Integrand size = 147, antiderivative size = 27 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=\frac {16}{x \left (2-\log \left (e^{\frac {2+x}{x}} x+x^2\right )\right )} \]
Time = 0.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=-\frac {16}{x \left (-2+\log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )\right )} \]
Integrate[(E^((2 + x)/x)*(-32 - 16*x) + (16*E^((2 + x)/x)*x + 16*x^2)*Log[ E^((2 + x)/x)*x + x^2])/(4*E^((2 + x)/x)*x^3 + 4*x^4 + (-4*E^((2 + x)/x)*x ^3 - 4*x^4)*Log[E^((2 + x)/x)*x + x^2] + (E^((2 + x)/x)*x^3 + x^4)*Log[E^( (2 + x)/x)*x + x^2]^2),x]
Time = 0.97 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7239, 27, 25, 7238}
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.
\(\displaystyle \int \frac {\left (16 x^2+16 e^{\frac {x+2}{x}} x\right ) \log \left (x^2+e^{\frac {x+2}{x}} x\right )+e^{\frac {x+2}{x}} (-16 x-32)}{4 x^4+4 e^{\frac {x+2}{x}} x^3+\left (x^4+e^{\frac {x+2}{x}} x^3\right ) \log ^2\left (x^2+e^{\frac {x+2}{x}} x\right )+\left (-4 x^4-4 e^{\frac {x+2}{x}} x^3\right ) \log \left (x^2+e^{\frac {x+2}{x}} x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {16 \left (x \left (x+e^{\frac {2}{x}+1}\right ) \log \left (x \left (x+e^{\frac {2}{x}+1}\right )\right )-e^{\frac {2}{x}+1} (x+2)\right )}{x^3 \left (x+e^{\frac {2}{x}+1}\right ) \left (2-\log \left (x \left (x+e^{\frac {2}{x}+1}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 16 \int -\frac {e^{1+\frac {2}{x}} (x+2)-x \left (x+e^{1+\frac {2}{x}}\right ) \log \left (x \left (x+e^{1+\frac {2}{x}}\right )\right )}{x^3 \left (x+e^{1+\frac {2}{x}}\right ) \left (2-\log \left (x \left (x+e^{1+\frac {2}{x}}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -16 \int \frac {e^{1+\frac {2}{x}} (x+2)-x \left (x+e^{1+\frac {2}{x}}\right ) \log \left (x \left (x+e^{1+\frac {2}{x}}\right )\right )}{x^3 \left (x+e^{1+\frac {2}{x}}\right ) \left (2-\log \left (x \left (x+e^{1+\frac {2}{x}}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 7238 |
\(\displaystyle \frac {16}{x \left (2-\log \left (x \left (x+e^{\frac {2}{x}+1}\right )\right )\right )}\) |
Int[(E^((2 + x)/x)*(-32 - 16*x) + (16*E^((2 + x)/x)*x + 16*x^2)*Log[E^((2 + x)/x)*x + x^2])/(4*E^((2 + x)/x)*x^3 + 4*x^4 + (-4*E^((2 + x)/x)*x^3 - 4 *x^4)*Log[E^((2 + x)/x)*x + x^2] + (E^((2 + x)/x)*x^3 + x^4)*Log[E^((2 + x )/x)*x + x^2]^2),x]
3.19.31.3.1 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[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y* z, u*z^(n - m), x]}, Simp[q*y^(m + 1)*(z^(m + 1)/(m + 1)), x] /; !FalseQ[q ]] /; FreeQ[{m, n}, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\frac {16}{x \left (\ln \left (x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )-2\right )}\) | \(26\) |
risch | \(-\frac {32 i}{x \left (\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )}^{2}+\pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )}^{3}+2 i \ln \left (x \right )+2 i \ln \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )-4 i\right )}\) | \(145\) |
int(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*ln(x*exp(1/2*(2+x)/x)^2+x^2)+(-16*x- 32)*exp(1/2*(2+x)/x)^2)/((x^3*exp(1/2*(2+x)/x)^2+x^4)*ln(x*exp(1/2*(2+x)/x )^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*ln(x*exp(1/2*(2+x)/x)^2+x^2)+ 4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=-\frac {16}{x \log \left (x^{2} + x e^{\left (\frac {x + 2}{x}\right )}\right ) - 2 \, x} \]
integrate(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*log(x*exp(1/2*(2+x)/x)^2+x^2)+ (-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3*exp(1/2*(2+x)/x)^2+x^4)*log(x*exp(1/2 *(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*log(x*exp(1/2*(2+x)/x )^2+x^2)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=- \frac {16}{x \log {\left (x^{2} + x e^{\frac {2 \left (\frac {x}{2} + 1\right )}{x}} \right )} - 2 x} \]
integrate(((16*x*exp(1/2*(2+x)/x)**2+16*x**2)*ln(x*exp(1/2*(2+x)/x)**2+x** 2)+(-16*x-32)*exp(1/2*(2+x)/x)**2)/((x**3*exp(1/2*(2+x)/x)**2+x**4)*ln(x*e xp(1/2*(2+x)/x)**2+x**2)**2+(-4*x**3*exp(1/2*(2+x)/x)**2-4*x**4)*ln(x*exp( 1/2*(2+x)/x)**2+x**2)+4*x**3*exp(1/2*(2+x)/x)**2+4*x**4),x)
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=-\frac {16}{x \log \left (x + e^{\left (\frac {2}{x} + 1\right )}\right ) + x \log \left (x\right ) - 2 \, x} \]
integrate(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*log(x*exp(1/2*(2+x)/x)^2+x^2)+ (-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3*exp(1/2*(2+x)/x)^2+x^4)*log(x*exp(1/2 *(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*log(x*exp(1/2*(2+x)/x )^2+x^2)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x, algorithm=\
Time = 0.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=-\frac {16}{x \log \left (x^{2} + x e^{\left (\frac {x + 2}{x}\right )}\right ) - 2 \, x} \]
integrate(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*log(x*exp(1/2*(2+x)/x)^2+x^2)+ (-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3*exp(1/2*(2+x)/x)^2+x^4)*log(x*exp(1/2 *(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*log(x*exp(1/2*(2+x)/x )^2+x^2)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x, algorithm=\
Time = 11.92 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {2+x}{x}} (-32-16 x)+\left (16 e^{\frac {2+x}{x}} x+16 x^2\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )}{4 e^{\frac {2+x}{x}} x^3+4 x^4+\left (-4 e^{\frac {2+x}{x}} x^3-4 x^4\right ) \log \left (e^{\frac {2+x}{x}} x+x^2\right )+\left (e^{\frac {2+x}{x}} x^3+x^4\right ) \log ^2\left (e^{\frac {2+x}{x}} x+x^2\right )} \, dx=-\frac {16}{x\,\left (\ln \left (x^2+x\,\mathrm {e}\,{\mathrm {e}}^{2/x}\right )-2\right )} \]
int(-(exp((2*(x/2 + 1))/x)*(16*x + 32) - log(x*exp((2*(x/2 + 1))/x) + x^2) *(16*x*exp((2*(x/2 + 1))/x) + 16*x^2))/(4*x^3*exp((2*(x/2 + 1))/x) - log(x *exp((2*(x/2 + 1))/x) + x^2)*(4*x^3*exp((2*(x/2 + 1))/x) + 4*x^4) + log(x* exp((2*(x/2 + 1))/x) + x^2)^2*(x^3*exp((2*(x/2 + 1))/x) + x^4) + 4*x^4),x)