Integrand size = 88, antiderivative size = 34 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=1-\frac {\log (x)}{-\frac {x}{5}+x^2}+\frac {2 \left (x-\frac {3}{\log (2 x)}\right )^2}{x} \end {dmath*}
Time = 0.46 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2 x-\frac {5 \log (x)}{x (-1+5 x)}+\frac {18}{x \log ^2(2 x)}-\frac {12}{\log (2 x)} \end {dmath*}
Integrate[(-36 + 360*x - 900*x^2 + (-18 + 192*x - 570*x^2 + 300*x^3)*Log[2 *x] + (5 - 25*x + 2*x^2 - 20*x^3 + 50*x^4 + (-5 + 50*x)*Log[x])*Log[2*x]^3 )/((x^2 - 10*x^3 + 25*x^4)*Log[2*x]^3),x]
Time = 1.84 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2026, 7277, 27, 7293, 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.
\(\displaystyle \int \frac {-900 x^2+\left (300 x^3-570 x^2+192 x-18\right ) \log (2 x)+\left (50 x^4-20 x^3+2 x^2-25 x+(50 x-5) \log (x)+5\right ) \log ^3(2 x)+360 x-36}{\left (25 x^4-10 x^3+x^2\right ) \log ^3(2 x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-900 x^2+\left (300 x^3-570 x^2+192 x-18\right ) \log (2 x)+\left (50 x^4-20 x^3+2 x^2-25 x+(50 x-5) \log (x)+5\right ) \log ^3(2 x)+360 x-36}{x^2 \left (25 x^2-10 x+1\right ) \log ^3(2 x)}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 100 \int -\frac {-\left (\left (50 x^4-20 x^3+2 x^2-25 x-5 (1-10 x) \log (x)+5\right ) \log ^3(2 x)\right )+6 \left (-50 x^3+95 x^2-32 x+3\right ) \log (2 x)+900 x^2-360 x+36}{100 (1-5 x)^2 x^2 \log ^3(2 x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {-\left (\left (50 x^4-20 x^3+2 x^2-25 x-5 (1-10 x) \log (x)+5\right ) \log ^3(2 x)\right )+6 \left (-50 x^3+95 x^2-32 x+3\right ) \log (2 x)+900 x^2-360 x+36}{(1-5 x)^2 x^2 \log ^3(2 x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (-\frac {6 (2 x-3)}{x^2 \log ^2(2 x)}+\frac {-50 x^4+20 x^3-2 x^2-50 \log (x) x+25 x+5 \log (x)-5}{x^2 (5 x-1)^2}+\frac {36}{x^2 \log ^3(2 x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 x+\frac {18}{x \log ^2(2 x)}+\frac {125 x \log (x)}{1-5 x}+25 \log (x)-\frac {12}{\log (2 x)}+\frac {5 \log (x)}{x}\) |
Int[(-36 + 360*x - 900*x^2 + (-18 + 192*x - 570*x^2 + 300*x^3)*Log[2*x] + (5 - 25*x + 2*x^2 - 20*x^3 + 50*x^4 + (-5 + 50*x)*Log[x])*Log[2*x]^3)/((x^ 2 - 10*x^3 + 25*x^4)*Log[2*x]^3),x]
3.14.21.3.1 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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.64 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38
method | result | size |
risch | \(-\frac {5 \ln \left (x \right )}{x \left (5 x -1\right )}+2 x -\frac {24 \left (-3+2 x \ln \left (2\right )+2 x \ln \left (x \right )\right )}{x \left (2 \ln \left (2\right )+2 \ln \left (x \right )\right )^{2}}\) | \(47\) |
parts | \(2 x +25 \ln \left (x \right )+\frac {18}{x \ln \left (2 x \right )^{2}}-\frac {12}{\ln \left (2 x \right )}+\frac {5 \ln \left (x \right )}{x}-\frac {125 \ln \left (x \right ) x}{5 x -1}\) | \(47\) |
parallelrisch | \(\frac {-540+2700 x -12 x \ln \left (2 x \right )^{2}-1800 x^{2} \ln \left (2 x \right )+300 x^{3} \ln \left (2 x \right )^{2}+360 x \ln \left (2 x \right )-150 \ln \left (2 x \right )^{2} \ln \left (x \right )}{30 x \ln \left (2 x \right )^{2} \left (5 x -1\right )}\) | \(70\) |
default | \(\frac {-18-60 x^{2} \ln \left (2\right )+\left (90-\frac {2 \ln \left (2\right )^{2}}{5}+12 \ln \left (2\right )\right ) x -\frac {2 x \ln \left (x \right )^{2}}{5}-60 x^{2} \ln \left (x \right )+\left (-\frac {4 \ln \left (2\right )}{5}+12\right ) \ln \left (x \right ) x -5 \ln \left (x \right )^{3}+10 x^{3} \ln \left (x \right )^{2}+10 x^{3} \ln \left (2\right )^{2}-10 \ln \left (2\right ) \ln \left (x \right )^{2}-5 \ln \left (2\right )^{2} \ln \left (x \right )+20 x^{3} \ln \left (2\right ) \ln \left (x \right )}{x \left (5 x -1\right ) \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}\) | \(115\) |
int((((50*x-5)*ln(x)+50*x^4-20*x^3+2*x^2-25*x+5)*ln(2*x)^3+(300*x^3-570*x^ 2+192*x-18)*ln(2*x)-900*x^2+360*x-36)/(25*x^4-10*x^3+x^2)/ln(2*x)^3,x,meth od=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (34) = 68\).
Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 4.21 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=\frac {2 \, {\left (5 \, x^{3} - x^{2}\right )} \log \left (2\right )^{2} + 2 \, {\left (5 \, x^{3} - x^{2} - 5 \, \log \left (2\right )\right )} \log \left (x\right )^{2} - 5 \, \log \left (x\right )^{3} - 12 \, {\left (5 \, x^{2} - x\right )} \log \left (2\right ) - {\left (60 \, x^{2} - 4 \, {\left (5 \, x^{3} - x^{2}\right )} \log \left (2\right ) + 5 \, \log \left (2\right )^{2} - 12 \, x\right )} \log \left (x\right ) + 90 \, x - 18}{{\left (5 \, x^{2} - x\right )} \log \left (2\right )^{2} + 2 \, {\left (5 \, x^{2} - x\right )} \log \left (2\right ) \log \left (x\right ) + {\left (5 \, x^{2} - x\right )} \log \left (x\right )^{2}} \end {dmath*}
integrate((((50*x-5)*log(x)+50*x^4-20*x^3+2*x^2-25*x+5)*log(2*x)^3+(300*x^ 3-570*x^2+192*x-18)*log(2*x)-900*x^2+360*x-36)/(25*x^4-10*x^3+x^2)/log(2*x )^3,x, algorithm=\
(2*(5*x^3 - x^2)*log(2)^2 + 2*(5*x^3 - x^2 - 5*log(2))*log(x)^2 - 5*log(x) ^3 - 12*(5*x^2 - x)*log(2) - (60*x^2 - 4*(5*x^3 - x^2)*log(2) + 5*log(2)^2 - 12*x)*log(x) + 90*x - 18)/((5*x^2 - x)*log(2)^2 + 2*(5*x^2 - x)*log(2)* log(x) + (5*x^2 - x)*log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2 x + \frac {- 12 x \log {\left (x \right )} - 12 x \log {\left (2 \right )} + 18}{x \log {\left (x \right )}^{2} + 2 x \log {\left (2 \right )} \log {\left (x \right )} + x \log {\left (2 \right )}^{2}} - \frac {5 \log {\left (x \right )}}{5 x^{2} - x} \end {dmath*}
integrate((((50*x-5)*ln(x)+50*x**4-20*x**3+2*x**2-25*x+5)*ln(2*x)**3+(300* x**3-570*x**2+192*x-18)*ln(2*x)-900*x**2+360*x-36)/(25*x**4-10*x**3+x**2)/ ln(2*x)**3,x)
2*x + (-12*x*log(x) - 12*x*log(2) + 18)/(x*log(x)**2 + 2*x*log(2)*log(x) + x*log(2)**2) - 5*log(x)/(5*x**2 - x)
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (34) = 68\).
Time = 0.31 (sec) , antiderivative size = 141, normalized size of antiderivative = 4.15 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=\frac {10 \, x^{3} \log \left (2\right )^{2} - 2 \, {\left (\log \left (2\right )^{2} + 30 \, \log \left (2\right )\right )} x^{2} + 2 \, {\left (5 \, x^{3} - x^{2} - 5 \, \log \left (2\right )\right )} \log \left (x\right )^{2} - 5 \, \log \left (x\right )^{3} + 6 \, x {\left (2 \, \log \left (2\right ) + 15\right )} + {\left (20 \, x^{3} \log \left (2\right ) - 4 \, x^{2} {\left (\log \left (2\right ) + 15\right )} - 5 \, \log \left (2\right )^{2} + 12 \, x\right )} \log \left (x\right ) - 18}{5 \, x^{2} \log \left (2\right )^{2} - x \log \left (2\right )^{2} + {\left (5 \, x^{2} - x\right )} \log \left (x\right )^{2} + 2 \, {\left (5 \, x^{2} \log \left (2\right ) - x \log \left (2\right )\right )} \log \left (x\right )} \end {dmath*}
integrate((((50*x-5)*log(x)+50*x^4-20*x^3+2*x^2-25*x+5)*log(2*x)^3+(300*x^ 3-570*x^2+192*x-18)*log(2*x)-900*x^2+360*x-36)/(25*x^4-10*x^3+x^2)/log(2*x )^3,x, algorithm=\
(10*x^3*log(2)^2 - 2*(log(2)^2 + 30*log(2))*x^2 + 2*(5*x^3 - x^2 - 5*log(2 ))*log(x)^2 - 5*log(x)^3 + 6*x*(2*log(2) + 15) + (20*x^3*log(2) - 4*x^2*(l og(2) + 15) - 5*log(2)^2 + 12*x)*log(x) - 18)/(5*x^2*log(2)^2 - x*log(2)^2 + (5*x^2 - x)*log(x)^2 + 2*(5*x^2*log(2) - x*log(2))*log(x))
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=-5 \, {\left (\frac {5}{5 \, x - 1} - \frac {1}{x}\right )} \log \left (x\right ) + 2 \, x - \frac {6 \, {\left (2 \, x \log \left (2\right ) + 2 \, x \log \left (x\right ) - 3\right )}}{x \log \left (2\right )^{2} + 2 \, x \log \left (2\right ) \log \left (x\right ) + x \log \left (x\right )^{2}} \end {dmath*}
integrate((((50*x-5)*log(x)+50*x^4-20*x^3+2*x^2-25*x+5)*log(2*x)^3+(300*x^ 3-570*x^2+192*x-18)*log(2*x)-900*x^2+360*x-36)/(25*x^4-10*x^3+x^2)/log(2*x )^3,x, algorithm=\
-5*(5/(5*x - 1) - 1/x)*log(x) + 2*x - 6*(2*x*log(2) + 2*x*log(x) - 3)/(x*l og(2)^2 + 2*x*log(2)*log(x) + x*log(x)^2)
Time = 17.97 (sec) , antiderivative size = 133, normalized size of antiderivative = 3.91 \begin {dmath*} \int \frac {-36+360 x-900 x^2+\left (-18+192 x-570 x^2+300 x^3\right ) \log (2 x)+\left (5-25 x+2 x^2-20 x^3+50 x^4+(-5+50 x) \log (x)\right ) \log ^3(2 x)}{\left (x^2-10 x^3+25 x^4\right ) \log ^3(2 x)} \, dx=2\,x+\frac {\ln \left (x\right )}{\frac {x}{5}-x^2}+\frac {\frac {3\,\left (3\,\ln \left (2\,x\right )-3\,\ln \left (x\right )-2\,x\,\left (\ln \left (2\,x\right )-\ln \left (x\right )\right )+6\right )}{x}-\frac {3\,\ln \left (x\right )\,\left (2\,x-3\right )}{x}}{2\,\ln \left (x\right )\,\left (\ln \left (2\,x\right )-\ln \left (x\right )\right )+{\ln \left (x\right )}^2+{\left (\ln \left (2\,x\right )-\ln \left (x\right )\right )}^2}-\frac {\frac {9\,\ln \left (x\right )}{x}+\frac {3\,\left (2\,x+3\,\ln \left (2\,x\right )-3\,\ln \left (x\right )+3\right )}{x}}{\ln \left (2\,x\right )}+\frac {9}{x} \end {dmath*}
int((360*x + log(2*x)*(192*x - 570*x^2 + 300*x^3 - 18) + log(2*x)^3*(log(x )*(50*x - 5) - 25*x + 2*x^2 - 20*x^3 + 50*x^4 + 5) - 900*x^2 - 36)/(log(2* x)^3*(x^2 - 10*x^3 + 25*x^4)),x)