3.7.31 \(\int \frac {2 x^2+12 x^3-4 x^5-24 x^6+2 x^8+12 x^9+e^{e^8} (-1+2 x^2+4 x^3+4 x^5)}{x^2-2 x^5+x^8} \, dx\)

Optimal. Leaf size=30 \[ 2 x+\left (1+2 x^2\right ) \left (3-\frac {e^{e^8}}{-x+x^4}\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 37, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {1594, 28, 1829, 1586, 14} \begin {gather*} \frac {e^{e^8} (x+2) x}{1-x^3}+6 x^2+2 x+\frac {e^{e^8}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 + 12*x^3 - 4*x^5 - 24*x^6 + 2*x^8 + 12*x^9 + E^E^8*(-1 + 2*x^2 + 4*x^3 + 4*x^5))/(x^2 - 2*x^5 + x^8
),x]

[Out]

E^E^8/x + 2*x + 6*x^2 + (E^E^8*x*(2 + x))/(1 - x^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 1586

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]

Rule 1594

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 1829

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)*Coeff[R, x, i]*x^(i - m))/a, {i, 0, n - 1}], x], x], x] - S
imp[(x*R*(a + b*x^n)^(p + 1))/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x^2+12 x^3-4 x^5-24 x^6+2 x^8+12 x^9+e^{e^8} \left (-1+2 x^2+4 x^3+4 x^5\right )}{x^2 \left (1-2 x^3+x^6\right )} \, dx\\ &=\int \frac {2 x^2+12 x^3-4 x^5-24 x^6+2 x^8+12 x^9+e^{e^8} \left (-1+2 x^2+4 x^3+4 x^5\right )}{x^2 \left (-1+x^3\right )^2} \, dx\\ &=\frac {e^{e^8} x (2+x)}{1-x^3}+\frac {1}{3} \int \frac {3 e^{e^8}-6 x^2-3 \left (12+e^{e^8}\right ) x^3+6 x^5+36 x^6}{x^2 \left (-1+x^3\right )} \, dx\\ &=\frac {e^{e^8} x (2+x)}{1-x^3}+\frac {1}{3} \int \frac {-3 e^{e^8}+6 x^2+36 x^3}{x^2} \, dx\\ &=\frac {e^{e^8} x (2+x)}{1-x^3}+\frac {1}{3} \int \left (6-\frac {3 e^{e^8}}{x^2}+36 x\right ) \, dx\\ &=\frac {e^{e^8}}{x}+2 x+6 x^2+\frac {e^{e^8} x (2+x)}{1-x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 40, normalized size = 1.33 \begin {gather*} \frac {e^{e^8}}{x}+2 x+6 x^2+\frac {e^{e^8} \left (-2 x-x^2\right )}{-1+x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + 12*x^3 - 4*x^5 - 24*x^6 + 2*x^8 + 12*x^9 + E^E^8*(-1 + 2*x^2 + 4*x^3 + 4*x^5))/(x^2 - 2*x^5
+ x^8),x]

[Out]

E^E^8/x + 2*x + 6*x^2 + (E^E^8*(-2*x - x^2))/(-1 + x^3)

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fricas [A]  time = 0.62, size = 43, normalized size = 1.43 \begin {gather*} \frac {6 \, x^{6} + 2 \, x^{5} - 6 \, x^{3} - 2 \, x^{2} - {\left (2 \, x^{2} + 1\right )} e^{\left (e^{8}\right )}}{x^{4} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+4*x^3+2*x^2-1)*exp(exp(4)^2)+12*x^9+2*x^8-24*x^6-4*x^5+12*x^3+2*x^2)/(x^8-2*x^5+x^2),x, algo
rithm="fricas")

[Out]

(6*x^6 + 2*x^5 - 6*x^3 - 2*x^2 - (2*x^2 + 1)*e^(e^8))/(x^4 - x)

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giac [A]  time = 0.43, size = 32, normalized size = 1.07 \begin {gather*} 6 \, x^{2} + 2 \, x - \frac {2 \, x^{2} e^{\left (e^{8}\right )} + e^{\left (e^{8}\right )}}{x^{4} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+4*x^3+2*x^2-1)*exp(exp(4)^2)+12*x^9+2*x^8-24*x^6-4*x^5+12*x^3+2*x^2)/(x^8-2*x^5+x^2),x, algo
rithm="giac")

[Out]

6*x^2 + 2*x - (2*x^2*e^(e^8) + e^(e^8))/(x^4 - x)

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maple [A]  time = 0.12, size = 35, normalized size = 1.17

method	result	size
risch	\(6 x^{2}+2 x +\frac {-2 \,{\mathrm e}^{{\mathrm e}^{8}} x^{2}-{\mathrm e}^{{\mathrm e}^{8}}}{x \left (x^{3}-1\right )}\)	\(35\)
default	\(2 x +6 x^{2}-\frac {{\mathrm e}^{{\mathrm e}^{8}}}{x -1}-\frac {{\mathrm e}^{{\mathrm e}^{8}}}{x^{2}+x +1}+\frac {{\mathrm e}^{{\mathrm e}^{8}}}{x}\)	\(40\)
norman	\(\frac {6 x^{6}+2 x^{5}-6 x^{3}+\left (-2 \,{\mathrm e}^{{\mathrm e}^{8}}-2\right ) x^{2}-{\mathrm e}^{{\mathrm e}^{8}}}{x \left (x^{3}-1\right )}\)	\(48\)
gosper	\(-\frac {-6 x^{6}-2 x^{5}+2 \,{\mathrm e}^{{\mathrm e}^{8}} x^{2}+6 x^{3}+2 x^{2}+{\mathrm e}^{{\mathrm e}^{8}}}{x \left (x^{3}-1\right )}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^5+4*x^3+2*x^2-1)*exp(exp(4)^2)+12*x^9+2*x^8-24*x^6-4*x^5+12*x^3+2*x^2)/(x^8-2*x^5+x^2),x,method=_RET
URNVERBOSE)

[Out]

6*x^2+2*x+(-2*exp(exp(8))*x^2-exp(exp(8)))/x/(x^3-1)

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maxima [A]  time = 0.50, size = 32, normalized size = 1.07 \begin {gather*} 6 \, x^{2} + 2 \, x - \frac {2 \, x^{2} e^{\left (e^{8}\right )} + e^{\left (e^{8}\right )}}{x^{4} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+4*x^3+2*x^2-1)*exp(exp(4)^2)+12*x^9+2*x^8-24*x^6-4*x^5+12*x^3+2*x^2)/(x^8-2*x^5+x^2),x, algo
rithm="maxima")

[Out]

6*x^2 + 2*x - (2*x^2*e^(e^8) + e^(e^8))/(x^4 - x)

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mupad [B]  time = 0.53, size = 31, normalized size = 1.03 \begin {gather*} 2\,x+6\,x^2+\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^8}\,x^2+{\mathrm {e}}^{{\mathrm {e}}^8}}{x-x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(8))*(2*x^2 + 4*x^3 + 4*x^5 - 1) + 2*x^2 + 12*x^3 - 4*x^5 - 24*x^6 + 2*x^8 + 12*x^9)/(x^2 - 2*x^5
+ x^8),x)

[Out]

2*x + 6*x^2 + (exp(exp(8)) + 2*x^2*exp(exp(8)))/(x - x^4)

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sympy [A]  time = 0.32, size = 29, normalized size = 0.97 \begin {gather*} 6 x^{2} + 2 x + \frac {- 2 x^{2} e^{e^{8}} - e^{e^{8}}}{x^{4} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**5+4*x**3+2*x**2-1)*exp(exp(4)**2)+12*x**9+2*x**8-24*x**6-4*x**5+12*x**3+2*x**2)/(x**8-2*x**5+
x**2),x)

[Out]

6*x**2 + 2*x + (-2*x**2*exp(exp(8)) - exp(exp(8)))/(x**4 - x)

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