\(\int \frac {75-100 x^3+e^{e^x+x} (45 x^2-30 x^5+5 x^8)}{9 x^2-6 x^5+x^8} \, dx\) [3617]

Optimal result

Integrand size = 48, antiderivative size = 22 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=8+5 \left (e^{e^x}+\frac {5}{x \left (-3+x^3\right )}\right ) \]

[Out]

8+5*exp(exp(x))+25/x/(x^3-3)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1608, 28, 6874, 2320, 2225, 460} \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=5 e^{e^x}-\frac {25}{x \left (3-x^3\right )} \]

[In]

Int[(75 - 100*x^3 + E^(E^x + x)*(45*x^2 - 30*x^5 + 5*x^8))/(9*x^2 - 6*x^5 + x^8),x]

[Out]

5*E^E^x - 25/(x*(3 - x^3))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 460

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{x^2 \left (9-6 x^3+x^6\right )} \, dx \\ & = \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{x^2 \left (-3+x^3\right )^2} \, dx \\ & = \int \left (5 e^{e^x+x}-\frac {25 \left (-3+4 x^3\right )}{x^2 \left (-3+x^3\right )^2}\right ) \, dx \\ & = 5 \int e^{e^x+x} \, dx-25 \int \frac {-3+4 x^3}{x^2 \left (-3+x^3\right )^2} \, dx \\ & = -\frac {25}{x \left (3-x^3\right )}+5 \text {Subst}\left (\int e^x \, dx,x,e^x\right ) \\ & = 5 e^{e^x}-\frac {25}{x \left (3-x^3\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=5 \left (e^{e^x}-\frac {5}{3 x}+\frac {5}{18} x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {x^3}{3}\right )-\frac {5}{6} x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},2,\frac {5}{3},\frac {x^3}{3}\right )\right ) \]

[In]

Integrate[(75 - 100*x^3 + E^(E^x + x)*(45*x^2 - 30*x^5 + 5*x^8))/(9*x^2 - 6*x^5 + x^8),x]

[Out]

5*(E^E^x - 5/(3*x) + (5*x^2*Hypergeometric2F1[2/3, 1, 5/3, x^3/3])/18 - (5*x^2*Hypergeometric2F1[2/3, 2, 5/3,
x^3/3])/6)

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {25}{x \left (x^{3}-3\right )}+5 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(19\)
parts	\(5 \,{\mathrm e}^{{\mathrm e}^{x}}-\frac {25}{3 x}+\frac {25 x^{2}}{3 \left (x^{3}-3\right )}\)	\(24\)
parallelrisch	\(\frac {5 x^{4} {\mathrm e}^{{\mathrm e}^{x}}+25-15 x \,{\mathrm e}^{{\mathrm e}^{x}}}{x \left (x^{3}-3\right )}\)	\(28\)

[In]

int(((5*x^8-30*x^5+45*x^2)*exp(x)*exp(exp(x))-100*x^3+75)/(x^8-6*x^5+9*x^2),x,method=_RETURNVERBOSE)

[Out]

25/x/(x^3-3)+5*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=\frac {5 \, {\left ({\left (x^{4} - 3 \, x\right )} e^{\left (x + e^{x}\right )} + 5 \, e^{x}\right )} e^{\left (-x\right )}}{x^{4} - 3 \, x} \]

[In]

integrate(((5*x^8-30*x^5+45*x^2)*exp(x)*exp(exp(x))-100*x^3+75)/(x^8-6*x^5+9*x^2),x, algorithm="fricas")

[Out]

5*((x^4 - 3*x)*e^(x + e^x) + 5*e^x)*e^(-x)/(x^4 - 3*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=5 e^{e^{x}} + \frac {25}{x^{4} - 3 x} \]

[In]

integrate(((5*x**8-30*x**5+45*x**2)*exp(x)*exp(exp(x))-100*x**3+75)/(x**8-6*x**5+9*x**2),x)

[Out]

5*exp(exp(x)) + 25/(x**4 - 3*x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=\frac {100 \, x^{2}}{9 \, {\left (x^{3} - 3\right )}} - \frac {25 \, {\left (4 \, x^{3} - 9\right )}}{9 \, {\left (x^{4} - 3 \, x\right )}} + 5 \, e^{\left (e^{x}\right )} \]

[In]

integrate(((5*x^8-30*x^5+45*x^2)*exp(x)*exp(exp(x))-100*x^3+75)/(x^8-6*x^5+9*x^2),x, algorithm="maxima")

[Out]

100/9*x^2/(x^3 - 3) - 25/9*(4*x^3 - 9)/(x^4 - 3*x) + 5*e^(e^x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=\frac {5 \, {\left (x^{4} e^{\left (x + e^{x}\right )} - 3 \, x e^{\left (x + e^{x}\right )} + 5 \, e^{x}\right )}}{x^{4} e^{x} - 3 \, x e^{x}} \]

[In]

integrate(((5*x^8-30*x^5+45*x^2)*exp(x)*exp(exp(x))-100*x^3+75)/(x^8-6*x^5+9*x^2),x, algorithm="giac")

[Out]

5*(x^4*e^(x + e^x) - 3*x*e^(x + e^x) + 5*e^x)/(x^4*e^x - 3*x*e^x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {75-100 x^3+e^{e^x+x} \left (45 x^2-30 x^5+5 x^8\right )}{9 x^2-6 x^5+x^8} \, dx=5\,{\mathrm {e}}^{{\mathrm {e}}^x}+\frac {25}{x\,\left (x^3-3\right )} \]

[In]

int((exp(exp(x))*exp(x)*(45*x^2 - 30*x^5 + 5*x^8) - 100*x^3 + 75)/(9*x^2 - 6*x^5 + x^8),x)

[Out]

5*exp(exp(x)) + 25/(x*(x^3 - 3))