Integrand size = 59, antiderivative size = 27 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=5-x-5 \log (x) \left (\frac {x}{4-x}-4 \log \left (4 x^2\right )\right ) \] Output:
5-x-ln(x)*(5*x/(4-x)-20*ln(4*x^2))
Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=-x+\frac {5 x \log (x)}{-4+x}+20 \log ^2(x)+5 \log ^2\left (4 x^2\right ) \] Input:
Integrate[(-36*x + 13*x^2 - x^3 + (640 - 340*x + 40*x^2)*Log[x] + (320 - 1 60*x + 20*x^2)*Log[4*x^2])/(16*x - 8*x^2 + x^3),x]
Output:
-x + (5*x*Log[x])/(-4 + x) + 20*Log[x]^2 + 5*Log[4*x^2]^2
Time = 0.69 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, 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.
\(\displaystyle \int \frac {-x^3+13 x^2+\left (40 x^2-340 x+640\right ) \log (x)+\left (20 x^2-160 x+320\right ) \log \left (4 x^2\right )-36 x}{x^3-8 x^2+16 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^3+13 x^2+\left (40 x^2-340 x+640\right ) \log (x)+\left (20 x^2-160 x+320\right ) \log \left (4 x^2\right )-36 x}{x \left (x^2-8 x+16\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {x^3-13 x^2+36 x-20 \left (2 x^2-17 x+32\right ) \log (x)-20 \left (x^2-8 x+16\right ) \log \left (4 x^2\right )}{4 (4-x)^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^3-13 x^2+36 x-20 \left (2 x^2-17 x+32\right ) \log (x)-20 \left (x^2-8 x+16\right ) \log \left (4 x^2\right )}{(4-x)^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {x^2}{(x-4)^2}-\frac {13 x}{(x-4)^2}+\frac {36}{(x-4)^2}-\frac {20 \left (2 x^2-17 x+32\right ) \log (x)}{(x-4)^2 x}-\frac {20 \log \left (4 x^2\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \log ^2\left (4 x^2\right )-x+20 \log ^2(x)-\frac {5 x \log (x)}{4-x}\) |
Input:
Int[(-36*x + 13*x^2 - x^3 + (640 - 340*x + 40*x^2)*Log[x] + (320 - 160*x + 20*x^2)*Log[4*x^2])/(16*x - 8*x^2 + x^3),x]
Output:
-x - (5*x*Log[x])/(4 - x) + 20*Log[x]^2 + 5*Log[4*x^2]^2
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 = 1.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
parts | \(5 \ln \left (4 x^{2}\right )^{2}-x +20 \ln \left (x \right )^{2}+\frac {5 x \ln \left (x \right )}{x -4}\) | \(31\) |
default | \(-x +40 \ln \left (2\right ) \ln \left (x \right )+5 \ln \left (x^{2}\right )^{2}+20 \ln \left (x \right )^{2}+\frac {5 x \ln \left (x \right )}{x -4}\) | \(35\) |
parallelrisch | \(\frac {16+20 x \ln \left (x \right ) \ln \left (4 x^{2}\right )-x^{2}+5 x \ln \left (x \right )-80 \ln \left (x \right ) \ln \left (4 x^{2}\right )}{x -4}\) | \(40\) |
norman | \(\frac {20 \ln \left (x \right )-10 \ln \left (4 x^{2}\right )-80 \ln \left (x \right ) \ln \left (4 x^{2}\right )+\frac {5 \ln \left (4 x^{2}\right ) x}{2}+20 x \ln \left (x \right ) \ln \left (4 x^{2}\right )-x^{2}+16}{x -4}\) | \(56\) |
risch | \(40 \ln \left (x \right )^{2}+\frac {20 \ln \left (x \right )}{x -4}-x -10 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+20 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-10 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+40 \ln \left (2\right ) \ln \left (x \right )+5 \ln \left (x \right )\) | \(85\) |
Input:
int(((20*x^2-160*x+320)*ln(4*x^2)+(40*x^2-340*x+640)*ln(x)-x^3+13*x^2-36*x )/(x^3-8*x^2+16*x),x,method=_RETURNVERBOSE)
Output:
5*ln(4*x^2)^2-x+20*ln(x)^2+5/(x-4)*x*ln(x)
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=\frac {40 \, {\left (x - 4\right )} \log \left (x\right )^{2} - x^{2} + 5 \, {\left (8 \, {\left (x - 4\right )} \log \left (2\right ) + x\right )} \log \left (x\right ) + 4 \, x}{x - 4} \] Input:
integrate(((20*x^2-160*x+320)*log(4*x^2)+(40*x^2-340*x+640)*log(x)-x^3+13* x^2-36*x)/(x^3-8*x^2+16*x),x, algorithm="fricas")
Output:
(40*(x - 4)*log(x)^2 - x^2 + 5*(8*(x - 4)*log(2) + x)*log(x) + 4*x)/(x - 4 )
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=- x + 40 \log {\left (x \right )}^{2} + 5 \cdot \left (1 + 8 \log {\left (2 \right )}\right ) \log {\left (x \right )} + \frac {20 \log {\left (x \right )}}{x - 4} \] Input:
integrate(((20*x**2-160*x+320)*ln(4*x**2)+(40*x**2-340*x+640)*ln(x)-x**3+1 3*x**2-36*x)/(x**3-8*x**2+16*x),x)
Output:
-x + 40*log(x)**2 + 5*(1 + 8*log(2))*log(x) + 20*log(x)/(x - 4)
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=5 \, {\left (8 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) - x + \frac {20 \, {\left (2 \, {\left (x - 4\right )} \log \left (x\right )^{2} + \log \left (x\right )\right )}}{x - 4} \] Input:
integrate(((20*x^2-160*x+320)*log(4*x^2)+(40*x^2-340*x+640)*log(x)-x^3+13* x^2-36*x)/(x^3-8*x^2+16*x),x, algorithm="maxima")
Output:
5*(8*log(2) + 1)*log(x) - x + 20*(2*(x - 4)*log(x)^2 + log(x))/(x - 4)
\[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=\int { -\frac {x^{3} - 13 \, x^{2} - 20 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (4 \, x^{2}\right ) - 20 \, {\left (2 \, x^{2} - 17 \, x + 32\right )} \log \left (x\right ) + 36 \, x}{x^{3} - 8 \, x^{2} + 16 \, x} \,d x } \] Input:
integrate(((20*x^2-160*x+320)*log(4*x^2)+(40*x^2-340*x+640)*log(x)-x^3+13* x^2-36*x)/(x^3-8*x^2+16*x),x, algorithm="giac")
Output:
integrate(-(x^3 - 13*x^2 - 20*(x^2 - 8*x + 16)*log(4*x^2) - 20*(2*x^2 - 17 *x + 32)*log(x) + 36*x)/(x^3 - 8*x^2 + 16*x), x)
Time = 2.62 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=5\,\ln \left (x\right )-x+20\,\ln \left (x^2\right )\,\ln \left (x\right )+40\,\ln \left (2\right )\,\ln \left (x\right )-\frac {20\,x^2\,\ln \left (x\right )}{4\,x^2-x^3} \] Input:
int((log(4*x^2)*(20*x^2 - 160*x + 320) - 36*x + log(x)*(40*x^2 - 340*x + 6 40) + 13*x^2 - x^3)/(16*x - 8*x^2 + x^3),x)
Output:
5*log(x) - x + 20*log(x^2)*log(x) + 40*log(2)*log(x) - (20*x^2*log(x))/(4* x^2 - x^3)
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {-36 x+13 x^2-x^3+\left (640-340 x+40 x^2\right ) \log (x)+\left (320-160 x+20 x^2\right ) \log \left (4 x^2\right )}{16 x-8 x^2+x^3} \, dx=\frac {5 \mathrm {log}\left (4 x^{2}\right )^{2} x -20 \mathrm {log}\left (4 x^{2}\right )^{2}+20 \mathrm {log}\left (x \right )^{2} x -80 \mathrm {log}\left (x \right )^{2}+5 \,\mathrm {log}\left (x \right ) x -x^{2}+4 x}{x -4} \] Input:
int(((20*x^2-160*x+320)*log(4*x^2)+(40*x^2-340*x+640)*log(x)-x^3+13*x^2-36 *x)/(x^3-8*x^2+16*x),x)
Output:
(5*log(4*x**2)**2*x - 20*log(4*x**2)**2 + 20*log(x)**2*x - 80*log(x)**2 + 5*log(x)*x - x**2 + 4*x)/(x - 4)