Integrand size = 76, antiderivative size = 19 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=4 x \left (1+\log \left (3 (3+2 x) \left (\frac {1}{x}+\log (2)\right )\right )\right ) \] Output:
4*(ln((1/x+ln(2))*(6*x+9))+1)*x
Leaf count is larger than twice the leaf count of optimal. \(44\) vs. \(2(19)=38\).
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.32 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=\frac {4 x \left (-\log (4)+\log (2) \log (8)+\log (2) (-2+\log (8)) \log \left (6+\frac {9}{x}+x \log (64)+\log (512)\right )\right )}{\log (2) (-2+\log (8))} \] Input:
Integrate[(8*x + (12*x + 16*x^2)*Log[2] + (12 + 8*x + (12*x + 8*x^2)*Log[2 ])*Log[(9 + 6*x + (9*x + 6*x^2)*Log[2])/x])/(3 + 2*x + (3*x + 2*x^2)*Log[2 ]),x]
Output:
(4*x*(-Log[4] + Log[2]*Log[8] + Log[2]*(-2 + Log[8])*Log[6 + 9/x + x*Log[6 4] + Log[512]]))/(Log[2]*(-2 + Log[8]))
Leaf count is larger than twice the leaf count of optimal. \(78\) vs. \(2(19)=38\).
Time = 0.47 (sec) , antiderivative size = 78, normalized size of antiderivative = 4.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {7239, 27, 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.
\(\displaystyle \int \frac {\left (\left (8 x^2+12 x\right ) \log (2)+8 x+12\right ) \log \left (\frac {\left (6 x^2+9 x\right ) \log (2)+6 x+9}{x}\right )+\left (16 x^2+12 x\right ) \log (2)+8 x}{\left (2 x^2+3 x\right ) \log (2)+2 x+3} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int 4 \left (\frac {x (x \log (16)+2+\log (8))}{(2 x+3) (x \log (2)+1)}+\log \left (\frac {3 (2 x+3) (x \log (2)+1)}{x}\right )\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \left (\frac {x (\log (16) x+\log (8)+2)}{(2 x+3) (\log (2) x+1)}+\log \left (\frac {3 (2 x+3) (\log (2) x+1)}{x}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (-x-\frac {\left (3 \log ^2(2)-\log (4)\right ) \log (x \log (2)+1)}{\log ^2(2) (2-\log (8))}+x \log \left (\frac {3 (2 x+3) (x \log (2)+1)}{x}\right )+\frac {x \log (16)}{\log (4)}-\frac {\log (x \log (2)+1)}{\log (2)}\right )\) |
Input:
Int[(8*x + (12*x + 16*x^2)*Log[2] + (12 + 8*x + (12*x + 8*x^2)*Log[2])*Log [(9 + 6*x + (9*x + 6*x^2)*Log[2])/x])/(3 + 2*x + (3*x + 2*x^2)*Log[2]),x]
Output:
4*(-x + (x*Log[16])/Log[4] - Log[1 + x*Log[2]]/Log[2] - ((3*Log[2]^2 - Log [4])*Log[1 + x*Log[2]])/(Log[2]^2*(2 - Log[8])) + x*Log[(3*(3 + 2*x)*(1 + x*Log[2]))/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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58
method | result | size |
norman | \(4 x +4 x \ln \left (\frac {\left (6 x^{2}+9 x \right ) \ln \left (2\right )+6 x +9}{x}\right )\) | \(30\) |
risch | \(4 x +4 x \ln \left (\frac {\left (6 x^{2}+9 x \right ) \ln \left (2\right )+6 x +9}{x}\right )\) | \(30\) |
default | \(4 x +4 x \ln \left (3\right )+4 x \ln \left (\frac {2 x^{2} \ln \left (2\right )+3 x \ln \left (2\right )+2 x +3}{x}\right )\) | \(35\) |
parts | \(4 x +4 x \ln \left (3\right )+4 x \ln \left (\frac {2 x^{2} \ln \left (2\right )+3 x \ln \left (2\right )+2 x +3}{x}\right )\) | \(35\) |
parallelrisch | \(-\frac {-16 \ln \left (\frac {\left (6 x^{2}+9 x \right ) \ln \left (2\right )+6 x +9}{x}\right ) x \ln \left (2\right )^{2}-16 x \ln \left (2\right )^{2}+48 \ln \left (2\right )^{2}+32 \ln \left (2\right )}{4 \ln \left (2\right )^{2}}\) | \(54\) |
Input:
int((((8*x^2+12*x)*ln(2)+8*x+12)*ln(((6*x^2+9*x)*ln(2)+6*x+9)/x)+(16*x^2+1 2*x)*ln(2)+8*x)/((2*x^2+3*x)*ln(2)+2*x+3),x,method=_RETURNVERBOSE)
Output:
4*x+4*x*ln(((6*x^2+9*x)*ln(2)+6*x+9)/x)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=4 \, x \log \left (\frac {3 \, {\left ({\left (2 \, x^{2} + 3 \, x\right )} \log \left (2\right ) + 2 \, x + 3\right )}}{x}\right ) + 4 \, x \] Input:
integrate((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+ (16*x^2+12*x)*log(2)+8*x)/((2*x^2+3*x)*log(2)+2*x+3),x, algorithm="fricas" )
Output:
4*x*log(3*((2*x^2 + 3*x)*log(2) + 2*x + 3)/x) + 4*x
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=4 x \log {\left (\frac {6 x + \left (6 x^{2} + 9 x\right ) \log {\left (2 \right )} + 9}{x} \right )} + 4 x \] Input:
integrate((((8*x**2+12*x)*ln(2)+8*x+12)*ln(((6*x**2+9*x)*ln(2)+6*x+9)/x)+( 16*x**2+12*x)*ln(2)+8*x)/((2*x**2+3*x)*ln(2)+2*x+3),x)
Output:
4*x*log((6*x + (6*x**2 + 9*x)*log(2) + 9)/x) + 4*x
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (19) = 38\).
Time = 0.18 (sec) , antiderivative size = 189, normalized size of antiderivative = 9.95 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=4 \, {\left (\frac {2 \, x}{\log \left (2\right )} + \frac {4 \, \log \left (x \log \left (2\right ) + 1\right )}{3 \, \log \left (2\right )^{3} - 2 \, \log \left (2\right )^{2}} - \frac {9 \, \log \left (2 \, x + 3\right )}{3 \, \log \left (2\right ) - 2}\right )} \log \left (2\right ) - 6 \, {\left (\frac {2 \, \log \left (x \log \left (2\right ) + 1\right )}{3 \, \log \left (2\right )^{2} - 2 \, \log \left (2\right )} - \frac {3 \, \log \left (2 \, x + 3\right )}{3 \, \log \left (2\right ) - 2}\right )} \log \left (2\right ) + \frac {2 \, {\left (2 \, x {\left (\log \left (3\right ) - 1\right )} \log \left (2\right ) - 2 \, x \log \left (2\right ) \log \left (x\right ) + 2 \, {\left (x \log \left (2\right ) + 1\right )} \log \left (x \log \left (2\right ) + 1\right ) + {\left (2 \, x \log \left (2\right ) + 3 \, \log \left (2\right )\right )} \log \left (2 \, x + 3\right )\right )}}{\log \left (2\right )} - \frac {8 \, \log \left (x \log \left (2\right ) + 1\right )}{3 \, \log \left (2\right )^{2} - 2 \, \log \left (2\right )} + \frac {12 \, \log \left (2 \, x + 3\right )}{3 \, \log \left (2\right ) - 2} \] Input:
integrate((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+ (16*x^2+12*x)*log(2)+8*x)/((2*x^2+3*x)*log(2)+2*x+3),x, algorithm="maxima" )
Output:
4*(2*x/log(2) + 4*log(x*log(2) + 1)/(3*log(2)^3 - 2*log(2)^2) - 9*log(2*x + 3)/(3*log(2) - 2))*log(2) - 6*(2*log(x*log(2) + 1)/(3*log(2)^2 - 2*log(2 )) - 3*log(2*x + 3)/(3*log(2) - 2))*log(2) + 2*(2*x*(log(3) - 1)*log(2) - 2*x*log(2)*log(x) + 2*(x*log(2) + 1)*log(x*log(2) + 1) + (2*x*log(2) + 3*l og(2))*log(2*x + 3))/log(2) - 8*log(x*log(2) + 1)/(3*log(2)^2 - 2*log(2)) + 12*log(2*x + 3)/(3*log(2) - 2)
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=4 \, x \log \left (6 \, x^{2} \log \left (2\right ) + 9 \, x \log \left (2\right ) + 6 \, x + 9\right ) - 4 \, x \log \left (x\right ) + 4 \, x \] Input:
integrate((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+ (16*x^2+12*x)*log(2)+8*x)/((2*x^2+3*x)*log(2)+2*x+3),x, algorithm="giac")
Output:
4*x*log(6*x^2*log(2) + 9*x*log(2) + 6*x + 9) - 4*x*log(x) + 4*x
Time = 3.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=4\,x\,\left (\ln \left (\frac {6\,x+\ln \left (2\right )\,\left (6\,x^2+9\,x\right )+9}{x}\right )+1\right ) \] Input:
int((8*x + log(2)*(12*x + 16*x^2) + log((6*x + log(2)*(9*x + 6*x^2) + 9)/x )*(8*x + log(2)*(12*x + 8*x^2) + 12))/(2*x + log(2)*(3*x + 2*x^2) + 3),x)
Output:
4*x*(log((6*x + log(2)*(9*x + 6*x^2) + 9)/x) + 1)
Time = 0.49 (sec) , antiderivative size = 78, normalized size of antiderivative = 4.11 \[ \int \frac {8 x+\left (12 x+16 x^2\right ) \log (2)+\left (12+8 x+\left (12 x+8 x^2\right ) \log (2)\right ) \log \left (\frac {9+6 x+\left (9 x+6 x^2\right ) \log (2)}{x}\right )}{3+2 x+\left (3 x+2 x^2\right ) \log (2)} \, dx=\frac {-4 \,\mathrm {log}\left (\mathrm {log}\left (2\right ) x +1\right )-4 \,\mathrm {log}\left (3+2 x \right )+4 \,\mathrm {log}\left (\frac {6 \,\mathrm {log}\left (2\right ) x^{2}+9 \,\mathrm {log}\left (2\right ) x +6 x +9}{x}\right ) \mathrm {log}\left (2\right ) x +4 \,\mathrm {log}\left (\frac {6 \,\mathrm {log}\left (2\right ) x^{2}+9 \,\mathrm {log}\left (2\right ) x +6 x +9}{x}\right )+4 \,\mathrm {log}\left (x \right )+4 \,\mathrm {log}\left (2\right ) x}{\mathrm {log}\left (2\right )} \] Input:
int((((8*x^2+12*x)*log(2)+8*x+12)*log(((6*x^2+9*x)*log(2)+6*x+9)/x)+(16*x^ 2+12*x)*log(2)+8*x)/((2*x^2+3*x)*log(2)+2*x+3),x)
Output:
(4*( - log(log(2)*x + 1) - log(2*x + 3) + log((6*log(2)*x**2 + 9*log(2)*x + 6*x + 9)/x)*log(2)*x + log((6*log(2)*x**2 + 9*log(2)*x + 6*x + 9)/x) + l og(x) + log(2)*x))/log(2)