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

Optimal. Leaf size=15 \[ -3+x^2+\left (2+e^{e^x}\right ) \log (x) \]

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Rubi [A]  time = 0.10, antiderivative size = 16, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {14, 2282, 2194, 2554} \begin {gather*} x^2+e^{e^x} \log (x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2 + 2*Log[x] + E^E^x*Log[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 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 2554

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2+e^{e^x}+2 x^2}{x}+e^{e^x+x} \log (x)\right ) \, dx\\ &=\int \frac {2+e^{e^x}+2 x^2}{x} \, dx+\int e^{e^x+x} \log (x) \, dx\\ &=e^{e^x} \log (x)-\int \frac {e^{e^x}}{x} \, dx+\int \left (\frac {e^{e^x}}{x}+\frac {2 \left (1+x^2\right )}{x}\right ) \, dx\\ &=e^{e^x} \log (x)+2 \int \frac {1+x^2}{x} \, dx\\ &=e^{e^x} \log (x)+2 \int \left (\frac {1}{x}+x\right ) \, dx\\ &=x^2+2 \log (x)+e^{e^x} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 16, normalized size = 1.07 \begin {gather*} x^2+2 \log (x)+e^{e^x} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2 + 2*Log[x] + E^E^x*Log[x]

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fricas [A]  time = 0.70, size = 14, normalized size = 0.93 \begin {gather*} x^{2} + e^{\left (e^{x}\right )} \log \relax (x) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + e^(e^x)*log(x) + 2*log(x)

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giac [A]  time = 0.17, size = 26, normalized size = 1.73 \begin {gather*} {\left (x^{2} e^{x} + e^{\left (x + e^{x}\right )} \log \relax (x) + 2 \, e^{x} \log \relax (x)\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 15, normalized size = 1.00

method	result	size
risch	\(x^{2}+2 \ln \relax (x )+\ln \relax (x ) {\mathrm e}^{{\mathrm e}^{x}}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2+2*ln(x)+ln(x)*exp(exp(x))

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maxima [A]  time = 0.40, size = 14, normalized size = 0.93 \begin {gather*} x^{2} + e^{\left (e^{x}\right )} \log \relax (x) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + e^(e^x)*log(x) + 2*log(x)

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mupad [B]  time = 8.00, size = 14, normalized size = 0.93 \begin {gather*} 2\,\ln \relax (x)+{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \relax (x)+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x) + exp(exp(x))*log(x) + x^2

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sympy [A]  time = 4.16, size = 15, normalized size = 1.00 \begin {gather*} x^{2} + e^{e^{x}} \log {\relax (x )} + 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2 + exp(exp(x))*log(x) + 2*log(x)

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