\(\int (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} (e^x+2 e^{\frac {1}{4} (1+8 x)})+\log (x)) \, dx\) [192]

Optimal result

Integrand size = 39, antiderivative size = 20 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=e^{e^x+e^{\frac {1}{4}+2 x}}+x \log (x) \]

[Out]

x*ln(x)+exp(exp(2*x+1/4)+exp(x))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2320, 2268, 2332} \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=e^{e^x+e^{2 x+\frac {1}{4}}}+x \log (x) \]

[In]

Int[1 + E^(E^x + E^((1 + 8*x)/4))*(E^x + 2*E^((1 + 8*x)/4)) + Log[x],x]

[Out]

E^(E^x + E^(1/4 + 2*x)) + x*Log[x]

Rule 2268

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = x+\int e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right ) \, dx+\int \log (x) \, dx \\ & = x \log (x)+\text {Subst}\left (\int e^{x+\sqrt [4]{e} x^2} \left (1+2 \sqrt [4]{e} x\right ) \, dx,x,e^x\right ) \\ & = e^{e^x+e^{\frac {1}{4}+2 x}}+x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=e^{e^x+e^{\frac {1}{4}+2 x}}+x \log (x) \]

[In]

Integrate[1 + E^(E^x + E^((1 + 8*x)/4))*(E^x + 2*E^((1 + 8*x)/4)) + Log[x],x]

[Out]

E^(E^x + E^(1/4 + 2*x)) + x*Log[x]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
default	\(x \ln \left (x \right )+{\mathrm e}^{{\mathrm e}^{2 x +\frac {1}{4}}+{\mathrm e}^{x}}\)	\(16\)
risch	\(x \ln \left (x \right )+{\mathrm e}^{{\mathrm e}^{2 x +\frac {1}{4}}+{\mathrm e}^{x}}\)	\(16\)
parallelrisch	\(x \ln \left (x \right )+{\mathrm e}^{{\mathrm e}^{2 x +\frac {1}{4}}+{\mathrm e}^{x}}\)	\(16\)
norman	\(x \ln \left (x \right )+{\mathrm e}^{{\mathrm e}^{\frac {1}{4}} {\mathrm e}^{2 x}+{\mathrm e}^{x}}\)	\(17\)

[In]

int((2*exp(2*x+1/4)+exp(x))*exp(exp(2*x+1/4)+exp(x))+ln(x)+1,x,method=_RETURNVERBOSE)

[Out]

x*ln(x)+exp(exp(2*x+1/4)+exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=x \log \left (x\right ) + e^{\left (e^{\left (2 \, x + \frac {1}{4}\right )} + e^{x}\right )} \]

[In]

integrate((2*exp(2*x+1/4)+exp(x))*exp(exp(2*x+1/4)+exp(x))+log(x)+1,x, algorithm="fricas")

[Out]

x*log(x) + e^(e^(2*x + 1/4) + e^x)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=x \log {\left (x \right )} + e^{e^{\frac {1}{4}} e^{2 x} + e^{x}} \]

[In]

integrate((2*exp(2*x+1/4)+exp(x))*exp(exp(2*x+1/4)+exp(x))+ln(x)+1,x)

[Out]

x*log(x) + exp(exp(1/4)*exp(2*x) + exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=x \log \left (x\right ) + e^{\left (e^{\left (2 \, x + \frac {1}{4}\right )} + e^{x}\right )} \]

[In]

integrate((2*exp(2*x+1/4)+exp(x))*exp(exp(2*x+1/4)+exp(x))+log(x)+1,x, algorithm="maxima")

[Out]

x*log(x) + e^(e^(2*x + 1/4) + e^x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx=x \log \left (x\right ) + e^{\left (e^{\left (2 \, x + \frac {1}{4}\right )} + e^{x}\right )} \]

[In]

integrate((2*exp(2*x+1/4)+exp(x))*exp(exp(2*x+1/4)+exp(x))+log(x)+1,x, algorithm="giac")

[Out]

x*log(x) + e^(e^(2*x + 1/4) + e^x)

Mupad [B] (verification not implemented)

Time = 8.85 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (1+e^{e^x+e^{\frac {1}{4} (1+8 x)}} \left (e^x+2 e^{\frac {1}{4} (1+8 x)}\right )+\log (x)\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{1/4}}+x\,\ln \left (x\right ) \]

[In]

int(log(x) + exp(exp(2*x + 1/4) + exp(x))*(2*exp(2*x + 1/4) + exp(x)) + 1,x)

[Out]

exp(exp(x))*exp(exp(2*x)*exp(1/4)) + x*log(x)