Integrand size = 101, antiderivative size = 23 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx=\left (e^x-x \log \left (-\frac {1}{5}+x\right )\right )^{e^{-1-e^2}} \] Output:
exp(ln(-x*ln(x-1/5)+exp(x))/exp(exp(2)+1))
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx=\left (e^x-x \log \left (-\frac {1}{5}+x\right )\right )^{e^{-1-e^2}} \] Input:
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]
Output:
(E^x - x*Log[-1/5 + x])^E^(-1 - E^2)
Time = 1.52 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {7292, 27, 7248, 15}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e^x-x \log \left (\frac {1}{5} (5 x-1)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(5 x-1) \log \left (\frac {1}{5} (5 x-1)\right )\right )}{e^{1+e^2} \left (5 x^2-x\right ) \log \left (\frac {1}{5} (5 x-1)\right )+e^{x+e^2+1} (1-5 x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-1-e^2} \left (e^x-x \log \left (\frac {1}{5} (5 x-1)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(5 x-1) \log \left (\frac {1}{5} (5 x-1)\right )\right )}{(1-5 x) \left (e^x-x \log \left (x-\frac {1}{5}\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{-1-e^2} \int \frac {\left (e^x (1-5 x)-\log \left (\frac {1}{5} (5 x-1)\right ) (1-5 x)+5 x\right ) \left (e^x-x \log \left (\frac {1}{5} (5 x-1)\right )\right )^{e^{-1-e^2}}}{(1-5 x) \left (e^x-x \log \left (x-\frac {1}{5}\right )\right )}dx\) |
\(\Big \downarrow \) 7248 |
\(\displaystyle e^{-1-e^2} \int \left (e^x-x \log \left (\frac {1}{5} (5 x-1)\right )\right )^{-1+e^{-1-e^2}}d\left (e^x-x \log \left (\frac {1}{5} (5 x-1)\right )\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \left (e^x-x \log \left (\frac {1}{5} (5 x-1)\right )\right )^{e^{-1-e^2}}\) |
Input:
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]
Output:
(E^x - x*Log[(-1 + 5*x)/5])^E^(-1 - E^2)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(v_)^(m_.)*((a_.) + (b_.)*(y_)^(n_))^(p_.), x_Symbol] :> Module[{ q, r}, Simp[q*r Subst[Int[x^m*(a + b*x^n)^p, x], x, y], x] /; !FalseQ[r = Divides[y^m, v^m, x]] && !FalseQ[q = DerivativeDivides[y, u, x]]] /; Fre eQ[{a, b, m, n, p}, x]
Time = 35.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
parallelrisch | \({\mathrm e}^{\ln \left (-x \ln \left (x -\frac {1}{5}\right )+{\mathrm e}^{x}\right ) {\mathrm e}^{-{\mathrm e}^{2}-1}}\) | \(21\) |
Input:
int(((5*x-1)*ln(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(ln(-x*ln(x-1/5)+exp(x))/ex p(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,method=_RETURNVERBOSE)
Output:
exp(ln(-x*ln(x-1/5)+exp(x))/exp(exp(2)+1))
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx=\left (-{\left (x e^{\left (e^{2} + 1\right )} \log \left (x - \frac {1}{5}\right ) - e^{\left (x + e^{2} + 1\right )}\right )} e^{\left (-e^{2} - 1\right )}\right )^{e^{\left (-e^{2} - 1\right )}} \] Input:
integrate(((5*x-1)*log(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(log(-x*log(x-1/5)+e xp(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")
Output:
(-(x*e^(e^2 + 1)*log(x - 1/5) - e^(x + e^2 + 1))*e^(-e^2 - 1))^e^(-e^2 - 1 )
Time = 18.98 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx=\left (- x \log {\left (x - \frac {1}{5} \right )} + e^{x}\right )^{e^{- e^{2} - 1}} \] Input:
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(exp(2)+1)*ln(x-1/5)+(-5*x+1)*exp(x)*exp (exp(2)+1)),x)
Output:
(-x*log(x - 1/5) + exp(x))**exp(-exp(2) - 1)
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx={\left (x {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )} + e^{x}\right )}^{e^{\left (-e^{2} - 1\right )}} \] Input:
integrate(((5*x-1)*log(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(log(-x*log(x-1/5)+e xp(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")
Output:
(x*(log(5) - log(5*x - 1)) + e^x)^e^(-e^2 - 1)
\[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx=\int { -\frac {{\left ({\left (5 \, x - 1\right )} e^{x} - {\left (5 \, x - 1\right )} \log \left (x - \frac {1}{5}\right ) - 5 \, x\right )} {\left (-x \log \left (x - \frac {1}{5}\right ) + e^{x}\right )}^{e^{\left (-e^{2} - 1\right )}}}{{\left (5 \, x^{2} - x\right )} e^{\left (e^{2} + 1\right )} \log \left (x - \frac {1}{5}\right ) - {\left (5 \, x - 1\right )} e^{\left (x + e^{2} + 1\right )}} \,d x } \] Input:
integrate(((5*x-1)*log(x-1/5)+(-5*x+1)*exp(x)+5*x)*exp(log(-x*log(x-1/5)+e xp(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")
Output:
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)
Time = 3.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx={\left ({\mathrm {e}}^x-x\,\ln \left (x-\frac {1}{5}\right )\right )}^{{\mathrm {e}}^{-{\mathrm {e}}^2-1}} \] Input:
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)
Output:
(exp(x) - x*log(x - 1/5))^exp(- exp(2) - 1)
Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\left (e^x-x \log \left (\frac {1}{5} (-1+5 x)\right )\right )^{e^{-1-e^2}} \left (e^x (1-5 x)+5 x+(-1+5 x) \log \left (\frac {1}{5} (-1+5 x)\right )\right )}{e^{1+e^2+x} (1-5 x)+e^{1+e^2} \left (-x+5 x^2\right ) \log \left (\frac {1}{5} (-1+5 x)\right )} \, dx=e^{\frac {\mathrm {log}\left (e^{x}-\mathrm {log}\left (x -\frac {1}{5}\right ) x \right )}{e^{e^{2}} e}} \] Input:
int(((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(ex p(2)+1)),x)
Output:
e**(log(e**x - log((5*x - 1)/5)*x)/(e**(e**2)*e))