Integrand size = 87, antiderivative size = 26 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=3+\frac {\log \left (2+e^x\right )}{2 e^2}+\log \left (\frac {1}{\log \left (-25+e^{2+x}\right )}\right ) \] Output:
3+ln(1/ln(exp(2+x)-25))+1/2*ln(2+exp(x))/exp(2)
Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=\frac {1}{2} \left (\frac {\log \left (2+e^x\right )}{e^2}-2 \log \left (\log \left (-25+e^{2+x}\right )\right )\right ) \] Input:
Integrate[(E^(2 + x)*(-4*E^2 - 2*E^(2 + x)) + (-25*E^x + E^(2 + 2*x))*Log[ -25 + E^(2 + x)])/((-100*E^2 - 50*E^(2 + x) + E^(2 + x)*(4*E^2 + 2*E^(2 + x)))*Log[-25 + E^(2 + x)]),x]
Output:
(Log[2 + E^x]/E^2 - 2*Log[Log[-25 + E^(2 + x)]])/2
Time = 0.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2720, 27, 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.
\(\displaystyle \int \frac {e^{x+2} \left (-2 e^{x+2}-4 e^2\right )+\left (e^{2 x+2}-25 e^x\right ) \log \left (e^{x+2}-25\right )}{\left (e^{x+2} \left (2 e^{x+2}+4 e^2\right )-50 e^{x+2}-100 e^2\right ) \log \left (e^{x+2}-25\right )} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int \frac {\frac {1}{e^x+2}+\frac {2 e^4}{\left (25-e^{x+2}\right ) \log \left (e^{x+2}-25\right )}}{2 e^2}de^x\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \left (\frac {2 e^4}{\left (25-e^{x+2}\right ) \log \left (-25+e^{x+2}\right )}+\frac {1}{2+e^x}\right )de^x}{2 e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log \left (e^x+2\right )-2 e^2 \log \left (\log \left (e^{x+2}-25\right )\right )}{2 e^2}\) |
Input:
Int[(E^(2 + x)*(-4*E^2 - 2*E^(2 + x)) + (-25*E^x + E^(2 + 2*x))*Log[-25 + E^(2 + x)])/((-100*E^2 - 50*E^(2 + x) + E^(2 + x)*(4*E^2 + 2*E^(2 + x)))*L og[-25 + E^(2 + x)]),x]
Output:
(Log[2 + E^x] - 2*E^2*Log[Log[-25 + E^(2 + x)]])/(2*E^2)
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 = 0.66 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\ln \left ({\mathrm e}^{x}+2\right ) {\mathrm e}^{-2}}{2}-\ln \left (\ln \left ({\mathrm e}^{2+x}-25\right )\right )\) | \(21\) |
norman | \(\frac {\ln \left ({\mathrm e}^{x}+2\right ) {\mathrm e}^{-2}}{2}-\ln \left (\ln \left ({\mathrm e}^{2} {\mathrm e}^{x}-25\right )\right )\) | \(24\) |
parallelrisch | \(-\frac {\left (2 \ln \left (\ln \left ({\mathrm e}^{2+x}-25\right )\right ) {\mathrm e}^{2}-\ln \left ({\mathrm e}^{x}+2\right )\right ) {\mathrm e}^{-2}}{2}\) | \(27\) |
default | \(\frac {{\mathrm e}^{2} \ln \left ({\mathrm e}^{x}+2\right )}{25 \,{\mathrm e}^{2}+2 \,{\mathrm e}^{4}}+\frac {25 \,{\mathrm e}^{4} {\mathrm e}^{-4} \ln \left ({\mathrm e}^{4} {\mathrm e}^{x}-25 \,{\mathrm e}^{2}\right )}{2 \left (25 \,{\mathrm e}^{2}+2 \,{\mathrm e}^{4}\right )}-\ln \left (\ln \left ({\mathrm e}^{2+x}-25\right )\right )+\frac {25 \ln \left ({\mathrm e}^{x}+2\right )}{2 \left (25 \,{\mathrm e}^{2}+2 \,{\mathrm e}^{4}\right )}-\frac {25 \ln \left ({\mathrm e}^{4} {\mathrm e}^{x}-25 \,{\mathrm e}^{2}\right )}{2 \left (25 \,{\mathrm e}^{2}+2 \,{\mathrm e}^{4}\right )}\) | \(105\) |
Input:
int(((exp(x)*exp(2+x)-25*exp(x))*ln(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp(2 ))*exp(2+x))/((2*exp(2)*exp(x)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-100*exp (2))/ln(exp(2+x)-25),x,method=_RETURNVERBOSE)
Output:
1/2*ln(exp(x)+2)*exp(-2)-ln(ln(exp(2+x)-25))
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=-\frac {1}{2} \, {\left (2 \, e^{2} \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) - \log \left (2 \, e^{2} + e^{\left (x + 2\right )}\right )\right )} e^{\left (-2\right )} \] Input:
integrate(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)- 4*exp(2))*exp(2+x))/((2*exp(2)*exp(x)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)- 100*exp(2))/log(exp(2+x)-25),x, algorithm="fricas")
Output:
-1/2*(2*e^2*log(log(e^(x + 2) - 25)) - log(2*e^2 + e^(x + 2)))*e^(-2)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=\frac {\log {\left (e^{x} + 2 \right )}}{2 e^{2}} - \log {\left (\log {\left (e^{2} e^{x} - 25 \right )} \right )} \] Input:
integrate(((exp(x)*exp(2+x)-25*exp(x))*ln(exp(2+x)-25)+(-2*exp(2)*exp(x)-4 *exp(2))*exp(2+x))/((2*exp(2)*exp(x)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-1 00*exp(2))/ln(exp(2+x)-25),x)
Output:
exp(-2)*log(exp(x) + 2)/2 - log(log(exp(2)*exp(x) - 25))
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=\frac {1}{2} \, e^{\left (-2\right )} \log \left (2 \, e^{2} + e^{\left (x + 2\right )}\right ) - \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) \] Input:
integrate(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)- 4*exp(2))*exp(2+x))/((2*exp(2)*exp(x)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)- 100*exp(2))/log(exp(2+x)-25),x, algorithm="maxima")
Output:
1/2*e^(-2)*log(2*e^2 + e^(x + 2)) - log(log(e^(x + 2) - 25))
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=-\frac {1}{2} \, {\left (2 \, e^{2} \log \left (\log \left (e^{\left (x + 2\right )} - 25\right )\right ) - \log \left (2 \, e^{2} + e^{\left (x + 2\right )}\right )\right )} e^{\left (-2\right )} \] Input:
integrate(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)- 4*exp(2))*exp(2+x))/((2*exp(2)*exp(x)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)- 100*exp(2))/log(exp(2+x)-25),x, algorithm="giac")
Output:
-1/2*(2*e^2*log(log(e^(x + 2) - 25)) - log(2*e^2 + e^(x + 2)))*e^(-2)
Time = 2.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=\frac {{\mathrm {e}}^{-2}\,\ln \left ({\mathrm {e}}^x+2\right )}{2}-\ln \left (\ln \left ({\mathrm {e}}^2\,{\mathrm {e}}^x-25\right )\right ) \] Input:
int((exp(x + 2)*(4*exp(2) + 2*exp(2)*exp(x)) + log(exp(x + 2) - 25)*(25*ex p(x) - exp(x + 2)*exp(x)))/(log(exp(x + 2) - 25)*(100*exp(2) - exp(x + 2)* (4*exp(2) + 2*exp(2)*exp(x)) + 50*exp(2)*exp(x))),x)
Output:
(exp(-2)*log(exp(x) + 2))/2 - log(log(exp(2)*exp(x) - 25))
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{2+x} \left (-4 e^2-2 e^{2+x}\right )+\left (-25 e^x+e^{2+2 x}\right ) \log \left (-25+e^{2+x}\right )}{\left (-100 e^2-50 e^{2+x}+e^{2+x} \left (4 e^2+2 e^{2+x}\right )\right ) \log \left (-25+e^{2+x}\right )} \, dx=\frac {-2 \,\mathrm {log}\left (\mathrm {log}\left (e^{x} e^{2}-25\right )\right ) e^{2}+\mathrm {log}\left (e^{x}+2\right )}{2 e^{2}} \] Input:
int(((exp(x)*exp(2+x)-25*exp(x))*log(exp(2+x)-25)+(-2*exp(2)*exp(x)-4*exp( 2))*exp(2+x))/((2*exp(2)*exp(x)+4*exp(2))*exp(2+x)-50*exp(2)*exp(x)-100*ex p(2))/log(exp(2+x)-25),x)
Output:
( - 2*log(log(e**x*e**2 - 25))*e**2 + log(e**x + 2))/(2*e**2)