3.100.2 \(\int \frac {-10 x \log (5) \log (e^{-x} x)+(-1+x) \log ^{\frac {2}{\log (5)}}(e^{-x} x)}{10 x \log (5) \log (e^{-x} x)} \, dx\)

Optimal. Leaf size=29 \[ 25-e^5-x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 6688, 6686} \begin {gather*} -x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x*Log[5]*Log[x/E^x] + (-1 + x)*Log[x/E^x]^(2/Log[5]))/(10*x*Log[5]*Log[x/E^x]),x]

[Out]

-x - Log[x/E^x]^(2/Log[5])/20

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{x \log \left (e^{-x} x\right )} \, dx}{10 \log (5)}\\ &=\frac {\int \left (-10 \log (5)+\frac {(-1+x) \log ^{-1+\frac {2}{\log (5)}}\left (e^{-x} x\right )}{x}\right ) \, dx}{10 \log (5)}\\ &=-x+\frac {\int \frac {(-1+x) \log ^{-1+\frac {2}{\log (5)}}\left (e^{-x} x\right )}{x} \, dx}{10 \log (5)}\\ &=-x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.79 \begin {gather*} -x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x*Log[5]*Log[x/E^x] + (-1 + x)*Log[x/E^x]^(2/Log[5]))/(10*x*Log[5]*Log[x/E^x]),x]

[Out]

-x - Log[x/E^x]^(2/Log[5])/20

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fricas [A]  time = 0.77, size = 20, normalized size = 0.69 \begin {gather*} -x - \frac {1}{20} \, \log \left (x e^{\left (-x\right )}\right )^{\frac {2}{\log \relax (5)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x-1)*exp(2*log(log(x/exp(x)))/log(5))-10*x*log(5)*log(x/exp(x)))/x/log(5)/log(x/exp(x)),x, al
gorithm="fricas")

[Out]

-x - 1/20*log(x*e^(-x))^(2/log(5))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {10 \, x \log \relax (5) \log \left (x e^{\left (-x\right )}\right ) - {\left (x - 1\right )} \log \left (x e^{\left (-x\right )}\right )^{\frac {2}{\log \relax (5)}}}{10 \, x \log \relax (5) \log \left (x e^{\left (-x\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x-1)*exp(2*log(log(x/exp(x)))/log(5))-10*x*log(5)*log(x/exp(x)))/x/log(5)/log(x/exp(x)),x, al
gorithm="giac")

[Out]

integrate(-1/10*(10*x*log(5)*log(x*e^(-x)) - (x - 1)*log(x*e^(-x))^(2/log(5)))/(x*log(5)*log(x*e^(-x))), x)

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maple [C]  time = 0.26, size = 72, normalized size = 2.48




method result size



risch \(-x -\frac {\left (\ln \relax (x )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right )^{\frac {2}{\ln \relax (5)}}}{20}\) \(72\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((x-1)*exp(2*ln(ln(x/exp(x)))/ln(5))-10*x*ln(5)*ln(x/exp(x)))/x/ln(5)/ln(x/exp(x)),x,method=_RETURNVE
RBOSE)

[Out]

-x-1/20*(ln(x)-ln(exp(x))-1/2*I*Pi*csgn(I*x*exp(-x))*(-csgn(I*x*exp(-x))+csgn(I*x))*(-csgn(I*x*exp(-x))+csgn(I
*exp(-x))))^(2/ln(5))

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maxima [A]  time = 0.48, size = 28, normalized size = 0.97 \begin {gather*} -\frac {20 \, x \log \relax (5) + {\left (-x + \log \relax (x)\right )}^{\frac {2}{\log \relax (5)}} \log \relax (5)}{20 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x-1)*exp(2*log(log(x/exp(x)))/log(5))-10*x*log(5)*log(x/exp(x)))/x/log(5)/log(x/exp(x)),x, al
gorithm="maxima")

[Out]

-1/20*(20*x*log(5) + (-x + log(x))^(2/log(5))*log(5))/log(5)

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mupad [B]  time = 5.67, size = 19, normalized size = 0.66 \begin {gather*} -x-\frac {{\left (\ln \relax (x)-x\right )}^{\frac {2}{\ln \relax (5)}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x*exp(-x))^(2/log(5))*(x - 1))/10 - x*log(5)*log(x*exp(-x)))/(x*log(5)*log(x*exp(-x))),x)

[Out]

- x - (log(x) - x)^(2/log(5))/20

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sympy [A]  time = 2.32, size = 24, normalized size = 0.83 \begin {gather*} \frac {- x \log {\relax (5 )} - \frac {\left (- x + \log {\relax (x )}\right )^{\frac {2}{\log {\relax (5 )}}} \log {\relax (5 )}}{20}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x-1)*exp(2*ln(ln(x/exp(x)))/ln(5))-10*x*ln(5)*ln(x/exp(x)))/x/ln(5)/ln(x/exp(x)),x)

[Out]

(-x*log(5) - (-x + log(x))**(2/log(5))*log(5)/20)/log(5)

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