Integrand size = 123, antiderivative size = 25 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=5 \left (3-x^2+\frac {x}{\log \left (\frac {x}{2}+x^2\right )}\right )^2 \] Output:
5*(x/ln(x^2+1/2*x)-x^2+3)^2
Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=5 x \left (-6 x+x^3+\frac {x}{\log ^2\left (\frac {x}{2}+x^2\right )}-\frac {2 \left (-3+x^2\right )}{\log \left (\frac {x}{2}+x^2\right )}\right ) \] Input:
Integrate[(-10*x - 40*x^2 + (-30 - 110*x + 30*x^2 + 40*x^3)*Log[(x + 2*x^2 )/2] + (30 + 60*x - 30*x^2 - 60*x^3)*Log[(x + 2*x^2)/2]^2 + (-60*x - 120*x ^2 + 20*x^3 + 40*x^4)*Log[(x + 2*x^2)/2]^3)/((1 + 2*x)*Log[(x + 2*x^2)/2]^ 3),x]
Output:
5*x*(-6*x + x^3 + x/Log[x/2 + x^2]^2 - (2*(-3 + x^2))/Log[x/2 + x^2])
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.
\(\displaystyle \int \frac {-40 x^2+\left (-60 x^3-30 x^2+60 x+30\right ) \log ^2\left (\frac {1}{2} \left (2 x^2+x\right )\right )+\left (40 x^3+30 x^2-110 x-30\right ) \log \left (\frac {1}{2} \left (2 x^2+x\right )\right )+\left (40 x^4+20 x^3-120 x^2-60 x\right ) \log ^3\left (\frac {1}{2} \left (2 x^2+x\right )\right )-10 x}{(2 x+1) \log ^3\left (\frac {1}{2} \left (2 x^2+x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-40 x^2+\left (-60 x^3-30 x^2+60 x+30\right ) \log ^2\left (\frac {1}{2} \left (2 x^2+x\right )\right )+\left (40 x^3+30 x^2-110 x-30\right ) \log \left (\frac {1}{2} \left (2 x^2+x\right )\right )+\left (40 x^4+20 x^3-120 x^2-60 x\right ) \log ^3\left (\frac {1}{2} \left (2 x^2+x\right )\right )-10 x}{(2 x+1) \log ^3\left (\frac {1}{2} x (2 x+1)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (20 \left (x^2-3\right ) x+\frac {30 \left (1-x^2\right )}{\log \left (x \left (x+\frac {1}{2}\right )\right )}+\frac {10 \left (4 x^3+3 x^2-11 x-3\right )}{(2 x+1) \log ^2\left (x \left (x+\frac {1}{2}\right )\right )}+\frac {10 (-4 x-1) x}{(2 x+1) \log ^3\left (x \left (x+\frac {1}{2}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 30 \int \frac {1-x^2}{\log \left (x \left (x+\frac {1}{2}\right )\right )}dx+10 \int \frac {4 x^3+3 x^2-11 x-3}{(2 x+1) \log ^2\left (x \left (x+\frac {1}{2}\right )\right )}dx+10 \int \frac {(-4 x-1) x}{(2 x+1) \log ^3\left (x \left (x+\frac {1}{2}\right )\right )}dx+5 x^4-30 x^2\) |
Input:
Int[(-10*x - 40*x^2 + (-30 - 110*x + 30*x^2 + 40*x^3)*Log[(x + 2*x^2)/2] + (30 + 60*x - 30*x^2 - 60*x^3)*Log[(x + 2*x^2)/2]^2 + (-60*x - 120*x^2 + 2 0*x^3 + 40*x^4)*Log[(x + 2*x^2)/2]^3)/((1 + 2*x)*Log[(x + 2*x^2)/2]^3),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(52\) vs. \(2(23)=46\).
Time = 6.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12
method | result | size |
risch | \(5 x^{4}-30 x^{2}+45-\frac {5 \left (2 \ln \left (x^{2}+\frac {1}{2} x \right ) x^{2}-x -6 \ln \left (x^{2}+\frac {1}{2} x \right )\right ) x}{\ln \left (x^{2}+\frac {1}{2} x \right )^{2}}\) | \(53\) |
norman | \(\frac {5 x^{2}+30 \ln \left (x^{2}+\frac {1}{2} x \right ) x -10 \ln \left (x^{2}+\frac {1}{2} x \right ) x^{3}-30 \ln \left (x^{2}+\frac {1}{2} x \right )^{2} x^{2}+5 \ln \left (x^{2}+\frac {1}{2} x \right )^{2} x^{4}}{\ln \left (x^{2}+\frac {1}{2} x \right )^{2}}\) | \(72\) |
parallelrisch | \(\frac {20 \ln \left (x^{2}+\frac {1}{2} x \right )^{2} x^{4}-40 \ln \left (x^{2}+\frac {1}{2} x \right ) x^{3}-120 \ln \left (x^{2}+\frac {1}{2} x \right )^{2} x^{2}+20 x^{2}+120 \ln \left (x^{2}+\frac {1}{2} x \right ) x +30 \ln \left (x^{2}+\frac {1}{2} x \right )^{2}}{4 \ln \left (x^{2}+\frac {1}{2} x \right )^{2}}\) | \(85\) |
Input:
int(((40*x^4+20*x^3-120*x^2-60*x)*ln(x^2+1/2*x)^3+(-60*x^3-30*x^2+60*x+30) *ln(x^2+1/2*x)^2+(40*x^3+30*x^2-110*x-30)*ln(x^2+1/2*x)-40*x^2-10*x)/(1+2* x)/ln(x^2+1/2*x)^3,x,method=_RETURNVERBOSE)
Output:
5*x^4-30*x^2+45-5*(2*ln(x^2+1/2*x)*x^2-x-6*ln(x^2+1/2*x))*x/ln(x^2+1/2*x)^ 2
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=\frac {5 \, {\left ({\left (x^{4} - 6 \, x^{2}\right )} \log \left (x^{2} + \frac {1}{2} \, x\right )^{2} + x^{2} - 2 \, {\left (x^{3} - 3 \, x\right )} \log \left (x^{2} + \frac {1}{2} \, x\right )\right )}}{\log \left (x^{2} + \frac {1}{2} \, x\right )^{2}} \] Input:
integrate(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+6 0*x+30)*log(x^2+1/2*x)^2+(40*x^3+30*x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10 *x)/(1+2*x)/log(x^2+1/2*x)^3,x, algorithm="fricas")
Output:
5*((x^4 - 6*x^2)*log(x^2 + 1/2*x)^2 + x^2 - 2*(x^3 - 3*x)*log(x^2 + 1/2*x) )/log(x^2 + 1/2*x)^2
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=5 x^{4} - 30 x^{2} + \frac {5 x^{2} + \left (- 10 x^{3} + 30 x\right ) \log {\left (x^{2} + \frac {x}{2} \right )}}{\log {\left (x^{2} + \frac {x}{2} \right )}^{2}} \] Input:
integrate(((40*x**4+20*x**3-120*x**2-60*x)*ln(x**2+1/2*x)**3+(-60*x**3-30* x**2+60*x+30)*ln(x**2+1/2*x)**2+(40*x**3+30*x**2-110*x-30)*ln(x**2+1/2*x)- 40*x**2-10*x)/(1+2*x)/ln(x**2+1/2*x)**3,x)
Output:
5*x**4 - 30*x**2 + (5*x**2 + (-10*x**3 + 30*x)*log(x**2 + x/2))/log(x**2 + x/2)**2
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 173, normalized size of antiderivative = 6.92 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=\frac {5 \, {\left (x^{4} \log \left (2\right )^{2} + 2 \, x^{3} \log \left (2\right ) - {\left (6 \, \log \left (2\right )^{2} - 1\right )} x^{2} + {\left (x^{4} - 6 \, x^{2}\right )} \log \left (2 \, x + 1\right )^{2} + {\left (x^{4} - 6 \, x^{2}\right )} \log \left (x\right )^{2} - 6 \, x \log \left (2\right ) - 2 \, {\left (x^{4} \log \left (2\right ) + x^{3} - 6 \, x^{2} \log \left (2\right ) - {\left (x^{4} - 6 \, x^{2}\right )} \log \left (x\right ) - 3 \, x\right )} \log \left (2 \, x + 1\right ) - 2 \, {\left (x^{4} \log \left (2\right ) + x^{3} - 6 \, x^{2} \log \left (2\right ) - 3 \, x\right )} \log \left (x\right )\right )}}{\log \left (2\right )^{2} - 2 \, {\left (\log \left (2\right ) - \log \left (x\right )\right )} \log \left (2 \, x + 1\right ) + \log \left (2 \, x + 1\right )^{2} - 2 \, \log \left (2\right ) \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:
integrate(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+6 0*x+30)*log(x^2+1/2*x)^2+(40*x^3+30*x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10 *x)/(1+2*x)/log(x^2+1/2*x)^3,x, algorithm="maxima")
Output:
5*(x^4*log(2)^2 + 2*x^3*log(2) - (6*log(2)^2 - 1)*x^2 + (x^4 - 6*x^2)*log( 2*x + 1)^2 + (x^4 - 6*x^2)*log(x)^2 - 6*x*log(2) - 2*(x^4*log(2) + x^3 - 6 *x^2*log(2) - (x^4 - 6*x^2)*log(x) - 3*x)*log(2*x + 1) - 2*(x^4*log(2) + x ^3 - 6*x^2*log(2) - 3*x)*log(x))/(log(2)^2 - 2*(log(2) - log(x))*log(2*x + 1) + log(2*x + 1)^2 - 2*log(2)*log(x) + log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=5 \, x^{4} - 30 \, x^{2} - \frac {5 \, {\left (2 \, x^{3} \log \left (x^{2} + \frac {1}{2} \, x\right ) - x^{2} - 6 \, x \log \left (x^{2} + \frac {1}{2} \, x\right )\right )}}{\log \left (x^{2} + \frac {1}{2} \, x\right )^{2}} \] Input:
integrate(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+6 0*x+30)*log(x^2+1/2*x)^2+(40*x^3+30*x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10 *x)/(1+2*x)/log(x^2+1/2*x)^3,x, algorithm="giac")
Output:
5*x^4 - 30*x^2 - 5*(2*x^3*log(x^2 + 1/2*x) - x^2 - 6*x*log(x^2 + 1/2*x))/l og(x^2 + 1/2*x)^2
Time = 0.62 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=\frac {5\,x^2-5\,x\,\ln \left (x^2+\frac {x}{2}\right )\,\left (2\,x^2-6\right )}{{\ln \left (x^2+\frac {x}{2}\right )}^2}-5\,x\,\left (6\,x-x^3\right ) \] Input:
int(-(10*x + log(x/2 + x^2)^3*(60*x + 120*x^2 - 20*x^3 - 40*x^4) + log(x/2 + x^2)*(110*x - 30*x^2 - 40*x^3 + 30) + 40*x^2 - log(x/2 + x^2)^2*(60*x - 30*x^2 - 60*x^3 + 30))/(log(x/2 + x^2)^3*(2*x + 1)),x)
Output:
(5*x^2 - 5*x*log(x/2 + x^2)*(2*x^2 - 6))/log(x/2 + x^2)^2 - 5*x*(6*x - x^3 )
Time = 0.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.60 \[ \int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx=\frac {5 x \left (\mathrm {log}\left (x^{2}+\frac {1}{2} x \right )^{2} x^{3}-6 \mathrm {log}\left (x^{2}+\frac {1}{2} x \right )^{2} x -2 \,\mathrm {log}\left (x^{2}+\frac {1}{2} x \right ) x^{2}+6 \,\mathrm {log}\left (x^{2}+\frac {1}{2} x \right )+x \right )}{\mathrm {log}\left (x^{2}+\frac {1}{2} x \right )^{2}} \] Input:
int(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+60*x+30 )*log(x^2+1/2*x)^2+(40*x^3+30*x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10*x)/(1 +2*x)/log(x^2+1/2*x)^3,x)
Output:
(5*x*(log((2*x**2 + x)/2)**2*x**3 - 6*log((2*x**2 + x)/2)**2*x - 2*log((2* x**2 + x)/2)*x**2 + 6*log((2*x**2 + x)/2) + x))/log((2*x**2 + x)/2)**2