Integrand size = 124, antiderivative size = 23 \[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\log \left (\log \left (5 \log \left (-\frac {x^2}{5+x}+(-3+x) x \log (3)\right )\right )\right ) \]
\[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx \]
Integrate[(-10*x - x^2 + (-75 + 20*x + 17*x^2 + 2*x^3)*Log[3])/((-5*x^2 - x^3 + (-75*x - 5*x^2 + 7*x^3 + x^4)*Log[3])*Log[(-x^2 + (-15*x + 2*x^2 + x ^3)*Log[3])/(5 + x)]*Log[5*Log[(-x^2 + (-15*x + 2*x^2 + x^3)*Log[3])/(5 + x)]]),x]
Integrate[(-10*x - x^2 + (-75 + 20*x + 17*x^2 + 2*x^3)*Log[3])/((-5*x^2 - x^3 + (-75*x - 5*x^2 + 7*x^3 + x^4)*Log[3])*Log[(-x^2 + (-15*x + 2*x^2 + x ^3)*Log[3])/(5 + x)]*Log[5*Log[(-x^2 + (-15*x + 2*x^2 + x^3)*Log[3])/(5 + x)]]), 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.
\(\displaystyle \int \frac {-x^2+\left (2 x^3+17 x^2+20 x-75\right ) \log (3)-10 x}{\left (-x^3-5 x^2+\left (x^4+7 x^3-5 x^2-75 x\right ) \log (3)\right ) \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right ) \log \left (5 \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right )\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^2+\left (2 x^3+17 x^2+20 x-75\right ) \log (3)-10 x}{x \left (x^3 \log (3)-x^2 (1-7 \log (3))-5 x (1+\log (3))-75 \log (3)\right ) \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right ) \log \left (5 \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right )\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {(x \log (3)-1-\log (27)) \left (-x^2+\left (2 x^3+17 x^2+20 x-75\right ) \log (3)-10 x\right )}{5 x \left (x^2 (-\log (3))+x (1-\log (9))+15 \log (3)\right ) \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right ) \log \left (5 \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right )\right )}+\frac {-x^2+\left (2 x^3+17 x^2+20 x-75\right ) \log (3)-10 x}{5 x (x+5) \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right ) \log \left (5 \log \left (\frac {\left (x^3+2 x^2-15 x\right ) \log (3)-x^2}{x+5}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} (1-\log (2187)) \int \frac {1}{\log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx-\frac {1}{5} (1-7 \log (3)) \int \frac {1}{\log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx+\int \frac {1}{(-x-5) \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx+(1+\log (27)) \int \frac {1}{x \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx-3 \log (3) \int \frac {1}{x \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx+\frac {1}{5} \left (1-\frac {1-\log (9)}{\sqrt {60 \log ^2(3)+(-1+\log (9))^2}}\right ) \left (50 \log ^2(3)-\log (3) (17 \log (9)+12 \log (27))-(2-\log (81)) \log (243)\right ) \int \frac {1}{\left (-2 \log (3) x-\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}-\log (9)+1\right ) \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx+\frac {1}{5} \left (1+\frac {1-\log (9)}{\sqrt {60 \log ^2(3)+(-1+\log (9))^2}}\right ) \left (50 \log ^2(3)-\log (3) (17 \log (9)+12 \log (27))-(2-\log (81)) \log (243)\right ) \int \frac {1}{\left (-2 \log (3) x+\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}-\log (9)+1\right ) \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx+\frac {2 \log (3) \left (36 \log ^2(3)+(1+\log (9)) (1+\log (27))-\log (3) (7+6 \log (9)+10 \log (27))\right ) \int \frac {1}{\left (-\log (9) x+\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}-\log (9)+1\right ) \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx}{\sqrt {60 \log ^2(3)+(-1+\log (9))^2}}+\frac {2 \log (3) \left (36 \log ^2(3)+(1+\log (9)) (1+\log (27))-\log (3) (7+6 \log (9)+10 \log (27))\right ) \int \frac {1}{\left (\log (9) x+\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}+\log (9)-1\right ) \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right ) \log \left (5 \log \left (\frac {x \left (\log (3) x^2-(1-\log (9)) x-15 \log (3)\right )}{x+5}\right )\right )}dx}{\sqrt {60 \log ^2(3)+(-1+\log (9))^2}}\) |
Int[(-10*x - x^2 + (-75 + 20*x + 17*x^2 + 2*x^3)*Log[3])/((-5*x^2 - x^3 + (-75*x - 5*x^2 + 7*x^3 + x^4)*Log[3])*Log[(-x^2 + (-15*x + 2*x^2 + x^3)*Lo g[3])/(5 + x)]*Log[5*Log[(-x^2 + (-15*x + 2*x^2 + x^3)*Log[3])/(5 + x)]]), x]
3.28.56.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 7.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43
method | result | size |
default | \(\ln \left (\ln \left (5\right )+\ln \left (\ln \left (\frac {x \left (x^{2} \ln \left (3\right )+2 x \ln \left (3\right )-15 \ln \left (3\right )-x \right )}{5+x}\right )\right )\right )\) | \(33\) |
parallelrisch | \(\ln \left (\ln \left (5 \ln \left (\frac {\left (x^{3}+2 x^{2}-15 x \right ) \ln \left (3\right )-x^{2}}{5+x}\right )\right )\right )\) | \(33\) |
int(((2*x^3+17*x^2+20*x-75)*ln(3)-x^2-10*x)/((x^4+7*x^3-5*x^2-75*x)*ln(3)- x^3-5*x^2)/ln(((x^3+2*x^2-15*x)*ln(3)-x^2)/(5+x))/ln(5*ln(((x^3+2*x^2-15*x )*ln(3)-x^2)/(5+x))),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\log \left (\log \left (5 \, \log \left (-\frac {x^{2} - {\left (x^{3} + 2 \, x^{2} - 15 \, x\right )} \log \left (3\right )}{x + 5}\right )\right )\right ) \]
integrate(((2*x^3+17*x^2+20*x-75)*log(3)-x^2-10*x)/((x^4+7*x^3-5*x^2-75*x) *log(3)-x^3-5*x^2)/log(((x^3+2*x^2-15*x)*log(3)-x^2)/(5+x))/log(5*log(((x^ 3+2*x^2-15*x)*log(3)-x^2)/(5+x))),x, algorithm=\
Time = 0.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\log {\left (\log {\left (5 \log {\left (\frac {- x^{2} + \left (x^{3} + 2 x^{2} - 15 x\right ) \log {\left (3 \right )}}{x + 5} \right )} \right )} \right )} \]
integrate(((2*x**3+17*x**2+20*x-75)*ln(3)-x**2-10*x)/((x**4+7*x**3-5*x**2- 75*x)*ln(3)-x**3-5*x**2)/ln(((x**3+2*x**2-15*x)*ln(3)-x**2)/(5+x))/ln(5*ln (((x**3+2*x**2-15*x)*ln(3)-x**2)/(5+x))),x)
Time = 0.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\log \left (\log \left (5\right ) + \log \left (\log \left (x^{2} \log \left (3\right ) + x {\left (2 \, \log \left (3\right ) - 1\right )} - 15 \, \log \left (3\right )\right ) - \log \left (x + 5\right ) + \log \left (x\right )\right )\right ) \]
integrate(((2*x^3+17*x^2+20*x-75)*log(3)-x^2-10*x)/((x^4+7*x^3-5*x^2-75*x) *log(3)-x^3-5*x^2)/log(((x^3+2*x^2-15*x)*log(3)-x^2)/(5+x))/log(5*log(((x^ 3+2*x^2-15*x)*log(3)-x^2)/(5+x))),x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\log \left (\log \left (5 \, \log \left (x^{3} \log \left (3\right ) + 2 \, x^{2} \log \left (3\right ) - x^{2} - 15 \, x \log \left (3\right )\right ) - 5 \, \log \left (x + 5\right )\right )\right ) \]
integrate(((2*x^3+17*x^2+20*x-75)*log(3)-x^2-10*x)/((x^4+7*x^3-5*x^2-75*x) *log(3)-x^3-5*x^2)/log(((x^3+2*x^2-15*x)*log(3)-x^2)/(5+x))/log(5*log(((x^ 3+2*x^2-15*x)*log(3)-x^2)/(5+x))),x, algorithm=\
Time = 14.43 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx=\ln \left (\ln \left (5\,\ln \left (\frac {\ln \left (3\right )\,\left (x^3+2\,x^2-15\,x\right )-x^2}{x+5}\right )\right )\right ) \]