\(\int \frac {e^{e^x} (45+x^2+x \log (3)+(3 x^2+2 x \log (3)) \log (2 x))+e^{e^x} (45 e^x x+e^x (x^3+x^2 \log (3)) \log (2 x)) \log (\frac {1}{9} (45 x+(x^3+x^2 \log (3)) \log (2 x)))}{45 x+(x^3+x^2 \log (3)) \log (2 x)} \, dx\) [7864]

Optimal result

Integrand size = 109, antiderivative size = 26 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=e^{e^x} \log \left (5 \left (x+\frac {1}{45} x^2 (x+\log (3)) \log (2 x)\right )\right ) \]

[Out]

exp(exp(x))*ln(5*x+1/9*ln(2*x)*x^2*(ln(3)+x))

Rubi [A] (verified)

Time = 5.66 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 25, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6874, 6820, 2320, 2225, 2635} \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=e^{e^x} \log \left (\frac {1}{9} x (x (x+\log (3)) \log (2 x)+45)\right ) \]

[In]

Int[(E^E^x*(45 + x^2 + x*Log[3] + (3*x^2 + 2*x*Log[3])*Log[2*x]) + E^E^x*(45*E^x*x + E^x*(x^3 + x^2*Log[3])*Lo
g[2*x])*Log[(45*x + (x^3 + x^2*Log[3])*Log[2*x])/9])/(45*x + (x^3 + x^2*Log[3])*Log[2*x]),x]

[Out]

E^E^x*Log[(x*(45 + x*(x + Log[3])*Log[2*x]))/9]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2635

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{e^x} \left (45+x^2+x \log (3)+3 x^2 \log (2 x)+x \log (9) \log (2 x)\right )}{x \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}+e^{e^x+x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )\right ) \, dx \\ & = \int \frac {e^{e^x} \left (45+x^2+x \log (3)+3 x^2 \log (2 x)+x \log (9) \log (2 x)\right )}{x \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx+\int e^{e^x+x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right ) \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )-\int \frac {e^{e^x} \left (45+x^2+x \log (3)+x (3 x+\log (9)) \log (2 x)\right )}{x (45+x (x+\log (3)) \log (2 x))} \, dx+\int \left (\frac {e^{e^x} (3 x+\log (9))}{x (x+\log (3))}+\frac {e^{e^x} \left (x^3-45 \log (3)-x \left (90-\log ^2(3)\right )+x^2 \log (9)\right )}{x (x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}\right ) \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )+\int \frac {e^{e^x} (3 x+\log (9))}{x (x+\log (3))} \, dx+\int \frac {e^{e^x} \left (x^3-45 \log (3)-x \left (90-\log ^2(3)\right )+x^2 \log (9)\right )}{x (x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx-\int \left (\frac {e^{e^x} (3 x+\log (9))}{x (x+\log (3))}+\frac {e^{e^x} \left (x^3-45 \log (3)-x \left (90-\log ^2(3)\right )+x^2 \log (9)\right )}{x (x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}\right ) \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )-\int \frac {e^{e^x} (3 x+\log (9))}{x (x+\log (3))} \, dx+\int \left (\frac {e^{e^x}}{x+\log (3)}+\frac {e^{e^x} \log (9)}{x \log (3)}\right ) \, dx-\int \frac {e^{e^x} \left (x^3-45 \log (3)-x \left (90-\log ^2(3)\right )+x^2 \log (9)\right )}{x (x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx+\int \frac {e^{e^x} \left (x^3-45 \log (3)-x \left (90-\log ^2(3)\right )+x^2 \log (9)\right )}{x (x+\log (3)) (45+x (x+\log (3)) \log (2 x))} \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )+\frac {\log (9) \int \frac {e^{e^x}}{x} \, dx}{\log (3)}+\int \frac {e^{e^x}}{x+\log (3)} \, dx-\int \left (\frac {e^{e^x}}{x+\log (3)}+\frac {e^{e^x} \log (9)}{x \log (3)}\right ) \, dx-\int \frac {e^{e^x} \left (x^3-45 \log (3)-x \left (90-\log ^2(3)\right )+x^2 \log (9)\right )}{x (x+\log (3)) (45+x (x+\log (3)) \log (2 x))} \, dx+\int \left (-\frac {45 e^{e^x}}{x \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}+\frac {e^{e^x} x}{45+x^2 \log (2 x)+x \log (3) \log (2 x)}-\frac {e^{e^x} \log (3) \left (1-\frac {\log (9)}{\log (3)}\right )}{45+x^2 \log (2 x)+x \log (3) \log (2 x)}+\frac {e^{e^x} \left (-45+2 \log ^2(3)-\log (3) \log (9)\right )}{(x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}\right ) \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )-45 \int \frac {e^{e^x}}{x \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx+\log (3) \int \frac {e^{e^x}}{45+x^2 \log (2 x)+x \log (3) \log (2 x)} \, dx+\left (-45+2 \log ^2(3)-\log (3) \log (9)\right ) \int \frac {e^{e^x}}{(x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx+\int \frac {e^{e^x} x}{45+x^2 \log (2 x)+x \log (3) \log (2 x)} \, dx-\int \left (-\frac {45 e^{e^x}}{x \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}+\frac {e^{e^x} x}{45+x^2 \log (2 x)+x \log (3) \log (2 x)}-\frac {e^{e^x} \log (3) \left (1-\frac {\log (9)}{\log (3)}\right )}{45+x^2 \log (2 x)+x \log (3) \log (2 x)}+\frac {e^{e^x} \left (-45+2 \log ^2(3)-\log (3) \log (9)\right )}{(x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )}\right ) \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right )+45 \int \frac {e^{e^x}}{x \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx-45 \int \frac {e^{e^x}}{x (45+x (x+\log (3)) \log (2 x))} \, dx-\log (3) \int \frac {e^{e^x}}{45+x^2 \log (2 x)+x \log (3) \log (2 x)} \, dx+\log (3) \int \frac {e^{e^x}}{45+x (x+\log (3)) \log (2 x)} \, dx-\left (-45+2 \log ^2(3)-\log (3) \log (9)\right ) \int \frac {e^{e^x}}{(x+\log (3)) \left (45+x^2 \log (2 x)+x \log (3) \log (2 x)\right )} \, dx+\left (-45+2 \log ^2(3)-\log (3) \log (9)\right ) \int \frac {e^{e^x}}{(x+\log (3)) (45+x (x+\log (3)) \log (2 x))} \, dx-\int \frac {e^{e^x} x}{45+x^2 \log (2 x)+x \log (3) \log (2 x)} \, dx+\int \frac {e^{e^x} x}{45+x (x+\log (3)) \log (2 x)} \, dx \\ & = e^{e^x} \log \left (\frac {1}{9} x (45+x (x+\log (3)) \log (2 x))\right ) \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=\int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx \]

[In]

Integrate[(E^E^x*(45 + x^2 + x*Log[3] + (3*x^2 + 2*x*Log[3])*Log[2*x]) + E^E^x*(45*E^x*x + E^x*(x^3 + x^2*Log[
3])*Log[2*x])*Log[(45*x + (x^3 + x^2*Log[3])*Log[2*x])/9])/(45*x + (x^3 + x^2*Log[3])*Log[2*x]),x]

[Out]

Integrate[(E^E^x*(45 + x^2 + x*Log[3] + (3*x^2 + 2*x*Log[3])*Log[2*x]) + E^E^x*(45*E^x*x + E^x*(x^3 + x^2*Log[
3])*Log[2*x])*Log[(45*x + (x^3 + x^2*Log[3])*Log[2*x])/9])/(45*x + (x^3 + x^2*Log[3])*Log[2*x]), x]

Maple [F]

\[\int \frac {\left (\left (x^{2} \ln \left (3\right )+x^{3}\right ) {\mathrm e}^{x} \ln \left (2 x \right )+45 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{{\mathrm e}^{x}} \ln \left (\frac {\left (x^{2} \ln \left (3\right )+x^{3}\right ) \ln \left (2 x \right )}{9}+5 x \right )+\left (\left (2 x \ln \left (3\right )+3 x^{2}\right ) \ln \left (2 x \right )+x \ln \left (3\right )+x^{2}+45\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\left (x^{2} \ln \left (3\right )+x^{3}\right ) \ln \left (2 x \right )+45 x}d x\]

[In]

int((((x^2*ln(3)+x^3)*exp(x)*ln(2*x)+45*exp(x)*x)*exp(exp(x))*ln(1/9*(x^2*ln(3)+x^3)*ln(2*x)+5*x)+((2*x*ln(3)+
3*x^2)*ln(2*x)+x*ln(3)+x^2+45)*exp(exp(x)))/((x^2*ln(3)+x^3)*ln(2*x)+45*x),x)

[Out]

int((((x^2*ln(3)+x^3)*exp(x)*ln(2*x)+45*exp(x)*x)*exp(exp(x))*ln(1/9*(x^2*ln(3)+x^3)*ln(2*x)+5*x)+((2*x*ln(3)+
3*x^2)*ln(2*x)+x*ln(3)+x^2+45)*exp(exp(x)))/((x^2*ln(3)+x^3)*ln(2*x)+45*x),x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=e^{\left (e^{x}\right )} \log \left (\frac {1}{9} \, {\left (x^{3} + x^{2} \log \left (3\right )\right )} \log \left (2 \, x\right ) + 5 \, x\right ) \]

[In]

integrate((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log(1/9*(x^2*log(3)+x^3)*log(2*x)+5*x)+(
(2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x^2+45)*exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x, algorithm="frica
s")

[Out]

e^(e^x)*log(1/9*(x^3 + x^2*log(3))*log(2*x) + 5*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=\text {Timed out} \]

[In]

integrate((((x**2*ln(3)+x**3)*exp(x)*ln(2*x)+45*exp(x)*x)*exp(exp(x))*ln(1/9*(x**2*ln(3)+x**3)*ln(2*x)+5*x)+((
2*x*ln(3)+3*x**2)*ln(2*x)+x*ln(3)+x**2+45)*exp(exp(x)))/((x**2*ln(3)+x**3)*ln(2*x)+45*x),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).

Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=-{\left (2 \, \log \left (3\right ) - \log \left (x\right )\right )} e^{\left (e^{x}\right )} + e^{\left (e^{x}\right )} \log \left (x^{2} \log \left (2\right ) + x \log \left (3\right ) \log \left (2\right ) + {\left (x^{2} + x \log \left (3\right )\right )} \log \left (x\right ) + 45\right ) \]

[In]

integrate((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log(1/9*(x^2*log(3)+x^3)*log(2*x)+5*x)+(
(2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x^2+45)*exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x, algorithm="maxim
a")

[Out]

-(2*log(3) - log(x))*e^(e^x) + e^(e^x)*log(x^2*log(2) + x*log(3)*log(2) + (x^2 + x*log(3))*log(x) + 45)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).

Time = 0.38 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=-{\left (2 \, e^{\left (x + e^{x}\right )} \log \left (3\right ) - e^{\left (x + e^{x}\right )} \log \left (x^{2} \log \left (2 \, x\right ) + x \log \left (3\right ) \log \left (2 \, x\right ) + 45\right ) - e^{\left (x + e^{x}\right )} \log \left (x\right )\right )} e^{\left (-x\right )} \]

[In]

integrate((((x^2*log(3)+x^3)*exp(x)*log(2*x)+45*exp(x)*x)*exp(exp(x))*log(1/9*(x^2*log(3)+x^3)*log(2*x)+5*x)+(
(2*x*log(3)+3*x^2)*log(2*x)+x*log(3)+x^2+45)*exp(exp(x)))/((x^2*log(3)+x^3)*log(2*x)+45*x),x, algorithm="giac"
)

[Out]

-(2*e^(x + e^x)*log(3) - e^(x + e^x)*log(x^2*log(2*x) + x*log(3)*log(2*x) + 45) - e^(x + e^x)*log(x))*e^(-x)

Mupad [B] (verification not implemented)

Time = 14.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x} \left (45+x^2+x \log (3)+\left (3 x^2+2 x \log (3)\right ) \log (2 x)\right )+e^{e^x} \left (45 e^x x+e^x \left (x^3+x^2 \log (3)\right ) \log (2 x)\right ) \log \left (\frac {1}{9} \left (45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)\right )\right )}{45 x+\left (x^3+x^2 \log (3)\right ) \log (2 x)} \, dx=\ln \left (5\,x+\frac {\ln \left (2\,x\right )\,\left (x^3+\ln \left (3\right )\,x^2\right )}{9}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x} \]

[In]

int((exp(exp(x))*(x*log(3) + log(2*x)*(2*x*log(3) + 3*x^2) + x^2 + 45) + log(5*x + (log(2*x)*(x^2*log(3) + x^3
))/9)*exp(exp(x))*(45*x*exp(x) + log(2*x)*exp(x)*(x^2*log(3) + x^3)))/(45*x + log(2*x)*(x^2*log(3) + x^3)),x)

[Out]

log(5*x + (log(2*x)*(x^2*log(3) + x^3))/9)*exp(exp(x))