\(\int \frac {-10 x \log (5) \log (e^{-x} x)+(-1+x) \log ^{\frac {2}{\log (5)}}(e^{-x} x)}{10 x \log (5) \log (e^{-x} x)} \, dx\) [9902]

Optimal result

Integrand size = 54, antiderivative size = 29 \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=25-e^5-x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right ) \]

[Out]

25-1/20*exp(2*ln(ln(x/exp(x)))/ln(5))-exp(5)-x

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 6820, 6818} \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=-x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right ) \]

[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 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]

Rule 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=-x-\frac {1}{20} \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right ) \]

[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

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76

method	result	size
parts	\(-x -\frac {{\mathrm e}^{\frac {2 \ln \left (\ln \left (x \,{\mathrm e}^{-x}\right )\right )}{\ln \left (5\right )}}}{20}\)	\(22\)
default	\(\frac {-\frac {{\mathrm e}^{\frac {2 \ln \left (\ln \left (x \,{\mathrm e}^{-x}\right )\right )}{\ln \left (5\right )}} \ln \left (5\right )}{2}-10 x \ln \left (5\right )}{10 \ln \left (5\right )}\)	\(32\)
parallelrisch	\(\frac {-\frac {{\mathrm e}^{\frac {2 \ln \left (\ln \left (x \,{\mathrm e}^{-x}\right )\right )}{\ln \left (5\right )}} \ln \left (5\right )}{2}-10 x \ln \left (5\right )}{10 \ln \left (5\right )}\)	\(32\)
risch	\(-x -\frac {{\left (\ln \left (x \right )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right )}^{\frac {2}{\ln \left (5\right )}}}{20}\)	\(72\)

[In]

int(1/10*((-1+x)*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=_RETURNV
ERBOSE)

[Out]

-x-1/20*exp(2*ln(ln(x/exp(x)))/ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=-x - \frac {1}{20} \, \log \left (x e^{\left (-x\right )}\right )^{\frac {2}{\log \left (5\right )}} \]

[In]

integrate(1/10*((-1+x)*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, a
lgorithm="fricas")

[Out]

-x - 1/20*log(x*e^(-x))^(2/log(5))

Sympy [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=\frac {- x \log {\left (5 \right )} - \frac {\log {\left (5 \right )} \log {\left (x e^{- x} \right )}^{\frac {2}{\log {\left (5 \right )}}}}{20}}{\log {\left (5 \right )}} \]

[In]

integrate(1/10*((-1+x)*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) - log(5)*log(x*exp(-x))**(2/log(5))/20)/log(5)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=-\frac {20 \, x \log \left (5\right ) + {\left (-x + \log \left (x\right )\right )}^{\frac {2}{\log \left (5\right )}} \log \left (5\right )}{20 \, \log \left (5\right )} \]

[In]

integrate(1/10*((-1+x)*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, a
lgorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=\int { -\frac {10 \, x \log \left (5\right ) \log \left (x e^{\left (-x\right )}\right ) - {\left (x - 1\right )} \log \left (x e^{\left (-x\right )}\right )^{\frac {2}{\log \left (5\right )}}}{10 \, x \log \left (5\right ) \log \left (x e^{\left (-x\right )}\right )} \,d x } \]

[In]

integrate(1/10*((-1+x)*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, a
lgorithm="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)

Mupad [B] (verification not implemented)

Time = 14.70 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {-10 x \log (5) \log \left (e^{-x} x\right )+(-1+x) \log ^{\frac {2}{\log (5)}}\left (e^{-x} x\right )}{10 x \log (5) \log \left (e^{-x} x\right )} \, dx=-x-\frac {{\left (\ln \left (x\right )-x\right )}^{\frac {2}{\ln \left (5\right )}}}{20} \]

[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