Integrand size = 86, antiderivative size = 19 \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=4 e^{\frac {1+x}{x}} \log \left (\left (-4+e^x\right ) x\right ) \] Output:
4*ln(x*(exp(x)-4))*exp((1+x)/x)
Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=4 e^{1+\frac {1}{x}} \log \left (\left (-4+e^x\right ) x\right ) \] Input:
Integrate[(-16*E^((1 + x)/x)*x + E^(x + (1 + x)/x)*(4*x + 4*x^2) + (16*E^( (1 + x)/x) - 4*E^(x + (1 + x)/x))*Log[-4*x + E^x*x])/(-4*x^2 + E^x*x^2),x]
Output:
4*E^(1 + x^(-1))*Log[(-4 + E^x)*x]
Time = 0.56 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7239, 27, 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.
\(\displaystyle \int \frac {e^{x+\frac {x+1}{x}} \left (4 x^2+4 x\right )-16 e^{\frac {x+1}{x}} x+\left (16 e^{\frac {x+1}{x}}-4 e^{x+\frac {x+1}{x}}\right ) \log \left (e^x x-4 x\right )}{e^x x^2-4 x^2} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 e^{\frac {1}{x}+1} \left (\left (e^x-4\right ) \log \left (\left (e^x-4\right ) x\right )-x \left (e^x (x+1)-4\right )\right )}{\left (4-e^x\right ) x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {e^{1+\frac {1}{x}} \left (x \left (4-e^x (x+1)\right )-\left (4-e^x\right ) \log \left (-\left (\left (4-e^x\right ) x\right )\right )\right )}{\left (4-e^x\right ) x^2}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle 4 e^{\frac {1}{x}+1} \log \left (-\left (\left (4-e^x\right ) x\right )\right )\) |
Input:
Int[(-16*E^((1 + x)/x)*x + E^(x + (1 + x)/x)*(4*x + 4*x^2) + (16*E^((1 + x )/x) - 4*E^(x + (1 + x)/x))*Log[-4*x + E^x*x])/(-4*x^2 + E^x*x^2),x]
Output:
4*E^(1 + x^(-1))*Log[-((4 - E^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[(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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.75 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(4 \ln \left (x \left ({\mathrm e}^{x}-4\right )\right ) {\mathrm e}^{\frac {1+x}{x}}\) | \(18\) |
risch | \(4 \,{\mathrm e}^{\frac {1+x}{x}} \ln \left ({\mathrm e}^{x}-4\right )+2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-4\right )\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right )}^{2} {\mathrm e}^{\frac {1+x}{x}}-2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-4\right )\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right ) \operatorname {csgn}\left (i x \right ) {\mathrm e}^{\frac {1+x}{x}}-2 i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right )}^{3} {\mathrm e}^{\frac {1+x}{x}}+2 i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right )}^{2} \operatorname {csgn}\left (i x \right ) {\mathrm e}^{\frac {1+x}{x}}+4 \,{\mathrm e}^{\frac {1+x}{x}} \ln \left (x \right )\) | \(145\) |
Input:
int(((-4*exp((1+x)/x)*exp(x)+16*exp((1+x)/x))*ln(exp(x)*x-4*x)+(4*x^2+4*x) *exp((1+x)/x)*exp(x)-16*x*exp((1+x)/x))/(exp(x)*x^2-4*x^2),x,method=_RETUR NVERBOSE)
Output:
4*ln(x*(exp(x)-4))*exp((1+x)/x)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=4 \, e^{\left (-x + \frac {x^{2} + x + 1}{x}\right )} \log \left (x e^{x} - 4 \, x\right ) \] Input:
integrate(((-4*exp((1+x)/x)*exp(x)+16*exp((1+x)/x))*log(exp(x)*x-4*x)+(4*x ^2+4*x)*exp((1+x)/x)*exp(x)-16*x*exp((1+x)/x))/(exp(x)*x^2-4*x^2),x, algor ithm="fricas")
Output:
4*e^(-x + (x^2 + x + 1)/x)*log(x*e^x - 4*x)
Timed out. \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=\text {Timed out} \] Input:
integrate(((-4*exp((1+x)/x)*exp(x)+16*exp((1+x)/x))*ln(exp(x)*x-4*x)+(4*x* *2+4*x)*exp((1+x)/x)*exp(x)-16*x*exp((1+x)/x))/(exp(x)*x**2-4*x**2),x)
Output:
Timed out
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=4 \, e^{\left (\frac {1}{x} + 1\right )} \log \left (x\right ) + 4 \, e^{\left (\frac {1}{x} + 1\right )} \log \left (e^{x} - 4\right ) \] Input:
integrate(((-4*exp((1+x)/x)*exp(x)+16*exp((1+x)/x))*log(exp(x)*x-4*x)+(4*x ^2+4*x)*exp((1+x)/x)*exp(x)-16*x*exp((1+x)/x))/(exp(x)*x^2-4*x^2),x, algor ithm="maxima")
Output:
4*e^(1/x + 1)*log(x) + 4*e^(1/x + 1)*log(e^x - 4)
\[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=\int { \frac {4 \, {\left ({\left (x^{2} + x\right )} e^{\left (x + \frac {x + 1}{x}\right )} - 4 \, x e^{\left (\frac {x + 1}{x}\right )} - {\left (e^{\left (x + \frac {x + 1}{x}\right )} - 4 \, e^{\left (\frac {x + 1}{x}\right )}\right )} \log \left (x e^{x} - 4 \, x\right )\right )}}{x^{2} e^{x} - 4 \, x^{2}} \,d x } \] Input:
integrate(((-4*exp((1+x)/x)*exp(x)+16*exp((1+x)/x))*log(exp(x)*x-4*x)+(4*x ^2+4*x)*exp((1+x)/x)*exp(x)-16*x*exp((1+x)/x))/(exp(x)*x^2-4*x^2),x, algor ithm="giac")
Output:
integrate(4*((x^2 + x)*e^(x + (x + 1)/x) - 4*x*e^((x + 1)/x) - (e^(x + (x + 1)/x) - 4*e^((x + 1)/x))*log(x*e^x - 4*x))/(x^2*e^x - 4*x^2), x)
Timed out. \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=\int \frac {\ln \left (x\,{\mathrm {e}}^x-4\,x\right )\,\left (16\,{\mathrm {e}}^{\frac {x+1}{x}}-4\,{\mathrm {e}}^{\frac {x+1}{x}}\,{\mathrm {e}}^x\right )-16\,x\,{\mathrm {e}}^{\frac {x+1}{x}}+{\mathrm {e}}^{\frac {x+1}{x}}\,{\mathrm {e}}^x\,\left (4\,x^2+4\,x\right )}{x^2\,{\mathrm {e}}^x-4\,x^2} \,d x \] Input:
int((log(x*exp(x) - 4*x)*(16*exp((x + 1)/x) - 4*exp((x + 1)/x)*exp(x)) - 1 6*x*exp((x + 1)/x) + exp((x + 1)/x)*exp(x)*(4*x + 4*x^2))/(x^2*exp(x) - 4* x^2),x)
Output:
int((log(x*exp(x) - 4*x)*(16*exp((x + 1)/x) - 4*exp((x + 1)/x)*exp(x)) - 1 6*x*exp((x + 1)/x) + exp((x + 1)/x)*exp(x)*(4*x + 4*x^2))/(x^2*exp(x) - 4* x^2), x)
Time = 2.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} \left (4 x+4 x^2\right )+\left (16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}\right ) \log \left (-4 x+e^x x\right )}{-4 x^2+e^x x^2} \, dx=4 e^{\frac {1}{x}} \mathrm {log}\left (e^{x} x -4 x \right ) e \] Input:
int(((-4*exp((1+x)/x)*exp(x)+16*exp((1+x)/x))*log(exp(x)*x-4*x)+(4*x^2+4*x )*exp((1+x)/x)*exp(x)-16*x*exp((1+x)/x))/(exp(x)*x^2-4*x^2),x)
Output:
4*e**(1/x)*log(e**x*x - 4*x)*e