Integrand size = 109, antiderivative size = 33 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\log \left (\frac {e^{x/5}}{-5+\log \left (e^{1-e^5}-e^{5+2 x}\right )}\right ) \]
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\frac {1}{5} \left (x-5 \log \left (4+e^5-\log \left (1-e^{4+e^5+2 x}\right )\right )\right ) \]
Integrate[(5 - 15*E^(4 + E^5 + 2*x) + (-1 + E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))])/(25 - 25*E^(4 + E^5 + 2*x) + (-5 + 5*E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))]),x]
Time = 0.99 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {2720, 27, 7292, 7293, 2009}
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.
\(\displaystyle \int \frac {-15 e^{2 x+e^5+4}+\left (e^{2 x+e^5+4}-1\right ) \log \left (e^{1-e^5} \left (1-e^{2 x+e^5+4}\right )\right )+5}{-25 e^{2 x+e^5+4}+\left (5 e^{2 x+e^5+4}-5\right ) \log \left (e^{1-e^5} \left (1-e^{2 x+e^5+4}\right )\right )+25} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} \int \frac {e^{-2 x-e^5-4} \left (-\log \left (1-e^{2 x+e^5+4}\right ) \left (1-e^{2 x+e^5+4}\right )+e^5 \left (1-e^{2 x+e^5+4}\right )-14 e^{2 x+e^5+4}+4\right )}{5 \left (1-e^{2 x+e^5+4}\right ) \left (-\log \left (1-e^{2 x+e^5+4}\right )+e^5+4\right )}de^{2 x+e^5+4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int \frac {e^{-2 x-e^5-4} \left (-\log \left (1-e^{2 x+e^5+4}\right ) \left (1-e^{2 x+e^5+4}\right )+e^5 \left (1-e^{2 x+e^5+4}\right )-14 e^{2 x+e^5+4}+4\right )}{\left (1-e^{2 x+e^5+4}\right ) \left (-\log \left (1-e^{2 x+e^5+4}\right )+e^5+4\right )}de^{2 x+e^5+4}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{10} \int \frac {e^{-2 x-e^5-4} \left (-\log \left (1-e^{2 x+e^5+4}\right ) \left (1-e^{2 x+e^5+4}\right )+e^5 \left (1-e^{2 x+e^5+4}\right )-14 e^{2 x+e^5+4}+4\right )}{\left (1-e^{2 x+e^5+4}\right ) \left (4 \left (1+\frac {e^5}{4}\right )-\log \left (1-e^{2 x+e^5+4}\right )\right )}de^{2 x+e^5+4}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{10} \int \left (e^{-2 x-e^5-4}+\frac {10}{\left (-1+e^{2 x+e^5+4}\right ) \left (4 \left (1+\frac {e^5}{4}\right )-\log \left (1-e^{2 x+e^5+4}\right )\right )}\right )de^{2 x+e^5+4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} \left (\log \left (e^{2 x+e^5+4}\right )-10 \log \left (-\log \left (1-e^{2 x+e^5+4}\right )+e^5+4\right )\right )\) |
Int[(5 - 15*E^(4 + E^5 + 2*x) + (-1 + E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*( 1 - E^(4 + E^5 + 2*x))])/(25 - 25*E^(4 + E^5 + 2*x) + (-5 + 5*E^(4 + E^5 + 2*x))*Log[E^(1 - E^5)*(1 - E^(4 + E^5 + 2*x))]),x]
3.28.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.79 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
method | result | size |
risch | \(\frac {x}{5}-\ln \left (\ln \left (\left (-{\mathrm e}^{{\mathrm e}^{5}+2 x +4}+1\right ) {\mathrm e}^{1-{\mathrm e}^{5}}\right )-5\right )\) | \(31\) |
norman | \(\frac {x}{5}-\ln \left (\ln \left (\left (-{\mathrm e}^{5+2 x} {\mathrm e}^{{\mathrm e}^{5}-1}+1\right ) {\mathrm e}^{1-{\mathrm e}^{5}}\right )-5\right )\) | \(34\) |
parallelrisch | \(-\ln \left (\ln \left (-\left ({\mathrm e}^{5+2 x} {\mathrm e}^{{\mathrm e}^{5}-1}-1\right ) {\mathrm e}^{1-{\mathrm e}^{5}}\right )-5\right )+\frac {x}{5}\) | \(34\) |
int(((exp(5+2*x)*exp(exp(5)-1)-1)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp (5)-1))-15*exp(5+2*x)*exp(exp(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1)-5)*ln( (-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp(5)-1)+2 5),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\frac {1}{5} \, x - \log \left (\log \left (-{\left (e^{\left (2 \, x + e^{5} + 4\right )} - 1\right )} e^{\left (-e^{5} + 1\right )}\right ) - 5\right ) \]
integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/ exp(exp(5)-1))-15*exp(5+2*x)*exp(exp(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1) -5)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp (5)-1)+25),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\frac {x}{5} - \log {\left (\log {\left (\frac {- e^{-1 + e^{5}} e^{2 x + 5} + 1}{e^{-1 + e^{5}}} \right )} - 5 \right )} \]
integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/e xp(exp(5)-1))-15*exp(5+2*x)*exp(exp(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1)- 5)*ln((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp(5 )-1)+25),x)
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\frac {1}{5} \, x + \frac {1}{10} \, e^{5} - \log \left (-e^{5} + \log \left (-e^{\left (2 \, x + e^{5} + 4\right )} + 1\right ) - 4\right ) + \frac {2}{5} \]
integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/ exp(exp(5)-1))-15*exp(5+2*x)*exp(exp(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1) -5)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp (5)-1)+25),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (27) = 54\).
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\frac {1}{5} \, x + \frac {1}{10} \, \log \left (e^{\left (2 \, x + e^{5} + 4\right )} - 1\right ) - \frac {1}{10} \, \log \left (-e^{\left (2 \, x + e^{5} + 4\right )} + 1\right ) - \log \left (\log \left ({\left (e^{5} - e^{\left (2 \, x + e^{5} + 9\right )}\right )} e^{\left (-e^{5} - 4\right )}\right ) - 5\right ) \]
integrate(((exp(5+2*x)*exp(exp(5)-1)-1)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/ exp(exp(5)-1))-15*exp(5+2*x)*exp(exp(5)-1)+5)/((5*exp(5+2*x)*exp(exp(5)-1) -5)*log((-exp(5+2*x)*exp(exp(5)-1)+1)/exp(exp(5)-1))-25*exp(5+2*x)*exp(exp (5)-1)+25),x, algorithm=\
1/5*x + 1/10*log(e^(2*x + e^5 + 4) - 1) - 1/10*log(-e^(2*x + e^5 + 4) + 1) - log(log((e^5 - e^(2*x + e^5 + 9))*e^(-e^5 - 4)) - 5)
Time = 1.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {5-15 e^{4+e^5+2 x}+\left (-1+e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )}{25-25 e^{4+e^5+2 x}+\left (-5+5 e^{4+e^5+2 x}\right ) \log \left (e^{1-e^5} \left (1-e^{4+e^5+2 x}\right )\right )} \, dx=\frac {x}{5}-\ln \left (\ln \left ({\mathrm {e}}^{-{\mathrm {e}}^5}\,\mathrm {e}-{\mathrm {e}}^{2\,x+5}\right )-5\right ) \]