3.34.67 \(\int \frac {160 x^2+6 x^3-80 x^4+10 x^6+e^2 (240-120 x^2+15 x^4)}{80 x^2-40 x^4+5 x^6} \, dx\)

Optimal. Leaf size=30 \[ x+\frac {x^2-3 \left (e^2+x+\frac {1}{5 \left (-\frac {4}{x}+x\right )}\right )}{x} \]

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Rubi [A]  time = 0.09, antiderivative size = 25, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1594, 28, 1805, 1253, 14} \begin {gather*} \frac {3}{5 \left (4-x^2\right )}+2 x-\frac {3 e^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(160*x^2 + 6*x^3 - 80*x^4 + 10*x^6 + E^2*(240 - 120*x^2 + 15*x^4))/(80*x^2 - 40*x^4 + 5*x^6),x]

[Out]

(-3*E^2)/x + 2*x + 3/(5*(4 - x^2))

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 1253

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[(f*x)^m*(d + e*x^2)^(q + p)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {160 x^2+6 x^3-80 x^4+10 x^6+e^2 \left (240-120 x^2+15 x^4\right )}{x^2 \left (80-40 x^2+5 x^4\right )} \, dx\\ &=5 \int \frac {160 x^2+6 x^3-80 x^4+10 x^6+e^2 \left (240-120 x^2+15 x^4\right )}{x^2 \left (-20+5 x^2\right )^2} \, dx\\ &=\frac {3}{5 \left (4-x^2\right )}+\frac {1}{8} \int \frac {-480 e^2-40 \left (8-3 e^2\right ) x^2+80 x^4}{x^2 \left (-20+5 x^2\right )} \, dx\\ &=\frac {3}{5 \left (4-x^2\right )}+\frac {1}{8} \int \frac {24 e^2+16 x^2}{x^2} \, dx\\ &=\frac {3}{5 \left (4-x^2\right )}+\frac {1}{8} \int \left (16+\frac {24 e^2}{x^2}\right ) \, dx\\ &=-\frac {3 e^2}{x}+2 x+\frac {3}{5 \left (4-x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.77 \begin {gather*} -\frac {3 e^2}{x}+2 x-\frac {3}{5 \left (-4+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(160*x^2 + 6*x^3 - 80*x^4 + 10*x^6 + E^2*(240 - 120*x^2 + 15*x^4))/(80*x^2 - 40*x^4 + 5*x^6),x]

[Out]

(-3*E^2)/x + 2*x - 3/(5*(-4 + x^2))

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fricas [A]  time = 0.45, size = 34, normalized size = 1.13 \begin {gather*} \frac {10 \, x^{4} - 40 \, x^{2} - 15 \, {\left (x^{2} - 4\right )} e^{2} - 3 \, x}{5 \, {\left (x^{3} - 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^4-120*x^2+240)*exp(2)+10*x^6-80*x^4+6*x^3+160*x^2)/(5*x^6-40*x^4+80*x^2),x, algorithm="fricas
")

[Out]

1/5*(10*x^4 - 40*x^2 - 15*(x^2 - 4)*e^2 - 3*x)/(x^3 - 4*x)

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giac [A]  time = 0.29, size = 28, normalized size = 0.93 \begin {gather*} 2 \, x - \frac {3 \, {\left (5 \, x^{2} e^{2} + x - 20 \, e^{2}\right )}}{5 \, {\left (x^{3} - 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^4-120*x^2+240)*exp(2)+10*x^6-80*x^4+6*x^3+160*x^2)/(5*x^6-40*x^4+80*x^2),x, algorithm="giac")

[Out]

2*x - 3/5*(5*x^2*e^2 + x - 20*e^2)/(x^3 - 4*x)

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maple [A]  time = 0.05, size = 26, normalized size = 0.87

method	result	size
default	\(2 x -\frac {3 \,{\mathrm e}^{2}}{x}-\frac {3}{20 \left (x -2\right )}+\frac {3}{20 \left (2+x \right )}\)	\(26\)
risch	\(2 x +\frac {-3 x^{2} {\mathrm e}^{2}+12 \,{\mathrm e}^{2}-\frac {3 x}{5}}{x \left (x^{2}-4\right )}\)	\(31\)
norman	\(\frac {-\frac {3 x}{5}+\left (-3 \,{\mathrm e}^{2}-8\right ) x^{2}+2 x^{4}+12 \,{\mathrm e}^{2}}{x \left (x^{2}-4\right )}\)	\(35\)
gosper	\(-\frac {-10 x^{4}+15 x^{2} {\mathrm e}^{2}+40 x^{2}-60 \,{\mathrm e}^{2}+3 x}{5 x \left (x^{2}-4\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x^4-120*x^2+240)*exp(2)+10*x^6-80*x^4+6*x^3+160*x^2)/(5*x^6-40*x^4+80*x^2),x,method=_RETURNVERBOSE)

[Out]

2*x-3*exp(2)/x-3/20/(x-2)+3/20/(2+x)

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maxima [A]  time = 0.38, size = 28, normalized size = 0.93 \begin {gather*} 2 \, x - \frac {3 \, {\left (5 \, x^{2} e^{2} + x - 20 \, e^{2}\right )}}{5 \, {\left (x^{3} - 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^4-120*x^2+240)*exp(2)+10*x^6-80*x^4+6*x^3+160*x^2)/(5*x^6-40*x^4+80*x^2),x, algorithm="maxima
")

[Out]

2*x - 3/5*(5*x^2*e^2 + x - 20*e^2)/(x^3 - 4*x)

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mupad [B]  time = 0.10, size = 31, normalized size = 1.03 \begin {gather*} 2\,x-\frac {3\,{\mathrm {e}}^2\,x^2+\frac {3\,x}{5}-12\,{\mathrm {e}}^2}{x\,\left (x^2-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)*(15*x^4 - 120*x^2 + 240) + 160*x^2 + 6*x^3 - 80*x^4 + 10*x^6)/(80*x^2 - 40*x^4 + 5*x^6),x)

[Out]

2*x - ((3*x)/5 - 12*exp(2) + 3*x^2*exp(2))/(x*(x^2 - 4))

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sympy [A]  time = 0.35, size = 27, normalized size = 0.90 \begin {gather*} 2 x + \frac {- 15 x^{2} e^{2} - 3 x + 60 e^{2}}{5 x^{3} - 20 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x**4-120*x**2+240)*exp(2)+10*x**6-80*x**4+6*x**3+160*x**2)/(5*x**6-40*x**4+80*x**2),x)

[Out]

2*x + (-15*x**2*exp(2) - 3*x + 60*exp(2))/(5*x**3 - 20*x)

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