\(\int \frac {-60+140 \log (\frac {\log (5)}{3})+85 \log ^2(\frac {\log (5)}{3})}{36-240 x+400 x^2+(-204+680 x) \log (\frac {\log (5)}{3})+289 \log ^2(\frac {\log (5)}{3})} \, dx\) [7690]

Optimal result

Integrand size = 59, antiderivative size = 26 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=-1+\frac {x}{\frac {17}{5}+\frac {4 (-2+x)}{2+\log \left (\frac {\log (5)}{3}\right )}} \]

[Out]

x/((-2+x)/(1/2+1/4*ln(1/3*ln(5)))+17/5)-1

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 2006, 27, 32} \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=-\frac {\left (6-17 \log \left (\frac {\log (5)}{3}\right )\right ) \left (2+\log \left (\frac {\log (5)}{3}\right )\right )}{4 \left (-20 x+6-17 \log \left (\frac {\log (5)}{3}\right )\right )} \]

[In]

Int[(-60 + 140*Log[Log[5]/3] + 85*Log[Log[5]/3]^2)/(36 - 240*x + 400*x^2 + (-204 + 680*x)*Log[Log[5]/3] + 289*
Log[Log[5]/3]^2),x]

[Out]

-1/4*((6 - 17*Log[Log[5]/3])*(2 + Log[Log[5]/3]))/(6 - 20*x - 17*Log[Log[5]/3])

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2006

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (5 \left (6-17 \log \left (\frac {\log (5)}{3}\right )\right ) \left (2+\log \left (\frac {\log (5)}{3}\right )\right )\right ) \int \frac {1}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx\right ) \\ & = -\left (\left (5 \left (6-17 \log \left (\frac {\log (5)}{3}\right )\right ) \left (2+\log \left (\frac {\log (5)}{3}\right )\right )\right ) \int \frac {1}{400 x^2-40 x \left (6-17 \log \left (\frac {\log (5)}{3}\right )\right )+\left (6-17 \log \left (\frac {\log (5)}{3}\right )\right )^2} \, dx\right ) \\ & = -\left (\left (5 \left (6-17 \log \left (\frac {\log (5)}{3}\right )\right ) \left (2+\log \left (\frac {\log (5)}{3}\right )\right )\right ) \int \frac {1}{\left (-6+20 x+17 \log \left (\frac {\log (5)}{3}\right )\right )^2} \, dx\right ) \\ & = -\frac {\left (6-17 \log \left (\frac {\log (5)}{3}\right )\right ) \left (2+\log \left (\frac {\log (5)}{3}\right )\right )}{4 \left (6-20 x-17 \log \left (\frac {\log (5)}{3}\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=-\frac {-12+28 \log \left (\frac {\log (5)}{3}\right )+17 \log ^2\left (\frac {\log (5)}{3}\right )}{4 \left (-6+20 x+17 \log \left (\frac {\log (5)}{3}\right )\right )} \]

[In]

Integrate[(-60 + 140*Log[Log[5]/3] + 85*Log[Log[5]/3]^2)/(36 - 240*x + 400*x^2 + (-204 + 680*x)*Log[Log[5]/3]
+ 289*Log[Log[5]/3]^2),x]

[Out]

-1/4*(-12 + 28*Log[Log[5]/3] + 17*Log[Log[5]/3]^2)/(-6 + 20*x + 17*Log[Log[5]/3])

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35

method	result	size
gosper	\(-\frac {17 \ln \left (\frac {\ln \left (5\right )}{3}\right )^{2}+28 \ln \left (\frac {\ln \left (5\right )}{3}\right )-12}{4 \left (17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+20 x -6\right )}\)	\(35\)
default	\(-\frac {85 \ln \left (\frac {\ln \left (5\right )}{3}\right )^{2}+140 \ln \left (\frac {\ln \left (5\right )}{3}\right )-60}{20 \left (17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+20 x -6\right )}\)	\(35\)
parallelrisch	\(-\frac {85 \ln \left (\frac {\ln \left (5\right )}{3}\right )^{2}+140 \ln \left (\frac {\ln \left (5\right )}{3}\right )-60}{20 \left (17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+20 x -6\right )}\)	\(35\)
norman	\(\frac {3-\frac {17 \ln \left (3\right )^{2}}{4}+\frac {17 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )}{2}-\frac {17 \ln \left (\ln \left (5\right )\right )^{2}}{4}+7 \ln \left (3\right )-7 \ln \left (\ln \left (5\right )\right )}{17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+20 x -6}\)	\(47\)
risch	\(\frac {\ln \left (3\right )^{2}}{4 \ln \left (3\right )-4 \ln \left (\ln \left (5\right )\right )-\frac {80 x}{17}+\frac {24}{17}}-\frac {\ln \left (3\right ) \ln \left (\ln \left (5\right )\right )}{2 \left (\ln \left (3\right )-\ln \left (\ln \left (5\right )\right )-\frac {20 x}{17}+\frac {6}{17}\right )}+\frac {\ln \left (\ln \left (5\right )\right )^{2}}{4 \ln \left (3\right )-4 \ln \left (\ln \left (5\right )\right )-\frac {80 x}{17}+\frac {24}{17}}-\frac {7 \ln \left (3\right )}{17 \left (\ln \left (3\right )-\ln \left (\ln \left (5\right )\right )-\frac {20 x}{17}+\frac {6}{17}\right )}+\frac {7 \ln \left (\ln \left (5\right )\right )}{17 \left (\ln \left (3\right )-\ln \left (\ln \left (5\right )\right )-\frac {20 x}{17}+\frac {6}{17}\right )}-\frac {3}{17 \left (\ln \left (3\right )-\ln \left (\ln \left (5\right )\right )-\frac {20 x}{17}+\frac {6}{17}\right )}\)	\(117\)
meijerg	\(\frac {3 x}{\left (\frac {17 \ln \left (\frac {\ln \left (5\right )}{3}\right )}{20}-\frac {3}{10}\right ) \left (-17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+6\right ) \left (1-\frac {20 x}{-17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+6}\right )}-\frac {17 \ln \left (\frac {\ln \left (5\right )}{3}\right )^{2} x}{4 \left (\frac {17 \ln \left (\frac {\ln \left (5\right )}{3}\right )}{20}-\frac {3}{10}\right ) \left (-17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+6\right ) \left (1-\frac {20 x}{-17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+6}\right )}-\frac {7 \ln \left (\frac {\ln \left (5\right )}{3}\right ) x}{\left (\frac {17 \ln \left (\frac {\ln \left (5\right )}{3}\right )}{20}-\frac {3}{10}\right ) \left (-17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+6\right ) \left (1-\frac {20 x}{-17 \ln \left (\frac {\ln \left (5\right )}{3}\right )+6}\right )}\)	\(143\)

[In]

int((85*ln(1/3*ln(5))^2+140*ln(1/3*ln(5))-60)/(289*ln(1/3*ln(5))^2+(680*x-204)*ln(1/3*ln(5))+400*x^2-240*x+36)
,x,method=_RETURNVERBOSE)

[Out]

-1/4*(17*ln(1/3*ln(5))^2+28*ln(1/3*ln(5))-12)/(17*ln(1/3*ln(5))+20*x-6)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=-\frac {17 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right )^{2} + 28 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right ) - 12}{4 \, {\left (20 \, x + 17 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right ) - 6\right )}} \]

[In]

integrate((85*log(1/3*log(5))^2+140*log(1/3*log(5))-60)/(289*log(1/3*log(5))^2+(680*x-204)*log(1/3*log(5))+400
*x^2-240*x+36),x, algorithm="fricas")

[Out]

-1/4*(17*log(1/3*log(5))^2 + 28*log(1/3*log(5)) - 12)/(20*x + 17*log(1/3*log(5)) - 6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).

Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=- \frac {- 140 \log {\left (3 \right )} - 170 \log {\left (3 \right )} \log {\left (\log {\left (5 \right )} \right )} - 60 + 85 \log {\left (\log {\left (5 \right )} \right )}^{2} + 140 \log {\left (\log {\left (5 \right )} \right )} + 85 \log {\left (3 \right )}^{2}}{400 x - 340 \log {\left (3 \right )} - 120 + 340 \log {\left (\log {\left (5 \right )} \right )}} \]

[In]

integrate((85*ln(1/3*ln(5))**2+140*ln(1/3*ln(5))-60)/(289*ln(1/3*ln(5))**2+(680*x-204)*ln(1/3*ln(5))+400*x**2-
240*x+36),x)

[Out]

-(-140*log(3) - 170*log(3)*log(log(5)) - 60 + 85*log(log(5))**2 + 140*log(log(5)) + 85*log(3)**2)/(400*x - 340
*log(3) - 120 + 340*log(log(5)))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=-\frac {17 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right )^{2} + 28 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right ) - 12}{4 \, {\left (20 \, x + 17 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right ) - 6\right )}} \]

[In]

integrate((85*log(1/3*log(5))^2+140*log(1/3*log(5))-60)/(289*log(1/3*log(5))^2+(680*x-204)*log(1/3*log(5))+400
*x^2-240*x+36),x, algorithm="maxima")

[Out]

-1/4*(17*log(1/3*log(5))^2 + 28*log(1/3*log(5)) - 12)/(20*x + 17*log(1/3*log(5)) - 6)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=-\frac {17 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right )^{2} + 28 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right ) - 12}{4 \, {\left (20 \, x + 17 \, \log \left (\frac {1}{3} \, \log \left (5\right )\right ) - 6\right )}} \]

[In]

integrate((85*log(1/3*log(5))^2+140*log(1/3*log(5))-60)/(289*log(1/3*log(5))^2+(680*x-204)*log(1/3*log(5))+400
*x^2-240*x+36),x, algorithm="giac")

[Out]

-1/4*(17*log(1/3*log(5))^2 + 28*log(1/3*log(5)) - 12)/(20*x + 17*log(1/3*log(5)) - 6)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {-60+140 \log \left (\frac {\log (5)}{3}\right )+85 \log ^2\left (\frac {\log (5)}{3}\right )}{36-240 x+400 x^2+(-204+680 x) \log \left (\frac {\log (5)}{3}\right )+289 \log ^2\left (\frac {\log (5)}{3}\right )} \, dx=\frac {7\,\ln \left (3\right )-7\,\ln \left (\ln \left (5\right )\right )-\frac {17\,{\ln \left (\ln \left (5\right )\right )}^2}{4}+\frac {17\,\ln \left (3\right )\,\ln \left (\ln \left (5\right )\right )}{2}-\frac {17\,{\ln \left (3\right )}^2}{4}+3}{20\,x+\ln \left (\frac {{\ln \left (5\right )}^{17}}{129140163}\right )-6} \]

[In]

int((140*log(log(5)/3) + 85*log(log(5)/3)^2 - 60)/(289*log(log(5)/3)^2 - 240*x + log(log(5)/3)*(680*x - 204) +
400*x^2 + 36),x)

[Out]

(7*log(3) - 7*log(log(5)) - (17*log(log(5))^2)/4 + (17*log(3)*log(log(5)))/2 - (17*log(3)^2)/4 + 3)/(20*x + lo
g(log(5)^17/129140163) - 6)