Integrand size = 74, antiderivative size = 27 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=\frac {5 \left (-3+x \left (-x+\log \left (5 x^2\right )\right )\right )}{(2-5 x) \log (x)} \] Output:
5*(x*(ln(5*x^2)-x)-3)/(-5*x+2)/ln(x)
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=2+\frac {5 \left (3+x^2-x \log \left (5 x^2\right )\right )}{(-2+5 x) \log (x)} \] Input:
Integrate[(30 - 75*x + 10*x^2 - 25*x^3 + (-55*x - 70*x^2 + 25*x^3)*Log[x] + (-10*x + 25*x^2 + 10*x*Log[x])*Log[5*x^2])/((4*x - 20*x^2 + 25*x^3)*Log[ x]^2),x]
Output:
2 + (5*(3 + x^2 - x*Log[5*x^2]))/((-2 + 5*x)*Log[x])
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.
\(\displaystyle \int \frac {-25 x^3+10 x^2+\left (25 x^2-10 x+10 x \log (x)\right ) \log \left (5 x^2\right )+\left (25 x^3-70 x^2-55 x\right ) \log (x)-75 x+30}{\left (25 x^3-20 x^2+4 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-25 x^3+10 x^2+\left (25 x^2-10 x+10 x \log (x)\right ) \log \left (5 x^2\right )+\left (25 x^3-70 x^2-55 x\right ) \log (x)-75 x+30}{x \left (25 x^2-20 x+4\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 100 \int \frac {-5 x^3+2 x^2-15 x-\left (-5 x^3+14 x^2+11 x\right ) \log (x)-\left (-5 x^2-2 \log (x) x+2 x\right ) \log \left (5 x^2\right )+6}{20 (2-5 x)^2 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int \frac {-5 x^3+2 x^2-15 x-\left (-5 x^3+14 x^2+11 x\right ) \log (x)-\left (-5 x^2-2 \log (x) x+2 x\right ) \log \left (5 x^2\right )+6}{(2-5 x)^2 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 5 \int \left (-\frac {5 x^2}{(5 x-2)^2 \log ^2(x)}+\frac {2 x}{(5 x-2)^2 \log ^2(x)}+\frac {(5 x+2 \log (x)-2) \log \left (5 x^2\right )}{(5 x-2)^2 \log ^2(x)}+\frac {5 x^2-14 x-11}{(5 x-2)^2 \log (x)}-\frac {15}{(5 x-2)^2 \log ^2(x)}+\frac {6}{(5 x-2)^2 \log ^2(x) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (-5 \int \frac {x^2}{(5 x-2)^2 \log ^2(x)}dx+\int \frac {\log \left (5 x^2\right )}{(5 x-2) \log ^2(x)}dx+\int \frac {5 x^2-14 x-11}{(5 x-2)^2 \log (x)}dx+2 \int \frac {\log \left (5 x^2\right )}{(5 x-2)^2 \log (x)}dx-15 \int \frac {1}{(5 x-2)^2 \log ^2(x)}dx+6 \int \frac {1}{x (5 x-2)^2 \log ^2(x)}dx+2 \int \frac {x}{(5 x-2)^2 \log ^2(x)}dx\right )\) |
Input:
Int[(30 - 75*x + 10*x^2 - 25*x^3 + (-55*x - 70*x^2 + 25*x^3)*Log[x] + (-10 *x + 25*x^2 + 10*x*Log[x])*Log[5*x^2])/((4*x - 20*x^2 + 25*x^3)*Log[x]^2), x]
Output:
$Aborted
Time = 3.98 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(\frac {1800-600 \ln \left (5 x^{2}\right ) x +600 x^{2}}{120 \ln \left (x \right ) \left (5 x -2\right )}\) | \(30\) |
risch | \(-\frac {4}{5 x -2}+\frac {\frac {5 i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-5 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+15+\frac {5 i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-5 x \ln \left (5\right )+5 x^{2}}{\ln \left (x \right ) \left (5 x -2\right )}\) | \(88\) |
Input:
int(((10*x*ln(x)+25*x^2-10*x)*ln(5*x^2)+(25*x^3-70*x^2-55*x)*ln(x)-25*x^3+ 10*x^2-75*x+30)/(25*x^3-20*x^2+4*x)/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
1/120*(1800-600*ln(5*x^2)*x+600*x^2)/ln(x)/(5*x-2)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=\frac {5 \, x^{2} - 5 \, x \log \left (5\right ) - 4 \, \log \left (x\right ) + 15}{{\left (5 \, x - 2\right )} \log \left (x\right )} \] Input:
integrate(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x )-25*x^3+10*x^2-75*x+30)/(25*x^3-20*x^2+4*x)/log(x)^2,x, algorithm="fricas ")
Output:
(5*x^2 - 5*x*log(5) - 4*log(x) + 15)/((5*x - 2)*log(x))
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=- \frac {20}{25 x - 10} + \frac {5 x^{2} - 5 x \log {\left (5 \right )} + 15}{\left (5 x - 2\right ) \log {\left (x \right )}} \] Input:
integrate(((10*x*ln(x)+25*x**2-10*x)*ln(5*x**2)+(25*x**3-70*x**2-55*x)*ln( x)-25*x**3+10*x**2-75*x+30)/(25*x**3-20*x**2+4*x)/ln(x)**2,x)
Output:
-20/(25*x - 10) + (5*x**2 - 5*x*log(5) + 15)/((5*x - 2)*log(x))
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=\frac {5 \, x^{2} - 5 \, x \log \left (5\right ) - 4 \, \log \left (x\right ) + 15}{{\left (5 \, x - 2\right )} \log \left (x\right )} \] Input:
integrate(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x )-25*x^3+10*x^2-75*x+30)/(25*x^3-20*x^2+4*x)/log(x)^2,x, algorithm="maxima ")
Output:
(5*x^2 - 5*x*log(5) - 4*log(x) + 15)/((5*x - 2)*log(x))
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=\frac {5 \, {\left (x^{2} - x \log \left (5\right ) + 3\right )}}{5 \, x \log \left (x\right ) - 2 \, \log \left (x\right )} - \frac {4}{5 \, x - 2} \] Input:
integrate(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x )-25*x^3+10*x^2-75*x+30)/(25*x^3-20*x^2+4*x)/log(x)^2,x, algorithm="giac")
Output:
5*(x^2 - x*log(5) + 3)/(5*x*log(x) - 2*log(x)) - 4/(5*x - 2)
Time = 3.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=\frac {5\,\left (x^2-x\,\ln \left (5\,x^2\right )+3\right )}{\ln \left (x\right )\,\left (5\,x-2\right )} \] Input:
int(-(75*x - log(5*x^2)*(10*x*log(x) - 10*x + 25*x^2) - 10*x^2 + 25*x^3 + log(x)*(55*x + 70*x^2 - 25*x^3) - 30)/(log(x)^2*(4*x - 20*x^2 + 25*x^3)),x )
Output:
(5*(x^2 - x*log(5*x^2) + 3))/(log(x)*(5*x - 2))
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {30-75 x+10 x^2-25 x^3+\left (-55 x-70 x^2+25 x^3\right ) \log (x)+\left (-10 x+25 x^2+10 x \log (x)\right ) \log \left (5 x^2\right )}{\left (4 x-20 x^2+25 x^3\right ) \log ^2(x)} \, dx=\frac {-5 \,\mathrm {log}\left (5 x^{2}\right ) x +5 x^{2}+15}{\mathrm {log}\left (x \right ) \left (5 x -2\right )} \] Input:
int(((10*x*log(x)+25*x^2-10*x)*log(5*x^2)+(25*x^3-70*x^2-55*x)*log(x)-25*x ^3+10*x^2-75*x+30)/(25*x^3-20*x^2+4*x)/log(x)^2,x)
Output:
(5*( - log(5*x**2)*x + x**2 + 3))/(log(x)*(5*x - 2))