/c) + ((I + 1)*sqrt(2)*sqrt(pi)*(erf(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(1/2*sqrt(-(4* 
I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*sin(1/4*(b^2 + 4*a*c)/c))*x - 
 4*(c*(-I*e^(1/4*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) + I*e^(-1/4*(4*I*c^2 
*x^2 - 4*I*b*c*x + I*b^2)/c))*cos(1/4*(b^2 + 4*a*c)/c) - c*(e^(1/4*(4*I*c^ 
2*x^2 - 4*I*b*c*x + I*b^2)/c) + e^(-1/4*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/ 
c))*sin(1/4*(b^2 + 4*a*c)/c))*sqrt((4*c^2*x^2 - 4*b*c*x + b^2)/c))/(c^2*sq 
rt((4*c^2*x^2 - 4*b*c*x + b^2)/c))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.48 \[ \int x \cos \left (a+b x-c x^2\right ) \, dx=-\frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + 2 i \, e^{\left (i \, c x^{2} - i \, b x - i \, a\right )}}{8 \, c} - \frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 i \, e^{\left (-i \, c x^{2} + i \, b x + i \, a\right )}}{8 \, c} \] Input:

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

Output:

-1/8*(sqrt(2)*sqrt(pi)*b*erf(-1/4*sqrt(2)*(2*x - b/c)*(-I*c/abs(c) + 1)*sq 
rt(abs(c)))*e^(-1/4*(I*b^2 + 4*I*a*c)/c)/((-I*c/abs(c) + 1)*sqrt(abs(c))) 
+ 2*I*e^(I*c*x^2 - I*b*x - I*a))/c - 1/8*(sqrt(2)*sqrt(pi)*b*erf(-1/4*sqrt 
(2)*(2*x - b/c)*(I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(-I*b^2 - 4*I*a*c)/ 
c)/((I*c/abs(c) + 1)*sqrt(abs(c))) - 2*I*e^(-I*c*x^2 + I*b*x + I*a))/c
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(cos(a + b*x - c*x**2),x)*b - sin(a + b*x - c*x**2))/(2*c)