Integrand size = 46, antiderivative size = 39 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=e^9 \left (x-\frac {2+\frac {1}{4} \left (-e^{\frac {1}{2} (4+x+\log (3))}+x\right )}{i \pi +\log (3)}\right ) \] Output:
exp(9)*(x-(2+1/4*x-1/4*exp(1/2*ln(3)+2+1/2*x))/(ln(3)+I*Pi))
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {e^9 \left (2 \sqrt {3} e^{2+\frac {x}{2}}-2 x+8 i \pi x+8 x \log (3)\right )}{8 (i \pi +\log (3))} \] Input:
Integrate[(-2*E^9 + E^(9 + (4 + x + Log[3])/2) + 8*E^9*(I*Pi + Log[3]))/(8 *(I*Pi + Log[3])),x]
Output:
(E^9*(2*Sqrt[3]*E^(2 + x/2) - 2*x + (8*I)*Pi*x + 8*x*Log[3]))/(8*(I*Pi + L og[3]))
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {27, 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.
\(\displaystyle \int \frac {e^{\frac {1}{2} (x+4+\log (3))+9}-2 e^9+8 e^9 (\log (3)+i \pi )}{8 (\log (3)+i \pi )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \left (e^{\frac {1}{2} (x+\log (3)+4)+9}-2 e^9 (1-4 i \pi -\log (81))\right )dx}{8 (\log (3)+i \pi )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^{\frac {1}{2} (x+4+\log (3))+9}-2 e^9 x (1-4 i \pi -\log (81))}{8 (\log (3)+i \pi )}\) |
Input:
Int[(-2*E^9 + E^(9 + (4 + x + Log[3])/2) + 8*E^9*(I*Pi + Log[3]))/(8*(I*Pi + Log[3])),x]
Output:
(2*E^(9 + (4 + x + Log[3])/2) - 2*E^9*x*(1 - (4*I)*Pi - Log[81]))/(8*(I*Pi + Log[3]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {2 \,{\mathrm e}^{9} {\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}+8 \,{\mathrm e}^{9} \left (\ln \left (3\right )+i \pi \right ) x -2 x \,{\mathrm e}^{9}}{8 \ln \left (3\right )+8 i \pi }\) | \(44\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{9} {\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}+\left (8 \,{\mathrm e}^{9} \left (\ln \left (3\right )+i \pi \right )-2 \,{\mathrm e}^{9}\right ) x}{8 \ln \left (3\right )+8 i \pi }\) | \(45\) |
derivativedivides | \(\frac {{\mathrm e}^{9} \left ({\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}+\left (8 i \pi +8 \ln \left (3\right )-2\right ) \ln \left ({\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}\right )\right )}{4 \ln \left (3\right )+4 i \pi }\) | \(47\) |
parts | \(\frac {x \,{\mathrm e}^{9} \left (4 i \pi +4 \ln \left (3\right )-1\right )}{4 \ln \left (3\right )+4 i \pi }+\frac {{\mathrm e}^{9} {\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}}{4 \ln \left (3\right )+4 i \pi }\) | \(49\) |
risch | \(\frac {i x \,{\mathrm e}^{9} \pi }{\ln \left (3\right )+i \pi }+\frac {x \,{\mathrm e}^{9} \ln \left (3\right )}{\ln \left (3\right )+i \pi }-\frac {x \,{\mathrm e}^{9}}{4 \left (\ln \left (3\right )+i \pi \right )}+\frac {\sqrt {3}\, {\mathrm e}^{11+\frac {x}{2}}}{4 \ln \left (3\right )+4 i \pi }\) | \(67\) |
orering | \(\frac {\left (2+x \right ) \left ({\mathrm e}^{9} {\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}+8 \,{\mathrm e}^{9} \left (\ln \left (3\right )+i \pi \right )-2 \,{\mathrm e}^{9}\right )}{8 \ln \left (3\right )+8 i \pi }-\frac {x \,{\mathrm e}^{9} {\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}}{8 \left (\ln \left (3\right )+i \pi \right )}\) | \(69\) |
norman | \(-\frac {{\mathrm e}^{9} \left (i \pi -\ln \left (3\right )\right ) {\mathrm e}^{\frac {\ln \left (3\right )}{2}+2+\frac {x}{2}}}{4 \left (\pi ^{2}+\ln \left (3\right )^{2}\right )}+\frac {{\mathrm e}^{9} \left (4 \pi ^{2}+i \pi +4 \ln \left (3\right )^{2}-\ln \left (3\right )\right ) x}{4 \pi ^{2}+4 \ln \left (3\right )^{2}}\) | \(70\) |
Input:
int(1/8*(exp(9)*exp(1/2*ln(3)+2+1/2*x)+8*exp(9)*(ln(3)+I*Pi)-2*exp(9))/(ln (3)+I*Pi),x,method=_RETURNVERBOSE)
Output:
1/8/(ln(3)+I*Pi)*(2*exp(9)*exp(1/2*ln(3)+2+1/2*x)+8*exp(9)*(ln(3)+I*Pi)*x- 2*x*exp(9))
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {{\left (4 i \, \pi - 1\right )} x e^{9} + 4 \, x e^{9} \log \left (3\right ) + e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, \log \left (3\right ) + 11\right )}}{4 i \, \pi + 4 \, \log \left (3\right )} \] Input:
integrate(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*pi)-2*exp (9))/(log(3)+I*pi),x, algorithm="fricas")
Output:
((4*I*pi - 1)*x*e^9 + 4*x*e^9*log(3) + e^(1/2*x + 1/2*log(3) + 11))/(4*I*p i + 4*log(3))
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {x \left (- e^{9} + 4 e^{9} \log {\left (3 \right )} + 4 i \pi e^{9}\right )}{4 \log {\left (3 \right )} + 4 i \pi } + \frac {\sqrt {3} e^{11} e^{\frac {x}{2}}}{4 \log {\left (3 \right )} + 4 i \pi } \] Input:
integrate(1/8*(exp(9)*exp(1/2*ln(3)+2+1/2*x)+8*exp(9)*(ln(3)+I*pi)-2*exp(9 ))/(ln(3)+I*pi),x)
Output:
x*(-exp(9) + 4*exp(9)*log(3) + 4*I*pi*exp(9))/(4*log(3) + 4*I*pi) + sqrt(3 )*exp(11)*exp(x/2)/(4*log(3) + 4*I*pi)
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {4 \, {\left (i \, \pi + \log \left (3\right )\right )} x e^{9} - x e^{9} + e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, \log \left (3\right ) + 11\right )}}{4 \, {\left (i \, \pi + \log \left (3\right )\right )}} \] Input:
integrate(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*pi)-2*exp (9))/(log(3)+I*pi),x, algorithm="maxima")
Output:
1/4*(4*(I*pi + log(3))*x*e^9 - x*e^9 + e^(1/2*x + 1/2*log(3) + 11))/(I*pi + log(3))
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {4 \, {\left (i \, \pi + \log \left (3\right )\right )} x e^{9} - x e^{9} + e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, \log \left (3\right ) + 11\right )}}{4 \, {\left (i \, \pi + \log \left (3\right )\right )}} \] Input:
integrate(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*pi)-2*exp (9))/(log(3)+I*pi),x, algorithm="giac")
Output:
1/4*(4*(I*pi + log(3))*x*e^9 - x*e^9 + e^(1/2*x + 1/2*log(3) + 11))/(I*pi + log(3))
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {{\mathrm {e}}^9\,\left (x\,1{}\mathrm {i}+4\,\Pi \,x-x\,\ln \left (3\right )\,4{}\mathrm {i}-\sqrt {3}\,{\mathrm {e}}^{x/2}\,{\mathrm {e}}^2\,1{}\mathrm {i}\right )}{4\,\left (\Pi -\ln \left (3\right )\,1{}\mathrm {i}\right )} \] Input:
int(((exp(9)*exp(x/2 + log(3)/2 + 2))/8 - exp(9)/4 + exp(9)*(Pi*1i + log(3 )))/(Pi*1i + log(3)),x)
Output:
(exp(9)*(x*1i + 4*Pi*x - x*log(3)*4i - 3^(1/2)*exp(x/2)*exp(2)*1i))/(4*(Pi - log(3)*1i))
Time = 0.52 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03 \[ \int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx=\frac {e^{9} \left (e^{\frac {x}{2}} \sqrt {3}\, e^{2}+4 \,\mathrm {log}\left (3\right ) x +4 i \pi x -x \right )}{4 \,\mathrm {log}\left (3\right )+4 i \pi } \] Input:
int(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*Pi)-2*exp(9))/( log(3)+I*Pi),x)
Output:
(e**9*(e**(x/2)*sqrt(3)*e**2 + 4*log(3)*x + 4*i*pi*x - x))/(4*(log(3) + i* pi))