3.1.70 \(\int \frac {e^{-125/x} (2 x-e^{125/x} x+(250+2 x-e^{125/x} x) \log (x))}{x} \, dx\)

Optimal. Leaf size=22 \[ -3+5 \log (3)-\left (x-2 e^{-125/x} x\right ) \log (x) \]

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Rubi [A]  time = 0.35, antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 8, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {6688, 2206, 2210, 2288, 2554} \begin {gather*} 2 e^{-125/x} x \log (x)-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - E^(125/x)*x + (250 + 2*x - E^(125/x)*x)*Log[x])/(E^(125/x)*x),x]

[Out]

-(x*Log[x]) + (2*x*Log[x])/E^(125/x)

Rule 2206

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

Rule 2210

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[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]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+2 e^{-125/x}+\left (-1+\frac {2 e^{-125/x} (125+x)}{x}\right ) \log (x)\right ) \, dx\\ &=-x+2 \int e^{-125/x} \, dx+\int \left (-1+\frac {2 e^{-125/x} (125+x)}{x}\right ) \log (x) \, dx\\ &=-x+2 e^{-125/x} x-x \log (x)+2 e^{-125/x} x \log (x)-250 \int \frac {e^{-125/x}}{x} \, dx-\int \left (-1+2 e^{-125/x}\right ) \, dx\\ &=2 e^{-125/x} x+250 \text {Ei}\left (-\frac {125}{x}\right )-x \log (x)+2 e^{-125/x} x \log (x)-2 \int e^{-125/x} \, dx\\ &=250 \text {Ei}\left (-\frac {125}{x}\right )-x \log (x)+2 e^{-125/x} x \log (x)+250 \int \frac {e^{-125/x}}{x} \, dx\\ &=-x \log (x)+2 e^{-125/x} x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 15, normalized size = 0.68 \begin {gather*} \left (-1+2 e^{-125/x}\right ) x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - E^(125/x)*x + (250 + 2*x - E^(125/x)*x)*Log[x])/(E^(125/x)*x),x]

[Out]

(-1 + 2/E^(125/x))*x*Log[x]

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fricas [A]  time = 0.67, size = 22, normalized size = 1.00 \begin {gather*} -{\left (x e^{\frac {125}{x}} - 2 \, x\right )} e^{\left (-\frac {125}{x}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(125/x)+2*x+250)*log(x)-x*exp(125/x)+2*x)/x/exp(125/x),x, algorithm="fricas")

[Out]

-(x*e^(125/x) - 2*x)*e^(-125/x)*log(x)

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giac [A]  time = 0.30, size = 17, normalized size = 0.77 \begin {gather*} 2 \, x e^{\left (-\frac {125}{x}\right )} \log \relax (x) - x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(125/x)+2*x+250)*log(x)-x*exp(125/x)+2*x)/x/exp(125/x),x, algorithm="giac")

[Out]

2*x*e^(-125/x)*log(x) - x*log(x)

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maple [A]  time = 0.04, size = 20, normalized size = 0.91

method	result	size
default	\(2 \ln \relax (x ) x \,{\mathrm e}^{-\frac {125}{x}}-x \ln \relax (x )\)	\(20\)
risch	\(-x \left ({\mathrm e}^{\frac {125}{x}}-2\right ) {\mathrm e}^{-\frac {125}{x}} \ln \relax (x )\)	\(20\)
norman	\(\left (2 x \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{\frac {125}{x}} x \right ) {\mathrm e}^{-\frac {125}{x}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(125/x)+2*x+250)*ln(x)-x*exp(125/x)+2*x)/x/exp(125/x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)*x/exp(125/x)-x*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, x e^{\left (-\frac {125}{x}\right )} \log \relax (x) - x \log \relax (x) + 250 \, \Gamma \left (-1, \frac {125}{x}\right ) - 2 \, \int e^{\left (-\frac {125}{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(125/x)+2*x+250)*log(x)-x*exp(125/x)+2*x)/x/exp(125/x),x, algorithm="maxima")

[Out]

2*x*e^(-125/x)*log(x) - x*log(x) + 250*gamma(-1, 125/x) - 2*integrate(e^(-125/x), x)

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mupad [B]  time = 0.33, size = 14, normalized size = 0.64 \begin {gather*} x\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^{-\frac {125}{x}}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-125/x)*(2*x - x*exp(125/x) + log(x)*(2*x - x*exp(125/x) + 250)))/x,x)

[Out]

x*log(x)*(2*exp(-125/x) - 1)

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sympy [A]  time = 0.30, size = 15, normalized size = 0.68 \begin {gather*} - x \log {\relax (x )} + 2 x e^{- \frac {125}{x}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(125/x)+2*x+250)*ln(x)-x*exp(125/x)+2*x)/x/exp(125/x),x)

[Out]

-x*log(x) + 2*x*exp(-125/x)*log(x)

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