3.92.59 \(\int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+(45-30 x+5 x^2) \log (25)} \, dx\)

Optimal. Leaf size=22 \[ -\frac {x \left (4+\frac {4}{-\frac {18}{5}+\log (25)}\right ) \log (x)}{-3+x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6688, 12, 6742, 2314, 31} \begin {gather*} \frac {4 x (13-5 \log (25)) \log (x)}{(3-x) (18-5 \log (25))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-156 + 52*x + (60 - 20*x)*Log[25] + (-156 + 60*Log[25])*Log[x])/(-162 + 108*x - 18*x^2 + (45 - 30*x + 5*x
^2)*Log[25]),x]

[Out]

(4*x*(13 - 5*Log[25])*Log[x])/((3 - x)*(18 - 5*Log[25]))

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

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 {4 (13-5 \log (25)) (3-x+3 \log (x))}{(3-x)^2 (18-5 \log (25))} \, dx\\ &=\frac {(4 (13-5 \log (25))) \int \frac {3-x+3 \log (x)}{(3-x)^2} \, dx}{18-5 \log (25)}\\ &=\frac {(4 (13-5 \log (25))) \int \left (\frac {1}{3-x}+\frac {3 \log (x)}{(-3+x)^2}\right ) \, dx}{18-5 \log (25)}\\ &=-\frac {4 (13-5 \log (25)) \log (3-x)}{18-5 \log (25)}+\frac {(12 (13-5 \log (25))) \int \frac {\log (x)}{(-3+x)^2} \, dx}{18-5 \log (25)}\\ &=-\frac {4 (13-5 \log (25)) \log (3-x)}{18-5 \log (25)}+\frac {4 x (13-5 \log (25)) \log (x)}{(3-x) (18-5 \log (25))}+\frac {(4 (13-5 \log (25))) \int \frac {1}{-3+x} \, dx}{18-5 \log (25)}\\ &=\frac {4 x (13-5 \log (25)) \log (x)}{(3-x) (18-5 \log (25))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 24, normalized size = 1.09 \begin {gather*} -\frac {4 x (-13+5 \log (25)) \log (x)}{(-3+x) (-18+5 \log (25))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-156 + 52*x + (60 - 20*x)*Log[25] + (-156 + 60*Log[25])*Log[x])/(-162 + 108*x - 18*x^2 + (45 - 30*x
+ 5*x^2)*Log[25]),x]

[Out]

(-4*x*(-13 + 5*Log[25])*Log[x])/((-3 + x)*(-18 + 5*Log[25]))

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 27, normalized size = 1.23 \begin {gather*} -\frac {2 \, {\left (10 \, x \log \relax (5) - 13 \, x\right )} \log \relax (x)}{5 \, {\left (x - 3\right )} \log \relax (5) - 9 \, x + 27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2-30*x+45)*log(5)-18*x^2+108*x-162),x
, algorithm="fricas")

[Out]

-2*(10*x*log(5) - 13*x)*log(x)/(5*(x - 3)*log(5) - 9*x + 27)

________________________________________________________________________________________

giac [B]  time = 0.13, size = 45, normalized size = 2.05 \begin {gather*} -\frac {6 \, {\left (10 \, \log \relax (5) - 13\right )} \log \relax (x)}{5 \, x \log \relax (5) - 9 \, x - 15 \, \log \relax (5) + 27} - \frac {2 \, {\left (10 \, \log \relax (5) - 13\right )} \log \relax (x)}{5 \, \log \relax (5) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2-30*x+45)*log(5)-18*x^2+108*x-162),x
, algorithm="giac")

[Out]

-6*(10*log(5) - 13)*log(x)/(5*x*log(5) - 9*x - 15*log(5) + 27) - 2*(10*log(5) - 13)*log(x)/(5*log(5) - 9)

________________________________________________________________________________________

maple [A]  time = 0.64, size = 25, normalized size = 1.14

method	result	size
norman	\(-\frac {2 \left (10 \ln \relax (5)-13\right ) x \ln \relax (x )}{\left (5 \ln \relax (5)-9\right ) \left (x -3\right )}\)	\(25\)
risch	\(-\frac {6 \left (10 \ln \relax (5)-13\right ) \ln \relax (x )}{5 x \ln \relax (5)-15 \ln \relax (5)-9 x +27}-\frac {20 \ln \relax (x ) \ln \relax (5)}{5 \ln \relax (5)-9}+\frac {26 \ln \relax (x )}{5 \ln \relax (5)-9}\)	\(54\)
default	\(-\frac {20 \ln \left (\left (5 \ln \relax (5)-9\right ) x -15 \ln \relax (5)+27\right ) \ln \relax (5)}{5 \ln \relax (5)-9}+\frac {26 \ln \left (\left (5 \ln \relax (5)-9\right ) x -15 \ln \relax (5)+27\right )}{5 \ln \relax (5)-9}+\frac {20 \ln \relax (5) \ln \left (x -3\right )}{5 \ln \relax (5)-9}-\frac {20 \ln \relax (5) \ln \relax (x ) x}{\left (5 \ln \relax (5)-9\right ) \left (x -3\right )}-\frac {26 \ln \left (x -3\right )}{5 \ln \relax (5)-9}+\frac {26 \ln \relax (x ) x}{\left (5 \ln \relax (5)-9\right ) \left (x -3\right )}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((120*ln(5)-156)*ln(x)+2*(-20*x+60)*ln(5)+52*x-156)/(2*(5*x^2-30*x+45)*ln(5)-18*x^2+108*x-162),x,method=_R
ETURNVERBOSE)

[Out]

-2*(10*ln(5)-13)/(5*ln(5)-9)*x*ln(x)/(x-3)

________________________________________________________________________________________

maxima [B]  time = 0.36, size = 141, normalized size = 6.41 \begin {gather*} 20 \, {\left (\frac {\log \left (x - 3\right )}{5 \, \log \relax (5) - 9} - \frac {\log \relax (x)}{5 \, \log \relax (5) - 9}\right )} \log \relax (5) - 20 \, {\left (\frac {\log \left (x - 3\right )}{5 \, \log \relax (5) - 9} - \frac {3}{x {\left (5 \, \log \relax (5) - 9\right )} - 15 \, \log \relax (5) + 27}\right )} \log \relax (5) - \frac {60 \, \log \relax (5) \log \relax (x)}{x {\left (5 \, \log \relax (5) - 9\right )} - 15 \, \log \relax (5) + 27} - \frac {60 \, \log \relax (5)}{x {\left (5 \, \log \relax (5) - 9\right )} - 15 \, \log \relax (5) + 27} + \frac {78 \, \log \relax (x)}{x {\left (5 \, \log \relax (5) - 9\right )} - 15 \, \log \relax (5) + 27} + \frac {26 \, \log \relax (x)}{5 \, \log \relax (5) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2-30*x+45)*log(5)-18*x^2+108*x-162),x
, algorithm="maxima")

[Out]

20*(log(x - 3)/(5*log(5) - 9) - log(x)/(5*log(5) - 9))*log(5) - 20*(log(x - 3)/(5*log(5) - 9) - 3/(x*(5*log(5)
- 9) - 15*log(5) + 27))*log(5) - 60*log(5)*log(x)/(x*(5*log(5) - 9) - 15*log(5) + 27) - 60*log(5)/(x*(5*log(5
) - 9) - 15*log(5) + 27) + 78*log(x)/(x*(5*log(5) - 9) - 15*log(5) + 27) + 26*log(x)/(5*log(5) - 9)

________________________________________________________________________________________

mupad [B]  time = 7.96, size = 24, normalized size = 1.09 \begin {gather*} -\frac {2\,x\,\ln \relax (x)\,\left (10\,\ln \relax (5)-13\right )}{\left (5\,\ln \relax (5)-9\right )\,\left (x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((52*x - 2*log(5)*(20*x - 60) + log(x)*(120*log(5) - 156) - 156)/(108*x + 2*log(5)*(5*x^2 - 30*x + 45) - 18
*x^2 - 162),x)

[Out]

-(2*x*log(x)*(10*log(5) - 13))/((5*log(5) - 9)*(x - 3))

________________________________________________________________________________________

sympy [B]  time = 0.21, size = 42, normalized size = 1.91 \begin {gather*} \frac {\left (26 - 20 \log {\relax (5 )}\right ) \log {\relax (x )}}{-9 + 5 \log {\relax (5 )}} + \frac {\left (78 - 60 \log {\relax (5 )}\right ) \log {\relax (x )}}{- 9 x + 5 x \log {\relax (5 )} - 15 \log {\relax (5 )} + 27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((120*ln(5)-156)*ln(x)+2*(-20*x+60)*ln(5)+52*x-156)/(2*(5*x**2-30*x+45)*ln(5)-18*x**2+108*x-162),x)

[Out]

(26 - 20*log(5))*log(x)/(-9 + 5*log(5)) + (78 - 60*log(5))*log(x)/(-9*x + 5*x*log(5) - 15*log(5) + 27)

________________________________________________________________________________________