Integrand size = 47, antiderivative size = 20 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=\log \left (\log \left (\frac {3 x}{(2+x-\log (3)) (5+\log (5))}\right )\right ) \]
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=-\frac {(2-\log (3)) \log \left (\log \left (\frac {3 x}{(2+x-\log (3)) (5+\log (5))}\right )\right )}{-2+\log (3)} \]
Integrate[(-2 + Log[3])/((-2*x - x^2 + x*Log[3])*Log[(-3*x)/(-10 - 5*x + 5 *Log[3] + (-2 - x + Log[3])*Log[5])]),x]
Time = 0.36 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6, 27, 25, 2026, 2971, 2962, 2739, 14}
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.
\(\displaystyle \int \frac {\log (3)-2}{\left (-x^2-2 x+x \log (3)\right ) \log \left (-\frac {3 x}{-5 x+\log (5) (-x-2+\log (3))-10+5 \log (3)}\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\log (3)-2}{\left (x (\log (3)-2)-x^2\right ) \log \left (-\frac {3 x}{-5 x+\log (5) (-x-2+\log (3))-10+5 \log (3)}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\left ((2-\log (3)) \int -\frac {1}{\left (x^2+(2-\log (3)) x\right ) \log \left (\frac {3 x}{5 x+(x-\log (3)+2) \log (5)+5 (2-\log (3))}\right )}dx\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle (2-\log (3)) \int \frac {1}{\left (x^2+(2-\log (3)) x\right ) \log \left (\frac {3 x}{5 x+(x-\log (3)+2) \log (5)+5 (2-\log (3))}\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle (2-\log (3)) \int \frac {1}{x (x-\log (3)+2) \log \left (\frac {3 x}{5 x+(x-\log (3)+2) \log (5)+5 (2-\log (3))}\right )}dx\) |
\(\Big \downarrow \) 2971 |
\(\displaystyle (2-\log (3)) \int \frac {1}{x (x-\log (3)+2) \log \left (\frac {3 x}{(5+\log (5)) x+(2-\log (3)) (5+\log (5))}\right )}dx\) |
\(\Big \downarrow \) 2962 |
\(\displaystyle \int \frac {(5+\log (5)) (x+2-\log (3))}{x \log \left (\frac {3 x}{(5+\log (5)) (x+2-\log (3))}\right )}d\frac {x}{(5+\log (5)) (x+2-\log (3))}\) |
\(\Big \downarrow \) 2739 |
\(\displaystyle \int \frac {(5+\log (5)) (x+2-\log (3))}{x}d\log \left (\frac {3 x}{(5+\log (5)) (x+2-\log (3))}\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \log \left (\log \left (\frac {3 x}{(5+\log (5)) (x+2-\log (3))}\right )\right )\) |
Int[(-2 + Log[3])/((-2*x - x^2 + x*Log[3])*Log[(-3*x)/(-10 - 5*x + 5*Log[3 ] + (-2 - x + Log[3])*Log[5])]),x]
3.1.49.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1/( b*n) Subst[Int[x^p, x], x, a + b*Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p} , x]
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Sy mbol] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q Subst[Int[x^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && EqQ[n + mn, 0] && IGt Q[n, 0] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && I ntegersQ[m, q]
Int[((A_.) + Log[(e_.)*((u_)/(v_))^(n_.)]*(B_.))^(p_.)*(w_)^(m_.)*(y_)^(q_. ), x_Symbol] :> Int[ExpandToSum[w, x]^m*ExpandToSum[y, x]^q*(A + B*Log[e*(E xpandToSum[u, x]/ExpandToSum[v, x])^n])^p, x] /; FreeQ[{e, A, B, m, n, p, q }, x] && LinearQ[{u, v, w, y}, x] && !LinearMatchQ[{u, v, w, y}, x]
Time = 0.44 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35
method | result | size |
norman | \(\ln \left (\ln \left (-\frac {3 x}{\left (\ln \left (3\right )-x -2\right ) \ln \left (5\right )+5 \ln \left (3\right )-5 x -10}\right )\right )\) | \(27\) |
risch | \(\ln \left (\ln \left (-\frac {3 x}{\left (\ln \left (3\right )-x -2\right ) \ln \left (5\right )+5 \ln \left (3\right )-5 x -10}\right )\right )\) | \(27\) |
parallelrisch | \(\ln \left (\ln \left (-\frac {3 x}{\ln \left (3\right ) \ln \left (5\right )-x \ln \left (5\right )+5 \ln \left (3\right )-2 \ln \left (5\right )-5 x -10}\right )\right )\) | \(31\) |
default | \(-\frac {3 \left (\ln \left (3\right )-2\right )^{2} \left (-\frac {5 \ln \left (\ln \left (\frac {3}{5+\ln \left (5\right )}+\frac {3 \ln \left (3\right )-6}{\left (2+x -\ln \left (3\right )\right ) \left (5+\ln \left (5\right )\right )}\right )\right )}{3 \left (\ln \left (3\right )^{2}-4 \ln \left (3\right )+4\right )}-\frac {\ln \left (5\right ) \ln \left (\ln \left (\frac {3}{5+\ln \left (5\right )}+\frac {3 \ln \left (3\right )-6}{\left (2+x -\ln \left (3\right )\right ) \left (5+\ln \left (5\right )\right )}\right )\right )}{3 \left (\ln \left (3\right )^{2}-4 \ln \left (3\right )+4\right )}\right )}{5+\ln \left (5\right )}\) | \(110\) |
derivativedivides | \(\frac {3 \left (2-\ln \left (3\right )\right ) \left (\ln \left (3\right )-2\right ) \left (-\frac {5 \ln \left (\ln \left (\frac {3}{5+\ln \left (5\right )}+\frac {3 \ln \left (3\right )-6}{\left (2+x -\ln \left (3\right )\right ) \left (5+\ln \left (5\right )\right )}\right )\right )}{3 \left (\ln \left (3\right )^{2}-4 \ln \left (3\right )+4\right )}-\frac {\ln \left (5\right ) \ln \left (\ln \left (\frac {3}{5+\ln \left (5\right )}+\frac {3 \ln \left (3\right )-6}{\left (2+x -\ln \left (3\right )\right ) \left (5+\ln \left (5\right )\right )}\right )\right )}{3 \left (\ln \left (3\right )^{2}-4 \ln \left (3\right )+4\right )}\right )}{5+\ln \left (5\right )}\) | \(114\) |
int((ln(3)-2)/(x*ln(3)-x^2-2*x)/ln(-3*x/((ln(3)-x-2)*ln(5)+5*ln(3)-5*x-10) ),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=\log \left (\log \left (\frac {3 \, x}{{\left (x - \log \left (3\right ) + 2\right )} \log \left (5\right ) + 5 \, x - 5 \, \log \left (3\right ) + 10}\right )\right ) \]
integrate((log(3)-2)/(x*log(3)-x^2-2*x)/log(-3*x/((log(3)-x-2)*log(5)+5*lo g(3)-5*x-10)),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=\log {\left (\log {\left (- \frac {3 x}{- 5 x + \left (- x - 2 + \log {\left (3 \right )}\right ) \log {\left (5 \right )} - 10 + 5 \log {\left (3 \right )}} \right )} \right )} \]
Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=\log \left (-\log \left (3\right ) + \log \left (x - \log \left (3\right ) + 2\right ) - \log \left (x\right ) + \log \left (\log \left (5\right ) + 5\right )\right ) \]
integrate((log(3)-2)/(x*log(3)-x^2-2*x)/log(-3*x/((log(3)-x-2)*log(5)+5*lo g(3)-5*x-10)),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=\log \left (\log \left (3\right ) - \log \left (x \log \left (5\right ) - \log \left (5\right ) \log \left (3\right ) + 5 \, x + 2 \, \log \left (5\right ) - 5 \, \log \left (3\right ) + 10\right ) + \log \left (x\right )\right ) \]
integrate((log(3)-2)/(x*log(3)-x^2-2*x)/log(-3*x/((log(3)-x-2)*log(5)+5*lo g(3)-5*x-10)),x, algorithm=\
Time = 12.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-2+\log (3)}{\left (-2 x-x^2+x \log (3)\right ) \log \left (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)}\right )} \, dx=\ln \left (\ln \left (\frac {3\,x}{5\,x-5\,\ln \left (3\right )+\ln \left (5\right )\,\left (x-\ln \left (3\right )+2\right )+10}\right )\right ) \]