3.482 \(\int \frac {-1-8 x+8 x^3}{(1+2 x-4 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=45 \[ -\frac {4 (x+1)}{15 \left (-4 x^2+2 x+1\right )^{3/2}}-\frac {122 x+7}{75 \sqrt {-4 x^2+2 x+1}} \]

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1660, 636} \[ -\frac {4 (x+1)}{15 \left (-4 x^2+2 x+1\right )^{3/2}}-\frac {122 x+7}{75 \sqrt {-4 x^2+2 x+1}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 - 8*x + 8*x^3)/(1 + 2*x - 4*x^2)^(5/2),x]

[Out]

(-4*(1 + x))/(15*(1 + 2*x - 4*x^2)^(3/2)) - (7 + 122*x)/(75*Sqrt[1 + 2*x - 4*x^2])

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {-1-8 x+8 x^3}{\left (1+2 x-4 x^2\right )^{5/2}} \, dx &=-\frac {4 (1+x)}{15 \left (1+2 x-4 x^2\right )^{3/2}}-\frac {1}{30} \int \frac {46+60 x}{\left (1+2 x-4 x^2\right )^{3/2}} \, dx\\ &=-\frac {4 (1+x)}{15 \left (1+2 x-4 x^2\right )^{3/2}}-\frac {7+122 x}{75 \sqrt {1+2 x-4 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 33, normalized size = 0.73 \[ -\frac {-488 x^3+216 x^2+156 x+27}{75 \left (-4 x^2+2 x+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 8*x + 8*x^3)/(1 + 2*x - 4*x^2)^(5/2),x]

[Out]

-1/75*(27 + 156*x + 216*x^2 - 488*x^3)/(1 + 2*x - 4*x^2)^(3/2)

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IntegrateAlgebraic [A]  time = 0.29, size = 45, normalized size = 1.00 \[ \frac {\sqrt {-4 x^2+2 x+1} \left (488 x^3-216 x^2-156 x-27\right )}{75 \left (4 x^2-2 x-1\right )^2} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 - 8*x + 8*x^3)/(1 + 2*x - 4*x^2)^(5/2),x]

[Out]

(Sqrt[1 + 2*x - 4*x^2]*(-27 - 156*x - 216*x^2 + 488*x^3))/(75*(-1 - 2*x + 4*x^2)^2)

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fricas [A]  time = 1.10, size = 73, normalized size = 1.62 \[ -\frac {432 \, x^{4} - 432 \, x^{3} - 108 \, x^{2} - {\left (488 \, x^{3} - 216 \, x^{2} - 156 \, x - 27\right )} \sqrt {-4 \, x^{2} + 2 \, x + 1} + 108 \, x + 27}{75 \, {\left (16 \, x^{4} - 16 \, x^{3} - 4 \, x^{2} + 4 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3-8*x-1)/(-4*x^2+2*x+1)^(5/2),x, algorithm="fricas")

[Out]

-1/75*(432*x^4 - 432*x^3 - 108*x^2 - (488*x^3 - 216*x^2 - 156*x - 27)*sqrt(-4*x^2 + 2*x + 1) + 108*x + 27)/(16
*x^4 - 16*x^3 - 4*x^2 + 4*x + 1)

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giac [A]  time = 0.66, size = 41, normalized size = 0.91 \[ \frac {{\left (4 \, {\left (2 \, {\left (61 \, x - 27\right )} x - 39\right )} x - 27\right )} \sqrt {-4 \, x^{2} + 2 \, x + 1}}{75 \, {\left (4 \, x^{2} - 2 \, x - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3-8*x-1)/(-4*x^2+2*x+1)^(5/2),x, algorithm="giac")

[Out]

1/75*(4*(2*(61*x - 27)*x - 39)*x - 27)*sqrt(-4*x^2 + 2*x + 1)/(4*x^2 - 2*x - 1)^2

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maple [A]  time = 0.40, size = 30, normalized size = 0.67

method	result	size
gosper	\(\frac {488 x^{3}-216 x^{2}-156 x -27}{75 \left (-4 x^{2}+2 x +1\right )^{\frac {3}{2}}}\)	\(30\)
trager	\(\frac {\left (488 x^{3}-216 x^{2}-156 x -27\right ) \sqrt {-4 x^{2}+2 x +1}}{75 \left (4 x^{2}-2 x -1\right )^{2}}\)	\(42\)
risch	\(-\frac {488 x^{3}-216 x^{2}-156 x -27}{75 \left (4 x^{2}-2 x -1\right ) \sqrt {-4 x^{2}+2 x +1}}\)	\(42\)
default	\(\frac {2 x^{2}}{\left (-4 x^{2}+2 x +1\right )^{\frac {3}{2}}}-\frac {x}{4 \left (-4 x^{2}+2 x +1\right )^{\frac {3}{2}}}-\frac {49}{48 \left (-4 x^{2}+2 x +1\right )^{\frac {3}{2}}}+\frac {\frac {61}{240}-\frac {61 x}{60}}{\left (-4 x^{2}+2 x +1\right )^{\frac {3}{2}}}+\frac {\frac {61}{150}-\frac {122 x}{75}}{\sqrt {-4 x^{2}+2 x +1}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3-8*x-1)/(-4*x^2+2*x+1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/75*(488*x^3-216*x^2-156*x-27)/(-4*x^2+2*x+1)^(3/2)

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maxima [B]  time = 0.56, size = 76, normalized size = 1.69 \[ -\frac {122 \, x}{75 \, \sqrt {-4 \, x^{2} + 2 \, x + 1}} + \frac {2 \, x^{2}}{{\left (-4 \, x^{2} + 2 \, x + 1\right )}^{\frac {3}{2}}} + \frac {61}{150 \, \sqrt {-4 \, x^{2} + 2 \, x + 1}} - \frac {19 \, x}{15 \, {\left (-4 \, x^{2} + 2 \, x + 1\right )}^{\frac {3}{2}}} - \frac {23}{30 \, {\left (-4 \, x^{2} + 2 \, x + 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3-8*x-1)/(-4*x^2+2*x+1)^(5/2),x, algorithm="maxima")

[Out]

-122/75*x/sqrt(-4*x^2 + 2*x + 1) + 2*x^2/(-4*x^2 + 2*x + 1)^(3/2) + 61/150/sqrt(-4*x^2 + 2*x + 1) - 19/15*x/(-
4*x^2 + 2*x + 1)^(3/2) - 23/30/(-4*x^2 + 2*x + 1)^(3/2)

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mupad [B]  time = 0.19, size = 29, normalized size = 0.64 \[ -\frac {-488\,x^3+216\,x^2+156\,x+27}{75\,{\left (-4\,x^2+2\,x+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x - 8*x^3 + 1)/(2*x - 4*x^2 + 1)^(5/2),x)

[Out]

-(156*x + 216*x^2 - 488*x^3 + 27)/(75*(2*x - 4*x^2 + 1)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {8 x^{3} - 8 x - 1}{\left (- 4 x^{2} + 2 x + 1\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**3-8*x-1)/(-4*x**2+2*x+1)**(5/2),x)

[Out]

Integral((8*x**3 - 8*x - 1)/(-4*x**2 + 2*x + 1)**(5/2), x)

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