\(\int \frac {5 x^2+(-50-270 x-100 x^2) \log (\frac {1}{4} (1+5 x))+(-x-5 x^2) \log (\frac {1}{4} (1+5 x)) \log (\log (\frac {1}{4} (1+5 x)))}{(x^3+5 x^4) \log (\frac {1}{4} (1+5 x))} \, dx\) [7924]

Optimal result

Integrand size = 82, antiderivative size = 25 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\left (2+\frac {5}{x}\right )^2+\frac {\log \left (\log \left (x+\frac {1+x}{4}\right )\right )}{x} \]

[Out]

(5/x+2)^2+ln(ln(1/4+5/4*x))/x

Rubi [F]

\[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx \]

[In]

Int[(5*x^2 + (-50 - 270*x - 100*x^2)*Log[(1 + 5*x)/4] + (-x - 5*x^2)*Log[(1 + 5*x)/4]*Log[Log[(1 + 5*x)/4]])/(
(x^3 + 5*x^4)*Log[(1 + 5*x)/4]),x]

[Out]

(5 + 2*x)^2/x^2 - 5*Log[Log[(1 + 5*x)/4]] + 5*Defer[Int][1/(x*Log[1/4 + (5*x)/4]), x] - Defer[Int][Log[Log[1/4
+ (5*x)/4]]/x^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x^3 (1+5 x) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx \\ & = \int \left (\frac {5}{x (1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )}-\frac {50+20 x+x \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x^3}\right ) \, dx \\ & = 5 \int \frac {1}{x (1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-\int \frac {50+20 x+x \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x^3} \, dx \\ & = 5 \int \left (\frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )}-\frac {5}{(1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )}\right ) \, dx-\int \left (\frac {10 (5+2 x)}{x^3}+\frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2}\right ) \, dx \\ & = 5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-10 \int \frac {5+2 x}{x^3} \, dx-25 \int \frac {1}{(1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx \\ & = \frac {(5+2 x)^2}{x^2}+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-20 \text {Subst}\left (\int \frac {1}{4 x \log (x)} \, dx,x,\frac {1}{4}+\frac {5 x}{4}\right )-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx \\ & = \frac {(5+2 x)^2}{x^2}+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\frac {1}{4}+\frac {5 x}{4}\right )-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx \\ & = \frac {(5+2 x)^2}{x^2}+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {1}{4} (1+5 x)\right )\right )-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx \\ & = \frac {(5+2 x)^2}{x^2}-5 \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {25}{x^2}+\frac {20}{x}+\frac {\log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x} \]

[In]

Integrate[(5*x^2 + (-50 - 270*x - 100*x^2)*Log[(1 + 5*x)/4] + (-x - 5*x^2)*Log[(1 + 5*x)/4]*Log[Log[(1 + 5*x)/
4]])/((x^3 + 5*x^4)*Log[(1 + 5*x)/4]),x]

[Out]

25/x^2 + 20/x + Log[Log[(1 + 5*x)/4]]/x

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\ln \left (\ln \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x}+\frac {20 x +25}{x^{2}}\)	\(23\)
parallelrisch	\(-\frac {-1250+2500 x^{2}-50 \ln \left (\ln \left (\frac {1}{4}+\frac {5 x}{4}\right )\right ) x -1000 x}{50 x^{2}}\)	\(26\)

[In]

int(((-5*x^2-x)*ln(1/4+5/4*x)*ln(ln(1/4+5/4*x))+(-100*x^2-270*x-50)*ln(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)/ln(1/4+5/
4*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/4+5/4*x))/x+5*(5+4*x)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {x \log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right ) + 20 \, x + 25}{x^{2}} \]

[In]

integrate(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-50)*log(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)
/log(1/4+5/4*x),x, algorithm="fricas")

[Out]

(x*log(log(5/4*x + 1/4)) + 20*x + 25)/x^2

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {\log {\left (\log {\left (\frac {5 x}{4} + \frac {1}{4} \right )} \right )}}{x} - \frac {- 20 x - 25}{x^{2}} \]

[In]

integrate(((-5*x**2-x)*ln(1/4+5/4*x)*ln(ln(1/4+5/4*x))+(-100*x**2-270*x-50)*ln(1/4+5/4*x)+5*x**2)/(5*x**4+x**3
)/ln(1/4+5/4*x),x)

[Out]

log(log(5*x/4 + 1/4))/x - (-20*x - 25)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {x \log \left (-2 \, \log \left (2\right ) + \log \left (5 \, x + 1\right )\right ) + 20 \, x + 25}{x^{2}} \]

[In]

integrate(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-50)*log(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)
/log(1/4+5/4*x),x, algorithm="maxima")

[Out]

(x*log(-2*log(2) + log(5*x + 1)) + 20*x + 25)/x^2

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {\log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right )}{x} + \frac {5 \, {\left (4 \, x + 5\right )}}{x^{2}} \]

[In]

integrate(((-5*x^2-x)*log(1/4+5/4*x)*log(log(1/4+5/4*x))+(-100*x^2-270*x-50)*log(1/4+5/4*x)+5*x^2)/(5*x^4+x^3)
/log(1/4+5/4*x),x, algorithm="giac")

[Out]

log(log(5/4*x + 1/4))/x + 5*(4*x + 5)/x^2

Mupad [B] (verification not implemented)

Time = 14.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx=\frac {x\,\left (\ln \left (\ln \left (\frac {5\,x}{4}+\frac {1}{4}\right )\right )+20\right )+25}{x^2} \]

[In]

int(-(log((5*x)/4 + 1/4)*(270*x + 100*x^2 + 50) - 5*x^2 + log(log((5*x)/4 + 1/4))*log((5*x)/4 + 1/4)*(x + 5*x^
2))/(log((5*x)/4 + 1/4)*(x^3 + 5*x^4)),x)

[Out]

(x*(log(log((5*x)/4 + 1/4)) + 20) + 25)/x^2