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

Optimal. Leaf size=23 \[ \left (e^x-x \log \left (-\frac {1}{5}+x\right )\right )^{e^{-1-e^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.62, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 101, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6688, 12, 6686} \begin {gather*} \left (e^x-x \log \left (x-\frac {1}{5}\right )\right )^{e^{-1-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x - x*Log[-1/5 + x])^E^(-1 - E^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]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 23, normalized size = 1.00 \begin {gather*} \left (e^x-x \log \left (-\frac {1}{5}+x\right )\right )^{e^{-1-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x - x*Log[-1/5 + x])^E^(-1 - E^2)

________________________________________________________________________________________

fricas [B]  time = 0.90, size = 37, normalized size = 1.61 \begin {gather*} \left (-{\left (x e^{\left (e^{2} + 1\right )} \log \left (x - \frac {1}{5}\right ) - e^{\left (x + e^{2} + 1\right )}\right )} e^{\left (-e^{2} - 1\right )}\right )^{e^{\left (-e^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(log(-x*log(x-1/5)+exp(x))/exp(exp(2)+1))/((5*x^2-x)*exp
(exp(2)+1)*log(x-1/5)+(-5*x+1)*exp(x)*exp(exp(2)+1)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(log(-x*log(x-1/5)+exp(x))/exp(exp(2)+1))/((5*x^2-x)*exp
(exp(2)+1)*log(x-1/5)+(-5*x+1)*exp(x)*exp(exp(2)+1)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (5 x -1\right ) \ln \left (x -\frac {1}{5}\right )+\left (-5 x +1\right ) {\mathrm e}^{x}+5 x \right ) {\mathrm e}^{\ln \left (-x \ln \left (x -\frac {1}{5}\right )+{\mathrm e}^{x}\right ) {\mathrm e}^{-{\mathrm e}^{2}-1}}}{\left (5 x^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{2}+1} \ln \left (x -\frac {1}{5}\right )+\left (-5 x +1\right ) {\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-1)*ln(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(ln(-x*ln(x-1/5)+exp(x))/exp(exp(2)+1))/((5*x^2-x)*exp(exp(2)+1
)*ln(x-1/5)+(-5*x+1)*exp(x)*exp(exp(2)+1)),x)

[Out]

int(((5*x-1)*ln(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(ln(-x*ln(x-1/5)+exp(x))/exp(exp(2)+1))/((5*x^2-x)*exp(exp(2)+1
)*ln(x-1/5)+(-5*x+1)*exp(x)*exp(exp(2)+1)),x)

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 24, normalized size = 1.04 \begin {gather*} {\left (x {\left (\log \relax (5) - \log \left (5 \, x - 1\right )\right )} + e^{x}\right )}^{e^{\left (-e^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(log(-x*log(x-1/5)+exp(x))/exp(exp(2)+1))/((5*x^2-x)*exp
(exp(2)+1)*log(x-1/5)+(-5*x+1)*exp(x)*exp(exp(2)+1)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.93, size = 18, normalized size = 0.78 \begin {gather*} {\left ({\mathrm {e}}^x-x\,\ln \left (x-\frac {1}{5}\right )\right )}^{{\mathrm {e}}^{-{\mathrm {e}}^2-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 33.39, size = 19, normalized size = 0.83 \begin {gather*} \left (- x \log {\left (x - \frac {1}{5} \right )} + e^{x}\right )^{e^{- e^{2} - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*ln(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(ln(-x*ln(x-1/5)+exp(x))/exp(exp(2)+1))/((5*x**2-x)*exp(e
xp(2)+1)*ln(x-1/5)+(-5*x+1)*exp(x)*exp(exp(2)+1)),x)

[Out]

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

________________________________________________________________________________________