Integrand size = 97, antiderivative size = 27 \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=2+\log (3) \left (1-x+\frac {\log (20 x)}{5-x-3 x \log (2)}\right ) \]
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=\log (3) \left (-x+\frac {\log (20 x)}{5-x (1+\log (8))}\right ) \]
Integrate[((5 - 26*x + 10*x^2 - x^3 + (-3*x + 30*x^2 - 6*x^3)*Log[2] - 9*x ^3*Log[2]^2)*Log[3] + (x + 3*x*Log[2])*Log[3]*Log[20*x])/(25*x - 10*x^2 + x^3 + (-30*x^2 + 6*x^3)*Log[2] + 9*x^3*Log[2]^2),x]
Time = 0.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6, 2026, 7277, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {\log (3) \left (-x^3-9 x^3 \log ^2(2)+10 x^2+\left (-6 x^3+30 x^2-3 x\right ) \log (2)-26 x+5\right )+\log (3) (x+3 x \log (2)) \log (20 x)}{x^3+9 x^3 \log ^2(2)-10 x^2+\left (6 x^3-30 x^2\right ) \log (2)+25 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\log (3) \left (-x^3-9 x^3 \log ^2(2)+10 x^2+\left (-6 x^3+30 x^2-3 x\right ) \log (2)-26 x+5\right )+\log (3) (x+3 x \log (2)) \log (20 x)}{x^3 \left (1+9 \log ^2(2)\right )-10 x^2+\left (6 x^3-30 x^2\right ) \log (2)+25 x}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\log (3) \left (-x^3-9 x^3 \log ^2(2)+10 x^2+\left (-6 x^3+30 x^2-3 x\right ) \log (2)-26 x+5\right )+\log (3) (x+3 x \log (2)) \log (20 x)}{x \left (x^2 (1+\log (8))^2-10 x (1+\log (8))+25\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 (1+\log (8))^2 \int \frac {\log (3) \left (-9 \log ^2(2) x^3-x^3+10 x^2-26 x-3 \left (2 x^3-10 x^2+x\right ) \log (2)+5\right )+x \log (3) (1+\log (8)) \log (20 x)}{4 x (1+\log (8))^2 (5-x (1+\log (8)))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\log (3) \left (-x^3-9 x^3 \log ^2(2)+10 x^2-3 \left (2 x^3-10 x^2+x\right ) \log (2)-26 x+5\right )+x \log (3) (1+\log (8)) \log (20 x)}{x (5-x (1+\log (8)))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log (3) \left (x^2 (1+\log (8))-5 x+1\right )}{x (5-x (1+\log (8)))}+\frac {\log (3) (1+\log (8)) \log (20 x)}{(5-x (1+\log (8)))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \log (3) (1+\log (8)) \log (20 x)}{5 (5-x (1+\log (8)))}-x \log (3)+\frac {1}{5} \log (3) \log (x)\) |
Int[((5 - 26*x + 10*x^2 - x^3 + (-3*x + 30*x^2 - 6*x^3)*Log[2] - 9*x^3*Log [2]^2)*Log[3] + (x + 3*x*Log[2])*Log[3]*Log[20*x])/(25*x - 10*x^2 + x^3 + (-30*x^2 + 6*x^3)*Log[2] + 9*x^3*Log[2]^2),x]
3.10.76.3.1 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[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-\frac {\ln \left (3\right ) \ln \left (20 x \right )}{3 x \ln \left (2\right )+x -5}-x \ln \left (3\right )\) | \(25\) |
norman | \(\frac {-\ln \left (3\right ) \ln \left (20 x \right )+\left (-3 \ln \left (2\right ) \ln \left (3\right )-\ln \left (3\right )\right ) x^{2}+5 x \ln \left (3\right )}{3 x \ln \left (2\right )+x -5}\) | \(41\) |
parallelrisch | \(\frac {-15 \ln \left (3\right ) \ln \left (2\right ) x^{2}-5 x^{2} \ln \left (3\right )+25 x \ln \left (3\right )-5 \ln \left (3\right ) \ln \left (20 x \right )}{15 x \ln \left (2\right )+5 x -25}\) | \(43\) |
parts | \(\left (3 \ln \left (2\right )+1\right ) \ln \left (3\right ) \left (\frac {\ln \left (-100+20 \left (3 \ln \left (2\right )+1\right ) x \right )}{15 \ln \left (2\right )+5}-\frac {4 \ln \left (20 x \right ) x}{60 x \ln \left (2\right )+20 x -100}\right )-\ln \left (3\right ) \left (x -\frac {\ln \left (x \right )}{5}+\frac {\left (\frac {3 \ln \left (2\right )}{5}+\frac {1}{5}\right ) \ln \left (3 x \ln \left (2\right )+x -5\right )}{3 \ln \left (2\right )+1}\right )\) | \(87\) |
derivativedivides | \(\frac {\ln \left (3\right ) \left (\left (1200 \ln \left (2\right )+400\right ) \left (\frac {\ln \left (-100+20 \left (3 \ln \left (2\right )+1\right ) x \right )}{300 \ln \left (2\right )+100}-\frac {\ln \left (20 x \right ) x}{5 \left (60 x \ln \left (2\right )+20 x -100\right )}\right )-20 x -\frac {\left (12 \ln \left (2\right )+4\right ) \ln \left (60 x \ln \left (2\right )+20 x -100\right )}{3 \ln \left (2\right )+1}+4 \ln \left (20 x \right )\right )}{20}\) | \(91\) |
default | \(\frac {\ln \left (3\right ) \left (\left (1200 \ln \left (2\right )+400\right ) \left (\frac {\ln \left (-100+20 \left (3 \ln \left (2\right )+1\right ) x \right )}{300 \ln \left (2\right )+100}-\frac {\ln \left (20 x \right ) x}{5 \left (60 x \ln \left (2\right )+20 x -100\right )}\right )-20 x -\frac {\left (12 \ln \left (2\right )+4\right ) \ln \left (60 x \ln \left (2\right )+20 x -100\right )}{3 \ln \left (2\right )+1}+4 \ln \left (20 x \right )\right )}{20}\) | \(91\) |
int(((3*x*ln(2)+x)*ln(3)*ln(20*x)+(-9*x^3*ln(2)^2+(-6*x^3+30*x^2-3*x)*ln(2 )-x^3+10*x^2-26*x+5)*ln(3))/(9*x^3*ln(2)^2+(6*x^3-30*x^2)*ln(2)+x^3-10*x^2 +25*x),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=-\frac {{\left (3 \, x^{2} \log \left (2\right ) + x^{2} - 5 \, x\right )} \log \left (3\right ) + \log \left (3\right ) \log \left (20 \, x\right )}{3 \, x \log \left (2\right ) + x - 5} \]
integrate(((3*x*log(2)+x)*log(3)*log(20*x)+(-9*x^3*log(2)^2+(-6*x^3+30*x^2 -3*x)*log(2)-x^3+10*x^2-26*x+5)*log(3))/(9*x^3*log(2)^2+(6*x^3-30*x^2)*log (2)+x^3-10*x^2+25*x),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=- x \log {\left (3 \right )} - \frac {\log {\left (3 \right )} \log {\left (20 x \right )}}{x + 3 x \log {\left (2 \right )} - 5} \]
integrate(((3*x*ln(2)+x)*ln(3)*ln(20*x)+(-9*x**3*ln(2)**2+(-6*x**3+30*x**2 -3*x)*ln(2)-x**3+10*x**2-26*x+5)*ln(3))/(9*x**3*ln(2)**2+(6*x**3-30*x**2)* ln(2)+x**3-10*x**2+25*x),x)
Leaf count of result is larger than twice the leaf count of optimal. 678 vs. \(2 (24) = 48\).
Time = 0.19 (sec) , antiderivative size = 678, normalized size of antiderivative = 25.11 \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=\text {Too large to display} \]
integrate(((3*x*log(2)+x)*log(3)*log(20*x)+(-9*x^3*log(2)^2+(-6*x^3+30*x^2 -3*x)*log(2)-x^3+10*x^2-26*x+5)*log(3))/(9*x^3*log(2)^2+(6*x^3-30*x^2)*log (2)+x^3-10*x^2+25*x),x, algorithm=\
-9*(x/(9*log(2)^2 + 6*log(2) + 1) + 10*log(x*(3*log(2) + 1) - 5)/(27*log(2 )^3 + 27*log(2)^2 + 9*log(2) + 1) + 25/(135*log(2)^3 - (81*log(2)^4 + 108* log(2)^3 + 54*log(2)^2 + 12*log(2) + 1)*x + 135*log(2)^2 + 45*log(2) + 5)) *log(3)*log(2)^2 - 6*(x/(9*log(2)^2 + 6*log(2) + 1) + 10*log(x*(3*log(2) + 1) - 5)/(27*log(2)^3 + 27*log(2)^2 + 9*log(2) + 1) + 25/(135*log(2)^3 - ( 81*log(2)^4 + 108*log(2)^3 + 54*log(2)^2 + 12*log(2) + 1)*x + 135*log(2)^2 + 45*log(2) + 5))*log(3)*log(2) + 30*(log(x*(3*log(2) + 1) - 5)/(9*log(2) ^2 + 6*log(2) + 1) - 5/((27*log(2)^3 + 27*log(2)^2 + 9*log(2) + 1)*x - 45* log(2)^2 - 30*log(2) - 5))*log(3)*log(2) + 3/5*(log(x*(3*log(2) + 1) - 5)/ (3*log(2) + 1) - log(x)/(3*log(2) + 1))*log(3)*log(2) - (x/(9*log(2)^2 + 6 *log(2) + 1) + 10*log(x*(3*log(2) + 1) - 5)/(27*log(2)^3 + 27*log(2)^2 + 9 *log(2) + 1) + 25/(135*log(2)^3 - (81*log(2)^4 + 108*log(2)^3 + 54*log(2)^ 2 + 12*log(2) + 1)*x + 135*log(2)^2 + 45*log(2) + 5))*log(3) + 10*(log(x*( 3*log(2) + 1) - 5)/(9*log(2)^2 + 6*log(2) + 1) - 5/((27*log(2)^3 + 27*log( 2)^2 + 9*log(2) + 1)*x - 45*log(2)^2 - 30*log(2) - 5))*log(3) + 1/5*(log(x *(3*log(2) + 1) - 5)/(3*log(2) + 1) - log(x)/(3*log(2) + 1))*log(3) - 1/5* (5/(x*(3*log(2) + 1) - 5) + log(x*(3*log(2) + 1) - 5) - log(x))*log(3) - 3 *log(3)*log(2)*log(20*x)/((9*log(2)^2 + 6*log(2) + 1)*x - 15*log(2) - 5) + 3*log(3)*log(2)/((9*log(2)^2 + 6*log(2) + 1)*x - 15*log(2) - 5) - log(3)* log(20*x)/((9*log(2)^2 + 6*log(2) + 1)*x - 15*log(2) - 5) + 26*log(3)/(...
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=-x \log \left (3\right ) - \frac {2 \, \log \left (3\right ) \log \left (2\right )}{3 \, x \log \left (2\right ) + x - 5} - \frac {\log \left (3\right ) \log \left (5 \, x\right )}{3 \, x \log \left (2\right ) + x - 5} \]
integrate(((3*x*log(2)+x)*log(3)*log(20*x)+(-9*x^3*log(2)^2+(-6*x^3+30*x^2 -3*x)*log(2)-x^3+10*x^2-26*x+5)*log(3))/(9*x^3*log(2)^2+(6*x^3-30*x^2)*log (2)+x^3-10*x^2+25*x),x, algorithm=\
Timed out. \[ \int \frac {\left (5-26 x+10 x^2-x^3+\left (-3 x+30 x^2-6 x^3\right ) \log (2)-9 x^3 \log ^2(2)\right ) \log (3)+(x+3 x \log (2)) \log (3) \log (20 x)}{25 x-10 x^2+x^3+\left (-30 x^2+6 x^3\right ) \log (2)+9 x^3 \log ^2(2)} \, dx=-\int \frac {\ln \left (3\right )\,\left (26\,x+9\,x^3\,{\ln \left (2\right )}^2+\ln \left (2\right )\,\left (6\,x^3-30\,x^2+3\,x\right )-10\,x^2+x^3-5\right )-\ln \left (20\,x\right )\,\ln \left (3\right )\,\left (x+3\,x\,\ln \left (2\right )\right )}{25\,x+9\,x^3\,{\ln \left (2\right )}^2-\ln \left (2\right )\,\left (30\,x^2-6\,x^3\right )-10\,x^2+x^3} \,d x \]
int(-(log(3)*(26*x + 9*x^3*log(2)^2 + log(2)*(3*x - 30*x^2 + 6*x^3) - 10*x ^2 + x^3 - 5) - log(20*x)*log(3)*(x + 3*x*log(2)))/(25*x + 9*x^3*log(2)^2 - log(2)*(30*x^2 - 6*x^3) - 10*x^2 + x^3),x)