Integrand size = 148, antiderivative size = 24 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {\left (x^2\right )^{-\left (-x+\log \left ((2+5 x)^2\right )\right )^2}}{x} \] Output:
exp(-(ln((2+5*x)^2)-x)^2*ln(x^2))/x
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {\left (x^2\right )^{-\left (x-\log \left ((2+5 x)^2\right )\right )^2}}{x} \] Input:
Integrate[(E^(-(x^2*Log[x^2]) + 2*x*Log[x^2]*Log[4 + 20*x + 25*x^2] - Log[ x^2]*Log[4 + 20*x + 25*x^2]^2)*(-2 - 5*x - 4*x^2 - 10*x^3 + (16*x^2 - 10*x ^3)*Log[x^2] + (8*x + 20*x^2 + (-16*x + 10*x^2)*Log[x^2])*Log[4 + 20*x + 2 5*x^2] + (-4 - 10*x)*Log[4 + 20*x + 25*x^2]^2))/(2*x^2 + 5*x^3),x]
Output:
1/(x*(x^2)^(x - Log[(2 + 5*x)^2])^2)
Leaf count is larger than twice the leaf count of optimal. \(247\) vs. \(2(24)=48\).
Time = 1.48 (sec) , antiderivative size = 247, normalized size of antiderivative = 10.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2026, 2726}
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.
| \(\displaystyle \int \frac {\left (-10 x^3-4 x^2+(-10 x-4) \log ^2\left (25 x^2+20 x+4\right )+\left (20 x^2+\left (10 x^2-16 x\right ) \log \left (x^2\right )+8 x\right ) \log \left (25 x^2+20 x+4\right )+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )-5 x-2\right ) \exp \left (-\log \left (x^2\right ) \log ^2\left (25 x^2+20 x+4\right )+x^2 \left (-\log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (25 x^2+20 x+4\right )\right )}{5 x^3+2 x^2} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (-10 x^3-4 x^2+(-10 x-4) \log ^2\left (25 x^2+20 x+4\right )+\left (20 x^2+\left (10 x^2-16 x\right ) \log \left (x^2\right )+8 x\right ) \log \left (25 x^2+20 x+4\right )+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )-5 x-2\right ) \exp \left (-\log \left (x^2\right ) \log ^2\left (25 x^2+20 x+4\right )+x^2 \left (-\log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (25 x^2+20 x+4\right )\right )}{x^2 (5 x+2)}dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle \frac {\left (x^2\right )^{-x^2-\log ^2\left (25 x^2+20 x+4\right )+2 x \log \left (25 x^2+20 x+4\right )-1} \left (5 x^3+2 x^2+(5 x+2) \log ^2\left (25 x^2+20 x+4\right )-\left (10 x^2-\left (8 x-5 x^2\right ) \log \left (x^2\right )+4 x\right ) \log \left (25 x^2+20 x+4\right )-\left (8 x^2-5 x^3\right ) \log \left (x^2\right )\right )}{(5 x+2) \left (\frac {\log ^2\left (25 x^2+20 x+4\right )}{x}+\frac {10 (5 x+2) \log \left (x^2\right ) \log \left (25 x^2+20 x+4\right )}{25 x^2+20 x+4}-\log \left (x^2\right ) \log \left (25 x^2+20 x+4\right )-2 \log \left (25 x^2+20 x+4\right )+x \log \left (x^2\right )-\frac {10 x (5 x+2) \log \left (x^2\right )}{25 x^2+20 x+4}+x\right )}\) |
Input:
Int[(E^(-(x^2*Log[x^2]) + 2*x*Log[x^2]*Log[4 + 20*x + 25*x^2] - Log[x^2]*L og[4 + 20*x + 25*x^2]^2)*(-2 - 5*x - 4*x^2 - 10*x^3 + (16*x^2 - 10*x^3)*Lo g[x^2] + (8*x + 20*x^2 + (-16*x + 10*x^2)*Log[x^2])*Log[4 + 20*x + 25*x^2] + (-4 - 10*x)*Log[4 + 20*x + 25*x^2]^2))/(2*x^2 + 5*x^3),x]
Output:
((x^2)^(-1 - x^2 + 2*x*Log[4 + 20*x + 25*x^2] - Log[4 + 20*x + 25*x^2]^2)* (2*x^2 + 5*x^3 - (8*x^2 - 5*x^3)*Log[x^2] - (4*x + 10*x^2 - (8*x - 5*x^2)* Log[x^2])*Log[4 + 20*x + 25*x^2] + (2 + 5*x)*Log[4 + 20*x + 25*x^2]^2))/(( 2 + 5*x)*(x + x*Log[x^2] - (10*x*(2 + 5*x)*Log[x^2])/(4 + 20*x + 25*x^2) - 2*Log[4 + 20*x + 25*x^2] - Log[x^2]*Log[4 + 20*x + 25*x^2] + (10*(2 + 5*x )*Log[x^2]*Log[4 + 20*x + 25*x^2])/(4 + 20*x + 25*x^2) + Log[4 + 20*x + 25 *x^2]^2/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[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 4.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79
| method | result | size |
| parallelrisch | \(\frac {{\mathrm e}^{-\ln \left (x^{2}\right ) \left (\ln \left (25 x^{2}+20 x +4\right )^{2}-2 \ln \left (25 x^{2}+20 x +4\right ) x +x^{2}\right )}}{x}\) | \(43\) |
| risch | \(\frac {{\mathrm e}^{-\frac {\left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+4 \ln \left (x \right )\right ) {\left (i \pi \operatorname {csgn}\left (i \left (x +\frac {2}{5}\right )^{2}\right )^{3}-2 i \pi \operatorname {csgn}\left (i \left (x +\frac {2}{5}\right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (x +\frac {2}{5}\right )\right )+i \pi \,\operatorname {csgn}\left (i \left (x +\frac {2}{5}\right )^{2}\right ) \operatorname {csgn}\left (i \left (x +\frac {2}{5}\right )\right )^{2}-4 \ln \left (x +\frac {2}{5}\right )+2 x \right )}^{2}}{8}}}{x}\) | \(133\) |
Input:
int(((-10*x-4)*ln(25*x^2+20*x+4)^2+((10*x^2-16*x)*ln(x^2)+20*x^2+8*x)*ln(2 5*x^2+20*x+4)+(-10*x^3+16*x^2)*ln(x^2)-10*x^3-4*x^2-5*x-2)*exp(-ln(x^2)*ln (25*x^2+20*x+4)^2+2*x*ln(x^2)*ln(25*x^2+20*x+4)-x^2*ln(x^2))/(5*x^3+2*x^2) ,x,method=_RETURNVERBOSE)
Output:
exp(-ln(x^2)*(ln(25*x^2+20*x+4)^2-2*ln(25*x^2+20*x+4)*x+x^2))/x
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {e^{\left (-x^{2} \log \left (x^{2}\right ) + 2 \, x \log \left (25 \, x^{2} + 20 \, x + 4\right ) \log \left (x^{2}\right ) - \log \left (25 \, x^{2} + 20 \, x + 4\right )^{2} \log \left (x^{2}\right )\right )}}{x} \] Input:
integrate(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8 *x)*log(25*x^2+20*x+4)+(-10*x^3+16*x^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(- log(x^2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^2) )/(5*x^3+2*x^2),x, algorithm="fricas")
Output:
e^(-x^2*log(x^2) + 2*x*log(25*x^2 + 20*x + 4)*log(x^2) - log(25*x^2 + 20*x + 4)^2*log(x^2))/x
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).
Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {e^{- x^{2} \log {\left (x^{2} \right )} + 2 x \log {\left (x^{2} \right )} \log {\left (25 x^{2} + 20 x + 4 \right )} - \log {\left (x^{2} \right )} \log {\left (25 x^{2} + 20 x + 4 \right )}^{2}}}{x} \] Input:
integrate(((-10*x-4)*ln(25*x**2+20*x+4)**2+((10*x**2-16*x)*ln(x**2)+20*x** 2+8*x)*ln(25*x**2+20*x+4)+(-10*x**3+16*x**2)*ln(x**2)-10*x**3-4*x**2-5*x-2 )*exp(-ln(x**2)*ln(25*x**2+20*x+4)**2+2*x*ln(x**2)*ln(25*x**2+20*x+4)-x**2 *ln(x**2))/(5*x**3+2*x**2),x)
Output:
exp(-x**2*log(x**2) + 2*x*log(x**2)*log(25*x**2 + 20*x + 4) - log(x**2)*lo g(25*x**2 + 20*x + 4)**2)/x
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {e^{\left (-2 \, x^{2} \log \left (x\right ) + 8 \, x \log \left (5 \, x + 2\right ) \log \left (x\right ) - 8 \, \log \left (5 \, x + 2\right )^{2} \log \left (x\right )\right )}}{x} \] Input:
integrate(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8 *x)*log(25*x^2+20*x+4)+(-10*x^3+16*x^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(- log(x^2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^2) )/(5*x^3+2*x^2),x, algorithm="maxima")
Output:
e^(-2*x^2*log(x) + 8*x*log(5*x + 2)*log(x) - 8*log(5*x + 2)^2*log(x))/x
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 1.71 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {e^{\left (-x^{2} \log \left (x^{2}\right ) + 2 \, x \log \left (25 \, x^{2} + 20 \, x + 4\right ) \log \left (x^{2}\right ) - \log \left (25 \, x^{2} + 20 \, x + 4\right )^{2} \log \left (x^{2}\right )\right )}}{x} \] Input:
integrate(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8 *x)*log(25*x^2+20*x+4)+(-10*x^3+16*x^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(- log(x^2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^2) )/(5*x^3+2*x^2),x, algorithm="giac")
Output:
e^(-x^2*log(x^2) + 2*x*log(25*x^2 + 20*x + 4)*log(x^2) - log(25*x^2 + 20*x + 4)^2*log(x^2))/x
Time = 8.38 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx=\frac {{\left (x^2\right )}^{2\,x\,\ln \left (25\,x^2+20\,x+4\right )}}{x\,{\left (x^2\right )}^{x^2}\,{\left (x^2\right )}^{{\ln \left (25\,x^2+20\,x+4\right )}^2}} \] Input:
int(-(exp(2*x*log(x^2)*log(20*x + 25*x^2 + 4) - log(x^2)*log(20*x + 25*x^2 + 4)^2 - x^2*log(x^2))*(5*x - log(x^2)*(16*x^2 - 10*x^3) - log(20*x + 25* x^2 + 4)*(8*x - log(x^2)*(16*x - 10*x^2) + 20*x^2) + log(20*x + 25*x^2 + 4 )^2*(10*x + 4) + 4*x^2 + 10*x^3 + 2))/(2*x^2 + 5*x^3),x)
Output:
(x^2)^(2*x*log(20*x + 25*x^2 + 4))/(x*(x^2)^(x^2)*(x^2)^(log(20*x + 25*x^2 + 4)^2))
\[ \int \frac {e^{-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )} \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{2 x^2+5 x^3} \, dx =\text {Too large to display} \] Input:
int(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8*x)*lo g(25*x^2+20*x+4)+(-10*x^3+16*x^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(-log(x^ 2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^2))/(5*x ^3+2*x^2),x)
Output:
- 2*int(x**(4*log(25*x**2 + 20*x + 4)*x)/(5*x**(2*log(25*x**2 + 20*x + 4) **2 + 2*x**2)*x**3 + 2*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x**2),x) - 5*int(x**(4*log(25*x**2 + 20*x + 4)*x)/(5*x**(2*log(25*x**2 + 20*x + 4) **2 + 2*x**2)*x**2 + 2*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x),x) - 4*int(x**(4*log(25*x**2 + 20*x + 4)*x)/(5*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x + 2*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)),x) - 4*int((x **(4*log(25*x**2 + 20*x + 4)*x)*log(25*x**2 + 20*x + 4)**2)/(5*x**(2*log(2 5*x**2 + 20*x + 4)**2 + 2*x**2)*x**3 + 2*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x**2),x) - 10*int((x**(4*log(25*x**2 + 20*x + 4)*x)*log(25*x**2 + 20*x + 4)**2)/(5*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x**2 + 2*x** (2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x),x) - 16*int((x**(4*log(25*x**2 + 20*x + 4)*x)*log(x**2)*log(25*x**2 + 20*x + 4))/(5*x**(2*log(25*x**2 + 2 0*x + 4)**2 + 2*x**2)*x**2 + 2*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)* x),x) + 10*int((x**(4*log(25*x**2 + 20*x + 4)*x)*log(x**2)*log(25*x**2 + 2 0*x + 4))/(5*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x + 2*x**(2*log(25 *x**2 + 20*x + 4)**2 + 2*x**2)),x) - 10*int((x**(4*log(25*x**2 + 20*x + 4) *x)*log(x**2)*x)/(5*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x + 2*x**(2 *log(25*x**2 + 20*x + 4)**2 + 2*x**2)),x) + 16*int((x**(4*log(25*x**2 + 20 *x + 4)*x)*log(x**2))/(5*x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)*x + 2* x**(2*log(25*x**2 + 20*x + 4)**2 + 2*x**2)),x) + 8*int((x**(4*log(25*x*...