Integrand size = 93, antiderivative size = 25 \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=\frac {16}{\left (1-e^{e^9}+9 \left (e^3-x\right )\right ) x} \] Output:
16/(-9*x+9*exp(3)-exp(exp(9))+1)/x
Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=\frac {16}{x+9 e^3 x-e^{e^9} x-9 x^2} \] Input:
Integrate[(-16 - 144*E^3 + 16*E^E^9 + 288*x)/(x^2 + 81*E^6*x^2 + E^(2*E^9) *x^2 - 18*x^3 + 81*x^4 + E^3*(18*x^2 - 162*x^3) + E^E^9*(-2*x^2 - 18*E^3*x ^2 + 18*x^3)),x]
Output:
16/(x + 9*E^3*x - E^E^9*x - 9*x^2)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6, 6, 2026, 1184, 27, 83}
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.
\(\displaystyle \int \frac {288 x+16 e^{e^9}-144 e^3-16}{81 x^4-18 x^3+e^{2 e^9} x^2+81 e^6 x^2+x^2+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (18 x^3-18 e^3 x^2-2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {288 x+16 e^{e^9}-144 e^3-16}{81 x^4-18 x^3+\left (1+81 e^6\right ) x^2+e^{2 e^9} x^2+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (18 x^3-18 e^3 x^2-2 x^2\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {288 x+16 e^{e^9}-144 e^3-16}{81 x^4-18 x^3+\left (1+81 e^6+e^{2 e^9}\right ) x^2+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (18 x^3-18 e^3 x^2-2 x^2\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {288 x+16 e^{e^9}-144 e^3-16}{x^2 \left (81 x^2-18 \left (1+9 e^3-e^{e^9}\right ) x+\left (-1-9 e^3+e^{e^9}\right )^2\right )}dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle 81 \int -\frac {16 \left (-18 x-e^{e^9}+9 e^3+1\right )}{81 \left (-9 x-e^{e^9}+9 e^3+1\right )^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -16 \int \frac {-18 x-e^{e^9}+9 e^3+1}{\left (-9 x-e^{e^9}+9 e^3+1\right )^2 x^2}dx\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {16}{\left (-9 x-e^{e^9}+9 e^3+1\right ) x}\) |
Input:
Int[(-16 - 144*E^3 + 16*E^E^9 + 288*x)/(x^2 + 81*E^6*x^2 + E^(2*E^9)*x^2 - 18*x^3 + 81*x^4 + E^3*(18*x^2 - 162*x^3) + E^E^9*(-2*x^2 - 18*E^3*x^2 + 1 8*x^3)),x]
Output:
16/((1 + 9*E^3 - E^E^9 - 9*x)*x)
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
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 = 1.59 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(-\frac {16}{x \left (9 x -1+{\mathrm e}^{{\mathrm e}^{9}}-9 \,{\mathrm e}^{3}\right )}\) | \(20\) |
norman | \(-\frac {16}{x \left (9 x -1+{\mathrm e}^{{\mathrm e}^{9}}-9 \,{\mathrm e}^{3}\right )}\) | \(20\) |
risch | \(-\frac {16}{x \left (9 x -1+{\mathrm e}^{{\mathrm e}^{9}}-9 \,{\mathrm e}^{3}\right )}\) | \(20\) |
parallelrisch | \(-\frac {16}{x \left (9 x -1+{\mathrm e}^{{\mathrm e}^{9}}-9 \,{\mathrm e}^{3}\right )}\) | \(20\) |
Input:
int((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2*exp(3 )+18*x^3-2*x^2)*exp(exp(9))+81*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3)+81*x^ 4-18*x^3+x^2),x,method=_RETURNVERBOSE)
Output:
-16/x/(9*x-1+exp(exp(9))-9*exp(3))
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=-\frac {16}{9 \, x^{2} - 9 \, x e^{3} + x e^{\left (e^{9}\right )} - x} \] Input:
integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2 *exp(3)+18*x^3-2*x^2)*exp(exp(9))+81*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3) +81*x^4-18*x^3+x^2),x, algorithm="fricas")
Output:
-16/(9*x^2 - 9*x*e^3 + x*e^(e^9) - x)
Time = 0.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=- \frac {16}{9 x^{2} + x \left (- 9 e^{3} - 1 + e^{e^{9}}\right )} \] Input:
integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x**2*exp(exp(9))**2+(-18*x **2*exp(3)+18*x**3-2*x**2)*exp(exp(9))+81*x**2*exp(3)**2+(-162*x**3+18*x** 2)*exp(3)+81*x**4-18*x**3+x**2),x)
Output:
-16/(9*x**2 + x*(-9*exp(3) - 1 + exp(exp(9))))
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=-\frac {16}{9 \, x^{2} - x {\left (9 \, e^{3} - e^{\left (e^{9}\right )} + 1\right )}} \] Input:
integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2 *exp(3)+18*x^3-2*x^2)*exp(exp(9))+81*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3) +81*x^4-18*x^3+x^2),x, algorithm="maxima")
Output:
-16/(9*x^2 - x*(9*e^3 - e^(e^9) + 1))
\[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=\int { \frac {16 \, {\left (18 \, x - 9 \, e^{3} + e^{\left (e^{9}\right )} - 1\right )}}{81 \, x^{4} - 18 \, x^{3} + 81 \, x^{2} e^{6} + x^{2} e^{\left (2 \, e^{9}\right )} + x^{2} - 18 \, {\left (9 \, x^{3} - x^{2}\right )} e^{3} + 2 \, {\left (9 \, x^{3} - 9 \, x^{2} e^{3} - x^{2}\right )} e^{\left (e^{9}\right )}} \,d x } \] Input:
integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2 *exp(3)+18*x^3-2*x^2)*exp(exp(9))+81*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3) +81*x^4-18*x^3+x^2),x, algorithm="giac")
Output:
undef
Timed out. \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=\text {Hanged} \] Input:
int((288*x - 144*exp(3) + 16*exp(exp(9)) - 16)/(x^2*exp(2*exp(9)) + exp(3) *(18*x^2 - 162*x^3) - exp(exp(9))*(18*x^2*exp(3) + 2*x^2 - 18*x^3) + 81*x^ 2*exp(6) + x^2 - 18*x^3 + 81*x^4),x)
Output:
\text{Hanged}
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx=-\frac {16}{x \left (e^{e^{9}}-9 e^{3}+9 x -1\right )} \] Input:
int((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2*exp(3 )+18*x^3-2*x^2)*exp(exp(9))+81*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3)+81*x^ 4-18*x^3+x^2),x)
Output:
( - 16)/(x*(e**(e**9) - 9*e**3 + 9*x - 1))