Integrand size = 78, antiderivative size = 27 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx=-5+e^{\frac {4 e^{2+4 e^{e^x}} (1+x)^2}{(3+x)^2}} \]
Time = 2.80 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx=e^{\frac {4 e^{2+4 e^{e^x}} (1+x)^2}{(3+x)^2}} \]
Integrate[(E^(2 + 4*E^E^x + (E^(2 + 4*E^E^x)*(2 + 2*x)^2)/(3 + x)^2)*(2 + 2*x)^2*(4 + E^(E^x + x)*(12 + 16*x + 4*x^2)))/((3 + x)^2*(3 + 4*x + x^2)), 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.
\(\displaystyle \int \frac {(2 x+2)^2 \left (e^{x+e^x} \left (4 x^2+16 x+12\right )+4\right ) \exp \left (\frac {e^{4 e^{e^x}+2} (2 x+2)^2}{(x+3)^2}+4 e^{e^x}+2\right )}{(x+3)^2 \left (x^2+4 x+3\right )} \, dx\) |
\(\Big \downarrow \) 2004 |
\(\displaystyle \int \frac {(2 x+2)^2 \left (e^{x+e^x} \left (4 x^2+16 x+12\right )+4\right ) \exp \left (\frac {e^{4 e^{e^x}+2} (2 x+2)^2}{(x+3)^2}+4 e^{e^x}+2\right )}{(x+1) (x+3)^3}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle 4 \int \frac {4 \exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+2\right ) (x+1) \left (e^{x+e^x} \left (x^2+4 x+3\right )+1\right )}{(x+3)^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 16 \int \frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+2\right ) (x+1) \left (e^{x+e^x} \left (x^2+4 x+3\right )+1\right )}{(x+3)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 16 \int \left (\frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+e^x+x+2\right ) (x+1)^2}{(x+3)^2}+\frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+2\right ) (x+1)}{(x+3)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 16 \left (\int \exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+e^x+x+2\right )dx-2 \int \frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+2\right )}{(x+3)^3}dx+\int \frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+2\right )}{(x+3)^2}dx+4 \int \frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+e^x+x+2\right )}{(x+3)^2}dx-4 \int \frac {\exp \left (\frac {4 e^{2+4 e^{e^x}} (x+1)^2}{(x+3)^2}+4 e^{e^x}+e^x+x+2\right )}{x+3}dx\right )\) |
Int[(E^(2 + 4*E^E^x + (E^(2 + 4*E^E^x)*(2 + 2*x)^2)/(3 + x)^2)*(2 + 2*x)^2 *(4 + E^(E^x + x)*(12 + 16*x + 4*x^2)))/((3 + x)^2*(3 + 4*x + x^2)),x]
3.16.4.3.1 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_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[(u_)*((d_) + (e_.)*(x_))^(q_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.) , x_Symbol] :> Int[u*(d + e*x)^(p + q)*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, b , c, d, e, q}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]
Time = 233.87 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {4 \left (1+x \right )^{2} {\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{x}}+2}}{\left (3+x \right )^{2}}}\) | \(23\) |
risch | \({\mathrm e}^{\frac {4 \left (1+x \right )^{2} {\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{x}}+2-i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{3+x}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{3+x}\right )^{2} \operatorname {csgn}\left (i \left (1+x \right )\right )+i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{3+x}\right )^{2} \operatorname {csgn}\left (\frac {i}{3+x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i \left (1+x \right )}{3+x}\right ) \operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (\frac {i}{3+x}\right )}}{\left (3+x \right )^{2}}}\) | \(124\) |
int(((4*x^2+16*x+12)*exp(x)*exp(exp(x))+4)*exp(2*exp(exp(x))+ln((2+2*x)/(3 +x))+1)^2*exp(exp(2*exp(exp(x))+ln((2+2*x)/(3+x))+1)^2)/(x^2+4*x+3),x,meth od=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (24) = 48\).
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.59 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx=e^{\left ({\left (2 \, e^{x} \log \left (\frac {2 \, {\left (x + 1\right )}}{x + 3}\right ) + e^{\left (2 \, {\left (e^{x} \log \left (\frac {2 \, {\left (x + 1\right )}}{x + 3}\right ) + 2 \, e^{\left (x + e^{x}\right )} + e^{x}\right )} e^{\left (-x\right )} + x\right )} + 4 \, e^{\left (x + e^{x}\right )} + 2 \, e^{x}\right )} e^{\left (-x\right )} - 2 \, {\left (e^{x} \log \left (\frac {2 \, {\left (x + 1\right )}}{x + 3}\right ) + 2 \, e^{\left (x + e^{x}\right )} + e^{x}\right )} e^{\left (-x\right )}\right )} \]
integrate(((4*x^2+16*x+12)*exp(x)*exp(exp(x))+4)*exp(2*exp(exp(x))+log((2+ 2*x)/(3+x))+1)^2*exp(exp(2*exp(exp(x))+log((2+2*x)/(3+x))+1)^2)/(x^2+4*x+3 ),x, algorithm=\
e^((2*e^x*log(2*(x + 1)/(x + 3)) + e^(2*(e^x*log(2*(x + 1)/(x + 3)) + 2*e^ (x + e^x) + e^x)*e^(-x) + x) + 4*e^(x + e^x) + 2*e^x)*e^(-x) - 2*(e^x*log( 2*(x + 1)/(x + 3)) + 2*e^(x + e^x) + e^x)*e^(-x))
Time = 1.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx=e^{\frac {\left (2 x + 2\right )^{2} e^{4 e^{e^{x}} + 2}}{\left (x + 3\right )^{2}}} \]
integrate(((4*x**2+16*x+12)*exp(x)*exp(exp(x))+4)*exp(2*exp(exp(x))+ln((2+ 2*x)/(3+x))+1)**2*exp(exp(2*exp(exp(x))+ln((2+2*x)/(3+x))+1)**2)/(x**2+4*x +3),x)
Time = 0.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx=e^{\left (\frac {16 \, e^{\left (4 \, e^{\left (e^{x}\right )} + 2\right )}}{x^{2} + 6 \, x + 9} - \frac {16 \, e^{\left (4 \, e^{\left (e^{x}\right )} + 2\right )}}{x + 3} + 4 \, e^{\left (4 \, e^{\left (e^{x}\right )} + 2\right )}\right )} \]
integrate(((4*x^2+16*x+12)*exp(x)*exp(exp(x))+4)*exp(2*exp(exp(x))+log((2+ 2*x)/(3+x))+1)^2*exp(exp(2*exp(exp(x))+log((2+2*x)/(3+x))+1)^2)/(x^2+4*x+3 ),x, algorithm=\
e^(16*e^(4*e^(e^x) + 2)/(x^2 + 6*x + 9) - 16*e^(4*e^(e^x) + 2)/(x + 3) + 4 *e^(4*e^(e^x) + 2))
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx=e^{\left (\frac {4 \, {\left (x^{2} e^{\left (4 \, e^{\left (e^{x}\right )} + 2\right )} + 2 \, x e^{\left (4 \, e^{\left (e^{x}\right )} + 2\right )} + e^{\left (4 \, e^{\left (e^{x}\right )} + 2\right )}\right )}}{x^{2} + 6 \, x + 9}\right )} \]
integrate(((4*x^2+16*x+12)*exp(x)*exp(exp(x))+4)*exp(2*exp(exp(x))+log((2+ 2*x)/(3+x))+1)^2*exp(exp(2*exp(exp(x))+log((2+2*x)/(3+x))+1)^2)/(x^2+4*x+3 ),x, algorithm=\
Time = 15.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \[ \int \frac {e^{2+4 e^{e^x}+\frac {e^{2+4 e^{e^x}} (2+2 x)^2}{(3+x)^2}} (2+2 x)^2 \left (4+e^{e^x+x} \left (12+16 x+4 x^2\right )\right )}{(3+x)^2 \left (3+4 x+x^2\right )} \, dx={\mathrm {e}}^{\frac {8\,x\,{\mathrm {e}}^{4\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\mathrm {e}}^2}{x^2+6\,x+9}}\,{\mathrm {e}}^{\frac {4\,x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\mathrm {e}}^2}{x^2+6\,x+9}}\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{4\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\mathrm {e}}^2}{x^2+6\,x+9}} \]
int((exp(2*log((2*x + 2)/(x + 3)) + 4*exp(exp(x)) + 2)*exp(exp(2*log((2*x + 2)/(x + 3)) + 4*exp(exp(x)) + 2))*(exp(exp(x))*exp(x)*(16*x + 4*x^2 + 12 ) + 4))/(4*x + x^2 + 3),x)