Integrand size = 105, antiderivative size = 43 \[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=-4+2^{\frac {25 \left (5+x^2 \log (x)\right )}{2-x}} \left (x^2\right )^{\frac {25 \left (5+x^2 \log (x)\right )}{2-x}} \]
\[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=\int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx \]
Integrate[(2^((-125 - 25*x^2*Log[x])/(-2 + x))*(x^2)^((-125 - 25*x^2*Log[x ])/(-2 + x))*(500 - 250*x + (100*x^2 - 50*x^3)*Log[x] + (125*x + 50*x^2 - 25*x^3 + (100*x^2 - 25*x^3)*Log[x])*Log[2*x^2]))/(4*x - 4*x^2 + x^3),x]
Integrate[(2^((-125 - 25*x^2*Log[x])/(-2 + x))*(x^2)^((-125 - 25*x^2*Log[x ])/(-2 + x))*(500 - 250*x + (100*x^2 - 50*x^3)*Log[x] + (125*x + 50*x^2 - 25*x^3 + (100*x^2 - 25*x^3)*Log[x])*Log[2*x^2]))/(4*x - 4*x^2 + x^3), 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.
\(\displaystyle \int \frac {2^{\frac {-25 x^2 \log (x)-125}{x-2}} \left (x^2\right )^{\frac {-25 x^2 \log (x)-125}{x-2}} \left (\left (100 x^2-50 x^3\right ) \log (x)+\left (-25 x^3+50 x^2+\left (100 x^2-25 x^3\right ) \log (x)+125 x\right ) \log \left (2 x^2\right )-250 x+500\right )}{x^3-4 x^2+4 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {2^{\frac {-25 x^2 \log (x)-125}{x-2}} \left (x^2\right )^{\frac {-25 x^2 \log (x)-125}{x-2}} \left (\left (100 x^2-50 x^3\right ) \log (x)+\left (-25 x^3+50 x^2+\left (100 x^2-25 x^3\right ) \log (x)+125 x\right ) \log \left (2 x^2\right )-250 x+500\right )}{x \left (x^2-4 x+4\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {25\ 2^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}-2} \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}} \left (-10 x+2 \left (2 x^2-x^3\right ) \log (x)+\left (-x^3+2 x^2+5 x+\left (4 x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+20\right )}{(2-x)^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 100 \int \frac {2^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}-2} \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}} \left (-10 x+2 \left (2 x^2-x^3\right ) \log (x)+\left (-x^3+2 x^2+5 x+\left (4 x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+20\right )}{(2-x)^2 x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle 100 \int \frac {2^{\frac {-25 \log (x) x^2-2 x-121}{x-2}} \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}} \left (-10 x+2 \left (2 x^2-x^3\right ) \log (x)+\left (-x^3+2 x^2+5 x+\left (4 x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+20\right )}{(2-x)^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 100 \int \left (-\frac {2^{\frac {-25 \log (x) x^2-2 x-121}{x-2}+1} \left (\log (x) x^2+5\right ) \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}}}{(x-2) x}-\frac {2^{\frac {-25 \log (x) x^2-2 x-121}{x-2}} \left (\log (x) x^2+x^2-4 \log (x) x-2 x-5\right ) \log \left (2 x^2\right ) \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}}}{(x-2)^2}\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle 100 \int \left (-\frac {2^{\frac {-25 \log (x) x^2-x-123}{x-2}} \left (\log (x) x^2+5\right ) \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}}}{(x-2) x}-\frac {2^{\frac {-25 \log (x) x^2-2 x-121}{x-2}} \left (\log (x) x^2+x^2-4 \log (x) x-2 x-5\right ) \log \left (2 x^2\right ) \left (x^2\right )^{\frac {25 \left (\log (x) x^2+5\right )}{2-x}}}{(x-2)^2}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 100 \int \left (\frac {2^{\frac {-25 \log (x) x^2-x-123}{x-2}} \left (\log (x) x^2+5\right ) \left (x^2\right )^{-\frac {25 \left (\log (x) x^2+5\right )}{x-2}}}{(2-x) x}+\frac {2^{\frac {-25 \log (x) x^2-2 x-121}{x-2}} \left (-\log (x) x^2-x^2+4 \log (x) x+2 x+5\right ) \log \left (2 x^2\right ) \left (x^2\right )^{-\frac {25 \left (\log (x) x^2+5\right )}{x-2}}}{(2-x)^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle 100 \int \left (\frac {2^{\frac {-25 \log (x) x^2-x-123}{x-2}} \left (\log (x) x^2+5\right ) \left (x^2\right )^{-\frac {25 \left (\log (x) x^2+5\right )}{x-2}}}{(2-x) x}+\frac {2^{\frac {-25 \log (x) x^2-2 x-121}{x-2}} \left (-\log (x) x^2-x^2+4 \log (x) x+2 x+5\right ) \log \left (2 x^2\right ) \left (x^2\right )^{-\frac {25 \left (\log (x) x^2+5\right )}{x-2}}}{(2-x)^2}\right )dx\) |
Int[(2^((-125 - 25*x^2*Log[x])/(-2 + x))*(x^2)^((-125 - 25*x^2*Log[x])/(-2 + x))*(500 - 250*x + (100*x^2 - 50*x^3)*Log[x] + (125*x + 50*x^2 - 25*x^3 + (100*x^2 - 25*x^3)*Log[x])*Log[2*x^2]))/(4*x - 4*x^2 + x^3),x]
3.23.87.3.1 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 = 3.76 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.53
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (-25 x^{2} \ln \left (x \right )-125\right ) \ln \left (2 x^{2}\right )}{-2+x}}\) | \(23\) |
risch | \({\mathrm e}^{-\frac {25 \left (5+x^{2} \ln \left (x \right )\right ) \left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+4 \ln \left (x \right )+2 \ln \left (2\right )\right )}{2 \left (-2+x \right )}}\) | \(75\) |
int((((-25*x^3+100*x^2)*ln(x)-25*x^3+50*x^2+125*x)*ln(2*x^2)+(-50*x^3+100* x^2)*ln(x)-250*x+500)*exp((-25*x^2*ln(x)-125)*ln(2*x^2)/(-2+x))/(x^3-4*x^2 +4*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=e^{\left (-\frac {25 \, {\left (2 \, x^{2} \log \left (x\right )^{2} + {\left (x^{2} \log \left (2\right ) + 10\right )} \log \left (x\right ) + 5 \, \log \left (2\right )\right )}}{x - 2}\right )} \]
integrate((((-25*x^3+100*x^2)*log(x)-25*x^3+50*x^2+125*x)*log(2*x^2)+(-50* x^3+100*x^2)*log(x)-250*x+500)*exp((-25*x^2*log(x)-125)*log(2*x^2)/(-2+x)) /(x^3-4*x^2+4*x),x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.56 \[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=e^{\frac {\left (- 25 x^{2} \log {\left (x \right )} - 125\right ) \left (2 \log {\left (x \right )} + \log {\left (2 \right )}\right )}{x - 2}} \]
integrate((((-25*x**3+100*x**2)*ln(x)-25*x**3+50*x**2+125*x)*ln(2*x**2)+(- 50*x**3+100*x**2)*ln(x)-250*x+500)*exp((-25*x**2*ln(x)-125)*ln(2*x**2)/(-2 +x))/(x**3-4*x**2+4*x),x)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (25) = 50\).
Time = 0.40 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.58 \[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=e^{\left (-25 \, x \log \left (2\right ) \log \left (x\right ) - 50 \, x \log \left (x\right )^{2} - 50 \, \log \left (2\right ) \log \left (x\right ) - 100 \, \log \left (x\right )^{2} - \frac {100 \, \log \left (2\right ) \log \left (x\right )}{x - 2} - \frac {200 \, \log \left (x\right )^{2}}{x - 2} - \frac {125 \, \log \left (2\right )}{x - 2} - \frac {250 \, \log \left (x\right )}{x - 2}\right )} \]
integrate((((-25*x^3+100*x^2)*log(x)-25*x^3+50*x^2+125*x)*log(2*x^2)+(-50* x^3+100*x^2)*log(x)-250*x+500)*exp((-25*x^2*log(x)-125)*log(2*x^2)/(-2+x)) /(x^3-4*x^2+4*x),x, algorithm=\
e^(-25*x*log(2)*log(x) - 50*x*log(x)^2 - 50*log(2)*log(x) - 100*log(x)^2 - 100*log(2)*log(x)/(x - 2) - 200*log(x)^2/(x - 2) - 125*log(2)/(x - 2) - 2 50*log(x)/(x - 2))
\[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=\int { -\frac {25 \, {\left ({\left (x^{3} - 2 \, x^{2} + {\left (x^{3} - 4 \, x^{2}\right )} \log \left (x\right ) - 5 \, x\right )} \log \left (2 \, x^{2}\right ) + 2 \, {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x\right ) + 10 \, x - 20\right )}}{{\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \left (2 \, x^{2}\right )^{\frac {25 \, {\left (x^{2} \log \left (x\right ) + 5\right )}}{x - 2}}} \,d x } \]
integrate((((-25*x^3+100*x^2)*log(x)-25*x^3+50*x^2+125*x)*log(2*x^2)+(-50* x^3+100*x^2)*log(x)-250*x+500)*exp((-25*x^2*log(x)-125)*log(2*x^2)/(-2+x)) /(x^3-4*x^2+4*x),x, algorithm=\
integrate(-25*((x^3 - 2*x^2 + (x^3 - 4*x^2)*log(x) - 5*x)*log(2*x^2) + 2*( x^3 - 2*x^2)*log(x) + 10*x - 20)/((x^3 - 4*x^2 + 4*x)*(2*x^2)^(25*(x^2*log (x) + 5)/(x - 2))), x)
Time = 14.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {2^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (x^2\right )^{\frac {-125-25 x^2 \log (x)}{-2+x}} \left (500-250 x+\left (100 x^2-50 x^3\right ) \log (x)+\left (125 x+50 x^2-25 x^3+\left (100 x^2-25 x^3\right ) \log (x)\right ) \log \left (2 x^2\right )\right )}{4 x-4 x^2+x^3} \, dx=\frac {{\left (\frac {1}{42535295865117307932921825928971026432\,x^{250}}\right )}^{\frac {1}{x-2}}}{x^{\frac {25\,\left (x^2\,\ln \left (x^2\right )+x^2\,\ln \left (2\right )\right )}{x-2}}} \]
int((exp(-(log(2*x^2)*(25*x^2*log(x) + 125))/(x - 2))*(log(x)*(100*x^2 - 5 0*x^3) - 250*x + log(2*x^2)*(125*x + log(x)*(100*x^2 - 25*x^3) + 50*x^2 - 25*x^3) + 500))/(4*x - 4*x^2 + x^3),x)