\(\int \frac {4335+289 x+(-289 x+5 x^2) \log (\frac {-289+5 x}{x}) \log (\log (\frac {-289+5 x}{x}))}{(-289 x+5 x^2) \log (\frac {-289+5 x}{x})} \, dx\) [602]

Optimal result

Integrand size = 60, antiderivative size = 13 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=(15+x) \log \left (\log \left (5-\frac {289}{x}\right )\right ) \]

[Out]

ln(ln(5-289/x))*(x+15)

Rubi [F]

\[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=\int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx \]

[In]

Int[(4335 + 289*x + (-289*x + 5*x^2)*Log[(-289 + 5*x)/x]*Log[Log[(-289 + 5*x)/x]])/((-289*x + 5*x^2)*Log[(-289
+ 5*x)/x]),x]

[Out]

289*Defer[Int][(15 + x)/(x*(-289 + 5*x)*Log[(-289 + 5*x)/x]), x] + Defer[Int][Log[Log[5 - 289/x]], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{x (-289+5 x) \log \left (\frac {-289+5 x}{x}\right )} \, dx \\ & = \int \left (\frac {289 (15+x)}{x (-289+5 x) \log \left (5-\frac {289}{x}\right )}+\log \left (\log \left (5-\frac {289}{x}\right )\right )\right ) \, dx \\ & = 289 \int \frac {15+x}{x (-289+5 x) \log \left (5-\frac {289}{x}\right )} \, dx+\int \log \left (\log \left (5-\frac {289}{x}\right )\right ) \, dx \\ & = 289 \int \frac {15+x}{x (-289+5 x) \log \left (\frac {-289+5 x}{x}\right )} \, dx+\int \log \left (\log \left (5-\frac {289}{x}\right )\right ) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.77 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=15 \log \left (\log \left (5-\frac {289}{x}\right )\right )+x \log \left (\log \left (5-\frac {289}{x}\right )\right ) \]

[In]

Integrate[(4335 + 289*x + (-289*x + 5*x^2)*Log[(-289 + 5*x)/x]*Log[Log[(-289 + 5*x)/x]])/((-289*x + 5*x^2)*Log
[(-289 + 5*x)/x]),x]

[Out]

15*Log[Log[5 - 289/x]] + x*Log[Log[5 - 289/x]]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(27\) vs. \(2(13)=26\).

Time = 0.53 (sec) , antiderivative size = 28, normalized size of antiderivative = 2.15

method	result	size
norman	\(x \ln \left (\ln \left (\frac {5 x -289}{x}\right )\right )+15 \ln \left (\ln \left (\frac {5 x -289}{x}\right )\right )\)	\(28\)
parallelrisch	\(x \ln \left (\ln \left (\frac {5 x -289}{x}\right )\right )+15 \ln \left (\ln \left (\frac {5 x -289}{x}\right )\right )\)	\(28\)

[In]

int(((5*x^2-289*x)*ln((5*x-289)/x)*ln(ln((5*x-289)/x))+289*x+4335)/(5*x^2-289*x)/ln((5*x-289)/x),x,method=_RET
URNVERBOSE)

[Out]

x*ln(ln((5*x-289)/x))+15*ln(ln((5*x-289)/x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx={\left (x + 15\right )} \log \left (\log \left (\frac {5 \, x - 289}{x}\right )\right ) \]

[In]

integrate(((5*x^2-289*x)*log((5*x-289)/x)*log(log((5*x-289)/x))+289*x+4335)/(5*x^2-289*x)/log((5*x-289)/x),x,
algorithm="fricas")

[Out]

(x + 15)*log(log((5*x - 289)/x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (10) = 20\).

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.08 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=\left (x - \frac {289}{30}\right ) \log {\left (\log {\left (\frac {5 x - 289}{x} \right )} \right )} + \frac {739 \log {\left (\log {\left (\frac {5 x - 289}{x} \right )} \right )}}{30} \]

[In]

integrate(((5*x**2-289*x)*ln((5*x-289)/x)*ln(ln((5*x-289)/x))+289*x+4335)/(5*x**2-289*x)/ln((5*x-289)/x),x)

[Out]

(x - 289/30)*log(log((5*x - 289)/x)) + 739*log(log((5*x - 289)/x))/30

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (13) = 26\).

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 2.23 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=x \log \left (\log \left (5 \, x - 289\right ) - \log \left (x\right )\right ) + 15 \, \log \left (\log \left (5 \, x - 289\right ) - \log \left (x\right )\right ) \]

[In]

integrate(((5*x^2-289*x)*log((5*x-289)/x)*log(log((5*x-289)/x))+289*x+4335)/(5*x^2-289*x)/log((5*x-289)/x),x,
algorithm="maxima")

[Out]

x*log(log(5*x - 289) - log(x)) + 15*log(log(5*x - 289) - log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (13) = 26\).

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 2.15 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=x \log \left (\log \left (\frac {5 \, x - 289}{x}\right )\right ) + 15 \, \log \left (\log \left (5 \, x - 289\right ) - \log \left (x\right )\right ) \]

[In]

integrate(((5*x^2-289*x)*log((5*x-289)/x)*log(log((5*x-289)/x))+289*x+4335)/(5*x^2-289*x)/log((5*x-289)/x),x,
algorithm="giac")

[Out]

x*log(log((5*x - 289)/x)) + 15*log(log(5*x - 289) - log(x))

Mupad [B] (verification not implemented)

Time = 8.62 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {4335+289 x+\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right ) \log \left (\log \left (\frac {-289+5 x}{x}\right )\right )}{\left (-289 x+5 x^2\right ) \log \left (\frac {-289+5 x}{x}\right )} \, dx=\ln \left (\ln \left (\frac {5\,x-289}{x}\right )\right )\,\left (x+15\right ) \]

[In]

int(-(289*x - log(log((5*x - 289)/x))*log((5*x - 289)/x)*(289*x - 5*x^2) + 4335)/(log((5*x - 289)/x)*(289*x -
5*x^2)),x)

[Out]

log(log((5*x - 289)/x))*(x + 15)