Integrand size = 15, antiderivative size = 144 \[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt [4]{-1} e^{\frac {1}{4} i \left (4 a+\frac {(1+i b)^2}{c}\right )} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (1+i b+2 i c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {\sqrt [4]{-1} e^{-i a+\frac {i (i+b)^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (1-i b-2 i c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c}} \] Output:
-1/4*(-1)^(1/4)*exp(1/4*I*(4*a+(1+I*b)^2/c))*Pi^(1/2)*erf(1/2*(-1)^(1/4)*( 1+I*b+2*I*c*x)/c^(1/2))/c^(1/2)+1/4*(-1)^(1/4)*exp(-I*a+1/4*I*(I+b)^2/c)*P i^(1/2)*erfi(1/2*(-1)^(1/4)*(1-I*b-2*I*c*x)/c^(1/2))/c^(1/2)
Time = 0.17 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.94 \[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=\frac {\sqrt [4]{-1} e^{-\frac {i \left (1-2 i b+b^2\right )}{4 c}} \sqrt {\pi } \left (-e^{\frac {i b^2}{2 c}} \text {erfi}\left (\frac {(-1)^{3/4} (i+b+2 c x)}{2 \sqrt {c}}\right ) (\cos (a)-i \sin (a))+e^{\left .\frac {i}{2}\right /c} \text {erfi}\left (\frac {\sqrt [4]{-1} (-i+b+2 c x)}{2 \sqrt {c}}\right ) (-i \cos (a)+\sin (a))\right )}{4 \sqrt {c}} \] Input:
Integrate[E^x*Cos[a + b*x + c*x^2],x]
Output:
((-1)^(1/4)*Sqrt[Pi]*(-(E^(((I/2)*b^2)/c)*Erfi[((-1)^(3/4)*(I + b + 2*c*x) )/(2*Sqrt[c])]*(Cos[a] - I*Sin[a])) + E^((I/2)/c)*Erfi[((-1)^(1/4)*(-I + b + 2*c*x))/(2*Sqrt[c])]*((-I)*Cos[a] + Sin[a])))/(4*Sqrt[c]*E^(((I/4)*(1 - (2*I)*b + b^2))/c))
Time = 0.37 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4976, 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.
| \(\displaystyle \int e^x \cos \left (a+b x+c x^2\right ) \, dx\) |
| \(\Big \downarrow \) 4976 |
| \(\displaystyle \int \left (\frac {1}{2} e^{-i a+(1-i b) x-i c x^2}+\frac {1}{2} e^{i a+(1+i b) x+i c x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [4]{-1} \sqrt {\pi } e^{\frac {i (b+i)^2}{4 c}-i a} \text {erfi}\left (\frac {\sqrt [4]{-1} (-i b-2 i c x+1)}{2 \sqrt {c}}\right )}{4 \sqrt {c}}-\frac {\sqrt [4]{-1} \sqrt {\pi } e^{\frac {1}{4} i \left (4 a+\frac {(1+i b)^2}{c}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (i b+2 i c x+1)}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\) |
Input:
Int[E^x*Cos[a + b*x + c*x^2],x]
Output:
-1/4*((-1)^(1/4)*E^((I/4)*(4*a + (1 + I*b)^2/c))*Sqrt[Pi]*Erf[((-1)^(1/4)* (1 + I*b + (2*I)*c*x))/(2*Sqrt[c])])/Sqrt[c] + ((-1)^(1/4)*E^((-I)*a + ((I /4)*(I + b)^2)/c)*Sqrt[Pi]*Erfi[((-1)^(1/4)*(1 - I*b - (2*I)*c*x))/(2*Sqrt [c])])/(4*Sqrt[c])
Int[Cos[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cos[v]^n , x], x] /; FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]
Time = 0.74 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81
| method | result | size |
| risch | \(\frac {\sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}-2 i b +1\right )}{4 c}} \operatorname {erf}\left (\sqrt {i c}\, x -\frac {-i b +1}{2 \sqrt {i c}}\right )}{4 \sqrt {i c}}-\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}+2 i b +1\right )}{4 c}} \operatorname {erf}\left (-\sqrt {-i c}\, x +\frac {i b +1}{2 \sqrt {-i c}}\right )}{4 \sqrt {-i c}}\) | \(117\) |
Input:
int(exp(x)*cos(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
1/4*Pi^(1/2)*exp(-1/4*I*(-b^2-2*I*b+4*a*c+1)/c)/(I*c)^(1/2)*erf((I*c)^(1/2 )*x-1/2*(-I*b+1)/(I*c)^(1/2))-1/4*Pi^(1/2)*exp(1/4*I*(-b^2+2*I*b+4*a*c+1)/ c)/(-I*c)^(1/2)*erf(-(-I*c)^(1/2)*x+1/2*(1+I*b)/(-I*c)^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (91) = 182\).
Time = 0.07 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.59 \[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=\frac {\sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {i \, b^{2} - 4 i \, a c - 2 \, b - i}{4 \, c}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b + i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {-i \, b^{2} + 4 i \, a c - 2 \, b + i}{4 \, c}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, c x + b - i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - i \, \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {i \, b^{2} - 4 i \, a c - 2 \, b - i}{4 \, c}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b + i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - i \, \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {-i \, b^{2} + 4 i \, a c - 2 \, b + i}{4 \, c}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, c x + b - i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right )}{4 \, c} \] Input:
integrate(exp(x)*cos(c*x^2+b*x+a),x, algorithm="fricas")
Output:
1/4*(sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(I*b^2 - 4*I*a*c - 2*b - I)/c)*fresnel_c os(1/2*sqrt(2)*(2*c*x + b + I)*sqrt(c/pi)/c) - sqrt(2)*pi*sqrt(c/pi)*e^(1/ 4*(-I*b^2 + 4*I*a*c - 2*b + I)/c)*fresnel_cos(-1/2*sqrt(2)*(2*c*x + b - I) *sqrt(c/pi)/c) - I*sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(I*b^2 - 4*I*a*c - 2*b - I )/c)*fresnel_sin(1/2*sqrt(2)*(2*c*x + b + I)*sqrt(c/pi)/c) - I*sqrt(2)*pi* sqrt(c/pi)*e^(1/4*(-I*b^2 + 4*I*a*c - 2*b + I)/c)*fresnel_sin(-1/2*sqrt(2) *(2*c*x + b - I)*sqrt(c/pi)/c))/c
\[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=\int e^{x} \cos {\left (a + b x + c x^{2} \right )}\, dx \] Input:
integrate(exp(x)*cos(c*x**2+b*x+a),x)
Output:
Integral(exp(x)*cos(a + b*x + c*x**2), x)
Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91 \[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i - 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right ) - \left (i + 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, c x + i \, b - 1\right )} \sqrt {i \, c}}{2 \, c}\right ) + {\left (\left (i + 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right ) + \left (i - 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, c x + i \, b + 1\right )} \sqrt {-i \, c}}{2 \, c}\right )\right )} e^{\left (-\frac {b}{2 \, c}\right )}}{8 \, \sqrt {c}} \] Input:
integrate(exp(x)*cos(c*x^2+b*x+a),x, algorithm="maxima")
Output:
-1/8*sqrt(2)*sqrt(pi)*((-(I - 1)*cos(-1/4*(b^2 - 4*a*c - 1)/c) - (I + 1)*s in(-1/4*(b^2 - 4*a*c - 1)/c))*erf(1/2*I*(2*I*c*x + I*b - 1)*sqrt(I*c)/c) + ((I + 1)*cos(-1/4*(b^2 - 4*a*c - 1)/c) + (I - 1)*sin(-1/4*(b^2 - 4*a*c - 1)/c))*erf(1/2*I*(2*I*c*x + I*b + 1)*sqrt(-I*c)/c))*e^(-1/2*b/c)/sqrt(c)
Time = 0.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02 \[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b - i}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c + 2 \, b - i}{4 \, c}\right )}}{4 \, {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b + i}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c + 2 \, b + i}{4 \, c}\right )}}{4 \, {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} \] Input:
integrate(exp(x)*cos(c*x^2+b*x+a),x, algorithm="giac")
Output:
-1/4*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*(2*x + (b - I)/c)*(-I*c/abs(c) + 1) *sqrt(abs(c)))*e^(-1/4*(I*b^2 - 4*I*a*c + 2*b - I)/c)/((-I*c/abs(c) + 1)*s qrt(abs(c))) - 1/4*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*(2*x + (b + I)/c)*(I* c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(-I*b^2 + 4*I*a*c + 2*b + I)/c)/((I*c/ abs(c) + 1)*sqrt(abs(c)))
Timed out. \[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=\int {\mathrm {e}}^x\,\cos \left (c\,x^2+b\,x+a\right ) \,d x \] Input:
int(exp(x)*cos(a + b*x + c*x^2),x)
Output:
int(exp(x)*cos(a + b*x + c*x^2), x)
\[ \int e^x \cos \left (a+b x+c x^2\right ) \, dx=\frac {2 e^{x} \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2} c x -2 e^{x} \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2} c +2 e^{x} \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )+2 e^{x} c x -2 e^{x} c -2 \left (\int \frac {e^{x} \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}-2 \left (\int \frac {e^{x} \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right )-4 \left (\int \frac {e^{x} x}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) \tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2} c -4 \left (\int \frac {e^{x} x}{\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1}d x \right ) c}{b \left (\tan \left (\frac {1}{2} c \,x^{2}+\frac {1}{2} b x +\frac {1}{2} a \right )^{2}+1\right )} \] Input:
int(exp(x)*cos(c*x^2+b*x+a),x)
Output:
(2*(e**x*tan((a + b*x + c*x**2)/2)**2*c*x - e**x*tan((a + b*x + c*x**2)/2) **2*c + e**x*tan((a + b*x + c*x**2)/2) + e**x*c*x - e**x*c - int((e**x*tan ((a + b*x + c*x**2)/2))/(tan((a + b*x + c*x**2)/2)**2 + 1),x)*tan((a + b*x + c*x**2)/2)**2 - int((e**x*tan((a + b*x + c*x**2)/2))/(tan((a + b*x + c* x**2)/2)**2 + 1),x) - 2*int((e**x*x)/(tan((a + b*x + c*x**2)/2)**2 + 1),x) *tan((a + b*x + c*x**2)/2)**2*c - 2*int((e**x*x)/(tan((a + b*x + c*x**2)/2 )**2 + 1),x)*c))/(b*(tan((a + b*x + c*x**2)/2)**2 + 1))