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

Optimal. Leaf size=26 \[ e^{e^x} \log \left (5 \left (x+\frac {1}{45} x^2 (x+\log (3)) \log (2 x)\right )\right ) \]

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Rubi [A]  time = 9.42, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 25, number of rules used = 5, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6742, 6688, 2282, 2194, 2555} \begin {gather*} e^{e^x} \log \left (\frac {1}{9} x (x (x+\log (3)) \log (2 x)+45)\right ) \end {gather*}

Antiderivative was successfully verified.

[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 2194

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 2282

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 2555

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [F]  time = 0.37, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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]

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fricas [A]  time = 1.10, size = 25, normalized size = 0.96 \begin {gather*} e^{\left (e^{x}\right )} \log \left (\frac {1}{9} \, {\left (x^{3} + x^{2} \log \relax (3)\right )} \log \left (2 \, x\right ) + 5 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.34, size = 51, normalized size = 1.96 \begin {gather*} -{\left (2 \, e^{\left (x + e^{x}\right )} \log \relax (3) - e^{\left (x + e^{x}\right )} \log \left (x^{2} \log \left (2 \, x\right ) + x \log \relax (3) \log \left (2 \, x\right ) + 45\right ) - e^{\left (x + e^{x}\right )} \log \relax (x)\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (x^{2} \ln \relax (3)+x^{3}\right ) {\mathrm e}^{x} \ln \left (2 x \right )+45 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{{\mathrm e}^{x}} \ln \left (\frac {\left (x^{2} \ln \relax (3)+x^{3}\right ) \ln \left (2 x \right )}{9}+5 x \right )+\left (\left (2 x \ln \relax (3)+3 x^{2}\right ) \ln \left (2 x \right )+x \ln \relax (3)+x^{2}+45\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\left (x^{2} \ln \relax (3)+x^{3}\right ) \ln \left (2 x \right )+45 x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 0.49, size = 45, normalized size = 1.73 \begin {gather*} -{\left (2 \, \log \relax (3) - \log \relax (x)\right )} e^{\left (e^{x}\right )} + e^{\left (e^{x}\right )} \log \left (x^{2} \log \relax (2) + x \log \relax (3) \log \relax (2) + {\left (x^{2} + x \log \relax (3)\right )} \log \relax (x) + 45\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 6.17, size = 25, normalized size = 0.96 \begin {gather*} \ln \left (5\,x+\frac {\ln \left (2\,x\right )\,\left (x^3+\ln \relax (3)\,x^2\right )}{9}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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