3.98.39 \(\int \frac {-e^2 \log (x \log (5))-x \log (x \log (5))}{x \log (x \log (5))} \, dx\)

Optimal. Leaf size=15 \[ 9-x-e^2 \log (x \log (5)) \]

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Rubi [B]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 2.33, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2158, 29} \begin {gather*} \log (x) \left (x-\frac {x \log (x \log (5))+e^2 \log (x \log (5))}{\log (x \log (5))}\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^2*Log[x*Log[5]]) - x*Log[x*Log[5]])/(x*Log[x*Log[5]]),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-x-\left (-x-\frac {-e^2 \log (x \log (5))-x \log (x \log (5))}{\log (x \log (5))}\right ) \int \frac {1}{x} \, dx\\ &=-x+\log (x) \left (x-\frac {e^2 \log (x \log (5))+x \log (x \log (5))}{\log (x \log (5))}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.73 \begin {gather*} -x-e^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^2*Log[x*Log[5]]) - x*Log[x*Log[5]])/(x*Log[x*Log[5]]),x]

[Out]

-x - E^2*Log[x]

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fricas [A]  time = 0.83, size = 10, normalized size = 0.67 \begin {gather*} -e^{2} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^2*log(x) - x

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giac [A]  time = 0.12, size = 10, normalized size = 0.67 \begin {gather*} -e^{2} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^2*log(x) - x

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maple [A]  time = 0.03, size = 11, normalized size = 0.73

method	result	size
risch	\(-x -{\mathrm e}^{2} \ln \relax (x )\)	\(11\)
norman	\(-{\mathrm e}^{2} \ln \left (x \ln \relax (5)\right )-x\)	\(14\)
default	\(-x +{\mathrm e}^{\ln \left (-\ln \left (x \ln \relax (5)\right )\right )+2}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x-exp(2)*ln(x)

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maxima [A]  time = 0.38, size = 10, normalized size = 0.67 \begin {gather*} -e^{2} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^2*log(x) - x

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mupad [B]  time = 5.89, size = 10, normalized size = 0.67 \begin {gather*} -x-{\mathrm {e}}^2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- x - exp(2)*log(x)

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sympy [A]  time = 0.13, size = 8, normalized size = 0.53 \begin {gather*} - x - e^{2} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - exp(2)*log(x)

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