3.68.54 \(\int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{e \left (x+\frac {225 x}{\log (x)}\right )} \]

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Rubi [A]  time = 0.27, antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 9, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6688, 12, 6742, 2306, 2309, 2178} \begin {gather*} \frac {1}{e x}-\frac {225}{e x (\log (x)+225)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(225 - 225*Log[x] - Log[x]^2)/(50625*E*x^2 + 450*E*x^2*Log[x] + E*x^2*Log[x]^2),x]

[Out]

1/(E*x) - 225/(E*x*(225 + Log[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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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} &=\int \frac {225-225 \log (x)-\log ^2(x)}{e x^2 (225+\log (x))^2} \, dx\\ &=\frac {\int \frac {225-225 \log (x)-\log ^2(x)}{x^2 (225+\log (x))^2} \, dx}{e}\\ &=\frac {\int \left (-\frac {1}{x^2}+\frac {225}{x^2 (225+\log (x))^2}+\frac {225}{x^2 (225+\log (x))}\right ) \, dx}{e}\\ &=\frac {1}{e x}+\frac {225 \int \frac {1}{x^2 (225+\log (x))^2} \, dx}{e}+\frac {225 \int \frac {1}{x^2 (225+\log (x))} \, dx}{e}\\ &=\frac {1}{e x}-\frac {225}{e x (225+\log (x))}-\frac {225 \int \frac {1}{x^2 (225+\log (x))} \, dx}{e}+\frac {225 \operatorname {Subst}\left (\int \frac {e^{-x}}{225+x} \, dx,x,\log (x)\right )}{e}\\ &=\frac {1}{e x}+225 e^{224} \text {Ei}(-225-\log (x))-\frac {225}{e x (225+\log (x))}-\frac {225 \operatorname {Subst}\left (\int \frac {e^{-x}}{225+x} \, dx,x,\log (x)\right )}{e}\\ &=\frac {1}{e x}-\frac {225}{e x (225+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 16, normalized size = 1.07 \begin {gather*} \frac {\log (x)}{e (225 x+x \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(225 - 225*Log[x] - Log[x]^2)/(50625*E*x^2 + 450*E*x^2*Log[x] + E*x^2*Log[x]^2),x]

[Out]

Log[x]/(E*(225*x + x*Log[x]))

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fricas [A]  time = 0.63, size = 17, normalized size = 1.13 \begin {gather*} \frac {\log \relax (x)}{x e \log \relax (x) + 225 \, x e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2-225*log(x)+225)/(x^2*exp(1)*log(x)^2+450*x^2*exp(1)*log(x)+50625*x^2*exp(1)),x, algorithm
="fricas")

[Out]

log(x)/(x*e*log(x) + 225*x*e)

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giac [A]  time = 0.16, size = 17, normalized size = 1.13 \begin {gather*} \frac {\log \relax (x)}{x e \log \relax (x) + 225 \, x e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2-225*log(x)+225)/(x^2*exp(1)*log(x)^2+450*x^2*exp(1)*log(x)+50625*x^2*exp(1)),x, algorithm
="giac")

[Out]

log(x)/(x*e*log(x) + 225*x*e)

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maple [A]  time = 0.05, size = 17, normalized size = 1.13

method	result	size
norman	\(\frac {\ln \relax (x ) {\mathrm e}^{-1}}{x \left (\ln \relax (x )+225\right )}\)	\(17\)
risch	\(\frac {{\mathrm e}^{-1}}{x}-\frac {225 \,{\mathrm e}^{-1}}{x \left (\ln \relax (x )+225\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)^2-225*ln(x)+225)/(x^2*exp(1)*ln(x)^2+450*x^2*exp(1)*ln(x)+50625*x^2*exp(1)),x,method=_RETURNVERBOS
E)

[Out]

ln(x)/exp(1)/x/(ln(x)+225)

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maxima [A]  time = 0.41, size = 17, normalized size = 1.13 \begin {gather*} \frac {\log \relax (x)}{x e \log \relax (x) + 225 \, x e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2-225*log(x)+225)/(x^2*exp(1)*log(x)^2+450*x^2*exp(1)*log(x)+50625*x^2*exp(1)),x, algorithm
="maxima")

[Out]

log(x)/(x*e*log(x) + 225*x*e)

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mupad [B]  time = 4.15, size = 14, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^{-1}\,\ln \relax (x)}{x\,\left (\ln \relax (x)+225\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(225*log(x) + log(x)^2 - 225)/(50625*x^2*exp(1) + 450*x^2*exp(1)*log(x) + x^2*exp(1)*log(x)^2),x)

[Out]

(exp(-1)*log(x))/(x*(log(x) + 225))

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sympy [A]  time = 0.10, size = 22, normalized size = 1.47 \begin {gather*} - \frac {225}{e x \log {\relax (x )} + 225 e x} + \frac {1}{e x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)**2-225*ln(x)+225)/(x**2*exp(1)*ln(x)**2+450*x**2*exp(1)*ln(x)+50625*x**2*exp(1)),x)

[Out]

-225/(E*x*log(x) + 225*E*x) + exp(-1)/x

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