Integrand size = 118, antiderivative size = 31 \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=e^{\frac {25 x^2 \left (5+(i \pi +\log (3)) \left (e^x+x-\log (x)\right )\right )}{\log (2)}} \]
Time = 0.76 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77 \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=e^{\frac {25 x^2 \left (5+i \pi x+x \log (3)+e^x (i \pi +\log (3))\right )}{\log (2)}} x^{-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}} \]
Integrate[(E^((125*x^2 + 25*E^x*x^2*(I*Pi + Log[3]) + 25*x^3*(I*Pi + Log[3 ]) - 25*x^2*(I*Pi + Log[3])*Log[x])/Log[2])*(250*x + E^x*(50*x + 25*x^2)*( I*Pi + Log[3]) + (-25*x + 75*x^2)*(I*Pi + Log[3]) - 50*x*(I*Pi + Log[3])*L og[x]))/Log[2],x]
E^((25*x^2*(5 + I*Pi*x + x*Log[3] + E^x*(I*Pi + Log[3])))/Log[2])/x^((25*x ^2*(I*Pi + Log[3]))/Log[2])
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.
| \(\displaystyle \int \frac {\left (e^x \left (25 x^2+50 x\right ) (\log (3)+i \pi )+\left (75 x^2-25 x\right ) (\log (3)+i \pi )+250 x-50 x (\log (3)+i \pi ) \log (x)\right ) \exp \left (\frac {25 x^3 (\log (3)+i \pi )+125 x^2-25 x^2 (\log (3)+i \pi ) \log (x)+25 e^x x^2 (\log (3)+i \pi )}{\log (2)}\right )}{\log (2)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int 25 \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}} \left (-2 (i \pi +\log (3)) \log (x) x+10 x-\left (x-3 x^2\right ) (i \pi +\log (3))+e^x \left (x^2+2 x\right ) (i \pi +\log (3))\right )dx}{\log (2)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {25 \int \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}} \left (-2 (i \pi +\log (3)) \log (x) x+10 x-\left (x-3 x^2\right ) (i \pi +\log (3))+e^x \left (x^2+2 x\right ) (i \pi +\log (3))\right )dx}{\log (2)}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {25 \int \left (10 \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}-2 i \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) (\pi -i \log (3)) \log (x) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}+\exp \left (x+\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) (x+2) (i \pi +\log (3)) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}+\exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) (3 x-1) (i \pi +\log (3)) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}\right )dx}{\log (2)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {25 \left (2 (\log (3)+i \pi ) \int \frac {\int \exp \left (\frac {25 x^2 \left (i \pi \left (1-\frac {i \log (3)}{\pi }\right ) x+e^x (i \pi +\log (3))+5\right )}{\log (2)}\right ) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx}{x}dx-2 (\log (3)+i \pi ) \log (x) \int \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx-(\log (3)+i \pi ) \int \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx+10 \int \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx+2 (\log (3)+i \pi ) \int \exp \left (x+\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{1-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx+3 (\log (3)+i \pi ) \int \exp \left (\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{2-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx+(\log (3)+i \pi ) \int \exp \left (x+\frac {25 \left ((i \pi +\log (3)) x^3+e^x (i \pi +\log (3)) x^2+5 x^2\right )}{\log (2)}\right ) x^{2-\frac {25 x^2 (i \pi +\log (3))}{\log (2)}}dx\right )}{\log (2)}\) |
Int[(E^((125*x^2 + 25*E^x*x^2*(I*Pi + Log[3]) + 25*x^3*(I*Pi + Log[3]) - 2 5*x^2*(I*Pi + Log[3])*Log[x])/Log[2])*(250*x + E^x*(50*x + 25*x^2)*(I*Pi + Log[3]) + (-25*x + 75*x^2)*(I*Pi + Log[3]) - 50*x*(I*Pi + Log[3])*Log[x]) )/Log[2],x]
3.18.40.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.39 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45
| method | result | size |
| risch | \({\mathrm e}^{\frac {25 x^{2} \left (-i \pi \ln \left (x \right )+i \pi \,{\mathrm e}^{x}+i x \pi -\ln \left (3\right ) \ln \left (x \right )+\ln \left (3\right ) {\mathrm e}^{x}+x \ln \left (3\right )+5\right )}{\ln \left (2\right )}}\) | \(45\) |
| parallelrisch | \({\mathrm e}^{\frac {-25 x^{2} \left (\ln \left (3\right )+i \pi \right ) \ln \left (x \right )+25 x^{2} \left (\ln \left (3\right )+i \pi \right ) {\mathrm e}^{x}+25 x^{3} \left (\ln \left (3\right )+i \pi \right )+125 x^{2}}{\ln \left (2\right )}}\) | \(53\) |
int((-50*x*(ln(3)+I*Pi)*ln(x)+(25*x^2+50*x)*(ln(3)+I*Pi)*exp(x)+(75*x^2-25 *x)*(ln(3)+I*Pi)+250*x)*exp((-25*x^2*(ln(3)+I*Pi)*ln(x)+25*x^2*(ln(3)+I*Pi )*exp(x)+25*x^3*(ln(3)+I*Pi)+125*x^2)/ln(2))/ln(2),x,method=_RETURNVERBOSE )
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65 \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=e^{\left (\frac {25 i \, \pi x^{3}}{\log \left (2\right )} + \frac {25 i \, \pi x^{2} e^{x}}{\log \left (2\right )} + \frac {25 \, x^{3} \log \left (3\right )}{\log \left (2\right )} + \frac {25 \, x^{2} e^{x} \log \left (3\right )}{\log \left (2\right )} - \frac {25 i \, \pi x^{2} \log \left (x\right )}{\log \left (2\right )} - \frac {25 \, x^{2} \log \left (3\right ) \log \left (x\right )}{\log \left (2\right )} + \frac {125 \, x^{2}}{\log \left (2\right )}\right )} \]
integrate((-50*x*(log(3)+I*pi)*log(x)+(25*x^2+50*x)*(log(3)+I*pi)*exp(x)+( 75*x^2-25*x)*(log(3)+I*pi)+250*x)*exp((-25*x^2*(log(3)+I*pi)*log(x)+25*x^2 *(log(3)+I*pi)*exp(x)+25*x^3*(log(3)+I*pi)+125*x^2)/log(2))/log(2),x, algo rithm=\
e^(25*I*pi*x^3/log(2) + 25*I*pi*x^2*e^x/log(2) + 25*x^3*log(3)/log(2) + 25 *x^2*e^x*log(3)/log(2) - 25*I*pi*x^2*log(x)/log(2) - 25*x^2*log(3)*log(x)/ log(2) + 125*x^2/log(2))
Timed out. \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=\text {Timed out} \]
integrate((-50*x*(ln(3)+I*pi)*ln(x)+(25*x**2+50*x)*(ln(3)+I*pi)*exp(x)+(75 *x**2-25*x)*(ln(3)+I*pi)+250*x)*exp((-25*x**2*(ln(3)+I*pi)*ln(x)+25*x**2*( ln(3)+I*pi)*exp(x)+25*x**3*(ln(3)+I*pi)+125*x**2)/ln(2))/ln(2),x)
Time = 0.53 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65 \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=e^{\left (\frac {25 i \, \pi x^{3}}{\log \left (2\right )} + \frac {25 i \, \pi x^{2} e^{x}}{\log \left (2\right )} + \frac {25 \, x^{3} \log \left (3\right )}{\log \left (2\right )} + \frac {25 \, x^{2} e^{x} \log \left (3\right )}{\log \left (2\right )} - \frac {25 i \, \pi x^{2} \log \left (x\right )}{\log \left (2\right )} - \frac {25 \, x^{2} \log \left (3\right ) \log \left (x\right )}{\log \left (2\right )} + \frac {125 \, x^{2}}{\log \left (2\right )}\right )} \]
integrate((-50*x*(log(3)+I*pi)*log(x)+(25*x^2+50*x)*(log(3)+I*pi)*exp(x)+( 75*x^2-25*x)*(log(3)+I*pi)+250*x)*exp((-25*x^2*(log(3)+I*pi)*log(x)+25*x^2 *(log(3)+I*pi)*exp(x)+25*x^3*(log(3)+I*pi)+125*x^2)/log(2))/log(2),x, algo rithm=\
e^(25*I*pi*x^3/log(2) + 25*I*pi*x^2*e^x/log(2) + 25*x^3*log(3)/log(2) + 25 *x^2*e^x*log(3)/log(2) - 25*I*pi*x^2*log(x)/log(2) - 25*x^2*log(3)*log(x)/ log(2) + 125*x^2/log(2))
Time = 0.41 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65 \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=e^{\left (\frac {25 i \, \pi x^{3}}{\log \left (2\right )} + \frac {25 i \, \pi x^{2} e^{x}}{\log \left (2\right )} + \frac {25 \, x^{3} \log \left (3\right )}{\log \left (2\right )} + \frac {25 \, x^{2} e^{x} \log \left (3\right )}{\log \left (2\right )} - \frac {25 i \, \pi x^{2} \log \left (x\right )}{\log \left (2\right )} - \frac {25 \, x^{2} \log \left (3\right ) \log \left (x\right )}{\log \left (2\right )} + \frac {125 \, x^{2}}{\log \left (2\right )}\right )} \]
integrate((-50*x*(log(3)+I*pi)*log(x)+(25*x^2+50*x)*(log(3)+I*pi)*exp(x)+( 75*x^2-25*x)*(log(3)+I*pi)+250*x)*exp((-25*x^2*(log(3)+I*pi)*log(x)+25*x^2 *(log(3)+I*pi)*exp(x)+25*x^3*(log(3)+I*pi)+125*x^2)/log(2))/log(2),x, algo rithm=\
e^(25*I*pi*x^3/log(2) + 25*I*pi*x^2*e^x/log(2) + 25*x^3*log(3)/log(2) + 25 *x^2*e^x*log(3)/log(2) - 25*I*pi*x^2*log(x)/log(2) - 25*x^2*log(3)*log(x)/ log(2) + 125*x^2/log(2))
Time = 11.84 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.52 \[ \int \frac {e^{\frac {125 x^2+25 e^x x^2 (i \pi +\log (3))+25 x^3 (i \pi +\log (3))-25 x^2 (i \pi +\log (3)) \log (x)}{\log (2)}} \left (250 x+e^x \left (50 x+25 x^2\right ) (i \pi +\log (3))+\left (-25 x+75 x^2\right ) (i \pi +\log (3))-50 x (i \pi +\log (3)) \log (x)\right )}{\log (2)} \, dx=\frac {{847288609443}^{\frac {x^2\,{\mathrm {e}}^x+x^3}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {125\,x^2}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {\Pi \,x^2\,{\mathrm {e}}^x\,25{}\mathrm {i}}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {\Pi \,x^3\,25{}\mathrm {i}}{\ln \left (2\right )}}}{x^{\frac {25\,x^2\,\ln \left (3\right )+\Pi \,x^2\,25{}\mathrm {i}}{\ln \left (2\right )}}} \]
int((exp((25*x^3*(Pi*1i + log(3)) + 125*x^2 + 25*x^2*exp(x)*(Pi*1i + log(3 )) - 25*x^2*log(x)*(Pi*1i + log(3)))/log(2))*(250*x - (25*x - 75*x^2)*(Pi* 1i + log(3)) + exp(x)*(50*x + 25*x^2)*(Pi*1i + log(3)) - 50*x*log(x)*(Pi*1 i + log(3))))/log(2),x)