Integrand size = 73, antiderivative size = 32 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=e^{-x} \left (e^{1+x}-\frac {\log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}\right ) \] Output:
(exp(1+x)-ln(4/x*ln(5)^2-ln(3))/x)/exp(x)
Time = 0.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x} \] Input:
Integrate[(4*Log[5]^2 + ((-x - x^2)*Log[3] + (4 + 4*x)*Log[5]^2)*Log[(-(x* Log[3]) + 4*Log[5]^2)/x])/(E^x*(-(x^3*Log[3]) + 4*x^2*Log[5]^2)),x]
Output:
-(Log[-Log[3] + (4*Log[5]^2)/x]/(E^x*x))
Time = 1.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2026, 7293, 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.
\(\displaystyle \int \frac {e^{-x} \left (\left (\left (-x^2-x\right ) \log (3)+(4 x+4) \log ^2(5)\right ) \log \left (\frac {4 \log ^2(5)-x \log (3)}{x}\right )+4 \log ^2(5)\right )}{4 x^2 \log ^2(5)-x^3 \log (3)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (\left (\left (-x^2-x\right ) \log (3)+(4 x+4) \log ^2(5)\right ) \log \left (\frac {4 \log ^2(5)-x \log (3)}{x}\right )+4 \log ^2(5)\right )}{x^2 \left (4 \log ^2(5)-x \log (3)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-x} (x+1) \log \left (\frac {4 \log ^2(5)}{x}-\log (3)\right )}{x^2}-\frac {4 e^{-x} \log ^2(5)}{x^2 \left (x \log (3)-4 \log ^2(5)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{-x} \log \left (\frac {4 \log ^2(5)}{x}-\log (3)\right )}{x}\) |
Input:
Int[(4*Log[5]^2 + ((-x - x^2)*Log[3] + (4 + 4*x)*Log[5]^2)*Log[(-(x*Log[3] ) + 4*Log[5]^2)/x])/(E^x*(-(x^3*Log[3]) + 4*x^2*Log[5]^2)),x]
Output:
-(Log[-Log[3] + (4*Log[5]^2)/x]/(E^x*x))
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
norman | \(-\frac {\ln \left (\frac {4 \ln \left (5\right )^{2}-x \ln \left (3\right )}{x}\right ) {\mathrm e}^{-x}}{x}\) | \(27\) |
parallelrisch | \(-\frac {\ln \left (\frac {4 \ln \left (5\right )^{2}-x \ln \left (3\right )}{x}\right ) {\mathrm e}^{-x}}{x}\) | \(27\) |
risch | \(-\frac {{\mathrm e}^{-x} \ln \left (\ln \left (5\right )^{2}-\frac {x \ln \left (3\right )}{4}\right )}{x}-\frac {\left (-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )}^{3}+4 \ln \left (2\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{2 x}\) | \(184\) |
Input:
int((((4+4*x)*ln(5)^2+(-x^2-x)*ln(3))*ln((4*ln(5)^2-x*ln(3))/x)+4*ln(5)^2) /(4*x^2*ln(5)^2-x^3*ln(3))/exp(x),x,method=_RETURNVERBOSE)
Output:
-ln((4*ln(5)^2-x*ln(3))/x)/exp(x)/x
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (-x\right )} \log \left (\frac {4 \, \log \left (5\right )^{2} - x \log \left (3\right )}{x}\right )}{x} \] Input:
integrate((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x) +4*log(5)^2)/(4*x^2*log(5)^2-x^3*log(3))/exp(x),x, algorithm="fricas")
Output:
-e^(-x)*log((4*log(5)^2 - x*log(3))/x)/x
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=- \frac {e^{- x} \log {\left (\frac {- x \log {\left (3 \right )} + 4 \log {\left (5 \right )}^{2}}{x} \right )}}{x} \] Input:
integrate((((4+4*x)*ln(5)**2+(-x**2-x)*ln(3))*ln((4*ln(5)**2-x*ln(3))/x)+4 *ln(5)**2)/(4*x**2*ln(5)**2-x**3*ln(3))/exp(x),x)
Output:
-exp(-x)*log((-x*log(3) + 4*log(5)**2)/x)/x
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (-x\right )} \log \left (4 \, \log \left (5\right )^{2} - x \log \left (3\right )\right ) - e^{\left (-x\right )} \log \left (x\right )}{x} \] Input:
integrate((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x) +4*log(5)^2)/(4*x^2*log(5)^2-x^3*log(3))/exp(x),x, algorithm="maxima")
Output:
-(e^(-x)*log(4*log(5)^2 - x*log(3)) - e^(-x)*log(x))/x
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (-x\right )} \log \left (4 \, \log \left (5\right )^{2} - x \log \left (3\right )\right ) - e^{\left (-x\right )} \log \left (x\right )}{x} \] Input:
integrate((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x) +4*log(5)^2)/(4*x^2*log(5)^2-x^3*log(3))/exp(x),x, algorithm="giac")
Output:
-(e^(-x)*log(4*log(5)^2 - x*log(3)) - e^(-x)*log(x))/x
Timed out. \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=\int \frac {{\mathrm {e}}^{-x}\,\left (\ln \left (-\frac {x\,\ln \left (3\right )-4\,{\ln \left (5\right )}^2}{x}\right )\,\left ({\ln \left (5\right )}^2\,\left (4\,x+4\right )-\ln \left (3\right )\,\left (x^2+x\right )\right )+4\,{\ln \left (5\right )}^2\right )}{4\,x^2\,{\ln \left (5\right )}^2-x^3\,\ln \left (3\right )} \,d x \] Input:
int((exp(-x)*(log(-(x*log(3) - 4*log(5)^2)/x)*(log(5)^2*(4*x + 4) - log(3) *(x + x^2)) + 4*log(5)^2))/(4*x^2*log(5)^2 - x^3*log(3)),x)
Output:
int((exp(-x)*(log(-(x*log(3) - 4*log(5)^2)/x)*(log(5)^2*(4*x + 4) - log(3) *(x + x^2)) + 4*log(5)^2))/(4*x^2*log(5)^2 - x^3*log(3)), x)
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {\mathrm {log}\left (\frac {4 \mathrm {log}\left (5\right )^{2}-\mathrm {log}\left (3\right ) x}{x}\right )}{e^{x} x} \] Input:
int((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x)+4*log (5)^2)/(4*x^2*log(5)^2-x^3*log(3))/exp(x),x)
Output:
( - log((4*log(5)**2 - log(3)*x)/x))/(e**x*x)