Integrand size = 100, antiderivative size = 29 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx=\left (-4+e^3\right ) (i \pi +\log (10)) \log \left (x+\frac {e^x}{\log \left (\frac {5}{x^2}\right )}\right ) \]
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx=\left (-4+e^3\right ) (i \pi +\log (10)) \left (-\log \left (\log \left (\frac {5}{x^2}\right )\right )+\log \left (e^x+x \log \left (\frac {5}{x^2}\right )\right )\right ) \]
Integrate[(E^x*(-8 + 2*E^3)*(I*Pi + Log[10]) + E^x*(-4*x + E^3*x)*(I*Pi + Log[10])*Log[5/x^2] + (-4*x + E^3*x)*(I*Pi + Log[10])*Log[5/x^2]^2)/(E^x*x *Log[5/x^2] + x^2*Log[5/x^2]^2),x]
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.
| \(\displaystyle \int \frac {\left (e^3 x-4 x\right ) (\log (10)+i \pi ) \log ^2\left (\frac {5}{x^2}\right )+e^x \left (e^3 x-4 x\right ) (\log (10)+i \pi ) \log \left (\frac {5}{x^2}\right )+\left (2 e^3-8\right ) e^x (\log (10)+i \pi )}{x^2 \log ^2\left (\frac {5}{x^2}\right )+e^x x \log \left (\frac {5}{x^2}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (4-e^3\right ) (\log (10)+i \pi ) \left (-x \log ^2\left (\frac {5}{x^2}\right )-e^x x \log \left (\frac {5}{x^2}\right )-2 e^x\right )}{x \log \left (\frac {5}{x^2}\right ) \left (x \log \left (\frac {5}{x^2}\right )+e^x\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \left (4-e^3\right ) (\log (10)+i \pi ) \int -\frac {x \log ^2\left (\frac {5}{x^2}\right )+e^x x \log \left (\frac {5}{x^2}\right )+2 e^x}{x \log \left (\frac {5}{x^2}\right ) \left (x \log \left (\frac {5}{x^2}\right )+e^x\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\left (\left (4-e^3\right ) (\log (10)+i \pi ) \int \frac {x \log ^2\left (\frac {5}{x^2}\right )+e^x x \log \left (\frac {5}{x^2}\right )+2 e^x}{x \log \left (\frac {5}{x^2}\right ) \left (x \log \left (\frac {5}{x^2}\right )+e^x\right )}dx\right )\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\left (\left (4-e^3\right ) (\log (10)+i \pi ) \int \left (\frac {x \log \left (\frac {5}{x^2}\right )+2}{x \log \left (\frac {5}{x^2}\right )}-\frac {x \log \left (\frac {5}{x^2}\right )-\log \left (\frac {5}{x^2}\right )+2}{x \log \left (\frac {5}{x^2}\right )+e^x}\right )dx\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\left (\left (4-e^3\right ) (\log (10)+i \pi ) \left (-2 \int \frac {1}{x \log \left (\frac {5}{x^2}\right )+e^x}dx+\int \frac {\log \left (\frac {5}{x^2}\right )}{x \log \left (\frac {5}{x^2}\right )+e^x}dx-\int \frac {x \log \left (\frac {5}{x^2}\right )}{x \log \left (\frac {5}{x^2}\right )+e^x}dx-\log \left (\log \left (\frac {5}{x^2}\right )\right )+x\right )\right )\) |
Int[(E^x*(-8 + 2*E^3)*(I*Pi + Log[10]) + E^x*(-4*x + E^3*x)*(I*Pi + Log[10 ])*Log[5/x^2] + (-4*x + E^3*x)*(I*Pi + Log[10])*Log[5/x^2]^2)/(E^x*x*Log[5 /x^2] + x^2*Log[5/x^2]^2),x]
3.21.51.3.1 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]]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26 ) = 52\).
Time = 0.69 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21
| method | result | size |
| norman | \(\left (-i \pi \,{\mathrm e}^{3}-\ln \left (10\right ) {\mathrm e}^{3}+4 i \pi +4 \ln \left (10\right )\right ) \ln \left (\ln \left (\frac {5}{x^{2}}\right )\right )+\left (i \pi \,{\mathrm e}^{3}+\ln \left (10\right ) {\mathrm e}^{3}-4 i \pi -4 \ln \left (10\right )\right ) \ln \left (x \ln \left (\frac {5}{x^{2}}\right )+{\mathrm e}^{x}\right )\) | \(64\) |
| default | \(\left (-i \pi \,{\mathrm e}^{3}-\ln \left (10\right ) {\mathrm e}^{3}+4 i \pi +4 \ln \left (10\right )\right ) \ln \left (-\ln \left (\frac {5}{x^{2}}\right )\right )+\left (i \pi \,{\mathrm e}^{3}+\ln \left (10\right ) {\mathrm e}^{3}-4 i \pi -4 \ln \left (10\right )\right ) \ln \left (2 x \ln \left (x \right )-x \left (\ln \left (\frac {5}{x^{2}}\right )+2 \ln \left (x \right )\right )-{\mathrm e}^{x}\right )\) | \(79\) |
| parallelrisch | \(-i \pi \,{\mathrm e}^{3} \ln \left (\ln \left (\frac {5}{x^{2}}\right )\right )-\ln \left (10\right ) {\mathrm e}^{3} \ln \left (\ln \left (\frac {5}{x^{2}}\right )\right )+4 i \pi \ln \left (\ln \left (\frac {5}{x^{2}}\right )\right )+4 \ln \left (10\right ) \ln \left (\ln \left (\frac {5}{x^{2}}\right )\right )+i \pi \,{\mathrm e}^{3} \ln \left (x \ln \left (\frac {5}{x^{2}}\right )+{\mathrm e}^{x}\right )+\ln \left (10\right ) {\mathrm e}^{3} \ln \left (x \ln \left (\frac {5}{x^{2}}\right )+{\mathrm e}^{x}\right )-4 i \pi \ln \left (x \ln \left (\frac {5}{x^{2}}\right )+{\mathrm e}^{x}\right )-4 \ln \left (10\right ) \ln \left (x \ln \left (\frac {5}{x^{2}}\right )+{\mathrm e}^{x}\right )\) | \(117\) |
| risch | \(\text {Expression too large to display}\) | \(856\) |
int(((x*exp(3)-4*x)*(ln(10)+I*Pi)*ln(5/x^2)^2+(x*exp(3)-4*x)*(ln(10)+I*Pi) *exp(x)*ln(5/x^2)+(2*exp(3)-8)*(ln(10)+I*Pi)*exp(x))/(x^2*ln(5/x^2)^2+x*ex p(x)*ln(5/x^2)),x,method=_RETURNVERBOSE)
(-I*Pi*exp(3)-ln(10)*exp(3)+4*I*Pi+4*ln(10))*ln(ln(5/x^2))+(I*Pi*exp(3)+ln (10)*exp(3)-4*I*Pi-4*ln(10))*ln(x*ln(5/x^2)+exp(x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.97 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx=-{\left (4 i \, \pi - i \, \pi e^{3} - {\left (e^{3} - 4\right )} \log \left (10\right )\right )} \log \left (\frac {x \log \left (\frac {5}{x^{2}}\right ) + e^{x}}{x}\right ) + \frac {1}{2} \, {\left (4 i \, \pi - i \, \pi e^{3} - {\left (e^{3} - 4\right )} \log \left (10\right )\right )} \log \left (\frac {5}{x^{2}}\right ) - {\left (-4 i \, \pi + i \, \pi e^{3} + {\left (e^{3} - 4\right )} \log \left (10\right )\right )} \log \left (\log \left (\frac {5}{x^{2}}\right )\right ) \]
integrate(((x*exp(3)-4*x)*(log(10)+I*pi)*log(5/x^2)^2+(x*exp(3)-4*x)*(log( 10)+I*pi)*exp(x)*log(5/x^2)+(2*exp(3)-8)*(log(10)+I*pi)*exp(x))/(x^2*log(5 /x^2)^2+x*exp(x)*log(5/x^2)),x, algorithm=\
-(4*I*pi - I*pi*e^3 - (e^3 - 4)*log(10))*log((x*log(5/x^2) + e^x)/x) + 1/2 *(4*I*pi - I*pi*e^3 - (e^3 - 4)*log(10))*log(5/x^2) - (-4*I*pi + I*pi*e^3 + (e^3 - 4)*log(10))*log(log(5/x^2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.88 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx=- \left (-4 + e^{3}\right ) \left (\log {\left (10 \right )} + i \pi \right ) \log {\left (\log {\left (\frac {1}{x^{2}} \right )} + \log {\left (5 \right )} \right )} + \left (-4 + e^{3}\right ) \left (\log {\left (10 \right )} + i \pi \right ) \log {\left (x \log {\left (\frac {1}{x^{2}} \right )} + x \log {\left (5 \right )} + e^{x} \right )} \]
integrate(((x*exp(3)-4*x)*(ln(10)+I*pi)*ln(5/x**2)**2+(x*exp(3)-4*x)*(ln(1 0)+I*pi)*exp(x)*ln(5/x**2)+(2*exp(3)-8)*(ln(10)+I*pi)*exp(x))/(x**2*ln(5/x **2)**2+x*exp(x)*ln(5/x**2)),x)
-(-4 + exp(3))*(log(10) + I*pi)*log(log(x**(-2)) + log(5)) + (-4 + exp(3)) *(log(10) + I*pi)*log(x*log(x**(-2)) + x*log(5) + exp(x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (25) = 50\).
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.55 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx={\left (-4 i \, \pi + {\left (i \, \pi + \log \left (5\right ) + \log \left (2\right )\right )} e^{3} - 4 \, \log \left (5\right ) - 4 \, \log \left (2\right )\right )} \log \left (x \log \left (5\right ) - 2 \, x \log \left (x\right ) + e^{x}\right ) + {\left (4 i \, \pi + {\left (-i \, \pi - \log \left (5\right ) - \log \left (2\right )\right )} e^{3} + 4 \, \log \left (5\right ) + 4 \, \log \left (2\right )\right )} \log \left (-\frac {1}{2} \, \log \left (5\right ) + \log \left (x\right )\right ) \]
integrate(((x*exp(3)-4*x)*(log(10)+I*pi)*log(5/x^2)^2+(x*exp(3)-4*x)*(log( 10)+I*pi)*exp(x)*log(5/x^2)+(2*exp(3)-8)*(log(10)+I*pi)*exp(x))/(x^2*log(5 /x^2)^2+x*exp(x)*log(5/x^2)),x, algorithm=\
(-4*I*pi + (I*pi + log(5) + log(2))*e^3 - 4*log(5) - 4*log(2))*log(x*log(5 ) - 2*x*log(x) + e^x) + (4*I*pi + (-I*pi - log(5) - log(2))*e^3 + 4*log(5) + 4*log(2))*log(-1/2*log(5) + log(x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (25) = 50\).
Time = 0.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 7.07 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx=i \, \pi e^{3} \log \left (x \log \left (5\right ) - x \log \left (x^{2}\right ) + e^{x}\right ) + e^{3} \log \left (5\right ) \log \left (x \log \left (5\right ) - x \log \left (x^{2}\right ) + e^{x}\right ) + e^{3} \log \left (2\right ) \log \left (x \log \left (5\right ) - x \log \left (x^{2}\right ) + e^{x}\right ) - i \, \pi e^{3} \log \left (\log \left (5\right ) - \log \left (x^{2}\right )\right ) - e^{3} \log \left (5\right ) \log \left (\log \left (5\right ) - \log \left (x^{2}\right )\right ) - e^{3} \log \left (2\right ) \log \left (\log \left (5\right ) - \log \left (x^{2}\right )\right ) - 4 i \, \pi \log \left (x \log \left (5\right ) - x \log \left (x^{2}\right ) + e^{x}\right ) - 4 \, \log \left (5\right ) \log \left (x \log \left (5\right ) - x \log \left (x^{2}\right ) + e^{x}\right ) - 4 \, \log \left (2\right ) \log \left (x \log \left (5\right ) - x \log \left (x^{2}\right ) + e^{x}\right ) + 4 i \, \pi \log \left (\log \left (5\right ) - \log \left (x^{2}\right )\right ) + 4 \, \log \left (5\right ) \log \left (\log \left (5\right ) - \log \left (x^{2}\right )\right ) + 4 \, \log \left (2\right ) \log \left (\log \left (5\right ) - \log \left (x^{2}\right )\right ) \]
integrate(((x*exp(3)-4*x)*(log(10)+I*pi)*log(5/x^2)^2+(x*exp(3)-4*x)*(log( 10)+I*pi)*exp(x)*log(5/x^2)+(2*exp(3)-8)*(log(10)+I*pi)*exp(x))/(x^2*log(5 /x^2)^2+x*exp(x)*log(5/x^2)),x, algorithm=\
I*pi*e^3*log(x*log(5) - x*log(x^2) + e^x) + e^3*log(5)*log(x*log(5) - x*lo g(x^2) + e^x) + e^3*log(2)*log(x*log(5) - x*log(x^2) + e^x) - I*pi*e^3*log (log(5) - log(x^2)) - e^3*log(5)*log(log(5) - log(x^2)) - e^3*log(2)*log(l og(5) - log(x^2)) - 4*I*pi*log(x*log(5) - x*log(x^2) + e^x) - 4*log(5)*log (x*log(5) - x*log(x^2) + e^x) - 4*log(2)*log(x*log(5) - x*log(x^2) + e^x) + 4*I*pi*log(log(5) - log(x^2)) + 4*log(5)*log(log(5) - log(x^2)) + 4*log( 2)*log(log(5) - log(x^2))
Time = 12.51 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {e^x \left (-8+2 e^3\right ) (i \pi +\log (10))+e^x \left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log \left (\frac {5}{x^2}\right )+\left (-4 x+e^3 x\right ) (i \pi +\log (10)) \log ^2\left (\frac {5}{x^2}\right )}{e^x x \log \left (\frac {5}{x^2}\right )+x^2 \log ^2\left (\frac {5}{x^2}\right )} \, dx=-\left (\ln \left (10\right )+\Pi \,1{}\mathrm {i}\right )\,\left ({\mathrm {e}}^3-4\right )\,\left (\ln \left (\ln \left (\frac {5}{x^2}\right )\right )-\ln \left ({\mathrm {e}}^x+x\,\ln \left (\frac {5}{x^2}\right )\right )\right ) \]