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

Optimal. Leaf size=20 \[ e^{e^x+e^{\frac {1}{4}+2 x}}+x \log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2282, 2236, 2295} \begin {gather*} e^{e^x+e^{2 x+\frac {1}{4}}}+x \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 2236

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 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 2295

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 {gather*} \begin {aligned} \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)+\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 20, normalized size = 1.00 \begin {gather*} e^{e^x+e^{\frac {1}{4}+2 x}}+x \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.98, size = 15, normalized size = 0.75 \begin {gather*} x \log \relax (x) + e^{\left (e^{\left (2 \, x + \frac {1}{4}\right )} + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.30, size = 15, normalized size = 0.75 \begin {gather*} x \log \relax (x) + e^{\left (e^{\left (2 \, x + \frac {1}{4}\right )} + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 16, normalized size = 0.80

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.70, size = 15, normalized size = 0.75 \begin {gather*} x \log \relax (x) + e^{\left (e^{\left (2 \, x + \frac {1}{4}\right )} + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.40, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{1/4}}+x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.36, size = 19, normalized size = 0.95 \begin {gather*} x \log {\relax (x )} + e^{e^{\frac {1}{4}} e^{2 x} + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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