Integrand size = 76, antiderivative size = 30 \[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=e^{-e^{-5+e^5-x}} x \left (4+4 (4-x)-x^2\right ) \] Output:
exp(ln((-x^2-4*x+20)*x)-exp(exp(5)-x-5))
Time = 1.51 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=-e^{-e^{-5+e^5-x}} x \left (-20+4 x+x^2\right ) \] Input:
Integrate[((20*x - 4*x^2 - x^3)*(-20 + 8*x + 3*x^2 + E^(-5 + E^5 - x)*(-20 *x + 4*x^2 + x^3)))/(E^E^(-5 + E^5 - x)*(-20*x + 4*x^2 + x^3)),x]
Output:
-((x*(-20 + 4*x + x^2))/E^E^(-5 + E^5 - x))
Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2019, 25, 25, 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.
\(\displaystyle \int \frac {e^{-e^{-x+e^5-5}} \left (-x^3-4 x^2+20 x\right ) \left (3 x^2+e^{-x+e^5-5} \left (x^3+4 x^2-20 x\right )+8 x-20\right )}{x^3+4 x^2-20 x} \, dx\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle \int -e^{-e^{-x+e^5-5}} \left (3 x^2+e^{-x+e^5-5} \left (x^3+4 x^2-20 x\right )+8 x-20\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -e^{-e^{-x+e^5-5}} \left (-3 x^2-8 x+e^{-x+e^5-5} \left (-x^3-4 x^2+20 x\right )+20\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int e^{-e^{-x+e^5-5}} \left (-3 x^2+e^{-x+e^5-5} \left (-x^3-4 x^2+20 x\right )-8 x+20\right )dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle e^{-e^{-x+e^5-5}} \left (-x^3-4 x^2+20 x\right )\) |
Input:
Int[((20*x - 4*x^2 - x^3)*(-20 + 8*x + 3*x^2 + E^(-5 + E^5 - x)*(-20*x + 4 *x^2 + x^3)))/(E^E^(-5 + E^5 - x)*(-20*x + 4*x^2 + x^3)),x]
Output:
(20*x - 4*x^2 - x^3)/E^E^(-5 + E^5 - x)
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
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]
Time = 1.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \({\mathrm e}^{\ln \left (-x^{3}-4 x^{2}+20 x \right )-{\mathrm e}^{{\mathrm e}^{5}-x -5}}\) | \(28\) |
risch | \(-x \left (x^{2}+4 x -20\right ) {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )}^{2} \operatorname {csgn}\left (i x \right )}{2}+\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+4 x -20\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x^{2}+4 x -20\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (x^{2}+4 x -20\right )\right )}{2}-i \pi {\operatorname {csgn}\left (i x \left (x^{2}+4 x -20\right )\right )}^{2}-{\mathrm e}^{{\mathrm e}^{5}-x -5}}\) | \(151\) |
Input:
int(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(ln(-x^3-4*x^2+20*x )-exp(exp(5)-x-5))/(x^3+4*x^2-20*x),x,method=_RETURNVERBOSE)
Output:
exp(ln(-x^3-4*x^2+20*x)-exp(exp(5)-x-5))
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=e^{\left (-e^{\left (-x + e^{5} - 5\right )} + \log \left (-x^{3} - 4 \, x^{2} + 20 \, x\right )\right )} \] Input:
integrate(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x ^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2-20*x),x, algorithm="fricas")
Output:
e^(-e^(-x + e^5 - 5) + log(-x^3 - 4*x^2 + 20*x))
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=\left (- x^{3} - 4 x^{2} + 20 x\right ) e^{- e^{- x - 5 + e^{5}}} \] Input:
integrate(((x**3+4*x**2-20*x)*exp(exp(5)-x-5)+3*x**2+8*x-20)*exp(ln(-x**3- 4*x**2+20*x)-exp(exp(5)-x-5))/(x**3+4*x**2-20*x),x)
Output:
(-x**3 - 4*x**2 + 20*x)*exp(-exp(-x - 5 + exp(5)))
\[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=\int { \frac {{\left (3 \, x^{2} + {\left (x^{3} + 4 \, x^{2} - 20 \, x\right )} e^{\left (-x + e^{5} - 5\right )} + 8 \, x - 20\right )} e^{\left (-e^{\left (-x + e^{5} - 5\right )} + \log \left (-x^{3} - 4 \, x^{2} + 20 \, x\right )\right )}}{x^{3} + 4 \, x^{2} - 20 \, x} \,d x } \] Input:
integrate(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x ^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2-20*x),x, algorithm="maxima")
Output:
integrate(-(3*x^2 + (x^3 + 4*x^2 - 20*x)*e^(-x + e^5 - 5) + 8*x - 20)*e^(- e^(-x + e^5 - 5)), x)
\[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=\int { \frac {{\left (3 \, x^{2} + {\left (x^{3} + 4 \, x^{2} - 20 \, x\right )} e^{\left (-x + e^{5} - 5\right )} + 8 \, x - 20\right )} e^{\left (-e^{\left (-x + e^{5} - 5\right )} + \log \left (-x^{3} - 4 \, x^{2} + 20 \, x\right )\right )}}{x^{3} + 4 \, x^{2} - 20 \, x} \,d x } \] Input:
integrate(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x ^2+20*x)-exp(exp(5)-x-5))/(x^3+4*x^2-20*x),x, algorithm="giac")
Output:
integrate((3*x^2 + (x^3 + 4*x^2 - 20*x)*e^(-x + e^5 - 5) + 8*x - 20)*e^(-e ^(-x + e^5 - 5) + log(-x^3 - 4*x^2 + 20*x))/(x^3 + 4*x^2 - 20*x), x)
Time = 1.83 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=-x\,{\mathrm {e}}^{-{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^5}}\,\left (x^2+4\,x-20\right ) \] Input:
int((exp(log(20*x - 4*x^2 - x^3) - exp(exp(5) - x - 5))*(8*x + exp(exp(5) - x - 5)*(4*x^2 - 20*x + x^3) + 3*x^2 - 20))/(4*x^2 - 20*x + x^3),x)
Output:
-x*exp(-exp(-x)*exp(-5)*exp(exp(5)))*(4*x + x^2 - 20)
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^{-5+e^5-x}} \left (20 x-4 x^2-x^3\right ) \left (-20+8 x+3 x^2+e^{-5+e^5-x} \left (-20 x+4 x^2+x^3\right )\right )}{-20 x+4 x^2+x^3} \, dx=\frac {x \left (-x^{2}-4 x +20\right )}{e^{\frac {e^{e^{5}}}{e^{x} e^{5}}}} \] Input:
int(((x^3+4*x^2-20*x)*exp(exp(5)-x-5)+3*x^2+8*x-20)*exp(log(-x^3-4*x^2+20* x)-exp(exp(5)-x-5))/(x^3+4*x^2-20*x),x)
Output:
(x*( - x**2 - 4*x + 20))/e**(e**(e**5)/(e**x*e**5))