Integrand size = 45, antiderivative size = 13 \begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx=\left (-3+e^{-1+x} x\right )^{e^3} \end {dmath*}
Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx=\left (-3+e^{-1+x} x\right )^{e^3} \end {dmath*}
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.
\(\displaystyle \int \frac {e^3 (-x-1) \left (e^{x-1} \left (x-3 e^{1-x}\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^3 \int -\frac {\left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} (x+1)}{3 e^{1-x}-x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e^3 \int \frac {\left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} (x+1)}{3 e^{1-x}-x}dx\) |
\(\Big \downarrow \) 7270 |
\(\displaystyle -e^3 \left (e^{x-1}\right )^{-e^3} \left (3 e^{1-x}-x\right )^{-e^3} \left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} \int \left (e^{x-1}\right )^{e^3} \left (3 e^{1-x}-x\right )^{-1+e^3} (x+1)dx\) |
\(\Big \downarrow \) 2717 |
\(\displaystyle -e^{e^3 (1-x)+3} \left (3 e^{1-x}-x\right )^{-e^3} \left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} \int e^{-e^3 (1-x)} \left (3 e^{1-x}-x\right )^{-1+e^3} (x+1)dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -e^{e^3 (1-x)+3} \left (3 e^{1-x}-x\right )^{-e^3} \left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} \int e^{e^3 x-e^3} \left (3 e^{1-x}-x\right )^{-1+e^3} (x+1)dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -e^{e^3 (1-x)+3} \left (3 e^{1-x}-x\right )^{-e^3} \left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} \int \left (e^{e^3 x-e^3} \left (3 e^{1-x}-x\right )^{-1+e^3}+e^{e^3 x-e^3} x \left (3 e^{1-x}-x\right )^{-1+e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^{e^3 (1-x)+3} \left (3 e^{1-x}-x\right )^{-e^3} \left (-e^{x-1} \left (3 e^{1-x}-x\right )\right )^{e^3} \left (\int e^{e^3 x-e^3} \left (3 e^{1-x}-x\right )^{-1+e^3}dx+\int e^{e^3 x-e^3} \left (3 e^{1-x}-x\right )^{-1+e^3} xdx\right )\) |
3.12.36.3.1 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_.)*(F_)^(v_))^(n_), x_Symbol] :> Simp[(a*F^v)^n/F^(n*v) Int [u*F^(n*v), x], x] /; FreeQ[{F, a, n}, x] && !IntegerQ[n]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
Time = 0.81 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.92
method | result | size |
norman | \({\mathrm e}^{{\mathrm e}^{3} \ln \left (\left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{-1+x}\right )}\) | \(25\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{3} \ln \left (-\left (3 \,{\mathrm e}^{1-x}-x \right ) {\mathrm e}^{-1+x}\right )}\) | \(28\) |
risch | \(\left ({\mathrm e}^{1-x}\right )^{-{\mathrm e}^{3}} \left (-3 \,{\mathrm e}^{1-x}+x \right )^{{\mathrm e}^{3}} {\mathrm e}^{-\frac {i \pi \,{\mathrm e}^{3} \operatorname {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{-1+x}\right ) \left (-\operatorname {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{-1+x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-1+x}\right )\right ) \left (-\operatorname {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{-1+x}\right )+\operatorname {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right )\right )\right )}{2}}\) | \(115\) |
int((-1-x)*exp(3)*exp(exp(3)*ln((-3*exp(1-x)+x)/exp(1-x)))/(3*exp(1-x)-x), x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx={\left (x e^{\left (x - 1\right )} - 3\right )}^{e^{3}} \end {dmath*}
integrate((-1-x)*exp(3)*exp(exp(3)*log((-3*exp(1-x)+x)/exp(1-x)))/(3*exp(1 -x)-x),x, algorithm=\
Time = 9.90 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx=\left (\left (x - 3 e^{1 - x}\right ) e^{x - 1}\right )^{e^{3}} \end {dmath*}
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx=e^{\left (e^{3} \log \left (x e^{x} - 3 \, e\right ) - e^{3}\right )} \end {dmath*}
integrate((-1-x)*exp(3)*exp(exp(3)*log((-3*exp(1-x)+x)/exp(1-x)))/(3*exp(1 -x)-x),x, algorithm=\
\begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx=\int { \frac {\left ({\left (x - 3 \, e^{\left (-x + 1\right )}\right )} e^{\left (x - 1\right )}\right )^{e^{3}} {\left (x + 1\right )} e^{3}}{x - 3 \, e^{\left (-x + 1\right )}} \,d x } \end {dmath*}
integrate((-1-x)*exp(3)*exp(exp(3)*log((-3*exp(1-x)+x)/exp(1-x)))/(3*exp(1 -x)-x),x, algorithm=\
Time = 15.88 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \begin {dmath*} \int \frac {e^3 (-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx={\left (x\,{\mathrm {e}}^{x-1}-3\right )}^{{\mathrm {e}}^3} \end {dmath*}