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

Optimal. Leaf size=25 \[ \left (2+\frac {5}{x}\right )^2+\frac {\log \left (\log \left (x+\frac {1+x}{4}\right )\right )}{x} \]

________________________________________________________________________________________

Rubi [F]  time = 0.45, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 \operatorname {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 \operatorname {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 \operatorname {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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 26, normalized size = 1.04 \begin {gather*} \frac {25}{x^2}+\frac {20}{x}+\frac {\log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 1.17, size = 18, normalized size = 0.72 \begin {gather*} \frac {x \log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right ) + 20 \, x + 25}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [A]  time = 0.18, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right )}{x} + \frac {5 \, {\left (4 \, x + 5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maple [A]  time = 0.22, size = 23, normalized size = 0.92

method	result	size
risch	\(\frac {\ln \left (\ln \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x}+\frac {20 x +25}{x^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[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*(4*x+5)/x^2

________________________________________________________________________________________

maxima [A]  time = 0.63, size = 23, normalized size = 0.92 \begin {gather*} \frac {x \log \left (-2 \, \log \relax (2) + \log \left (5 \, x + 1\right )\right ) + 20 \, x + 25}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 5.80, size = 17, normalized size = 0.68 \begin {gather*} \frac {x\,\left (\ln \left (\ln \left (\frac {5\,x}{4}+\frac {1}{4}\right )\right )+20\right )+25}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.37, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log {\left (\log {\left (\frac {5 x}{4} + \frac {1}{4} \right )} \right )}}{x} - \frac {- 20 x - 25}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________