\(\int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+(9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+(-9 x-x^2) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))) \log (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x})}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} (9 x^2+2 x^3)} \, dx\) [8309]

Optimal result

Integrand size = 156, antiderivative size = 24 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log (x) \log \left (1+\frac {9}{e^{5 e^{2 x}}+x}\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 7.72 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 39, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6874, 2634, 12, 14, 6820, 2340, 2631, 2341, 30, 2637} \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )}{x} \]

[In]

Int[(-9*x*Log[x] - 90*E^(5*E^(2*x) + 2*x)*x*Log[x] + (9*x + x^2 + E^(10*E^(2*x))*(1 - Log[x]) + (-9*x - x^2)*L
og[x] + E^(5*E^(2*x))*(9 + 2*x + (-9 - 2*x)*Log[x]))*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/(E^(10*
E^(2*x))*x^2 + 9*x^3 + x^4 + E^(5*E^(2*x))*(9*x^2 + 2*x^3)),x]

[Out]

(Log[x]*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2340

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[b*(d*x)^(m + 1)*(Log[c*x^n]/(d
*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 2634

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

Rule 2637

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[z*Log[w]*(D[v, x]/v), x], x] - Int[SimplifyIntegrand[z*Log[v]*(D[w, x]/w), x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

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 {90 e^{5 e^{2 x}+2 x} \log (x)}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )}-\frac {9 x \log (x)-9 e^{5 e^{2 x}} \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-e^{10 e^{2 x}} \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-9 x \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-2 e^{5 e^{2 x}} x \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-x^2 \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+9 e^{5 e^{2 x}} \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+e^{10 e^{2 x}} \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+9 x \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+2 e^{5 e^{2 x}} x \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+x^2 \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2 \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx \\ & = -\left (90 \int \frac {e^{5 e^{2 x}+2 x} \log (x)}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx\right )-\int \frac {9 x \log (x)-9 e^{5 e^{2 x}} \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-e^{10 e^{2 x}} \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-9 x \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-2 e^{5 e^{2 x}} x \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )-x^2 \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+9 e^{5 e^{2 x}} \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+e^{10 e^{2 x}} \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+9 x \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+2 e^{5 e^{2 x}} x \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )+x^2 \log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2 \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx \\ & = 90 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{9 x} \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {-\left (\left (e^{10 e^{2 x}}+x (9+x)+e^{5 e^{2 x}} (9+2 x)\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )\right )+\log (x) \left (9 x+\left (e^{10 e^{2 x}}+x (9+x)+e^{5 e^{2 x}} (9+2 x)\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )\right )}{x^2 \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx \\ & = 10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \left (\frac {\log (x)}{x \left (e^{5 e^{2 x}}+x\right )}-\frac {\log (x)}{x \left (9+e^{5 e^{2 x}}+x\right )}+\frac {(-1+\log (x)) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2}\right ) \, dx \\ & = 10 \int \left (\frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x}-\frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x}\right ) \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {\log (x)}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {\log (x)}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {(-1+\log (x)) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2} \, dx \\ & = 10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\log (x) \int \frac {1}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+\log (x) \int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \left (-\frac {\log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2}+\frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2}\right ) \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx \\ & = 10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\log (x) \int \frac {1}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+\log (x) \int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {\log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2} \, dx-\int \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx \\ & = -\frac {\log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+\frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\log (x) \int \frac {1}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+\log (x) \int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {9 \left (-1-10 e^{5 e^{2 x}+2 x}\right )}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {9 \left (1+10 e^{5 e^{2 x}+2 x}\right ) \log (x)}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {\log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x^2} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+9 \int \frac {-1-10 e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx+9 \int \frac {\left (1+10 e^{5 e^{2 x}+2 x}\right ) \log (x)}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\log (x) \int \frac {1}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+\log (x) \int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+(10 \log (x)) \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {9 \left (-1-10 e^{5 e^{2 x}+2 x}\right )}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}-9 \int \frac {-1-10 e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx+9 \int \left (-\frac {1}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )}-\frac {10 e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx-9 \int \frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{9 x} \, dx+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}-9 \int \frac {1}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx-9 \int \left (-\frac {1}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )}-\frac {10 e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-90 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx-\int \frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+9 \int \frac {1}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx-9 \int \left (\frac {1}{9 x \left (e^{5 e^{2 x}}+x\right )}-\frac {1}{9 x \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+90 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right ) \left (9+e^{5 e^{2 x}}+x\right )} \, dx-90 \int \left (\frac {e^{5 e^{2 x}+2 x}}{9 x \left (e^{5 e^{2 x}}+x\right )}-\frac {e^{5 e^{2 x}+2 x}}{9 x \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+\int \frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x} \, dx-\int \left (\frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x}+\frac {\int \frac {1}{e^{5 e^{2 x}} x+x^2} \, dx}{x}\right ) \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+9 \int \left (\frac {1}{9 x \left (e^{5 e^{2 x}}+x\right )}-\frac {1}{9 x \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx-10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx+90 \int \left (\frac {e^{5 e^{2 x}+2 x}}{9 x \left (e^{5 e^{2 x}}+x\right )}-\frac {e^{5 e^{2 x}+2 x}}{9 x \left (9+e^{5 e^{2 x}}+x\right )}\right ) \, dx-\int \frac {1}{x \left (e^{5 e^{2 x}}+x\right )} \, dx+\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx-10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \left (\frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x}-\frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x}\right ) \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx-\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x}+10 \int \frac {\int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x} \, dx-\int \left (\frac {10 \int \frac {e^{5 e^{2 x}+2 x}}{x \left (e^{5 e^{2 x}}+x\right )} \, dx}{x}-\frac {\int \frac {1}{x \left (9+e^{5 e^{2 x}}+x\right )} \, dx}{x}\right ) \, dx \\ & = \frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x} \]

[In]

Integrate[(-9*x*Log[x] - 90*E^(5*E^(2*x) + 2*x)*x*Log[x] + (9*x + x^2 + E^(10*E^(2*x))*(1 - Log[x]) + (-9*x -
x^2)*Log[x] + E^(5*E^(2*x))*(9 + 2*x + (-9 - 2*x)*Log[x]))*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/(
E^(10*E^(2*x))*x^2 + 9*x^3 + x^4 + E^(5*E^(2*x))*(9*x^2 + 2*x^3)),x]

[Out]

(Log[x]*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/x

Maple [A] (verified)

Time = 49.65 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
parallelrisch	\(\frac {\ln \left (\frac {{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right ) \ln \left (x \right )}{x}\)	\(30\)
risch	\(\frac {\ln \left (x \right ) \ln \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{x}-\frac {\ln \left (x \right ) \left (i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )}^{3}+2 \ln \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x \right )\right )}{2 x}\)	\(219\)

[In]

int((((1-ln(x))*exp(5*exp(x)^2)^2+((-2*x-9)*ln(x)+2*x+9)*exp(5*exp(x)^2)+(-x^2-9*x)*ln(x)+x^2+9*x)*ln((exp(5*e
xp(x)^2)+x+9)/(exp(5*exp(x)^2)+x))-90*x*exp(x)^2*ln(x)*exp(5*exp(x)^2)-9*x*ln(x))/(x^2*exp(5*exp(x)^2)^2+(2*x^
3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (22) = 44\).

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log \left (x\right ) \log \left (\frac {{\left (x + 9\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x + 5 \, e^{\left (2 \, x\right )}\right )}}{x e^{\left (2 \, x\right )} + e^{\left (2 \, x + 5 \, e^{\left (2 \, x\right )}\right )}}\right )}{x} \]

[In]

integrate((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp(x)^2)+(-x^2-9*x)*log(x)+x^2+9*x)*lo
g((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^2)+x))-90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x
)^2)^2+(2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log {\left (x \right )} \log {\left (\frac {x + e^{5 e^{2 x}} + 9}{x + e^{5 e^{2 x}}} \right )}}{x} \]

[In]

integrate((((1-ln(x))*exp(5*exp(x)**2)**2+((-2*x-9)*ln(x)+2*x+9)*exp(5*exp(x)**2)+(-x**2-9*x)*ln(x)+x**2+9*x)*
ln((exp(5*exp(x)**2)+x+9)/(exp(5*exp(x)**2)+x))-90*x*exp(x)**2*ln(x)*exp(5*exp(x)**2)-9*x*ln(x))/(x**2*exp(5*e
xp(x)**2)**2+(2*x**3+9*x**2)*exp(5*exp(x)**2)+x**4+9*x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )} + 9\right ) \log \left (x\right ) - \log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )}\right ) \log \left (x\right )}{x} \]

[In]

integrate((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp(x)^2)+(-x^2-9*x)*log(x)+x^2+9*x)*lo
g((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^2)+x))-90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x
)^2)^2+(2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )} + 9\right ) \log \left (x\right ) - \log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )}\right ) \log \left (x\right )}{x} \]

[In]

integrate((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp(x)^2)+(-x^2-9*x)*log(x)+x^2+9*x)*lo
g((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^2)+x))-90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x
)^2)^2+(2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.75 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\ln \left (\frac {x+{\mathrm {e}}^{5\,{\mathrm {e}}^{2\,x}}+9}{x+{\mathrm {e}}^{5\,{\mathrm {e}}^{2\,x}}}\right )\,\ln \left (x\right )}{x} \]

[In]

int(-(9*x*log(x) - log((x + exp(5*exp(2*x)) + 9)/(x + exp(5*exp(2*x))))*(9*x + exp(5*exp(2*x))*(2*x - log(x)*(
2*x + 9) + 9) - log(x)*(9*x + x^2) - exp(10*exp(2*x))*(log(x) - 1) + x^2) + 90*x*exp(5*exp(2*x))*exp(2*x)*log(
x))/(exp(5*exp(2*x))*(9*x^2 + 2*x^3) + x^2*exp(10*exp(2*x)) + 9*x^3 + x^4),x)

[Out]

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