3.87.14 \(\int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ 1+\frac {2}{x \left (2+\frac {1}{-3+\log (x)}\right )} \]

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Rubi [A]  time = 0.28, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {6688, 12, 6742, 2306, 2309, 2178} \begin {gather*} \frac {1}{x}+\frac {1}{x (5-2 \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-28 + 22*Log[x] - 4*Log[x]^2)/(25*x^2 - 20*x^2*Log[x] + 4*x^2*Log[x]^2),x]

[Out]

x^(-1) + 1/(x*(5 - 2*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 {2 \left (-14+11 \log (x)-2 \log ^2(x)\right )}{x^2 (5-2 \log (x))^2} \, dx\\ &=2 \int \frac {-14+11 \log (x)-2 \log ^2(x)}{x^2 (5-2 \log (x))^2} \, dx\\ &=2 \int \left (-\frac {1}{2 x^2}+\frac {1}{x^2 (-5+2 \log (x))^2}+\frac {1}{2 x^2 (-5+2 \log (x))}\right ) \, dx\\ &=\frac {1}{x}+2 \int \frac {1}{x^2 (-5+2 \log (x))^2} \, dx+\int \frac {1}{x^2 (-5+2 \log (x))} \, dx\\ &=\frac {1}{x}+\frac {1}{x (5-2 \log (x))}-\int \frac {1}{x^2 (-5+2 \log (x))} \, dx+\operatorname {Subst}\left (\int \frac {e^{-x}}{-5+2 x} \, dx,x,\log (x)\right )\\ &=\frac {1}{x}+\frac {\text {Ei}\left (\frac {1}{2} (5-2 \log (x))\right )}{2 e^{5/2}}+\frac {1}{x (5-2 \log (x))}-\operatorname {Subst}\left (\int \frac {e^{-x}}{-5+2 x} \, dx,x,\log (x)\right )\\ &=\frac {1}{x}+\frac {1}{x (5-2 \log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 17, normalized size = 1.00 \begin {gather*} \frac {1}{x}-\frac {1}{x (-5+2 \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-28 + 22*Log[x] - 4*Log[x]^2)/(25*x^2 - 20*x^2*Log[x] + 4*x^2*Log[x]^2),x]

[Out]

x^(-1) - 1/(x*(-5 + 2*Log[x]))

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fricas [A]  time = 0.80, size = 17, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (\log \relax (x) - 3\right )}}{2 \, x \log \relax (x) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)^2+22*log(x)-28)/(4*x^2*log(x)^2-20*x^2*log(x)+25*x^2),x, algorithm="fricas")

[Out]

2*(log(x) - 3)/(2*x*log(x) - 5*x)

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giac [A]  time = 0.19, size = 17, normalized size = 1.00 \begin {gather*} -\frac {1}{2 \, x \log \relax (x) - 5 \, x} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)^2+22*log(x)-28)/(4*x^2*log(x)^2-20*x^2*log(x)+25*x^2),x, algorithm="giac")

[Out]

-1/(2*x*log(x) - 5*x) + 1/x

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maple [A]  time = 0.03, size = 18, normalized size = 1.06

method	result	size
risch	\(\frac {1}{x}-\frac {1}{x \left (2 \ln \relax (x )-5\right )}\)	\(18\)
norman	\(\frac {2 \ln \relax (x )-6}{x \left (2 \ln \relax (x )-5\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(x)^2+22*ln(x)-28)/(4*x^2*ln(x)^2-20*x^2*ln(x)+25*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.38, size = 17, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (\log \relax (x) - 3\right )}}{2 \, x \log \relax (x) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)^2+22*log(x)-28)/(4*x^2*log(x)^2-20*x^2*log(x)+25*x^2),x, algorithm="maxima")

[Out]

2*(log(x) - 3)/(2*x*log(x) - 5*x)

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mupad [B]  time = 5.61, size = 17, normalized size = 1.00 \begin {gather*} \frac {2\,\left (\ln \relax (x)-3\right )}{x\,\left (2\,\ln \relax (x)-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(x)^2 - 22*log(x) + 28)/(4*x^2*log(x)^2 - 20*x^2*log(x) + 25*x^2),x)

[Out]

(2*(log(x) - 3))/(x*(2*log(x) - 5))

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sympy [A]  time = 0.10, size = 14, normalized size = 0.82 \begin {gather*} - \frac {1}{2 x \log {\relax (x )} - 5 x} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(x)**2+22*ln(x)-28)/(4*x**2*ln(x)**2-20*x**2*ln(x)+25*x**2),x)

[Out]

-1/(2*x*log(x) - 5*x) + 1/x

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