Integrand size = 75, antiderivative size = 26 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=-2+x^5 (1+4 x)^2+\frac {20 x}{-x+4 \log (3)} \] Output:
(1+4*x)^2*x^5+20/(4*ln(3)-x)*x-2
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(26)=52\).
Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.12 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=\frac {16 x^8+x^7 (8-64 \log (3))+x^6 (1-32 \log (3))-4 x^5 \log (3)-1024 x \log ^5(3) (1+16 \log (3))^2+16 \log (3) \left (-5+256 \log ^5(3)+8192 \log ^6(3)+65536 \log ^7(3)\right )}{x-4 \log (3)} \] Input:
Integrate[(5*x^6 + 48*x^7 + 112*x^8 + (80 - 40*x^5 - 384*x^6 - 896*x^7)*Lo g[3] + (80*x^4 + 768*x^5 + 1792*x^6)*Log[3]^2)/(x^2 - 8*x*Log[3] + 16*Log[ 3]^2),x]
Output:
(16*x^8 + x^7*(8 - 64*Log[3]) + x^6*(1 - 32*Log[3]) - 4*x^5*Log[3] - 1024* x*Log[3]^5*(1 + 16*Log[3])^2 + 16*Log[3]*(-5 + 256*Log[3]^5 + 8192*Log[3]^ 6 + 65536*Log[3]^7))/(x - 4*Log[3])
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2188, 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.
\(\displaystyle \int \frac {112 x^8+48 x^7+5 x^6+\left (-896 x^7-384 x^6-40 x^5+80\right ) \log (3)+\left (1792 x^6+768 x^5+80 x^4\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle \int \left (112 x^6+48 x^5+5 x^4+\frac {80 \log (3)}{x^2-8 x \log (3)+16 \log ^2(3)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 16 x^7+8 x^6+x^5-\frac {80 \log (3)}{x-4 \log (3)}\) |
Input:
Int[(5*x^6 + 48*x^7 + 112*x^8 + (80 - 40*x^5 - 384*x^6 - 896*x^7)*Log[3] + (80*x^4 + 768*x^5 + 1792*x^6)*Log[3]^2)/(x^2 - 8*x*Log[3] + 16*Log[3]^2), x]
Output:
x^5 + 8*x^6 + 16*x^7 - (80*Log[3])/(x - 4*Log[3])
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq , x] && IGtQ[p, -2]
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
default | \(16 x^{7}+8 x^{6}+x^{5}-\frac {80 \ln \left (3\right )}{-4 \ln \left (3\right )+x}\) | \(27\) |
risch | \(16 x^{7}+8 x^{6}+x^{5}+\frac {20 \ln \left (3\right )}{\ln \left (3\right )-\frac {x}{4}}\) | \(27\) |
norman | \(\frac {-16 x^{8}+\left (64 \ln \left (3\right )-8\right ) x^{7}+\left (32 \ln \left (3\right )-1\right ) x^{6}+4 x^{5} \ln \left (3\right )+80 \ln \left (3\right )}{4 \ln \left (3\right )-x}\) | \(49\) |
gosper | \(\frac {64 \ln \left (3\right ) x^{7}-16 x^{8}+32 \ln \left (3\right ) x^{6}-8 x^{7}+4 x^{5} \ln \left (3\right )-x^{6}+80 \ln \left (3\right )}{4 \ln \left (3\right )-x}\) | \(53\) |
parallelrisch | \(\frac {64 \ln \left (3\right ) x^{7}-16 x^{8}+32 \ln \left (3\right ) x^{6}-8 x^{7}+4 x^{5} \ln \left (3\right )-x^{6}+80 \ln \left (3\right )}{4 \ln \left (3\right )-x}\) | \(53\) |
meijerg | \(-1835008 \ln \left (3\right )^{7} \left (-\frac {x \left (-\frac {45 x^{7}}{16384 \ln \left (3\right )^{7}}-\frac {15 x^{6}}{1024 \ln \left (3\right )^{6}}-\frac {21 x^{5}}{256 \ln \left (3\right )^{5}}-\frac {63 x^{4}}{128 \ln \left (3\right )^{4}}-\frac {105 x^{3}}{32 \ln \left (3\right )^{3}}-\frac {105 x^{2}}{4 \ln \left (3\right )^{2}}-\frac {315 x}{\ln \left (3\right )}+2520\right )}{1260 \ln \left (3\right ) \left (1-\frac {x}{4 \ln \left (3\right )}\right )}-8 \ln \left (1-\frac {x}{4 \ln \left (3\right )}\right )\right )+65536 \ln \left (3\right )^{6} \left (-56 \ln \left (3\right )+3\right ) \left (\frac {x \left (-\frac {5 x^{6}}{1024 \ln \left (3\right )^{6}}-\frac {7 x^{5}}{256 \ln \left (3\right )^{5}}-\frac {21 x^{4}}{128 \ln \left (3\right )^{4}}-\frac {35 x^{3}}{32 \ln \left (3\right )^{3}}-\frac {35 x^{2}}{4 \ln \left (3\right )^{2}}-\frac {105 x}{\ln \left (3\right )}+840\right )}{480 \ln \left (3\right ) \left (1-\frac {x}{4 \ln \left (3\right )}\right )}+7 \ln \left (1-\frac {x}{4 \ln \left (3\right )}\right )\right )-16384 \ln \left (3\right )^{5} \left (112 \ln \left (3\right )^{2}-24 \ln \left (3\right )+\frac {5}{16}\right ) \left (-\frac {x \left (-\frac {7 x^{5}}{512 \ln \left (3\right )^{5}}-\frac {21 x^{4}}{256 \ln \left (3\right )^{4}}-\frac {35 x^{3}}{64 \ln \left (3\right )^{3}}-\frac {35 x^{2}}{8 \ln \left (3\right )^{2}}-\frac {105 x}{2 \ln \left (3\right )}+420\right )}{280 \ln \left (3\right ) \left (1-\frac {x}{4 \ln \left (3\right )}\right )}-6 \ln \left (1-\frac {x}{4 \ln \left (3\right )}\right )\right )+4096 \ln \left (3\right )^{4} \left (48 \ln \left (3\right )^{2}-\frac {5 \ln \left (3\right )}{2}\right ) \left (\frac {x \left (-\frac {3 x^{4}}{256 \ln \left (3\right )^{4}}-\frac {5 x^{3}}{64 \ln \left (3\right )^{3}}-\frac {5 x^{2}}{8 \ln \left (3\right )^{2}}-\frac {15 x}{2 \ln \left (3\right )}+60\right )}{48 \ln \left (3\right ) \left (1-\frac {x}{4 \ln \left (3\right )}\right )}+5 \ln \left (1-\frac {x}{4 \ln \left (3\right )}\right )\right )-5120 \ln \left (3\right )^{5} \left (-\frac {x \left (-\frac {5 x^{3}}{64 \ln \left (3\right )^{3}}-\frac {5 x^{2}}{8 \ln \left (3\right )^{2}}-\frac {15 x}{2 \ln \left (3\right )}+60\right )}{60 \ln \left (3\right ) \left (1-\frac {x}{4 \ln \left (3\right )}\right )}-4 \ln \left (1-\frac {x}{4 \ln \left (3\right )}\right )\right )+\frac {5 x}{\ln \left (3\right ) \left (1-\frac {x}{4 \ln \left (3\right )}\right )}\) | \(459\) |
Input:
int(((1792*x^6+768*x^5+80*x^4)*ln(3)^2+(-896*x^7-384*x^6-40*x^5+80)*ln(3)+ 112*x^8+48*x^7+5*x^6)/(16*ln(3)^2-8*x*ln(3)+x^2),x,method=_RETURNVERBOSE)
Output:
16*x^7+8*x^6+x^5-80*ln(3)/(-4*ln(3)+x)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=\frac {16 \, x^{8} + 8 \, x^{7} + x^{6} - 4 \, {\left (16 \, x^{7} + 8 \, x^{6} + x^{5} + 20\right )} \log \left (3\right )}{x - 4 \, \log \left (3\right )} \] Input:
integrate(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80) *log(3)+112*x^8+48*x^7+5*x^6)/(16*log(3)^2-8*x*log(3)+x^2),x, algorithm="f ricas")
Output:
(16*x^8 + 8*x^7 + x^6 - 4*(16*x^7 + 8*x^6 + x^5 + 20)*log(3))/(x - 4*log(3 ))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=16 x^{7} + 8 x^{6} + x^{5} - \frac {80 \log {\left (3 \right )}}{x - 4 \log {\left (3 \right )}} \] Input:
integrate(((1792*x**6+768*x**5+80*x**4)*ln(3)**2+(-896*x**7-384*x**6-40*x* *5+80)*ln(3)+112*x**8+48*x**7+5*x**6)/(16*ln(3)**2-8*x*ln(3)+x**2),x)
Output:
16*x**7 + 8*x**6 + x**5 - 80*log(3)/(x - 4*log(3))
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=16 \, x^{7} + 8 \, x^{6} + x^{5} - \frac {80 \, \log \left (3\right )}{x - 4 \, \log \left (3\right )} \] Input:
integrate(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80) *log(3)+112*x^8+48*x^7+5*x^6)/(16*log(3)^2-8*x*log(3)+x^2),x, algorithm="m axima")
Output:
16*x^7 + 8*x^6 + x^5 - 80*log(3)/(x - 4*log(3))
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=16 \, x^{7} + 8 \, x^{6} + x^{5} - \frac {80 \, \log \left (3\right )}{x - 4 \, \log \left (3\right )} \] Input:
integrate(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80) *log(3)+112*x^8+48*x^7+5*x^6)/(16*log(3)^2-8*x*log(3)+x^2),x, algorithm="g iac")
Output:
16*x^7 + 8*x^6 + x^5 - 80*log(3)/(x - 4*log(3))
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=x^5-\frac {80\,\ln \left (3\right )}{x-4\,\ln \left (3\right )}+8\,x^6+16\,x^7 \] Input:
int((log(3)^2*(80*x^4 + 768*x^5 + 1792*x^6) - log(3)*(40*x^5 + 384*x^6 + 8 96*x^7 - 80) + 5*x^6 + 48*x^7 + 112*x^8)/(16*log(3)^2 - 8*x*log(3) + x^2), x)
Output:
x^5 - (80*log(3))/(x - 4*log(3)) + 8*x^6 + 16*x^7
Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {5 x^6+48 x^7+112 x^8+\left (80-40 x^5-384 x^6-896 x^7\right ) \log (3)+\left (80 x^4+768 x^5+1792 x^6\right ) \log ^2(3)}{x^2-8 x \log (3)+16 \log ^2(3)} \, dx=\frac {x \left (64 \,\mathrm {log}\left (3\right ) x^{6}+32 \,\mathrm {log}\left (3\right ) x^{5}+4 \,\mathrm {log}\left (3\right ) x^{4}-16 x^{7}-8 x^{6}-x^{5}+20\right )}{4 \,\mathrm {log}\left (3\right )-x} \] Input:
int(((1792*x^6+768*x^5+80*x^4)*log(3)^2+(-896*x^7-384*x^6-40*x^5+80)*log(3 )+112*x^8+48*x^7+5*x^6)/(16*log(3)^2-8*x*log(3)+x^2),x)
Output:
(x*(64*log(3)*x**6 + 32*log(3)*x**5 + 4*log(3)*x**4 - 16*x**7 - 8*x**6 - x **5 + 20))/(4*log(3) - x)