Integrand size = 112, antiderivative size = 31 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=\frac {e^{e^3-x} \left (1-e+\frac {5}{x}-\log (2)\right )}{x+x^2} \] Output:
exp(ln((1-ln(2)-exp(1)+5/x)/(x^2+x))-x+exp(3))
Time = 5.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=\frac {e^{e^3-x} (5-x (-1+e+\log (2)))}{x^2 (1+x)} \] Input:
Integrate[(E^(E^3 - x)*(5 + x - E*x - x*Log[2])*(10 + 21*x + 8*x^2 + x^3 + E*(-x - 3*x^2 - x^3) + (-x - 3*x^2 - x^3)*Log[2]))/((x^2 + x^3)*(-5*x - 6 *x^2 - x^3 + E*(x^2 + x^3) + (x^2 + x^3)*Log[2])),x]
Output:
(E^(E^3 - x)*(5 - x*(-1 + E + Log[2])))/(x^2*(1 + x))
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.61 (sec) , antiderivative size = 116, normalized size of antiderivative = 3.74, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6, 6, 6, 6, 2019, 3, 2026, 9, 25, 7292, 7293, 2009}
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.
| \(\displaystyle \int \frac {e^{e^3-x} (-e x+x+x (-\log (2))+5) \left (x^3+8 x^2+e \left (-x^3-3 x^2-x\right )+\left (-x^3-3 x^2-x\right ) \log (2)+21 x+10\right )}{\left (x^3+x^2\right ) \left (-x^3-6 x^2+e \left (x^3+x^2\right )+\left (x^3+x^2\right ) \log (2)-5 x\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{e^3-x} ((1-e) x+x (-\log (2))+5) \left (x^3+8 x^2+e \left (-x^3-3 x^2-x\right )+\left (-x^3-3 x^2-x\right ) \log (2)+21 x+10\right )}{\left (x^3+x^2\right ) \left (-x^3-6 x^2+e \left (x^3+x^2\right )+\left (x^3+x^2\right ) \log (2)-5 x\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{e^3-x} (x (1-e-\log (2))+5) \left (x^3+8 x^2+e \left (-x^3-3 x^2-x\right )+\left (-x^3-3 x^2-x\right ) \log (2)+21 x+10\right )}{\left (x^3+x^2\right ) \left (-x^3-6 x^2+e \left (x^3+x^2\right )+\left (x^3+x^2\right ) \log (2)-5 x\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{e^3-x} (x (1-e-\log (2))+5) \left (x^3+8 x^2+\left (-x^3-3 x^2-x\right ) (e+\log (2))+21 x+10\right )}{\left (x^3+x^2\right ) \left (-x^3-6 x^2+e \left (x^3+x^2\right )+\left (x^3+x^2\right ) \log (2)-5 x\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{e^3-x} (x (1-e-\log (2))+5) \left (x^3+8 x^2+\left (-x^3-3 x^2-x\right ) (e+\log (2))+21 x+10\right )}{\left (x^3+x^2\right ) \left (-x^3-6 x^2+\left (x^3+x^2\right ) (e+\log (2))-5 x\right )}dx\) |
| \(\Big \downarrow \) 2019 |
| \(\displaystyle \int \frac {e^{e^3-x} \left (x^3+8 x^2+\left (-x^3-3 x^2-x\right ) (e+\log (2))+21 x+10\right )}{\left (x^3+x^2\right ) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {25 \log (2)}{(1-e-\log (2))^3}-\frac {5}{1-e-\log (2)}-\frac {5 e}{(1-e-\log (2))^2}+\frac {30}{(1-e-\log (2))^2}+\frac {25 e}{(1-e-\log (2))^3}-\frac {25}{(1-e-\log (2))^3}\right )}dx\) |
| \(\Big \downarrow \) 3 |
| \(\displaystyle \int \frac {e^{e^3-x} \left (x^3+8 x^2+\left (-x^3-3 x^2-x\right ) (e+\log (2))+21 x+10\right )}{\left (x^3+x^2\right ) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{e^3-x} \left (x^3+8 x^2+\left (-x^3-3 x^2-x\right ) (e+\log (2))+21 x+10\right )}{x^2 (x+1) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )}dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int -\frac {e^{e^3-x} \left (x^3+8 x^2-\left (x^3+3 x^2+x\right ) (e+\log (2))+21 x+10\right )}{x^3 (x+1)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {e^{e^3-x} \left (x^3+8 x^2+21 x-\left (x^3+3 x^2+x\right ) (e+\log (2))+10\right )}{x^3 (x+1)^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\int \frac {e^{e^3-x} \left ((1-e-\log (2)) x^3+(8-3 e-\log (8)) x^2+(21-e-\log (2)) x+10\right )}{x^3 (x+1)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\int \left (\frac {e^{e^3-x} (4+e+\log (2))}{x+1}+\frac {e^{e^3-x} (4+e+\log (2))}{(x+1)^2}+\frac {e^{e^3-x} (-4-e-\log (2))}{x}+\frac {e^{e^3-x} (1-e-\log (2))}{x^2}+\frac {10 e^{e^3-x}}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -5 e^{e^3} \operatorname {ExpIntegralEi}(-x)+e^{e^3} (4+e+\log (2)) \operatorname {ExpIntegralEi}(-x)+e^{e^3} (1-e-\log (2)) \operatorname {ExpIntegralEi}(-x)+\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}+\frac {e^{e^3-x} (4+e+\log (2))}{x+1}+\frac {e^{e^3-x} (1-e-\log (2))}{x}\) |
Input:
Int[(E^(E^3 - x)*(5 + x - E*x - x*Log[2])*(10 + 21*x + 8*x^2 + x^3 + E*(-x - 3*x^2 - x^3) + (-x - 3*x^2 - x^3)*Log[2]))/((x^2 + x^3)*(-5*x - 6*x^2 - x^3 + E*(x^2 + x^3) + (x^2 + x^3)*Log[2])),x]
Output:
(5*E^(E^3 - x))/x^2 - (5*E^(E^3 - x))/x - 5*E^E^3*ExpIntegralEi[-x] + (E^( E^3 - x)*(1 - E - Log[2]))/x + E^E^3*ExpIntegralEi[-x]*(1 - E - Log[2]) + (E^(E^3 - x)*(4 + E + Log[2]))/(1 + x) + E^E^3*ExpIntegralEi[-x]*(4 + E + Log[2])
Int[(u_.)*((a_) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n + c*x^(2*n))^p, x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[a, 0]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, 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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 1.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(\frac {\left (-x \ln \left (2\right )-x \,{\mathrm e}+5+x \right ) {\mathrm e}^{-x +{\mathrm e}^{3}}}{x^{3}+x^{2}}\) | \(31\) |
| gosper | \({\mathrm e}^{\ln \left (-\frac {x \ln \left (2\right )+x \,{\mathrm e}-x -5}{x^{2} \left (1+x \right )}\right )-x +{\mathrm e}^{3}}\) | \(32\) |
| norman | \({\mathrm e}^{\ln \left (\frac {-x \ln \left (2\right )-x \,{\mathrm e}+5+x}{x^{3}+x^{2}}\right )-x +{\mathrm e}^{3}}\) | \(32\) |
| parallelrisch | \({\mathrm e}^{\ln \left (-\frac {x \ln \left (2\right )+x \,{\mathrm e}-x -5}{x^{2} \left (1+x \right )}\right )-x +{\mathrm e}^{3}}\) | \(32\) |
Input:
int(((-x^3-3*x^2-x)*ln(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)*exp(ln( (-x*ln(2)-x*exp(1)+5+x)/(x^3+x^2))-x+exp(3))/((x^3+x^2)*ln(2)+(x^3+x^2)*ex p(1)-x^3-6*x^2-5*x),x,method=_RETURNVERBOSE)
Output:
(-x*ln(2)-x*exp(1)+5+x)/(x^3+x^2)*exp(-x+exp(3))
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=e^{\left (-x + e^{3} + \log \left (-\frac {x e + x \log \left (2\right ) - x - 5}{x^{3} + x^{2}}\right )\right )} \] Input:
integrate(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)* exp(log((-x*log(2)-exp(1)*x+5+x)/(x^3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x ^3+x^2)*exp(1)-x^3-6*x^2-5*x),x, algorithm="fricas")
Output:
e^(-x + e^3 + log(-(x*e + x*log(2) - x - 5)/(x^3 + x^2)))
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=\frac {\left (- e x - x \log {\left (2 \right )} + x + 5\right ) e^{- x + e^{3}}}{x^{3} + x^{2}} \] Input:
integrate(((-x**3-3*x**2-x)*ln(2)+(-x**3-3*x**2-x)*exp(1)+x**3+8*x**2+21*x +10)*exp(ln((-x*ln(2)-exp(1)*x+5+x)/(x**3+x**2))-x+exp(3))/((x**3+x**2)*ln (2)+(x**3+x**2)*exp(1)-x**3-6*x**2-5*x),x)
Output:
(-E*x - x*log(2) + x + 5)*exp(-x + exp(3))/(x**3 + x**2)
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=-\frac {{\left ({\left ({\left (\log \left (2\right ) - 1\right )} e^{\left (e^{3}\right )} + e^{\left (e^{3} + 1\right )}\right )} x - 5 \, e^{\left (e^{3}\right )}\right )} e^{\left (-x\right )}}{x^{3} + x^{2}} \] Input:
integrate(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)* exp(log((-x*log(2)-exp(1)*x+5+x)/(x^3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x ^3+x^2)*exp(1)-x^3-6*x^2-5*x),x, algorithm="maxima")
Output:
-(((log(2) - 1)*e^(e^3) + e^(e^3 + 1))*x - 5*e^(e^3))*e^(-x)/(x^3 + x^2)
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (31) = 62\).
Time = 0.19 (sec) , antiderivative size = 277, normalized size of antiderivative = 8.94 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=-\frac {x e^{\left (-x + e^{3}\right )} \log \left (2\right )^{3} + 3 \, x e^{\left (-x + e^{3} + 1\right )} \log \left (2\right )^{2} + 7 \, x e^{\left (-x + e^{3}\right )} \log \left (2\right )^{2} + 3 \, x e^{\left (-x + e^{3} + 2\right )} \log \left (2\right ) + 14 \, x e^{\left (-x + e^{3} + 1\right )} \log \left (2\right ) + 8 \, x e^{\left (-x + e^{3}\right )} \log \left (2\right ) - 5 \, e^{\left (-x + e^{3}\right )} \log \left (2\right )^{2} + x e^{\left (-x + e^{3} + 3\right )} + 7 \, x e^{\left (-x + e^{3} + 2\right )} + 8 \, x e^{\left (-x + e^{3} + 1\right )} - 16 \, x e^{\left (-x + e^{3}\right )} - 10 \, e^{\left (-x + e^{3} + 1\right )} \log \left (2\right ) - 40 \, e^{\left (-x + e^{3}\right )} \log \left (2\right ) - 5 \, e^{\left (-x + e^{3} + 2\right )} - 40 \, e^{\left (-x + e^{3} + 1\right )} - 80 \, e^{\left (-x + e^{3}\right )}}{2 \, x^{3} e \log \left (2\right ) + x^{3} \log \left (2\right )^{2} + x^{3} e^{2} + 8 \, x^{3} e + 8 \, x^{3} \log \left (2\right ) + 2 \, x^{2} e \log \left (2\right ) + x^{2} \log \left (2\right )^{2} + 16 \, x^{3} + x^{2} e^{2} + 8 \, x^{2} e + 8 \, x^{2} \log \left (2\right ) + 16 \, x^{2}} \] Input:
integrate(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)* exp(log((-x*log(2)-exp(1)*x+5+x)/(x^3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x ^3+x^2)*exp(1)-x^3-6*x^2-5*x),x, algorithm="giac")
Output:
-(x*e^(-x + e^3)*log(2)^3 + 3*x*e^(-x + e^3 + 1)*log(2)^2 + 7*x*e^(-x + e^ 3)*log(2)^2 + 3*x*e^(-x + e^3 + 2)*log(2) + 14*x*e^(-x + e^3 + 1)*log(2) + 8*x*e^(-x + e^3)*log(2) - 5*e^(-x + e^3)*log(2)^2 + x*e^(-x + e^3 + 3) + 7*x*e^(-x + e^3 + 2) + 8*x*e^(-x + e^3 + 1) - 16*x*e^(-x + e^3) - 10*e^(-x + e^3 + 1)*log(2) - 40*e^(-x + e^3)*log(2) - 5*e^(-x + e^3 + 2) - 40*e^(- x + e^3 + 1) - 80*e^(-x + e^3))/(2*x^3*e*log(2) + x^3*log(2)^2 + x^3*e^2 + 8*x^3*e + 8*x^3*log(2) + 2*x^2*e*log(2) + x^2*log(2)^2 + 16*x^3 + x^2*e^2 + 8*x^2*e + 8*x^2*log(2) + 16*x^2)
Time = 4.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.52 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=\frac {5\,{\mathrm {e}}^{{\mathrm {e}}^3-x}}{x^3+x^2}+\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x}}{x^3+x^2}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x+1}}{x^3+x^2}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x}\,\ln \left (2\right )}{x^3+x^2} \] Input:
int(-(exp(exp(3) - x + log((x - x*exp(1) - x*log(2) + 5)/(x^2 + x^3)))*(21 *x - exp(1)*(x + 3*x^2 + x^3) - log(2)*(x + 3*x^2 + x^3) + 8*x^2 + x^3 + 1 0))/(5*x - exp(1)*(x^2 + x^3) + 6*x^2 + x^3 - log(2)*(x^2 + x^3)),x)
Output:
(5*exp(exp(3) - x))/(x^2 + x^3) + (x*exp(exp(3) - x))/(x^2 + x^3) - (x*exp (exp(3) - x + 1))/(x^2 + x^3) - (x*exp(exp(3) - x)*log(2))/(x^2 + x^3)
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^3-x} (5+x-e x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx=\frac {e^{e^{3}} \left (-\mathrm {log}\left (2\right ) x -e x +x +5\right )}{e^{x} x^{2} \left (x +1\right )} \] Input:
int(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)*exp(lo g((-x*log(2)-exp(1)*x+5+x)/(x^3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x^3+x^2 )*exp(1)-x^3-6*x^2-5*x),x)
Output:
(e**(e**3)*( - log(2)*x - e*x + x + 5))/(e**x*x**2*(x + 1))