Integrand size = 92, antiderivative size = 29 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=\frac {4}{2-x+x \left (4+\log \left (e^{2 e^{4 x}-2 x} x\right )\right )} \] Output:
4/((4+ln(x*exp(exp(4*x))^2/exp(x)^2))*x+2-x)
Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=\frac {4}{2+3 x+x \log \left (e^{2 e^{4 x}-2 x} x\right )} \] Input:
Integrate[(-16 + 8*x - 32*E^(4*x)*x - 4*Log[E^(2*E^(4*x) - 2*x)*x])/(4 + 1 2*x + 9*x^2 + (4*x + 6*x^2)*Log[E^(2*E^(4*x) - 2*x)*x] + x^2*Log[E^(2*E^(4 *x) - 2*x)*x]^2),x]
Output:
4/(2 + 3*x + x*Log[E^(2*E^(4*x) - 2*x)*x])
Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {7292, 27, 25, 7237}
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.
\(\displaystyle \int \frac {-32 e^{4 x} x+8 x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )-16}{9 x^2+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )+\left (6 x^2+4 x\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+12 x+4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 \left (-8 e^{4 x} x+2 x-\log \left (e^{2 e^{4 x}-2 x} x\right )-4\right )}{\left (3 x+x \log \left (e^{2 e^{4 x}-2 x} x\right )+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {8 e^{4 x} x-2 x+\log \left (e^{2 e^{4 x}-2 x} x\right )+4}{\left (\log \left (e^{2 e^{4 x}-2 x} x\right ) x+3 x+2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {8 e^{4 x} x-2 x+\log \left (e^{2 e^{4 x}-2 x} x\right )+4}{\left (\log \left (e^{2 e^{4 x}-2 x} x\right ) x+3 x+2\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {4}{3 x+x \log \left (e^{2 e^{4 x}-2 x} x\right )+2}\) |
Input:
Int[(-16 + 8*x - 32*E^(4*x)*x - 4*Log[E^(2*E^(4*x) - 2*x)*x])/(4 + 12*x + 9*x^2 + (4*x + 6*x^2)*Log[E^(2*E^(4*x) - 2*x)*x] + x^2*Log[E^(2*E^(4*x) - 2*x)*x]^2),x]
Output:
4/(2 + 3*x + x*Log[E^(2*E^(4*x) - 2*x)*x])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 1.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {4}{x \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}} {\mathrm e}^{-2 x}\right )+3 x +2}\) | \(26\) |
risch | \(-\frac {8 i}{\pi x \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )-2 \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+\pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-\pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )+\pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2}-\pi x \operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{4 x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )+2 \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{4 x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )^{2}-\pi x \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )^{3}+\pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2}-\pi x \operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{3}+\pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2}-\pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right ) \operatorname {csgn}\left (i x \right )-\pi x \operatorname {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{3}+\pi x \operatorname {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2} \operatorname {csgn}\left (i x \right )-4 i x \ln \left ({\mathrm e}^{{\mathrm e}^{4 x}}\right )-6 i x +4 i x \ln \left ({\mathrm e}^{x}\right )-2 i x \ln \left (x \right )-4 i}\) | \(399\) |
Input:
int((-4*ln(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*ln(x*exp (exp(4*x))^2/exp(x)^2)^2+(6*x^2+4*x)*ln(x*exp(exp(4*x))^2/exp(x)^2)+9*x^2+ 12*x+4),x,method=_RETURNVERBOSE)
Output:
4/(x*ln(x*exp(exp(4*x))^2/exp(x)^2)+3*x+2)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=\frac {4}{x \log \left (x e^{\left (-2 \, x + 2 \, e^{\left (4 \, x\right )}\right )}\right ) + 3 \, x + 2} \] Input:
integrate((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*l og(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^ 2)+9*x^2+12*x+4),x, algorithm="fricas")
Output:
4/(x*log(x*e^(-2*x + 2*e^(4*x))) + 3*x + 2)
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=\frac {4}{x \log {\left (x e^{- 2 x} e^{2 e^{4 x}} \right )} + 3 x + 2} \] Input:
integrate((-4*ln(x*exp(exp(4*x))**2/exp(x)**2)-32*x*exp(4*x)+8*x-16)/(x**2 *ln(x*exp(exp(4*x))**2/exp(x)**2)**2+(6*x**2+4*x)*ln(x*exp(exp(4*x))**2/ex p(x)**2)+9*x**2+12*x+4),x)
Output:
4/(x*log(x*exp(-2*x)*exp(2*exp(4*x))) + 3*x + 2)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=-\frac {4}{2 \, x^{2} - 2 \, x e^{\left (4 \, x\right )} - x \log \left (x\right ) - 3 \, x - 2} \] Input:
integrate((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*l og(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^ 2)+9*x^2+12*x+4),x, algorithm="maxima")
Output:
-4/(2*x^2 - 2*x*e^(4*x) - x*log(x) - 3*x - 2)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=-\frac {4}{2 \, x^{2} - 2 \, x e^{\left (4 \, x\right )} - x \log \left (x\right ) - 3 \, x - 2} \] Input:
integrate((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*l og(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^ 2)+9*x^2+12*x+4),x, algorithm="giac")
Output:
-4/(2*x^2 - 2*x*e^(4*x) - x*log(x) - 3*x - 2)
Time = 2.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=\frac {4}{x\,\left (2\,{\mathrm {e}}^{4\,x}+\ln \left (x\right )+3\right )-2\,x^2+2} \] Input:
int(-(4*log(x*exp(2*exp(4*x))*exp(-2*x)) - 8*x + 32*x*exp(4*x) + 16)/(12*x + x^2*log(x*exp(2*exp(4*x))*exp(-2*x))^2 + log(x*exp(2*exp(4*x))*exp(-2*x ))*(4*x + 6*x^2) + 9*x^2 + 4),x)
Output:
4/(x*(2*exp(4*x) + log(x) + 3) - 2*x^2 + 2)
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {-16+8 x-32 e^{4 x} x-4 \log \left (e^{2 e^{4 x}-2 x} x\right )}{4+12 x+9 x^2+\left (4 x+6 x^2\right ) \log \left (e^{2 e^{4 x}-2 x} x\right )+x^2 \log ^2\left (e^{2 e^{4 x}-2 x} x\right )} \, dx=\frac {2 x \left (-\mathrm {log}\left (\frac {e^{2 e^{4 x}} x}{e^{2 x}}\right )-3\right )}{\mathrm {log}\left (\frac {e^{2 e^{4 x}} x}{e^{2 x}}\right ) x +3 x +2} \] Input:
int((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*log(x*e xp(exp(4*x))^2/exp(x)^2)^2+(6*x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^2)+9*x ^2+12*x+4),x)
Output:
(2*x*( - log((e**(2*e**(4*x))*x)/e**(2*x)) - 3))/(log((e**(2*e**(4*x))*x)/ e**(2*x))*x + 3*x + 2)