Integrand size = 133, antiderivative size = 24 \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\frac {x}{2-\log (2)+\frac {1}{4} x (3+x+\log (3 (2+x)))} \]
Time = 10.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\frac {4 x}{8+3 x+x^2-4 \log (2)+x \log (3 (2+x))} \]
Integrate[(64 + 32*x - 12*x^2 - 4*x^3 + (-32 - 16*x)*Log[2])/(128 + 160*x + 98*x^2 + 37*x^3 + 8*x^4 + x^5 + (-128 - 112*x - 40*x^2 - 8*x^3)*Log[2] + (32 + 16*x)*Log[2]^2 + (32*x + 28*x^2 + 10*x^3 + 2*x^4 + (-16*x - 8*x^2)* Log[2])*Log[6 + 3*x] + (2*x^2 + x^3)*Log[6 + 3*x]^2),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.
\(\displaystyle \int \frac {-4 x^3-12 x^2+32 x+(-16 x-32) \log (2)+64}{x^5+8 x^4+37 x^3+98 x^2+\left (x^3+2 x^2\right ) \log ^2(3 x+6)+\left (-8 x^3-40 x^2-112 x-128\right ) \log (2)+\left (2 x^4+10 x^3+28 x^2+\left (-8 x^2-16 x\right ) \log (2)+32 x\right ) \log (3 x+6)+160 x+(16 x+32) \log ^2(2)+128} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 \left (-x^3-3 x^2+4 x (2-\log (2))+8 (2-\log (2))\right )}{(x+2) \left (x^2+3 x+x \log (3 (x+2))+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {x^3+3 x^2-4 (2-\log (2)) x-8 (2-\log (2))}{(x+2) \left (x^2+\log (3 (x+2)) x+3 x+4 (2-\log (2))\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {x^3+3 x^2-4 (2-\log (2)) x-8 (2-\log (2))}{(x+2) \left (x^2+\log (3 (x+2)) x+3 x+4 (2-\log (2))\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (\frac {x^2}{\left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}+\frac {x}{\left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}+\frac {4}{(x+2) \left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}+\frac {-10+\log (16)}{\left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (-\left ((10-\log (16)) \int \frac {1}{\left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}dx\right )+\int \frac {x}{\left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}dx+\int \frac {x^2}{\left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}dx+4 \int \frac {1}{(x+2) \left (x^2+\log (3 (x+2)) x+3 x+8 \left (1-\frac {\log (2)}{2}\right )\right )^2}dx\right )\) |
Int[(64 + 32*x - 12*x^2 - 4*x^3 + (-32 - 16*x)*Log[2])/(128 + 160*x + 98*x ^2 + 37*x^3 + 8*x^4 + x^5 + (-128 - 112*x - 40*x^2 - 8*x^3)*Log[2] + (32 + 16*x)*Log[2]^2 + (32*x + 28*x^2 + 10*x^3 + 2*x^4 + (-16*x - 8*x^2)*Log[2] )*Log[6 + 3*x] + (2*x^2 + x^3)*Log[6 + 3*x]^2),x]
3.24.74.3.1 Defintions of rubi rules used
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 = 1.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
norman | \(-\frac {4 x}{-\ln \left (6+3 x \right ) x -x^{2}+4 \ln \left (2\right )-3 x -8}\) | \(29\) |
risch | \(-\frac {4 x}{-\ln \left (6+3 x \right ) x -x^{2}+4 \ln \left (2\right )-3 x -8}\) | \(29\) |
parallelrisch | \(-\frac {4 x}{-\ln \left (6+3 x \right ) x -x^{2}+4 \ln \left (2\right )-3 x -8}\) | \(29\) |
derivativedivides | \(-\frac {36 x}{-\left (6+3 x \right )^{2}-3 \ln \left (6+3 x \right ) \left (6+3 x \right )+36 \ln \left (2\right )-36+9 x +18 \ln \left (6+3 x \right )}\) | \(45\) |
default | \(-\frac {36 x}{-\left (6+3 x \right )^{2}-3 \ln \left (6+3 x \right ) \left (6+3 x \right )+36 \ln \left (2\right )-36+9 x +18 \ln \left (6+3 x \right )}\) | \(45\) |
int(((-16*x-32)*ln(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*ln(6+3*x)^2+((-8* x^2-16*x)*ln(2)+2*x^4+10*x^3+28*x^2+32*x)*ln(6+3*x)+(16*x+32)*ln(2)^2+(-8* x^3-40*x^2-112*x-128)*ln(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+128),x,method=_R ETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\frac {4 \, x}{x^{2} + x \log \left (3 \, x + 6\right ) + 3 \, x - 4 \, \log \left (2\right ) + 8} \]
integrate(((-16*x-32)*log(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*log(6+3*x) ^2+((-8*x^2-16*x)*log(2)+2*x^4+10*x^3+28*x^2+32*x)*log(6+3*x)+(16*x+32)*lo g(2)^2+(-8*x^3-40*x^2-112*x-128)*log(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+128) ,x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\frac {4 x}{x^{2} + x \log {\left (3 x + 6 \right )} + 3 x - 4 \log {\left (2 \right )} + 8} \]
integrate(((-16*x-32)*ln(2)-4*x**3-12*x**2+32*x+64)/((x**3+2*x**2)*ln(6+3* x)**2+((-8*x**2-16*x)*ln(2)+2*x**4+10*x**3+28*x**2+32*x)*ln(6+3*x)+(16*x+3 2)*ln(2)**2+(-8*x**3-40*x**2-112*x-128)*ln(2)+x**5+8*x**4+37*x**3+98*x**2+ 160*x+128),x)
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\frac {4 \, x}{x^{2} + x {\left (\log \left (3\right ) + 3\right )} + x \log \left (x + 2\right ) - 4 \, \log \left (2\right ) + 8} \]
integrate(((-16*x-32)*log(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*log(6+3*x) ^2+((-8*x^2-16*x)*log(2)+2*x^4+10*x^3+28*x^2+32*x)*log(6+3*x)+(16*x+32)*lo g(2)^2+(-8*x^3-40*x^2-112*x-128)*log(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+128) ,x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\frac {4 \, x}{x^{2} + x \log \left (3 \, x + 6\right ) + 3 \, x - 4 \, \log \left (2\right ) + 8} \]
integrate(((-16*x-32)*log(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*log(6+3*x) ^2+((-8*x^2-16*x)*log(2)+2*x^4+10*x^3+28*x^2+32*x)*log(6+3*x)+(16*x+32)*lo g(2)^2+(-8*x^3-40*x^2-112*x-128)*log(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+128) ,x, algorithm=\
Timed out. \[ \int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx=\int -\frac {\ln \left (2\right )\,\left (16\,x+32\right )-32\,x+12\,x^2+4\,x^3-64}{160\,x+\ln \left (3\,x+6\right )\,\left (32\,x-\ln \left (2\right )\,\left (8\,x^2+16\,x\right )+28\,x^2+10\,x^3+2\,x^4\right )+{\ln \left (2\right )}^2\,\left (16\,x+32\right )-\ln \left (2\right )\,\left (8\,x^3+40\,x^2+112\,x+128\right )+{\ln \left (3\,x+6\right )}^2\,\left (x^3+2\,x^2\right )+98\,x^2+37\,x^3+8\,x^4+x^5+128} \,d x \]
int(-(log(2)*(16*x + 32) - 32*x + 12*x^2 + 4*x^3 - 64)/(160*x + log(3*x + 6)*(32*x - log(2)*(16*x + 8*x^2) + 28*x^2 + 10*x^3 + 2*x^4) + log(2)^2*(16 *x + 32) - log(2)*(112*x + 40*x^2 + 8*x^3 + 128) + log(3*x + 6)^2*(2*x^2 + x^3) + 98*x^2 + 37*x^3 + 8*x^4 + x^5 + 128),x)