Integrand size = 64, antiderivative size = 24 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=\frac {\left (-1-\frac {25}{16 x^4}+\frac {4}{x}\right ) x}{\log \left (-3+\frac {1}{x}\right )} \] Output:
(4/x-25/16/x^4-1)/ln(1/x-3)*x
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=\frac {-25+64 x^3-16 x^4}{16 x^3 \log \left (-3+\frac {1}{x}\right )} \] Input:
Integrate[(25 - 64*x^3 + 16*x^4 + (-75 + 225*x + 16*x^4 - 48*x^5)*Log[(1 - 3*x)/x])/((-16*x^4 + 48*x^5)*Log[(1 - 3*x)/x]^2),x]
Output:
(-25 + 64*x^3 - 16*x^4)/(16*x^3*Log[-3 + x^(-1)])
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {16 x^4-64 x^3+\left (-48 x^5+16 x^4+225 x-75\right ) \log \left (\frac {1-3 x}{x}\right )+25}{\left (48 x^5-16 x^4\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {16 x^4-64 x^3+\left (-48 x^5+16 x^4+225 x-75\right ) \log \left (\frac {1-3 x}{x}\right )+25}{x^4 (48 x-16) \log ^2\left (\frac {1-3 x}{x}\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-16 x^4+64 x^3-\left (-48 x^5+16 x^4+225 x-75\right ) \log \left (\frac {1-3 x}{x}\right )-25}{(16-48 x) x^4 \log ^2\left (\frac {1}{x}-3\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {75-16 x^4}{16 x^4 \log \left (\frac {1}{x}-3\right )}+\frac {16 x^4-64 x^3+25}{16 x^4 (3 x-1) \log ^2\left (\frac {1}{x}-3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \text {Subst}\left (\int \frac {1}{x^2 \log (x-3)}dx,x,\frac {1}{x}\right )+\frac {1}{16} \int \frac {16 x^4-64 x^3+25}{x^4 (3 x-1) \log ^2\left (\frac {1-3 x}{x}\right )}dx-\frac {225}{8} \operatorname {ExpIntegralEi}\left (2 \log \left (\frac {1}{x}-3\right )\right )-\frac {75}{16} \operatorname {ExpIntegralEi}\left (3 \log \left (\frac {1}{x}-3\right )\right )-\frac {675 \operatorname {LogIntegral}\left (\frac {1}{x}-3\right )}{16}\) |
Input:
Int[(25 - 64*x^3 + 16*x^4 + (-75 + 225*x + 16*x^4 - 48*x^5)*Log[(1 - 3*x)/ x])/((-16*x^4 + 48*x^5)*Log[(1 - 3*x)/x]^2),x]
Output:
$Aborted
Time = 5.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
norman | \(\frac {-\frac {25}{16}+4 x^{3}-x^{4}}{x^{3} \ln \left (\frac {1-3 x}{x}\right )}\) | \(29\) |
risch | \(-\frac {16 x^{4}-64 x^{3}+25}{16 x^{3} \ln \left (\frac {1-3 x}{x}\right )}\) | \(30\) |
parallelrisch | \(\frac {-96 x^{4}+384 x^{3}-150}{96 x^{3} \ln \left (-\frac {-1+3 x}{x}\right )}\) | \(31\) |
derivativedivides | \(-\frac {x}{\ln \left (\frac {1}{x}-3\right )}-\frac {611}{16 \ln \left (\frac {1}{x}-3\right )}-\frac {675 \left (\frac {1}{x}-3\right )}{16 \ln \left (\frac {1}{x}-3\right )}-\frac {225 \left (\frac {1}{x}-3\right )^{2}}{16 \ln \left (\frac {1}{x}-3\right )}-\frac {25 \left (\frac {1}{x}-3\right )^{3}}{16 \ln \left (\frac {1}{x}-3\right )}\) | \(72\) |
default | \(-\frac {x}{\ln \left (\frac {1}{x}-3\right )}-\frac {611}{16 \ln \left (\frac {1}{x}-3\right )}-\frac {675 \left (\frac {1}{x}-3\right )}{16 \ln \left (\frac {1}{x}-3\right )}-\frac {225 \left (\frac {1}{x}-3\right )^{2}}{16 \ln \left (\frac {1}{x}-3\right )}-\frac {25 \left (\frac {1}{x}-3\right )^{3}}{16 \ln \left (\frac {1}{x}-3\right )}\) | \(72\) |
Input:
int(((-48*x^5+16*x^4+225*x-75)*ln((1-3*x)/x)+16*x^4-64*x^3+25)/(48*x^5-16* x^4)/ln((1-3*x)/x)^2,x,method=_RETURNVERBOSE)
Output:
(-25/16+4*x^3-x^4)/x^3/ln((1-3*x)/x)
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=-\frac {16 \, x^{4} - 64 \, x^{3} + 25}{16 \, x^{3} \log \left (-\frac {3 \, x - 1}{x}\right )} \] Input:
integrate(((-48*x^5+16*x^4+225*x-75)*log((1-3*x)/x)+16*x^4-64*x^3+25)/(48* x^5-16*x^4)/log((1-3*x)/x)^2,x, algorithm="fricas")
Output:
-1/16*(16*x^4 - 64*x^3 + 25)/(x^3*log(-(3*x - 1)/x))
Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=\frac {- 16 x^{4} + 64 x^{3} - 25}{16 x^{3} \log {\left (\frac {1 - 3 x}{x} \right )}} \] Input:
integrate(((-48*x**5+16*x**4+225*x-75)*ln((1-3*x)/x)+16*x**4-64*x**3+25)/( 48*x**5-16*x**4)/ln((1-3*x)/x)**2,x)
Output:
(-16*x**4 + 64*x**3 - 25)/(16*x**3*log((1 - 3*x)/x))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=\frac {16 \, x^{4} - 64 \, x^{3} + 25}{16 \, {\left (x^{3} \log \left (x\right ) - x^{3} \log \left (-3 \, x + 1\right )\right )}} \] Input:
integrate(((-48*x^5+16*x^4+225*x-75)*log((1-3*x)/x)+16*x^4-64*x^3+25)/(48* x^5-16*x^4)/log((1-3*x)/x)^2,x, algorithm="maxima")
Output:
1/16*(16*x^4 - 64*x^3 + 25)/(x^3*log(x) - x^3*log(-3*x + 1))
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (23) = 46\).
Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.58 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=\frac {\frac {25 \, {\left (3 \, x - 1\right )}^{4}}{x^{4}} - \frac {300 \, {\left (3 \, x - 1\right )}^{3}}{x^{3}} + \frac {1350 \, {\left (3 \, x - 1\right )}^{2}}{x^{2}} - \frac {2636 \, {\left (3 \, x - 1\right )}}{x} + 1849}{16 \, {\left (\frac {{\left (3 \, x - 1\right )} \log \left (-\frac {3 \, x - 1}{x}\right )}{x} - 3 \, \log \left (-\frac {3 \, x - 1}{x}\right )\right )}} \] Input:
integrate(((-48*x^5+16*x^4+225*x-75)*log((1-3*x)/x)+16*x^4-64*x^3+25)/(48* x^5-16*x^4)/log((1-3*x)/x)^2,x, algorithm="giac")
Output:
1/16*(25*(3*x - 1)^4/x^4 - 300*(3*x - 1)^3/x^3 + 1350*(3*x - 1)^2/x^2 - 26 36*(3*x - 1)/x + 1849)/((3*x - 1)*log(-(3*x - 1)/x)/x - 3*log(-(3*x - 1)/x ))
Time = 2.82 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=-\frac {x^4-4\,x^3+\frac {25}{16}}{x^3\,\ln \left (-\frac {3\,x-1}{x}\right )} \] Input:
int(-(16*x^4 - 64*x^3 + log(-(3*x - 1)/x)*(225*x + 16*x^4 - 48*x^5 - 75) + 25)/(log(-(3*x - 1)/x)^2*(16*x^4 - 48*x^5)),x)
Output:
-(x^4 - 4*x^3 + 25/16)/(x^3*log(-(3*x - 1)/x))
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {25-64 x^3+16 x^4+\left (-75+225 x+16 x^4-48 x^5\right ) \log \left (\frac {1-3 x}{x}\right )}{\left (-16 x^4+48 x^5\right ) \log ^2\left (\frac {1-3 x}{x}\right )} \, dx=\frac {-16 x^{4}+64 x^{3}-25}{16 \,\mathrm {log}\left (\frac {-3 x +1}{x}\right ) x^{3}} \] Input:
int(((-48*x^5+16*x^4+225*x-75)*log((1-3*x)/x)+16*x^4-64*x^3+25)/(48*x^5-16 *x^4)/log((1-3*x)/x)^2,x)
Output:
( - 16*x**4 + 64*x**3 - 25)/(16*log(( - 3*x + 1)/x)*x**3)