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

3.1.7.1 Optimal result
3.1.7.2 Mathematica [A] (verified)
3.1.7.3 Rubi [A] (verified)
3.1.7.4 Maple [A] (verified)
3.1.7.5 Fricas [A] (verification not implemented)
3.1.7.6 Sympy [F]
3.1.7.7 Maxima [C] (verification not implemented)
3.1.7.8 Giac [C] (verification not implemented)
3.1.7.9 Mupad [F(-1)]

3.1.7.1 Optimal result

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

output
1/2*cos(-c*x^2+b*x+a)/c+1/4*b*cos(a+1/4*b^2/c)*FresnelS(1/2*(-2*c*x+b)/c^( 
1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/c^(3/2)-1/4*b*FresnelC(1/2*(-2*c*x 
+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a+1/4*b^2/c)*2^(1/2)*Pi^(1/2)/c^(3/2)
 
3.1.7.2 Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.94 \[ \int x \sin \left (a+b x-c x^2\right ) \, dx=\frac {2 \sqrt {c} \cos (a+x (b-c x))-b \sqrt {2 \pi } \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+b \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right )}{4 c^{3/2}} \]

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

Time = 0.32 (sec) , antiderivative size = 126, 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 = {3942, 3928, 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.

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

\(\Big \downarrow \) 3942

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

\(\Big \downarrow \) 3928

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

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

3.1.7.3.1 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 3928
Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[Cos[(b^2 - 4* 
a*c)/(4*c)]   Int[Sin[(b + 2*c*x)^2/(4*c)], x], x] - Simp[Sin[(b^2 - 4*a*c) 
/(4*c)]   Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] && Ne 
Q[b^2 - 4*a*c, 0]
 

rule 3942
Int[((d_.) + (e_.)*(x_))*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] 
:> Simp[(-e)*(Cos[a + b*x + c*x^2]/(2*c)), x] + Simp[(2*c*d - b*e)/(2*c) 
Int[Sin[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d 
- b*e, 0]
 
3.1.7.4 Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.88

method result size
default \(\frac {\cos \left (-c \,x^{2}+b x +a \right )}{2 c}+\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {S}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {C}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )\right )}{4 c \sqrt {-c}}\) \(109\)
risch \(\frac {i b \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c +b^{2}\right )}{4 c}} \operatorname {erf}\left (-\sqrt {i c}\, x +\frac {i b}{2 \sqrt {i c}}\right )}{8 c \sqrt {i c}}+\frac {i b \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c +b^{2}\right )}{4 c}} \operatorname {erf}\left (\sqrt {-i c}\, x +\frac {i b}{2 \sqrt {-i c}}\right )}{8 c \sqrt {-i c}}+\frac {\cos \left (-c \,x^{2}+b x +a \right )}{2 c}\) \(122\)
parts \(\frac {\sqrt {2}\, \sqrt {\pi }\, x \cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {S}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )}{2 \sqrt {-c}}+\frac {\sqrt {2}\, \sqrt {\pi }\, x \sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {C}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )}{2 \sqrt {-c}}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (-\frac {\cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-c}\, \left (\operatorname {S}\left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right ) \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )+\frac {\cos \left (\frac {\pi \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )^{2}}{2}\right )}{\pi }\right )}{2 c}-\frac {\sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-c}\, \left (\operatorname {C}\left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right ) \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )-\frac {\sin \left (\frac {\pi \left (-\frac {\sqrt {2}\, c x}{\sqrt {\pi }\, \sqrt {-c}}+\frac {\sqrt {2}\, b}{2 \sqrt {\pi }\, \sqrt {-c}}\right )^{2}}{2}\right )}{\pi }\right )}{2 c}\right )}{2 \sqrt {-c}}\) \(381\)

input
int(x*sin(-c*x^2+b*x+a),x,method=_RETURNVERBOSE)
 
output
1/2*cos(-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)*FresnelS(2^(1/2)/Pi^(1/2)/(-c)^(1/2)*(-c*x+1/2*b))+sin((1/4*b^2+a* 
c)/c)*FresnelC(2^(1/2)/Pi^(1/2)/(-c)^(1/2)*(-c*x+1/2*b)))
 
3.1.7.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02 \[ \int x \sin \left (a+b x-c x^2\right ) \, dx=-\frac {\sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - 2 \, c \cos \left (c x^{2} - b x - a\right )}{4 \, c^{2}} \]

input
integrate(x*sin(-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_sin(1/2*sqr 
t(2)*(2*c*x - b)*sqrt(c/pi)/c) - sqrt(2)*pi*b*sqrt(c/pi)*fresnel_cos(1/2*s 
qrt(2)*(2*c*x - b)*sqrt(c/pi)/c)*sin(1/4*(b^2 + 4*a*c)/c) - 2*c*cos(c*x^2 
- b*x - a))/c^2
 
3.1.7.6 Sympy [F]

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

input
integrate(x*sin(-c*x**2+b*x+a),x)
 
output
Integral(x*sin(a + b*x - c*x**2), x)
 
3.1.7.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 580, normalized size of antiderivative = 4.68 \[ \int x \sin \left (a+b x-c x^2\right ) \, dx=\frac {{\left (\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b^{2} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b^{2} \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - 2 \, {\left ({\left (\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b c \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b c \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} x + 4 \, {\left (c {\left (e^{\left (\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{4 \, c}\right )} + e^{\left (-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{4 \, c}\right )}\right )} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - c {\left (i \, e^{\left (\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{4 \, c}\right )} - i \, e^{\left (-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{4 \, c}\right )}\right )} \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {4 \, c^{2} x^{2} - 4 \, b c x + b^{2}}{c}}}{16 \, c^{2} \sqrt {\frac {4 \, c^{2} x^{2} - 4 \, b c x + b^{2}}{c}}} \]

input
integrate(x*sin(-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*(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))*cos(1/4*(b^2 + 4*a*c)/c) - 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))* 
sin(1/4*(b^2 + 4*a*c)/c))*sqrt((4*c^2*x^2 - 4*b*c*x + b^2)/c))/(c^2*sqrt(( 
4*c^2*x^2 - 4*b*c*x + b^2)/c))
 
3.1.7.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.48 \[ \int x \sin \left (a+b x-c x^2\right ) \, dx=-\frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} i \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 \, e^{\left (i \, c x^{2} - i \, b x - i \, a\right )}}{8 \, c} - \frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (\frac {1}{4} i \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 \, e^{\left (-i \, c x^{2} + i \, b x + i \, a\right )}}{8 \, c} \]

input
integrate(x*sin(-c*x^2+b*x+a),x, algorithm="giac")
 
output
-1/8*(sqrt(2)*sqrt(pi)*b*erf(-1/4*I*sqrt(2)*(2*x - b/c)*(I*c/abs(c) + 1)*s 
qrt(abs(c)))*e^(-1/4*(I*b^2 + 4*I*a*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) 
- 2*e^(I*c*x^2 - I*b*x - I*a))/c - 1/8*(sqrt(2)*sqrt(pi)*b*erf(1/4*I*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*e^(-I*c*x^2 + I*b*x + I*a))/c
 
3.1.7.9 Mupad [F(-1)]

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

input
int(x*sin(a + b*x - c*x^2),x)
 
output
int(x*sin(a + b*x - c*x^2), x)