\(\int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{(e^5-x) \log ^2(e^{10}-2 e^5 x+x^2)} \, dx\) [9621]

Optimal result

Integrand size = 45, antiderivative size = 34 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=5-\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (\left (e^5-x\right )^2\right )} \]

[Out]

5-ln(-ln(ln(ln(2))-5)+5)/ln((exp(5)-x)^2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {12, 6818} \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=-\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (x^2-2 e^5 x+e^{10}\right )} \]

[In]

Int[(-2*Log[5 - I*Pi - Log[5 - Log[Log[2]]]])/((E^5 - x)*Log[E^10 - 2*E^5*x + x^2]^2),x]

[Out]

-(Log[5 - I*Pi - Log[5 - Log[Log[2]]]]/Log[E^10 - 2*E^5*x + x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\left ((2 \log (5-i \pi -\log (5-\log (\log (2))))) \int \frac {1}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx\right ) \\ & = -\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (e^{10}-2 e^5 x+x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=-\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (\left (e^5-x\right )^2\right )} \]

[In]

Integrate[(-2*Log[5 - I*Pi - Log[5 - Log[Log[2]]]])/((E^5 - x)*Log[E^10 - 2*E^5*x + x^2]^2),x]

[Out]

-(Log[5 - I*Pi - Log[5 - Log[Log[2]]]]/Log[(E^5 - x)^2])

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82

method	result	size
risch	\(-\frac {\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )-5\right )+5\right )}{\ln \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\)	\(28\)
derivativedivides	\(-\frac {\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )-5\right )+5\right )}{\ln \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\)	\(30\)
default	\(-\frac {\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )-5\right )+5\right )}{\ln \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\)	\(30\)
norman	\(-\frac {\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )-5\right )+5\right )}{\ln \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\)	\(30\)
parallelrisch	\(-\frac {\ln \left (-\ln \left (\ln \left (\ln \left (2\right )\right )-5\right )+5\right )}{\ln \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\)	\(30\)

[In]

int(-2*ln(-ln(ln(ln(2))-5)+5)/(exp(5)-x)/ln(exp(5)^2-2*x*exp(5)+x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-ln(-ln(ln(ln(2))-5)+5)/ln(exp(10)-2*x*exp(5)+x^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=-\frac {\log \left (-\log \left (\log \left (\log \left (2\right )\right ) - 5\right ) + 5\right )}{\log \left (x^{2} - 2 \, x e^{5} + e^{10}\right )} \]

[In]

integrate(-2*log(-log(log(log(2))-5)+5)/(exp(5)-x)/log(exp(5)^2-2*x*exp(5)+x^2)^2,x, algorithm="fricas")

[Out]

-log(-log(log(log(2)) - 5) + 5)/log(x^2 - 2*x*e^5 + e^10)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=- \frac {\log {\left (- \log {\left (5 - \log {\left (\log {\left (2 \right )} \right )} \right )} + 5 - i \pi \right )}}{\log {\left (x^{2} - 2 x e^{5} + e^{10} \right )}} \]

[In]

integrate(-2*ln(-ln(ln(ln(2))-5)+5)/(exp(5)-x)/ln(exp(5)**2-2*x*exp(5)+x**2)**2,x)

[Out]

-log(-log(5 - log(log(2))) + 5 - I*pi)/log(x**2 - 2*x*exp(5) + exp(10))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=-\frac {\log \left (-\log \left (\log \left (\log \left (2\right )\right ) - 5\right ) + 5\right )}{2 \, \log \left (x - e^{5}\right )} \]

[In]

integrate(-2*log(-log(log(log(2))-5)+5)/(exp(5)-x)/log(exp(5)^2-2*x*exp(5)+x^2)^2,x, algorithm="maxima")

[Out]

-1/2*log(-log(log(log(2)) - 5) + 5)/log(x - e^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=-\frac {\log \left (-\log \left (\log \left (\log \left (2\right )\right ) - 5\right ) + 5\right )}{\log \left (x^{2} - 2 \, x e^{5} + e^{10}\right )} \]

[In]

integrate(-2*log(-log(log(log(2))-5)+5)/(exp(5)-x)/log(exp(5)^2-2*x*exp(5)+x^2)^2,x, algorithm="giac")

[Out]

-log(-log(log(log(2)) - 5) + 5)/log(x^2 - 2*x*e^5 + e^10)

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int -\frac {2 \log (5-i \pi -\log (5-\log (\log (2))))}{\left (e^5-x\right ) \log ^2\left (e^{10}-2 e^5 x+x^2\right )} \, dx=-\frac {\ln \left (5-\ln \left (\ln \left (\ln \left (2\right )\right )-5\right )\right )}{\ln \left (x^2-2\,{\mathrm {e}}^5\,x+{\mathrm {e}}^{10}\right )} \]

[In]

int((2*log(5 - log(log(log(2)) - 5)))/(log(exp(10) - 2*x*exp(5) + x^2)^2*(x - exp(5))),x)

[Out]

-log(5 - log(log(log(2)) - 5))/log(exp(10) - 2*x*exp(5) + x^2)