\(\int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx\) [8614]

Optimal result

Integrand size = 37, antiderivative size = 17 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=1+\frac {2}{x \left (2+\frac {1}{-3+\log (x)}\right )} \]

[Out]

2/(2+1/(ln(x)-3))/x+1

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {6820, 12, 6874, 2343, 2346, 2209} \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=\frac {1}{x}+\frac {1}{x (5-2 \log (x))} \]

[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 2209

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

Rule 2343

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 2346

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 6820

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

Rule 6874

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

Rubi steps \begin{align*} \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+\text {Subst}\left (\int \frac {e^{-x}}{-5+2 x} \, dx,x,\log (x)\right ) \\ & = \frac {1}{x}+\frac {\operatorname {ExpIntegralEi}\left (\frac {1}{2} (5-2 \log (x))\right )}{2 e^{5/2}}+\frac {1}{x (5-2 \log (x))}-\text {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{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=\frac {1}{x}-\frac {1}{x (-5+2 \log (x))} \]

[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]))

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {2 \ln \left (x \right )-6}{x \left (2 \ln \left (x \right )-5\right )}\)	\(18\)
risch	\(\frac {1}{x}-\frac {1}{x \left (2 \ln \left (x \right )-5\right )}\)	\(18\)
norman	\(\frac {2 \ln \left (x \right )-6}{x \left (2 \ln \left (x \right )-5\right )}\)	\(19\)
parallelrisch	\(\frac {-12+4 \ln \left (x \right )}{2 x \left (2 \ln \left (x \right )-5\right )}\)	\(20\)

[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]

2*(ln(x)-3)/(2*ln(x)-5)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=\frac {2 \, {\left (\log \left (x\right ) - 3\right )}}{2 \, x \log \left (x\right ) - 5 \, x} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=- \frac {1}{2 x \log {\left (x \right )} - 5 x} + \frac {1}{x} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=\frac {2 \, {\left (\log \left (x\right ) - 3\right )}}{2 \, x \log \left (x\right ) - 5 \, x} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=-\frac {1}{2 \, x \log \left (x\right ) - 5 \, x} + \frac {1}{x} \]

[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

Mupad [B] (verification not implemented)

Time = 13.86 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-28+22 \log (x)-4 \log ^2(x)}{25 x^2-20 x^2 \log (x)+4 x^2 \log ^2(x)} \, dx=\frac {2\,\left (\ln \left (x\right )-3\right )}{x\,\left (2\,\ln \left (x\right )-5\right )} \]

[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))