Integrand size = 85, antiderivative size = 26 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=5+\left (x+\left (3+\log ^2\left (\left (5-e^{e^x}\right )^2\right )\right )^2\right ) \log (\log (3)) \] Output:
(x+(ln((5-exp(exp(x)))^2)^2+3)^2)*ln(ln(3))+5
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=\left (x+6 \log ^2\left (\left (-5+e^{e^x}\right )^2\right )+\log ^4\left (\left (-5+e^{e^x}\right )^2\right )\right ) \log (\log (3)) \] Input:
Integrate[(-5*Log[Log[3]] + E^E^x*Log[Log[3]] + 24*E^(E^x + x)*Log[25 - 10 *E^E^x + E^(2*E^x)]*Log[Log[3]] + 8*E^(E^x + x)*Log[25 - 10*E^E^x + E^(2*E ^x)]^3*Log[Log[3]])/(-5 + E^E^x),x]
Output:
(x + 6*Log[(-5 + E^E^x)^2]^2 + Log[(-5 + E^E^x)^2]^4)*Log[Log[3]]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.20 (sec) , antiderivative size = 285, normalized size of antiderivative = 10.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2720, 27, 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.
| \(\displaystyle \int \frac {8 e^{x+e^x} \log (\log (3)) \log ^3\left (-10 e^{e^x}+e^{2 e^x}+25\right )+24 e^{x+e^x} \log (\log (3)) \log \left (-10 e^{e^x}+e^{2 e^x}+25\right )+e^{e^x} \log (\log (3))-5 \log (\log (3))}{e^{e^x}-5} \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \int \frac {e^{-x} \log (\log (3)) \left (-e^{e^x}-8 e^{x+e^x} \log ^3\left (\left (e^{e^x}-5\right )^2\right )-24 e^{x+e^x} \log \left (\left (e^{e^x}-5\right )^2\right )+5\right )}{5-e^{e^x}}de^x\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \log (\log (3)) \int \frac {e^{-x} \left (-8 e^{x+e^x} \log ^3\left (\left (-5+e^{e^x}\right )^2\right )-24 e^{x+e^x} \log \left (\left (-5+e^{e^x}\right )^2\right )-e^{e^x}+5\right )}{5-e^{e^x}}de^x\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \log (\log (3)) \int \left (\frac {40 \log \left (\left (-5+e^{e^x}\right )^2\right ) \left (\log ^2\left (\left (-5+e^{e^x}\right )^2\right )+3\right )}{-5+e^{e^x}}+e^{-x} \left (8 e^x \log ^3\left (\left (-5+e^{e^x}\right )^2\right )+24 e^x \log \left (\left (-5+e^{e^x}\right )^2\right )+1\right )\right )de^x\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \log (\log (3)) \left (48 \operatorname {PolyLog}\left (2,\frac {5}{5-e^{e^x}}\right )+48 \operatorname {PolyLog}\left (2,1-\frac {e^{e^x}}{5}\right )+384 \operatorname {PolyLog}\left (4,\frac {5}{5-e^{e^x}}\right )+384 \operatorname {PolyLog}\left (4,1-\frac {e^{e^x}}{5}\right )+48 \operatorname {PolyLog}\left (2,\frac {5}{5-e^{e^x}}\right ) \log ^2\left (\left (e^{e^x}-5\right )^2\right )+48 \operatorname {PolyLog}\left (2,1-\frac {e^{e^x}}{5}\right ) \log ^2\left (\left (e^{e^x}-5\right )^2\right )+192 \operatorname {PolyLog}\left (3,\frac {5}{5-e^{e^x}}\right ) \log \left (\left (e^{e^x}-5\right )^2\right )-192 \operatorname {PolyLog}\left (3,1-\frac {e^{e^x}}{5}\right ) \log \left (\left (e^{e^x}-5\right )^2\right )+8 \log \left (\frac {e^{e^x}}{5}\right ) \log ^3\left (\left (e^{e^x}-5\right )^2\right )-8 \log \left (1-\frac {5}{5-e^{e^x}}\right ) \log ^3\left (\left (e^{e^x}-5\right )^2\right )+24 \log \left (\frac {e^{e^x}}{5}\right ) \log \left (\left (e^{e^x}-5\right )^2\right )-24 \log \left (1-\frac {5}{5-e^{e^x}}\right ) \log \left (\left (e^{e^x}-5\right )^2\right )+\log \left (e^x\right )\right )\) |
Input:
Int[(-5*Log[Log[3]] + E^E^x*Log[Log[3]] + 24*E^(E^x + x)*Log[25 - 10*E^E^x + E^(2*E^x)]*Log[Log[3]] + 8*E^(E^x + x)*Log[25 - 10*E^E^x + E^(2*E^x)]^3 *Log[Log[3]])/(-5 + E^E^x),x]
Output:
Log[Log[3]]*(Log[E^x] + 24*Log[E^E^x/5]*Log[(-5 + E^E^x)^2] + 8*Log[E^E^x/ 5]*Log[(-5 + E^E^x)^2]^3 - 24*Log[(-5 + E^E^x)^2]*Log[1 - 5/(5 - E^E^x)] - 8*Log[(-5 + E^E^x)^2]^3*Log[1 - 5/(5 - E^E^x)] + 48*PolyLog[2, 5/(5 - E^E ^x)] + 48*Log[(-5 + E^E^x)^2]^2*PolyLog[2, 5/(5 - E^E^x)] + 48*PolyLog[2, 1 - E^E^x/5] + 48*Log[(-5 + E^E^x)^2]^2*PolyLog[2, 1 - E^E^x/5] + 192*Log[ (-5 + E^E^x)^2]*PolyLog[3, 5/(5 - E^E^x)] - 192*Log[(-5 + E^E^x)^2]*PolyLo g[3, 1 - E^E^x/5] + 384*PolyLog[4, 5/(5 - E^E^x)] + 384*PolyLog[4, 1 - E^E ^x/5])
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 = 1.49 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15
| method | result | size |
| parts | \(\ln \left (\ln \left (3\right )\right ) x +\ln \left (\ln \left (3\right )\right ) {\left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}-10 \,{\mathrm e}^{{\mathrm e}^{x}}+25\right )^{2}+3\right )}^{2}\) | \(30\) |
| parallelrisch | \(\ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}-10 \,{\mathrm e}^{{\mathrm e}^{x}}+25\right )^{4}+6 \ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}-10 \,{\mathrm e}^{{\mathrm e}^{x}}+25\right )^{2}+\ln \left (\ln \left (3\right )\right ) x\) | \(46\) |
| derivativedivides | \(\left (24 \ln \left (\ln \left (3\right )\right ) {\left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )}^{2}+24 \ln \left (\ln \left (3\right )\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}+\left (8 \ln \left (\ln \left (3\right )\right ) {\left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )}^{3}+24 \ln \left (\ln \left (3\right )\right ) \left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )+16 \ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{4}+32 \ln \left (\ln \left (3\right )\right ) \left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{3}+\ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{x}\right )\) | \(145\) |
| default | \(\left (24 \ln \left (\ln \left (3\right )\right ) {\left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )}^{2}+24 \ln \left (\ln \left (3\right )\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}+\left (8 \ln \left (\ln \left (3\right )\right ) {\left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )}^{3}+24 \ln \left (\ln \left (3\right )\right ) \left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )+16 \ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{4}+32 \ln \left (\ln \left (3\right )\right ) \left (\ln \left (\left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right ) \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{3}+\ln \left (\ln \left (3\right )\right ) \ln \left ({\mathrm e}^{x}\right )\) | \(145\) |
| risch | \(\text {Expression too large to display}\) | \(611\) |
Input:
int((8*exp(x)*ln(ln(3))*exp(exp(x))*ln(exp(exp(x))^2-10*exp(exp(x))+25)^3+ 24*exp(x)*ln(ln(3))*exp(exp(x))*ln(exp(exp(x))^2-10*exp(exp(x))+25)+ln(ln( 3))*exp(exp(x))-5*ln(ln(3)))/(exp(exp(x))-5),x,method=_RETURNVERBOSE)
Output:
ln(ln(3))*x+ln(ln(3))*(ln(exp(exp(x))^2-10*exp(exp(x))+25)^2+3)^2
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (22) = 44\).
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.12 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=\log \left ({\left (25 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + 2 \, e^{x}\right )} - 10 \, e^{\left (2 \, x + e^{x}\right )}\right )} e^{\left (-2 \, x\right )}\right )^{4} \log \left (\log \left (3\right )\right ) + 6 \, \log \left ({\left (25 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + 2 \, e^{x}\right )} - 10 \, e^{\left (2 \, x + e^{x}\right )}\right )} e^{\left (-2 \, x\right )}\right )^{2} \log \left (\log \left (3\right )\right ) + x \log \left (\log \left (3\right )\right ) \] Input:
integrate((8*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x ))+25)^3+24*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x) )+25)+log(log(3))*exp(exp(x))-5*log(log(3)))/(exp(exp(x))-5),x, algorithm= "fricas")
Output:
log((25*e^(2*x) + e^(2*x + 2*e^x) - 10*e^(2*x + e^x))*e^(-2*x))^4*log(log( 3)) + 6*log((25*e^(2*x) + e^(2*x + 2*e^x) - 10*e^(2*x + e^x))*e^(-2*x))^2* log(log(3)) + x*log(log(3))
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=x \log {\left (\log {\left (3 \right )} \right )} + \log {\left (e^{2 e^{x}} - 10 e^{e^{x}} + 25 \right )}^{4} \log {\left (\log {\left (3 \right )} \right )} + 6 \log {\left (e^{2 e^{x}} - 10 e^{e^{x}} + 25 \right )}^{2} \log {\left (\log {\left (3 \right )} \right )} \] Input:
integrate((8*exp(x)*ln(ln(3))*exp(exp(x))*ln(exp(exp(x))**2-10*exp(exp(x)) +25)**3+24*exp(x)*ln(ln(3))*exp(exp(x))*ln(exp(exp(x))**2-10*exp(exp(x))+2 5)+ln(ln(3))*exp(exp(x))-5*ln(ln(3)))/(exp(exp(x))-5),x)
Output:
x*log(log(3)) + log(exp(2*exp(x)) - 10*exp(exp(x)) + 25)**4*log(log(3)) + 6*log(exp(2*exp(x)) - 10*exp(exp(x)) + 25)**2*log(log(3))
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=16 \, \log \left (e^{\left (e^{x}\right )} - 5\right )^{4} \log \left (\log \left (3\right )\right ) + 24 \, \log \left (e^{\left (e^{x}\right )} - 5\right )^{2} \log \left (\log \left (3\right )\right ) + x \log \left (\log \left (3\right )\right ) \] Input:
integrate((8*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x ))+25)^3+24*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x) )+25)+log(log(3))*exp(exp(x))-5*log(log(3)))/(exp(exp(x))-5),x, algorithm= "maxima")
Output:
16*log(e^(e^x) - 5)^4*log(log(3)) + 24*log(e^(e^x) - 5)^2*log(log(3)) + x* log(log(3))
\[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=\int { \frac {8 \, e^{\left (x + e^{x}\right )} \log \left (e^{\left (2 \, e^{x}\right )} - 10 \, e^{\left (e^{x}\right )} + 25\right )^{3} \log \left (\log \left (3\right )\right ) + 24 \, e^{\left (x + e^{x}\right )} \log \left (e^{\left (2 \, e^{x}\right )} - 10 \, e^{\left (e^{x}\right )} + 25\right ) \log \left (\log \left (3\right )\right ) + e^{\left (e^{x}\right )} \log \left (\log \left (3\right )\right ) - 5 \, \log \left (\log \left (3\right )\right )}{e^{\left (e^{x}\right )} - 5} \,d x } \] Input:
integrate((8*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x ))+25)^3+24*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x) )+25)+log(log(3))*exp(exp(x))-5*log(log(3)))/(exp(exp(x))-5),x, algorithm= "giac")
Output:
integrate((8*e^(x + e^x)*log(e^(2*e^x) - 10*e^(e^x) + 25)^3*log(log(3)) + 24*e^(x + e^x)*log(e^(2*e^x) - 10*e^(e^x) + 25)*log(log(3)) + e^(e^x)*log( log(3)) - 5*log(log(3)))/(e^(e^x) - 5), x)
Time = 3.60 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=\ln \left (\ln \left (3\right )\right )\,\left ({\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^x}-10\,{\mathrm {e}}^{{\mathrm {e}}^x}+25\right )}^4+6\,{\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^x}-10\,{\mathrm {e}}^{{\mathrm {e}}^x}+25\right )}^2+x\right ) \] Input:
int((exp(exp(x))*log(log(3)) - 5*log(log(3)) + 24*log(exp(2*exp(x)) - 10*e xp(exp(x)) + 25)*exp(exp(x))*exp(x)*log(log(3)) + 8*log(exp(2*exp(x)) - 10 *exp(exp(x)) + 25)^3*exp(exp(x))*exp(x)*log(log(3)))/(exp(exp(x)) - 5),x)
Output:
log(log(3))*(x + 6*log(exp(2*exp(x)) - 10*exp(exp(x)) + 25)^2 + log(exp(2* exp(x)) - 10*exp(exp(x)) + 25)^4)
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {-5 \log (\log (3))+e^{e^x} \log (\log (3))+24 e^{e^x+x} \log \left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))+8 e^{e^x+x} \log ^3\left (25-10 e^{e^x}+e^{2 e^x}\right ) \log (\log (3))}{-5+e^{e^x}} \, dx=\mathrm {log}\left (\mathrm {log}\left (3\right )\right ) \left (\mathrm {log}\left (e^{2 e^{x}}-10 e^{e^{x}}+25\right )^{4}+6 \mathrm {log}\left (e^{2 e^{x}}-10 e^{e^{x}}+25\right )^{2}+x \right ) \] Input:
int((8*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x))+25) ^3+24*exp(x)*log(log(3))*exp(exp(x))*log(exp(exp(x))^2-10*exp(exp(x))+25)+ log(log(3))*exp(exp(x))-5*log(log(3)))/(exp(exp(x))-5),x)
Output:
log(log(3))*(log(e**(2*e**x) - 10*e**(e**x) + 25)**4 + 6*log(e**(2*e**x) - 10*e**(e**x) + 25)**2 + x)