3.29.79 \(\int \frac {5696-3328 x+2560 x^2-512 x^3+256 x^4}{81-36 x+40 x^2-8 x^3+4 x^4} \, dx\)

Optimal. Leaf size=25 \[ 4 x \left (16+\log \left (e^{\frac {32}{x \left (\frac {9}{2}-x+x^2\right )}}\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 19, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {1680, 12, 1814, 21, 8} \begin {gather*} 64 x+\frac {512}{4 \left (x-\frac {1}{2}\right )^2+17} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5696 - 3328*x + 2560*x^2 - 512*x^3 + 256*x^4)/(81 - 36*x + 40*x^2 - 8*x^3 + 4*x^4),x]

[Out]

512/(17 + 4*(-1/2 + x)^2) + 64*x

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[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[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {64 \left (289-64 x+136 x^2+16 x^4\right )}{\left (17+4 x^2\right )^2} \, dx,x,-\frac {1}{2}+x\right )\\ &=64 \operatorname {Subst}\left (\int \frac {289-64 x+136 x^2+16 x^4}{\left (17+4 x^2\right )^2} \, dx,x,-\frac {1}{2}+x\right )\\ &=\frac {512}{17+(-1+2 x)^2}-\frac {32}{17} \operatorname {Subst}\left (\int \frac {-578-136 x^2}{17+4 x^2} \, dx,x,-\frac {1}{2}+x\right )\\ &=\frac {512}{17+(-1+2 x)^2}+64 \operatorname {Subst}\left (\int 1 \, dx,x,-\frac {1}{2}+x\right )\\ &=64 x+\frac {512}{17+(-1+2 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.72 \begin {gather*} 64 \left (x+\frac {4}{9-2 x+2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5696 - 3328*x + 2560*x^2 - 512*x^3 + 256*x^4)/(81 - 36*x + 40*x^2 - 8*x^3 + 4*x^4),x]

[Out]

64*(x + 4/(9 - 2*x + 2*x^2))

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fricas [A]  time = 0.67, size = 29, normalized size = 1.16 \begin {gather*} \frac {64 \, {\left (2 \, x^{3} - 2 \, x^{2} + 9 \, x + 4\right )}}{2 \, x^{2} - 2 \, x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x^4-512*x^3+2560*x^2-3328*x+5696)/(4*x^4-8*x^3+40*x^2-36*x+81),x, algorithm="fricas")

[Out]

64*(2*x^3 - 2*x^2 + 9*x + 4)/(2*x^2 - 2*x + 9)

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giac [A]  time = 0.24, size = 18, normalized size = 0.72 \begin {gather*} 64 \, x + \frac {256}{2 \, x^{2} - 2 \, x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x^4-512*x^3+2560*x^2-3328*x+5696)/(4*x^4-8*x^3+40*x^2-36*x+81),x, algorithm="giac")

[Out]

64*x + 256/(2*x^2 - 2*x + 9)

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maple [A]  time = 0.02, size = 17, normalized size = 0.68

method	result	size
risch	\(64 x +\frac {128}{x^{2}+\frac {9}{2}-x}\)	\(17\)
default	\(64 x +\frac {256}{2 x^{2}-2 x +9}\)	\(19\)
norman	\(\frac {128 x^{3}+448 x +832}{2 x^{2}-2 x +9}\)	\(24\)
gosper	\(\frac {128 x^{3}+448 x +832}{2 x^{2}-2 x +9}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((256*x^4-512*x^3+2560*x^2-3328*x+5696)/(4*x^4-8*x^3+40*x^2-36*x+81),x,method=_RETURNVERBOSE)

[Out]

64*x+128/(x^2+9/2-x)

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maxima [A]  time = 0.34, size = 18, normalized size = 0.72 \begin {gather*} 64 \, x + \frac {256}{2 \, x^{2} - 2 \, x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x^4-512*x^3+2560*x^2-3328*x+5696)/(4*x^4-8*x^3+40*x^2-36*x+81),x, algorithm="maxima")

[Out]

64*x + 256/(2*x^2 - 2*x + 9)

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mupad [B]  time = 0.04, size = 16, normalized size = 0.64 \begin {gather*} 64\,x+\frac {128}{x^2-x+\frac {9}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2560*x^2 - 3328*x - 512*x^3 + 256*x^4 + 5696)/(40*x^2 - 36*x - 8*x^3 + 4*x^4 + 81),x)

[Out]

64*x + 128/(x^2 - x + 9/2)

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sympy [A]  time = 0.09, size = 14, normalized size = 0.56 \begin {gather*} 64 x + \frac {256}{2 x^{2} - 2 x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x**4-512*x**3+2560*x**2-3328*x+5696)/(4*x**4-8*x**3+40*x**2-36*x+81),x)

[Out]

64*x + 256/(2*x**2 - 2*x + 9)

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