3.79.44 \(\int \frac {e^{-6 e^{8+2 x}} (75 \log ^2(x)-900 e^{8+2 x} x \log ^3(x)+e^{6 e^{8+2 x}} (3 x-3 x \log (x)))}{3 x \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {-x+25 e^{-\frac {2}{3} e^{2 (4+x+\log (3))}} \log ^2(x)}{\log (x)} \]

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Rubi [A]  time = 1.45, antiderivative size = 23, normalized size of antiderivative = 0.77, number of steps used = 10, number of rules used = 8, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6742, 2282, 2194, 2554, 6688, 2297, 2298} \begin {gather*} 25 e^{-6 e^{2 x+8}} \log (x)-\frac {x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75*Log[x]^2 - 900*E^(8 + 2*x)*x*Log[x]^3 + E^(6*E^(8 + 2*x))*(3*x - 3*x*Log[x]))/(3*E^(6*E^(8 + 2*x))*x*L
og[x]^2),x]

[Out]

-(x/Log[x]) + (25*Log[x])/E^(6*E^(8 + 2*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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]]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{-6 e^{8+2 x}} \left (75 \log ^2(x)-900 e^{8+2 x} x \log ^3(x)+e^{6 e^{8+2 x}} (3 x-3 x \log (x))\right )}{x \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (-900 e^{8-6 e^{8+2 x}+2 x} \log (x)-\frac {3 e^{-6 e^{8+2 x}} \left (-e^{6 e^{8+2 x}} x+e^{6 e^{8+2 x}} x \log (x)-25 \log ^2(x)\right )}{x \log ^2(x)}\right ) \, dx\\ &=-\left (300 \int e^{8-6 e^{8+2 x}+2 x} \log (x) \, dx\right )-\int \frac {e^{-6 e^{8+2 x}} \left (-e^{6 e^{8+2 x}} x+e^{6 e^{8+2 x}} x \log (x)-25 \log ^2(x)\right )}{x \log ^2(x)} \, dx\\ &=25 e^{-6 e^{8+2 x}} \log (x)+300 \int -\frac {e^{-6 e^{8+2 x}}}{12 x} \, dx-\int \left (-\frac {25 e^{-6 e^{8+2 x}}}{x}-\frac {1}{\log ^2(x)}+\frac {1}{\log (x)}\right ) \, dx\\ &=25 e^{-6 e^{8+2 x}} \log (x)+\int \frac {1}{\log ^2(x)} \, dx-\int \frac {1}{\log (x)} \, dx\\ &=-\frac {x}{\log (x)}+25 e^{-6 e^{8+2 x}} \log (x)-\text {li}(x)+\int \frac {1}{\log (x)} \, dx\\ &=-\frac {x}{\log (x)}+25 e^{-6 e^{8+2 x}} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 23, normalized size = 0.77 \begin {gather*} -\frac {x}{\log (x)}+25 e^{-6 e^{8+2 x}} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*Log[x]^2 - 900*E^(8 + 2*x)*x*Log[x]^3 + E^(6*E^(8 + 2*x))*(3*x - 3*x*Log[x]))/(3*E^(6*E^(8 + 2*x
))*x*Log[x]^2),x]

[Out]

-(x/Log[x]) + (25*Log[x])/E^(6*E^(8 + 2*x))

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fricas [A]  time = 0.83, size = 41, normalized size = 1.37 \begin {gather*} -\frac {{\left (x e^{\left (\frac {2}{3} \, e^{\left (2 \, x + 2 \, \log \relax (3) + 8\right )}\right )} - 25 \, \log \relax (x)^{2}\right )} e^{\left (-\frac {2}{3} \, e^{\left (2 \, x + 2 \, \log \relax (3) + 8\right )}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-3*x*log(x)+3*x)*exp(1/3*exp(2*log(3)+2*x+8))^2-100*x*exp(2*log(3)+2*x+8)*log(x)^3+75*log(x)^2
)/x/log(x)^2/exp(1/3*exp(2*log(3)+2*x+8))^2,x, algorithm="fricas")

[Out]

-(x*e^(2/3*e^(2*x + 2*log(3) + 8)) - 25*log(x)^2)*e^(-2/3*e^(2*x + 2*log(3) + 8))/log(x)

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giac [A]  time = 0.17, size = 41, normalized size = 1.37 \begin {gather*} \frac {{\left (25 \, e^{\left (2 \, x - 6 \, e^{\left (2 \, x + 8\right )} + 8\right )} \log \relax (x)^{2} - x e^{\left (2 \, x + 8\right )}\right )} e^{\left (-2 \, x - 8\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-3*x*log(x)+3*x)*exp(1/3*exp(2*log(3)+2*x+8))^2-100*x*exp(2*log(3)+2*x+8)*log(x)^3+75*log(x)^2
)/x/log(x)^2/exp(1/3*exp(2*log(3)+2*x+8))^2,x, algorithm="giac")

[Out]

(25*e^(2*x - 6*e^(2*x + 8) + 8)*log(x)^2 - x*e^(2*x + 8))*e^(-2*x - 8)/log(x)

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maple [A]  time = 0.05, size = 22, normalized size = 0.73

method	result	size
risch	\(-\frac {x}{\ln \relax (x )}+25 \ln \relax (x ) {\mathrm e}^{-6 \,{\mathrm e}^{2 x +8}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-3*x*ln(x)+3*x)*exp(1/3*exp(2*ln(3)+2*x+8))^2-100*x*exp(2*ln(3)+2*x+8)*ln(x)^3+75*ln(x)^2)/x/ln(x)^2
/exp(1/3*exp(2*ln(3)+2*x+8))^2,x,method=_RETURNVERBOSE)

[Out]

-x/ln(x)+25*ln(x)*exp(-6*exp(2*x+8))

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maxima [A]  time = 0.40, size = 24, normalized size = 0.80 \begin {gather*} \frac {25 \, e^{\left (-6 \, e^{\left (2 \, x + 8\right )}\right )} \log \relax (x)^{2} - x}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-3*x*log(x)+3*x)*exp(1/3*exp(2*log(3)+2*x+8))^2-100*x*exp(2*log(3)+2*x+8)*log(x)^3+75*log(x)^2
)/x/log(x)^2/exp(1/3*exp(2*log(3)+2*x+8))^2,x, algorithm="maxima")

[Out]

(25*e^(-6*e^(2*x + 8))*log(x)^2 - x)/log(x)

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mupad [B]  time = 5.49, size = 21, normalized size = 0.70 \begin {gather*} 25\,{\mathrm {e}}^{-6\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^8}\,\ln \relax (x)-\frac {x}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*exp(2*x + 2*log(3) + 8))/3)*(25*log(x)^2 + (exp((2*exp(2*x + 2*log(3) + 8))/3)*(3*x - 3*x*log(x))
)/3 - (100*x*exp(2*x + 2*log(3) + 8)*log(x)^3)/3))/(x*log(x)^2),x)

[Out]

25*exp(-6*exp(2*x)*exp(8))*log(x) - x/log(x)

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sympy [A]  time = 29.52, size = 19, normalized size = 0.63 \begin {gather*} - \frac {x}{\log {\relax (x )}} + 25 e^{- 6 e^{2 x + 8}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-3*x*ln(x)+3*x)*exp(1/3*exp(2*ln(3)+2*x+8))**2-100*x*exp(2*ln(3)+2*x+8)*ln(x)**3+75*ln(x)**2)/
x/ln(x)**2/exp(1/3*exp(2*ln(3)+2*x+8))**2,x)

[Out]

-x/log(x) + 25*exp(-6*exp(2*x + 8))*log(x)

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