Integrand size = 283, antiderivative size = 30 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=x+\frac {6 (4-x)^2}{x \left (x+\log \left (\left (1-\frac {16}{x}+\log (3)\right )^2\right )\right )} \]
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=x+\frac {6 (-4+x)^2}{x \left (x+\log \left (\frac {(-16+x+x \log (3))^2}{x^2}\right )\right )} \]
Integrate[(-3072 + 4608*x - 1152*x^2 + 48*x^3 - 16*x^4 + x^5 + (-192*x^2 + 48*x^3 + x^5)*Log[3] + (1536 - 96*x - 96*x^2 - 26*x^3 + 2*x^4 + (-96*x + 6*x^3 + 2*x^4)*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)*Log[3] + x^ 2*Log[3]^2)/x^2] + (-16*x^2 + x^3 + x^3*Log[3])*Log[(256 - 32*x + x^2 + (- 32*x + 2*x^2)*Log[3] + x^2*Log[3]^2)/x^2]^2)/(-16*x^4 + x^5 + x^5*Log[3] + (-32*x^3 + 2*x^4 + 2*x^4*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)* Log[3] + x^2*Log[3]^2)/x^2] + (-16*x^2 + x^3 + x^3*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)*Log[3] + x^2*Log[3]^2)/x^2]^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.
| \(\displaystyle \int \frac {x^5-16 x^4+48 x^3-1152 x^2+\left (x^3+x^3 \log (3)-16 x^2\right ) \log ^2\left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )+\left (x^5+48 x^3-192 x^2\right ) \log (3)+\left (2 x^4-26 x^3-96 x^2+\left (2 x^4+6 x^3-96 x\right ) \log (3)-96 x+1536\right ) \log \left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )+4608 x-3072}{x^5+x^5 \log (3)-16 x^4+\left (x^3+x^3 \log (3)-16 x^2\right ) \log ^2\left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )+\left (2 x^4+2 x^4 \log (3)-32 x^3\right ) \log \left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {x^5-16 x^4+48 x^3-1152 x^2+\left (x^3+x^3 \log (3)-16 x^2\right ) \log ^2\left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )+\left (x^5+48 x^3-192 x^2\right ) \log (3)+\left (2 x^4-26 x^3-96 x^2+\left (2 x^4+6 x^3-96 x\right ) \log (3)-96 x+1536\right ) \log \left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )+4608 x-3072}{x^5 (1+\log (3))-16 x^4+\left (x^3+x^3 \log (3)-16 x^2\right ) \log ^2\left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )+\left (2 x^4+2 x^4 \log (3)-32 x^3\right ) \log \left (\frac {x^2+x^2 \log ^2(3)+\left (2 x^2-32 x\right ) \log (3)-32 x+256}{x^2}\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-\left (x^5 (1+\log (3))\right )+16 x^4-48 x^3 (1+\log (3))-x^2 (x+x \log (3)-16) \log ^2\left (\frac {(x+x \log (3)-16)^2}{x^2}\right )+192 x^2 (6+\log (3))-2 \left (x^3+3 x^2-48\right ) (x+x \log (3)-16) \log \left (\frac {(x+x \log (3)-16)^2}{x^2}\right )-4608 x+3072}{x^2 (16-x (1+\log (3))) \left (\log \left (\frac {(x+x \log (3)-16)^2}{x^2}\right )+x\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {6 (4-x)^2 \left (x^2 (1+\log (3))-16 x+32\right )}{x^2 (16-x (1+\log (3))) \left (\log \left (\frac {(x+x \log (3)-16)^2}{x^2}\right )+x\right )^2}+\frac {6 \left (x^2-16\right )}{x^2 \left (\log \left (\frac {(x+x \log (3)-16)^2}{x^2}\right )+x\right )}+1\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 48 \int \frac {1}{\left (x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )\right )^2}dx+192 \int \frac {1}{x^2 \left (x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )\right )^2}dx-6 (30-\log (9)) \int \frac {1}{x \left (x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )\right )^2}dx-6 \int \frac {x}{\left (x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )\right )^2}dx+12 (3-\log (3))^2 \int \frac {1}{(16-x (1+\log (3))) \left (x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )\right )^2}dx+6 \int \frac {1}{x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )}dx-96 \int \frac {1}{x^2 \left (x+\log \left (\frac {(\log (3) x+x-16)^2}{x^2}\right )\right )}dx+x\) |
Int[(-3072 + 4608*x - 1152*x^2 + 48*x^3 - 16*x^4 + x^5 + (-192*x^2 + 48*x^ 3 + x^5)*Log[3] + (1536 - 96*x - 96*x^2 - 26*x^3 + 2*x^4 + (-96*x + 6*x^3 + 2*x^4)*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)*Log[3] + x^2*Log[ 3]^2)/x^2] + (-16*x^2 + x^3 + x^3*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)*Log[3] + x^2*Log[3]^2)/x^2]^2)/(-16*x^4 + x^5 + x^5*Log[3] + (-32* x^3 + 2*x^4 + 2*x^4*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)*Log[3] + x^2*Log[3]^2)/x^2] + (-16*x^2 + x^3 + x^3*Log[3])*Log[(256 - 32*x + x^2 + (-32*x + 2*x^2)*Log[3] + x^2*Log[3]^2)/x^2]^2),x]
3.10.20.3.1 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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 7.46 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77
| method | result | size |
| risch | \(x +\frac {6 x^{2}-48 x +96}{x \left (x +\ln \left (\frac {x^{2} \ln \left (3\right )^{2}+\left (2 x^{2}-32 x \right ) \ln \left (3\right )+x^{2}-32 x +256}{x^{2}}\right )\right )}\) | \(53\) |
| parallelrisch | \(-\frac {-24576-256 x^{3}-256 x^{2} \ln \left (\frac {x^{2} \ln \left (3\right )^{2}+\left (2 x^{2}-32 x \right ) \ln \left (3\right )+x^{2}-32 x +256}{x^{2}}\right )-1536 x^{2}+12288 x}{256 x \left (x +\ln \left (\frac {x^{2} \ln \left (3\right )^{2}+\left (2 x^{2}-32 x \right ) \ln \left (3\right )+x^{2}-32 x +256}{x^{2}}\right )\right )}\) | \(96\) |
| norman | \(\frac {96+x^{3}-6 x \ln \left (\frac {x^{2} \ln \left (3\right )^{2}+\left (2 x^{2}-32 x \right ) \ln \left (3\right )+x^{2}-32 x +256}{x^{2}}\right )+x^{2} \ln \left (\frac {x^{2} \ln \left (3\right )^{2}+\left (2 x^{2}-32 x \right ) \ln \left (3\right )+x^{2}-32 x +256}{x^{2}}\right )-48 x}{x \left (x +\ln \left (\frac {x^{2} \ln \left (3\right )^{2}+\left (2 x^{2}-32 x \right ) \ln \left (3\right )+x^{2}-32 x +256}{x^{2}}\right )\right )}\) | \(123\) |
int(((x^3*ln(3)+x^3-16*x^2)*ln((x^2*ln(3)^2+(2*x^2-32*x)*ln(3)+x^2-32*x+25 6)/x^2)^2+((2*x^4+6*x^3-96*x)*ln(3)+2*x^4-26*x^3-96*x^2-96*x+1536)*ln((x^2 *ln(3)^2+(2*x^2-32*x)*ln(3)+x^2-32*x+256)/x^2)+(x^5+48*x^3-192*x^2)*ln(3)+ x^5-16*x^4+48*x^3-1152*x^2+4608*x-3072)/((x^3*ln(3)+x^3-16*x^2)*ln((x^2*ln (3)^2+(2*x^2-32*x)*ln(3)+x^2-32*x+256)/x^2)^2+(2*x^4*ln(3)+2*x^4-32*x^3)*l n((x^2*ln(3)^2+(2*x^2-32*x)*ln(3)+x^2-32*x+256)/x^2)+x^5*ln(3)+x^5-16*x^4) ,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (30) = 60\).
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.00 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=\frac {x^{3} + x^{2} \log \left (\frac {x^{2} \log \left (3\right )^{2} + x^{2} + 2 \, {\left (x^{2} - 16 \, x\right )} \log \left (3\right ) - 32 \, x + 256}{x^{2}}\right ) + 6 \, x^{2} - 48 \, x + 96}{x^{2} + x \log \left (\frac {x^{2} \log \left (3\right )^{2} + x^{2} + 2 \, {\left (x^{2} - 16 \, x\right )} \log \left (3\right ) - 32 \, x + 256}{x^{2}}\right )} \]
integrate(((x^3*log(3)+x^3-16*x^2)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x ^2-32*x+256)/x^2)^2+((2*x^4+6*x^3-96*x)*log(3)+2*x^4-26*x^3-96*x^2-96*x+15 36)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2)+(x^5+48*x^3-1 92*x^2)*log(3)+x^5-16*x^4+48*x^3-1152*x^2+4608*x-3072)/((x^3*log(3)+x^3-16 *x^2)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2)^2+(2*x^4*lo g(3)+2*x^4-32*x^3)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2 )+x^5*log(3)+x^5-16*x^4),x, algorithm=\
(x^3 + x^2*log((x^2*log(3)^2 + x^2 + 2*(x^2 - 16*x)*log(3) - 32*x + 256)/x ^2) + 6*x^2 - 48*x + 96)/(x^2 + x*log((x^2*log(3)^2 + x^2 + 2*(x^2 - 16*x) *log(3) - 32*x + 256)/x^2))
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=x + \frac {6 x^{2} - 48 x + 96}{x^{2} + x \log {\left (\frac {x^{2} + x^{2} \log {\left (3 \right )}^{2} - 32 x + \left (2 x^{2} - 32 x\right ) \log {\left (3 \right )} + 256}{x^{2}} \right )}} \]
integrate(((x**3*ln(3)+x**3-16*x**2)*ln((x**2*ln(3)**2+(2*x**2-32*x)*ln(3) +x**2-32*x+256)/x**2)**2+((2*x**4+6*x**3-96*x)*ln(3)+2*x**4-26*x**3-96*x** 2-96*x+1536)*ln((x**2*ln(3)**2+(2*x**2-32*x)*ln(3)+x**2-32*x+256)/x**2)+(x **5+48*x**3-192*x**2)*ln(3)+x**5-16*x**4+48*x**3-1152*x**2+4608*x-3072)/(( x**3*ln(3)+x**3-16*x**2)*ln((x**2*ln(3)**2+(2*x**2-32*x)*ln(3)+x**2-32*x+2 56)/x**2)**2+(2*x**4*ln(3)+2*x**4-32*x**3)*ln((x**2*ln(3)**2+(2*x**2-32*x) *ln(3)+x**2-32*x+256)/x**2)+x**5*ln(3)+x**5-16*x**4),x)
x + (6*x**2 - 48*x + 96)/(x**2 + x*log((x**2 + x**2*log(3)**2 - 32*x + (2* x**2 - 32*x)*log(3) + 256)/x**2))
Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=\frac {x^{3} + 2 \, x^{2} \log \left (x {\left (\log \left (3\right ) + 1\right )} - 16\right ) - 2 \, x^{2} \log \left (x\right ) + 6 \, x^{2} - 48 \, x + 96}{x^{2} + 2 \, x \log \left (x {\left (\log \left (3\right ) + 1\right )} - 16\right ) - 2 \, x \log \left (x\right )} \]
integrate(((x^3*log(3)+x^3-16*x^2)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x ^2-32*x+256)/x^2)^2+((2*x^4+6*x^3-96*x)*log(3)+2*x^4-26*x^3-96*x^2-96*x+15 36)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2)+(x^5+48*x^3-1 92*x^2)*log(3)+x^5-16*x^4+48*x^3-1152*x^2+4608*x-3072)/((x^3*log(3)+x^3-16 *x^2)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2)^2+(2*x^4*lo g(3)+2*x^4-32*x^3)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2 )+x^5*log(3)+x^5-16*x^4),x, algorithm=\
(x^3 + 2*x^2*log(x*(log(3) + 1) - 16) - 2*x^2*log(x) + 6*x^2 - 48*x + 96)/ (x^2 + 2*x*log(x*(log(3) + 1) - 16) - 2*x*log(x))
Time = 1.00 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=x + \frac {6 \, {\left (x^{2} - 8 \, x + 16\right )}}{x^{2} + x \log \left (x^{2} \log \left (3\right )^{2} + 2 \, x^{2} \log \left (3\right ) + x^{2} - 32 \, x \log \left (3\right ) - 32 \, x + 256\right ) - x \log \left (x^{2}\right )} \]
integrate(((x^3*log(3)+x^3-16*x^2)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x ^2-32*x+256)/x^2)^2+((2*x^4+6*x^3-96*x)*log(3)+2*x^4-26*x^3-96*x^2-96*x+15 36)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2)+(x^5+48*x^3-1 92*x^2)*log(3)+x^5-16*x^4+48*x^3-1152*x^2+4608*x-3072)/((x^3*log(3)+x^3-16 *x^2)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2)^2+(2*x^4*lo g(3)+2*x^4-32*x^3)*log((x^2*log(3)^2+(2*x^2-32*x)*log(3)+x^2-32*x+256)/x^2 )+x^5*log(3)+x^5-16*x^4),x, algorithm=\
x + 6*(x^2 - 8*x + 16)/(x^2 + x*log(x^2*log(3)^2 + 2*x^2*log(3) + x^2 - 32 *x*log(3) - 32*x + 256) - x*log(x^2))
Time = 12.92 (sec) , antiderivative size = 5211, normalized size of antiderivative = 173.70 \[ \int \frac {-3072+4608 x-1152 x^2+48 x^3-16 x^4+x^5+\left (-192 x^2+48 x^3+x^5\right ) \log (3)+\left (1536-96 x-96 x^2-26 x^3+2 x^4+\left (-96 x+6 x^3+2 x^4\right ) \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )}{-16 x^4+x^5+x^5 \log (3)+\left (-32 x^3+2 x^4+2 x^4 \log (3)\right ) \log \left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )+\left (-16 x^2+x^3+x^3 \log (3)\right ) \log ^2\left (\frac {256-32 x+x^2+\left (-32 x+2 x^2\right ) \log (3)+x^2 \log ^2(3)}{x^2}\right )} \, dx=\text {Too large to display} \]
int((4608*x + log(3)*(48*x^3 - 192*x^2 + x^5) - log((x^2*log(3)^2 - 32*x - log(3)*(32*x - 2*x^2) + x^2 + 256)/x^2)*(96*x - log(3)*(6*x^3 - 96*x + 2* x^4) + 96*x^2 + 26*x^3 - 2*x^4 - 1536) + log((x^2*log(3)^2 - 32*x - log(3) *(32*x - 2*x^2) + x^2 + 256)/x^2)^2*(x^3*log(3) - 16*x^2 + x^3) - 1152*x^2 + 48*x^3 - 16*x^4 + x^5 - 3072)/(x^5*log(3) + log((x^2*log(3)^2 - 32*x - log(3)*(32*x - 2*x^2) + x^2 + 256)/x^2)*(2*x^4*log(3) - 32*x^3 + 2*x^4) + log((x^2*log(3)^2 - 32*x - log(3)*(32*x - 2*x^2) + x^2 + 256)/x^2)^2*(x^3* log(3) - 16*x^2 + x^3) - 16*x^4 + x^5),x)
x + 3072/(32*x*log(1/x^2) + 32*x*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2 *x^2*log(3) + x^2 + 256) - 16*x^2*log(1/x^2) + x^3*log(1/x^2) - 16*x^2*log (x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) + x^3*log(x ^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) + x^4*log(3) + 32*x^2 - 16*x^3 + x^4 + x^3*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^ 2*log(3) + x^2 + 256)*log(3) + x^3*log(1/x^2)*log(3)) + (1152*x^2)/(32*x*l og(1/x^2) + 32*x*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^ 2 + 256) - 16*x^2*log(1/x^2) + x^3*log(1/x^2) - 16*x^2*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) + x^3*log(x^2*log(3)^2 - 32 *x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) + x^4*log(3) + 32*x^2 - 16*x^ 3 + x^4 + x^3*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256)*log(3) + x^3*log(1/x^2)*log(3)) - (48*x^3)/(32*x*log(1/x^2) + 32*x*l og(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) - 16*x^2* log(1/x^2) + x^3*log(1/x^2) - 16*x^2*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) + x^3*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) + x^4*log(3) + 32*x^2 - 16*x^3 + x^4 + x^3*log( x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256)*log(3) + x^3 *log(1/x^2)*log(3)) - (1536*log(1/x^2))/(32*x*log(1/x^2) + 32*x*log(x^2*lo g(3)^2 - 32*x - 32*x*log(3) + 2*x^2*log(3) + x^2 + 256) - 16*x^2*log(1/x^2 ) + x^3*log(1/x^2) - 16*x^2*log(x^2*log(3)^2 - 32*x - 32*x*log(3) + 2*x...