3.1.33 \(\int \frac {e^{2 e^{2 x}-4 e^x \log (-e^{e^x}+\log (4))+2 \log ^2(-e^{e^x}+\log (4))} (-100 e^{2 x} \log (4)+100 e^x \log (4) \log (-e^{e^x}+\log (4)))}{e^{e^x}-\log (4)} \, dx\) [33]

3.1.33.1 Optimal result

Integrand size = 82, antiderivative size = 25 \begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 e^{2 \left (-e^x+\log \left (-e^{e^x}+\log (4)\right )\right )^2} \end {dmath*}

output

25*exp((ln(-exp(exp(x))+2*ln(2))-exp(x))^2)^2
 

3.1.33.2 Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 e^{2 e^{2 x}+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-e^{e^x}+\log (4)\right )^{-4 e^x} \end {dmath*}

input

Integrate[(E^(2*E^(2*x) - 4*E^x*Log[-E^E^x + Log[4]] + 2*Log[-E^E^x + Log[ 
4]]^2)*(-100*E^(2*x)*Log[4] + 100*E^x*Log[4]*Log[-E^E^x + Log[4]]))/(E^E^x 
 - Log[4]),x]
 

output

(25*E^(2*E^(2*x) + 2*Log[-E^E^x + Log[4]]^2))/(-E^E^x + Log[4])^(4*E^x)
 

3.1.33.3 Rubi [F]

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.

\(\displaystyle \int \frac {\left (100 e^x \log (4) \log \left (\log (4)-e^{e^x}\right )-100 e^{2 x} \log (4)\right ) \exp \left (2 e^{2 x}+2 \log ^2\left (\log (4)-e^{e^x}\right )-4 e^x \log \left (\log (4)-e^{e^x}\right )\right )}{e^{e^x}-\log (4)} \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \int 100 \log (4) e^{2 \left (e^{2 x}+\log ^2\left (\log (4)-e^{e^x}\right )\right )} \left (\log (4)-e^{e^x}\right )^{-4 e^x-1} \left (e^x-\log \left (\log (4)-e^{e^x}\right )\right )de^x\)

\(\Big \downarrow \) 27

\(\displaystyle 100 \log (4) \int e^{2 \left (\log ^2\left (-e^{e^x}+\log (4)\right )+e^{2 x}\right )} \left (-e^{e^x}+\log (4)\right )^{-1-4 e^x} \left (e^x-\log \left (-e^{e^x}+\log (4)\right )\right )de^x\)

\(\Big \downarrow \) 7293

\(\displaystyle 100 \log (4) \int \left (e^{x+2 \left (\log ^2\left (-e^{e^x}+\log (4)\right )+e^{2 x}\right )} \left (-e^{e^x}+\log (4)\right )^{-1-4 e^x}-e^{2 \left (\log ^2\left (-e^{e^x}+\log (4)\right )+e^{2 x}\right )} \left (-e^{e^x}+\log (4)\right )^{-1-4 e^x} \log \left (-e^{e^x}+\log (4)\right )\right )de^x\)

\(\Big \downarrow \) 2009

\(\displaystyle 100 \log (4) \left (\int e^{x+2 \left (\log ^2\left (-e^{e^x}+\log (4)\right )+e^{2 x}\right )} \left (-e^{e^x}+\log (4)\right )^{-1-4 e^x}de^x-\int e^{2 \left (\log ^2\left (-e^{e^x}+\log (4)\right )+e^{2 x}\right )} \left (-e^{e^x}+\log (4)\right )^{-1-4 e^x} \log \left (-e^{e^x}+\log (4)\right )de^x\right )\)

input

Int[(E^(2*E^(2*x) - 4*E^x*Log[-E^E^x + Log[4]] + 2*Log[-E^E^x + Log[4]]^2) 
*(-100*E^(2*x)*Log[4] + 100*E^x*Log[4]*Log[-E^E^x + Log[4]]))/(E^E^x - Log 
[4]),x]
 

output

$Aborted
 

3.1.33.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

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]]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.1.33.4 Maple [A] (verified)

Time = 4.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56

method	result	size
parallelrisch	\(25 \,{\mathrm e}^{2 \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )^{2}-4 \,{\mathrm e}^{x} \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )+2 \,{\mathrm e}^{2 x}}\)	\(39\)
risch	\(25 \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )^{-4 \,{\mathrm e}^{x}} {\mathrm e}^{2 \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )^{2}+2 \,{\mathrm e}^{2 x}}\)	\(43\)

input

int((200*ln(2)*exp(x)*ln(-exp(exp(x))+2*ln(2))-200*ln(2)*exp(x)^2)*exp(ln( 
-exp(exp(x))+2*ln(2))^2-2*exp(x)*ln(-exp(exp(x))+2*ln(2))+exp(x)^2)^2/(exp 
(exp(x))-2*ln(2)),x,method=_RETURNVERBOSE)
 

output

25*exp(ln(-exp(exp(x))+2*ln(2))^2-2*exp(x)*ln(-exp(exp(x))+2*ln(2))+exp(x) 
^2)^2
 

3.1.33.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 \, e^{\left (-4 \, e^{x} \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) + 2 \, \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right )^{2} + 2 \, e^{\left (2 \, x\right )}\right )} \end {dmath*}

input

integrate((200*log(2)*exp(x)*log(-exp(exp(x))+2*log(2))-200*log(2)*exp(x)^ 
2)*exp(log(-exp(exp(x))+2*log(2))^2-2*exp(x)*log(-exp(exp(x))+2*log(2))+ex 
p(x)^2)^2/(exp(exp(x))-2*log(2)),x, algorithm=\
 

output

25*e^(-4*e^x*log(-e^(e^x) + 2*log(2)) + 2*log(-e^(e^x) + 2*log(2))^2 + 2*e 
^(2*x))
 

3.1.33.6 Sympy [F(-1)]

Timed out. \begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\text {Timed out} \end {dmath*}

input

integrate((200*ln(2)*exp(x)*ln(-exp(exp(x))+2*ln(2))-200*ln(2)*exp(x)**2)* 
exp(ln(-exp(exp(x))+2*ln(2))**2-2*exp(x)*ln(-exp(exp(x))+2*ln(2))+exp(x)** 
2)**2/(exp(exp(x))-2*ln(2)),x)
 

output

Timed out
 

3.1.33.7 Maxima [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 \, e^{\left (-4 \, e^{x} \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) + 2 \, \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right )^{2} + 2 \, e^{\left (2 \, x\right )}\right )} \end {dmath*}

input

integrate((200*log(2)*exp(x)*log(-exp(exp(x))+2*log(2))-200*log(2)*exp(x)^ 
2)*exp(log(-exp(exp(x))+2*log(2))^2-2*exp(x)*log(-exp(exp(x))+2*log(2))+ex 
p(x)^2)^2/(exp(exp(x))-2*log(2)),x, algorithm=\
 

output

25*e^(-4*e^x*log(-e^(e^x) + 2*log(2)) + 2*log(-e^(e^x) + 2*log(2))^2 + 2*e 
^(2*x))
 

3.1.33.8 Giac [F]

\begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\int { \frac {200 \, {\left (e^{x} \log \left (2\right ) \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) - e^{\left (2 \, x\right )} \log \left (2\right )\right )} e^{\left (-4 \, e^{x} \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) + 2 \, \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right )^{2} + 2 \, e^{\left (2 \, x\right )}\right )}}{e^{\left (e^{x}\right )} - 2 \, \log \left (2\right )} \,d x } \end {dmath*}

input

integrate((200*log(2)*exp(x)*log(-exp(exp(x))+2*log(2))-200*log(2)*exp(x)^ 
2)*exp(log(-exp(exp(x))+2*log(2))^2-2*exp(x)*log(-exp(exp(x))+2*log(2))+ex 
p(x)^2)^2/(exp(exp(x))-2*log(2)),x, algorithm=\
 

output

integrate(200*(e^x*log(2)*log(-e^(e^x) + 2*log(2)) - e^(2*x)*log(2))*e^(-4 
*e^x*log(-e^(e^x) + 2*log(2)) + 2*log(-e^(e^x) + 2*log(2))^2 + 2*e^(2*x))/ 
(e^(e^x) - 2*log(2)), x)
 

3.1.33.9 Mupad [B] (verification not implemented)

Time = 13.45 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \begin {dmath*} \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\frac {25\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,{\ln \left (\ln \left (4\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\right )}^2}}{{\left (2\,\ln \left (2\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\right )}^{4\,{\mathrm {e}}^x}} \end {dmath*}

input

int(-(exp(2*exp(2*x) - 4*log(2*log(2) - exp(exp(x)))*exp(x) + 2*log(2*log( 
2) - exp(exp(x)))^2)*(200*exp(2*x)*log(2) - 200*log(2*log(2) - exp(exp(x)) 
)*exp(x)*log(2)))/(exp(exp(x)) - 2*log(2)),x)
 

output

(25*exp(2*exp(2*x))*exp(2*log(log(4) - exp(exp(x)))^2))/(2*log(2) - exp(ex 
p(x)))^(4*exp(x))