3.63.1 \(\int \frac {e^{\frac {1}{12} e^{-5+2 x}} (-2-\frac {1}{3} e^{-5+2 x} x \log (x))}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {2}{-5+e^{\frac {1}{12} e^{-5+2 x}} \log (x)} \]

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Rubi [A]  time = 1.56, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6741, 6708, 32} \begin {gather*} -\frac {2}{5-e^{\frac {1}{12} e^{2 x-5}} \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^(-5 + 2*x)/12)*(-2 - (E^(-5 + 2*x)*x*Log[x])/3))/(25*x - 10*E^(E^(-5 + 2*x)/12)*x*Log[x] + E^(E^(-5
+ 2*x)/6)*x*Log[x]^2),x]

[Out]

-2/(5 - E^(E^(-5 + 2*x)/12)*Log[x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6708

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{x \left (5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)\right )^2} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{(5-x)^2} \, dx,x,e^{\frac {1}{12} e^{-5+2 x}} \log (x)\right )\right )\\ &=-\frac {2}{5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 23, normalized size = 1.05 \begin {gather*} -\frac {2}{5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(-5 + 2*x)/12)*(-2 - (E^(-5 + 2*x)*x*Log[x])/3))/(25*x - 10*E^(E^(-5 + 2*x)/12)*x*Log[x] + E^(
E^(-5 + 2*x)/6)*x*Log[x]^2),x]

[Out]

-2/(5 - E^(E^(-5 + 2*x)/12)*Log[x])

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fricas [A]  time = 0.65, size = 20, normalized size = 0.91 \begin {gather*} \frac {2}{e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \relax (x) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x*log(x)^2*exp(exp(-log(12)+2*x-5))^2-
10*x*log(x)*exp(exp(-log(12)+2*x-5))+25*x),x, algorithm="fricas")

[Out]

2/(e^(e^(2*x - log(12) - 5))*log(x) - 5)

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giac [B]  time = 0.28, size = 65, normalized size = 2.95 \begin {gather*} \frac {2 \, {\left (x e^{\left (2 \, x\right )} \log \relax (x) + 6 \, e^{5}\right )}}{x e^{\left (2 \, x + \frac {1}{12} \, e^{\left (2 \, x - 5\right )}\right )} \log \relax (x)^{2} - 5 \, x e^{\left (2 \, x\right )} \log \relax (x) + 6 \, e^{\left (\frac {1}{12} \, e^{\left (2 \, x - 5\right )} + 5\right )} \log \relax (x) - 30 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x*log(x)^2*exp(exp(-log(12)+2*x-5))^2-
10*x*log(x)*exp(exp(-log(12)+2*x-5))+25*x),x, algorithm="giac")

[Out]

2*(x*e^(2*x)*log(x) + 6*e^5)/(x*e^(2*x + 1/12*e^(2*x - 5))*log(x)^2 - 5*x*e^(2*x)*log(x) + 6*e^(1/12*e^(2*x -
5) + 5)*log(x) - 30*e^5)

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maple [A]  time = 0.06, size = 19, normalized size = 0.86

method	result	size
risch	\(\frac {2}{\ln \relax (x ) {\mathrm e}^{\frac {{\mathrm e}^{2 x -5}}{12}}-5}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x*exp(-ln(12)+2*x-5)*ln(x)-2)*exp(exp(-ln(12)+2*x-5))/(x*ln(x)^2*exp(exp(-ln(12)+2*x-5))^2-10*x*ln(x)*
exp(exp(-ln(12)+2*x-5))+25*x),x,method=_RETURNVERBOSE)

[Out]

2/(ln(x)*exp(1/12*exp(2*x-5))-5)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left (x e^{\left (2 \, x - 5\right )} \log \relax (x) + 6\right )} e^{\left (\frac {1}{12} \, e^{\left (2 \, x - 5\right )}\right )}}{6 \, {\left (x e^{\left (\frac {1}{6} \, e^{\left (2 \, x - 5\right )}\right )} \log \relax (x)^{2} - 10 \, x e^{\left (\frac {1}{12} \, e^{\left (2 \, x - 5\right )}\right )} \log \relax (x) + 25 \, x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x*log(x)^2*exp(exp(-log(12)+2*x-5))^2-
10*x*log(x)*exp(exp(-log(12)+2*x-5))+25*x),x, algorithm="maxima")

[Out]

-2*integrate(1/6*(x*e^(2*x - 5)*log(x) + 6)*e^(1/12*e^(2*x - 5))/(x*e^(1/6*e^(2*x - 5))*log(x)^2 - 10*x*e^(1/1
2*e^(2*x - 5))*log(x) + 25*x), x)

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mupad [B]  time = 4.16, size = 18, normalized size = 0.82 \begin {gather*} \frac {2}{{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-5}}{12}}\,\ln \relax (x)-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(2*x - log(12) - 5))*(4*x*exp(2*x - log(12) - 5)*log(x) + 2))/(25*x + x*exp(2*exp(2*x - log(12) -
5))*log(x)^2 - 10*x*exp(exp(2*x - log(12) - 5))*log(x)),x)

[Out]

2/(exp((exp(2*x)*exp(-5))/12)*log(x) - 5)

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sympy [A]  time = 0.30, size = 15, normalized size = 0.68 \begin {gather*} \frac {2}{e^{\frac {e^{2 x - 5}}{12}} \log {\relax (x )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*exp(-ln(12)+2*x-5)*ln(x)-2)*exp(exp(-ln(12)+2*x-5))/(x*ln(x)**2*exp(exp(-ln(12)+2*x-5))**2-10*
x*ln(x)*exp(exp(-ln(12)+2*x-5))+25*x),x)

[Out]

2/(exp(exp(2*x - 5)/12)*log(x) - 5)

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