Integrand size = 47, antiderivative size = 26 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx=\frac {x^2+4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)} \] Output:
(x^2+4*ln(x^2/ln(5)^2))/exp(5)/ln(x)/x
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx=\frac {x^2+4 \log \left (x^2\right )-8 \log (\log (5))}{e^5 x \log (x)} \] Input:
Integrate[(-x^2 + Log[x]*(8 + x^2 - 4*Log[x^2/Log[5]^2]) - 4*Log[x^2/Log[5 ]^2])/(E^5*x^2*Log[x]^2),x]
Output:
(x^2 + 4*Log[x^2] - 8*Log[Log[5]])/(E^5*x*Log[x])
Time = 1.65 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {-x^2+\log (x) \left (x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )+8\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {x^2-\log (x) \left (x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )+8\right )+4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)}dx}{e^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {x^2-\log (x) \left (x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )+8\right )+4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)}dx}{e^5}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {-\log (x) x^2+x^2-8 \log (x)}{x^2 \log ^2(x)}+\frac {4 (\log (x)+1) \log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)}\right )dx}{e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{x \log (x)}-\frac {x}{\log (x)}}{e^5}\) |
Input:
Int[(-x^2 + Log[x]*(8 + x^2 - 4*Log[x^2/Log[5]^2]) - 4*Log[x^2/Log[5]^2])/ (E^5*x^2*Log[x]^2),x]
Output:
-((-(x/Log[x]) - (4*Log[x^2/Log[5]^2])/(x*Log[x]))/E^5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.80 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{-5} \left (-4 x^{2}-16 \ln \left (\frac {x^{2}}{\ln \left (5\right )^{2}}\right )\right )}{4 x \ln \left (x \right )}\) | \(31\) |
risch | \(\frac {8 \,{\mathrm e}^{-5}}{x}-\frac {{\mathrm e}^{-5} \left (2 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-4 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-x^{2}+8 \ln \left (\ln \left (5\right )\right )\right )}{x \ln \left (x \right )}\) | \(80\) |
default | \({\mathrm e}^{-5} \left (-4 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )\right )+8 \ln \left (\ln \left (5\right )\right ) \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )\right )+\frac {x}{\ln \left (x \right )}+\frac {8}{x}+4 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )-8 \ln \left (\ln \left (5\right )\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )\right )\) | \(86\) |
parts | \({\mathrm e}^{-5} \left (-4 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )\right )+8 \,\operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )+8 \ln \left (\ln \left (5\right )\right ) \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )\right )+\frac {x}{\ln \left (x \right )}+\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )\right )+{\mathrm e}^{-5} \left (-\operatorname {expIntegral}_{1}\left (-\ln \left (x \right )\right )+\frac {8}{x}-8 \,\operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )+4 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )-8 \ln \left (\ln \left (5\right )\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \right )\right )\right )\) | \(119\) |
Input:
int(((-4*ln(x^2/ln(5)^2)+x^2+8)*ln(x)-4*ln(x^2/ln(5)^2)-x^2)/x^2/exp(5)/ln (x)^2,x,method=_RETURNVERBOSE)
Output:
-1/4/exp(5)/x*(-4*x^2-16*ln(x^2/ln(5)^2))/ln(x)
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx=\frac {2 \, {\left (x^{2} + 4 \, \log \left (\frac {x^{2}}{\log \left (5\right )^{2}}\right )\right )}}{x e^{5} \log \left (\log \left (5\right )^{2}\right ) + x e^{5} \log \left (\frac {x^{2}}{\log \left (5\right )^{2}}\right )} \] Input:
integrate(((-4*log(x^2/log(5)^2)+x^2+8)*log(x)-4*log(x^2/log(5)^2)-x^2)/x^ 2/exp(5)/log(x)^2,x, algorithm="fricas")
Output:
2*(x^2 + 4*log(x^2/log(5)^2))/(x*e^5*log(log(5)^2) + x*e^5*log(x^2/log(5)^ 2))
Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx=\frac {x^{2} - 8 \log {\left (\log {\left (5 \right )} \right )}}{x e^{5} \log {\left (x \right )}} + \frac {8}{x e^{5}} \] Input:
integrate(((-4*ln(x**2/ln(5)**2)+x**2+8)*ln(x)-4*ln(x**2/ln(5)**2)-x**2)/x **2/exp(5)/ln(x)**2,x)
Output:
(x**2 - 8*log(log(5)))*exp(-5)/(x*log(x)) + 8*exp(-5)/x
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx={\left (8 \, {\rm Ei}\left (-\log \left (x\right )\right ) \log \left (x\right ) - 8 \, \Gamma \left (-1, \log \left (x\right )\right ) \log \left (x\right ) - 4 \, {\rm Ei}\left (-\log \left (x\right )\right ) \log \left (\frac {x^{2}}{\log \left (5\right )^{2}}\right ) + 4 \, \Gamma \left (-1, \log \left (x\right )\right ) \log \left (\frac {x^{2}}{\log \left (5\right )^{2}}\right ) + \frac {8}{x} + {\rm Ei}\left (\log \left (x\right )\right ) - \Gamma \left (-1, -\log \left (x\right )\right )\right )} e^{\left (-5\right )} \] Input:
integrate(((-4*log(x^2/log(5)^2)+x^2+8)*log(x)-4*log(x^2/log(5)^2)-x^2)/x^ 2/exp(5)/log(x)^2,x, algorithm="maxima")
Output:
(8*Ei(-log(x))*log(x) - 8*gamma(-1, log(x))*log(x) - 4*Ei(-log(x))*log(x^2 /log(5)^2) + 4*gamma(-1, log(x))*log(x^2/log(5)^2) + 8/x + Ei(log(x)) - ga mma(-1, -log(x)))*e^(-5)
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx={\left (\frac {8}{x} + \frac {x^{2} - 8 \, \log \left (\log \left (5\right )\right )}{x \log \left (x\right )}\right )} e^{\left (-5\right )} \] Input:
integrate(((-4*log(x^2/log(5)^2)+x^2+8)*log(x)-4*log(x^2/log(5)^2)-x^2)/x^ 2/exp(5)/log(x)^2,x, algorithm="giac")
Output:
(8/x + (x^2 - 8*log(log(5)))/(x*log(x)))*e^(-5)
Time = 1.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-5}\,\left (4\,\ln \left (x^2\right )-8\,\ln \left (\ln \left (5\right )\right )+x^2\right )}{x\,\ln \left (x\right )} \] Input:
int(-(exp(-5)*(4*log(x^2/log(5)^2) - log(x)*(x^2 - 4*log(x^2/log(5)^2) + 8 ) + x^2))/(x^2*log(x)^2),x)
Output:
(exp(-5)*(4*log(x^2) - 8*log(log(5)) + x^2))/(x*log(x))
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x^2 \log ^2(x)} \, dx=\frac {4 \,\mathrm {log}\left (\frac {x^{2}}{\mathrm {log}\left (5\right )^{2}}\right )+x^{2}}{\mathrm {log}\left (x \right ) e^{5} x} \] Input:
int(((-4*log(x^2/log(5)^2)+x^2+8)*log(x)-4*log(x^2/log(5)^2)-x^2)/x^2/exp( 5)/log(x)^2,x)
Output:
(4*log(x**2/log(5)**2) + x**2)/(log(x)*e**5*x)