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

Optimal. Leaf size=34 \[ 5-\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (\left (e^5-x\right )^2\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 36, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {12, 6686} \begin {gather*} -\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (x^2-2 e^5 x+e^{10}\right )} \end {gather*}

Antiderivative was successfully verified.

[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 6686

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.94 \begin {gather*} -\frac {\log (5-i \pi -\log (5-\log (\log (2))))}{\log \left (\left (e^5-x\right )^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.61, size = 27, normalized size = 0.79 \begin {gather*} -\frac {\log \left (-\log \left (\log \left (\log \relax (2)\right ) - 5\right ) + 5\right )}{\log \left (x^{2} - 2 \, x e^{5} + e^{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 27, normalized size = 0.79 \begin {gather*} -\frac {\log \left (-\log \left (\log \left (\log \relax (2)\right ) - 5\right ) + 5\right )}{\log \left (x^{2} - 2 \, x e^{5} + e^{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.10, size = 28, normalized size = 0.82

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.35, size = 22, normalized size = 0.65 \begin {gather*} -\frac {\log \left (-\log \left (\log \left (\log \relax (2)\right ) - 5\right ) + 5\right )}{2 \, \log \left (x - e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.61, size = 27, normalized size = 0.79 \begin {gather*} -\frac {\ln \left (5-\ln \left (\ln \left (\ln \relax (2)\right )-5\right )\right )}{\ln \left (x^2-2\,{\mathrm {e}}^5\,x+{\mathrm {e}}^{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.12, size = 31, normalized size = 0.91 \begin {gather*} - \frac {\log {\left (- \log {\left (5 - \log {\left (\log {\relax (2 )} \right )} \right )} + 5 - i \pi \right )}}{\log {\left (x^{2} - 2 x e^{5} + e^{10} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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