Integrand size = 94, antiderivative size = 20 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\log (2)}{6+\log (25 x (2-4 x-5 \log (x)))} \] Output:
ln(2)/(ln(25*x*(2-4*x-5*ln(x)))+6)
Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\log (2)}{6+\log (-25 x (-2+4 x+5 \log (x)))} \] Input:
Integrate[((-3 - 8*x)*Log[2] - 5*Log[2]*Log[x])/(-72*x + 144*x^2 + 180*x*L og[x] + (-24*x + 48*x^2 + 60*x*Log[x])*Log[50*x - 100*x^2 - 125*x*Log[x]] + (-2*x + 4*x^2 + 5*x*Log[x])*Log[50*x - 100*x^2 - 125*x*Log[x]]^2),x]
Output:
Log[2]/(6 + Log[-25*x*(-2 + 4*x + 5*Log[x])])
Time = 0.59 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {7239, 27, 7237}
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.
\(\displaystyle \int \frac {(-8 x-3) \log (2)-5 \log (2) \log (x)}{144 x^2+\left (4 x^2-2 x+5 x \log (x)\right ) \log ^2\left (-100 x^2+50 x-125 x \log (x)\right )+\left (48 x^2-24 x+60 x \log (x)\right ) \log \left (-100 x^2+50 x-125 x \log (x)\right )-72 x+180 x \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log (2) (8 x+5 \log (x)+3)}{x (-4 x-5 \log (x)+2) (\log (-25 x (4 x+5 \log (x)-2))+6)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (2) \int \frac {8 x+5 \log (x)+3}{x (-4 x-5 \log (x)+2) (\log (25 x (-4 x-5 \log (x)+2))+6)^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {\log (2)}{\log (25 x (-4 x-5 \log (x)+2))+6}\) |
Input:
Int[((-3 - 8*x)*Log[2] - 5*Log[2]*Log[x])/(-72*x + 144*x^2 + 180*x*Log[x] + (-24*x + 48*x^2 + 60*x*Log[x])*Log[50*x - 100*x^2 - 125*x*Log[x]] + (-2* x + 4*x^2 + 5*x*Log[x])*Log[50*x - 100*x^2 - 125*x*Log[x]]^2),x]
Output:
Log[2]/(6 + Log[25*x*(2 - 4*x - 5*Log[x])])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
parallelrisch | \(\frac {\ln \left (2\right )}{\ln \left (-25 x \left (5 \ln \left (x \right )+4 x -2\right )\right )+6}\) | \(21\) |
default | \(\frac {\ln \left (2\right )}{2 \ln \left (5\right )+\ln \left (-5 x \ln \left (x \right )-4 x^{2}+2 x \right )+6}\) | \(27\) |
risch | \(-\frac {2 i \ln \left (2\right )}{-2 \pi \operatorname {csgn}\left (i x \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i x \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i x \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right )^{3}+\pi \operatorname {csgn}\left (i x \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )\right )^{2} \operatorname {csgn}\left (i x \right )+2 \pi -4 i \ln \left (5\right )-2 i \ln \left (x \right )-2 i \ln \left (x +\frac {5 \ln \left (x \right )}{4}-\frac {1}{2}\right )-12 i}\) | \(147\) |
Input:
int((-5*ln(2)*ln(x)+(-8*x-3)*ln(2))/((5*x*ln(x)+4*x^2-2*x)*ln(-125*x*ln(x) -100*x^2+50*x)^2+(60*x*ln(x)+48*x^2-24*x)*ln(-125*x*ln(x)-100*x^2+50*x)+18 0*x*ln(x)+144*x^2-72*x),x,method=_RETURNVERBOSE)
Output:
ln(2)/(ln(-25*x*(5*ln(x)+4*x-2))+6)
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\log \left (2\right )}{\log \left (-100 \, x^{2} - 125 \, x \log \left (x\right ) + 50 \, x\right ) + 6} \] Input:
integrate((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(- 125*x*log(x)-100*x^2+50*x)^2+(60*x*log(x)+48*x^2-24*x)*log(-125*x*log(x)-1 00*x^2+50*x)+180*x*log(x)+144*x^2-72*x),x, algorithm="fricas")
Output:
log(2)/(log(-100*x^2 - 125*x*log(x) + 50*x) + 6)
Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\log {\left (2 \right )}}{\log {\left (- 100 x^{2} - 125 x \log {\left (x \right )} + 50 x \right )} + 6} \] Input:
integrate((-5*ln(2)*ln(x)+(-8*x-3)*ln(2))/((5*x*ln(x)+4*x**2-2*x)*ln(-125* x*ln(x)-100*x**2+50*x)**2+(60*x*ln(x)+48*x**2-24*x)*ln(-125*x*ln(x)-100*x* *2+50*x)+180*x*ln(x)+144*x**2-72*x),x)
Output:
log(2)/(log(-100*x**2 - 125*x*log(x) + 50*x) + 6)
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\log \left (2\right )}{i \, \pi + 2 \, \log \left (5\right ) + \log \left (4 \, x + 5 \, \log \left (x\right ) - 2\right ) + \log \left (x\right ) + 6} \] Input:
integrate((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(- 125*x*log(x)-100*x^2+50*x)^2+(60*x*log(x)+48*x^2-24*x)*log(-125*x*log(x)-1 00*x^2+50*x)+180*x*log(x)+144*x^2-72*x),x, algorithm="maxima")
Output:
log(2)/(I*pi + 2*log(5) + log(4*x + 5*log(x) - 2) + log(x) + 6)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\log \left (2\right )}{\log \left (-100 \, x^{2} - 125 \, x \log \left (x\right ) + 50 \, x\right ) + 6} \] Input:
integrate((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(- 125*x*log(x)-100*x^2+50*x)^2+(60*x*log(x)+48*x^2-24*x)*log(-125*x*log(x)-1 00*x^2+50*x)+180*x*log(x)+144*x^2-72*x),x, algorithm="giac")
Output:
log(2)/(log(-100*x^2 - 125*x*log(x) + 50*x) + 6)
Time = 1.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=\frac {\ln \left (2\right )}{\ln \left (50\,x-125\,x\,\ln \left (x\right )-100\,x^2\right )+6} \] Input:
int(-(log(2)*(8*x + 3) + 5*log(2)*log(x))/(log(50*x - 125*x*log(x) - 100*x ^2)*(60*x*log(x) - 24*x + 48*x^2) - 72*x + log(50*x - 125*x*log(x) - 100*x ^2)^2*(5*x*log(x) - 2*x + 4*x^2) + 180*x*log(x) + 144*x^2),x)
Output:
log(2)/(log(50*x - 125*x*log(x) - 100*x^2) + 6)
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+\left (-24 x+48 x^2+60 x \log (x)\right ) \log \left (50 x-100 x^2-125 x \log (x)\right )+\left (-2 x+4 x^2+5 x \log (x)\right ) \log ^2\left (50 x-100 x^2-125 x \log (x)\right )} \, dx=-\frac {\mathrm {log}\left (-125 \,\mathrm {log}\left (x \right ) x -100 x^{2}+50 x \right ) \mathrm {log}\left (2\right )}{6 \,\mathrm {log}\left (-125 \,\mathrm {log}\left (x \right ) x -100 x^{2}+50 x \right )+36} \] Input:
int((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(-125*x* log(x)-100*x^2+50*x)^2+(60*x*log(x)+48*x^2-24*x)*log(-125*x*log(x)-100*x^2 +50*x)+180*x*log(x)+144*x^2-72*x),x)
Output:
( - log( - 125*log(x)*x - 100*x**2 + 50*x)*log(2))/(6*(log( - 125*log(x)*x - 100*x**2 + 50*x) + 6))