3.801 \(\int \frac{1}{(8+24 x+8 x^2-15 x^3+8 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=434 \[ -\frac{\left (172-7 \left (\frac{4}{x}+3\right )^2\right ) x^2}{208 \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}+\frac{\left (50896-2455 \left (\frac{4}{x}+3\right )^2\right ) \left (\frac{4}{x}+3\right ) x^2}{322608 \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}+\frac{2455 \left (\left (\frac{4}{x}+3\right )^4-38 \left (\frac{4}{x}+3\right )^2+517\right ) \left (\frac{4}{x}+3\right ) x^2}{322608 \left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right ) \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}+\frac{\left (4910-203 \sqrt{517}\right ) \left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right ) \sqrt{\frac{\left (\frac{4}{x}+3\right )^4-38 \left (\frac{4}{x}+3\right )^2+517}{\left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac{3 x+4}{\sqrt [4]{517} x}\right )|\frac{517+19 \sqrt{517}}{1034}\right )}{2496\ 517^{3/4} \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}-\frac{2455 \left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right ) \sqrt{\frac{\left (\frac{4}{x}+3\right )^4-38 \left (\frac{4}{x}+3\right )^2+517}{\left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right )^2}} x^2 E\left (2 \tan ^{-1}\left (\frac{3 x+4}{\sqrt [4]{517} x}\right )|\frac{517+19 \sqrt{517}}{1034}\right )}{624\ 517^{3/4} \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}} \]

[Out]

-((172 - 7*(3 + 4/x)^2)*x^2)/(208*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) + ((50896 - 2455*(3 + 4/x)^2)*(3 +
4/x)*x^2)/(322608*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) + (2455*(517 - 38*(3 + 4/x)^2 + (3 + 4/x)^4)*(3 + 4
/x)*x^2)/(322608*(Sqrt[517] + (3 + 4/x)^2)*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) - (2455*(Sqrt[517] + (3 +
4/x)^2)*Sqrt[(517 - 38*(3 + 4/x)^2 + (3 + 4/x)^4)/(Sqrt[517] + (3 + 4/x)^2)^2]*x^2*EllipticE[2*ArcTan[(4 + 3*x
)/(517^(1/4)*x)], (517 + 19*Sqrt[517])/1034])/(624*517^(3/4)*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) + ((4910
- 203*Sqrt[517])*(Sqrt[517] + (3 + 4/x)^2)*Sqrt[(517 - 38*(3 + 4/x)^2 + (3 + 4/x)^4)/(Sqrt[517] + (3 + 4/x)^2
)^2]*x^2*EllipticF[2*ArcTan[(4 + 3*x)/(517^(1/4)*x)], (517 + 19*Sqrt[517])/1034])/(2496*517^(3/4)*Sqrt[8 + 24*
x + 8*x^2 - 15*x^3 + 8*x^4])

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Rubi [A]  time = 0.52758, antiderivative size = 434, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2069, 12, 6719, 1673, 1678, 1197, 1103, 1195, 1247, 636} \[ -\frac{\left (172-7 \left (\frac{4}{x}+3\right )^2\right ) x^2}{208 \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}+\frac{\left (50896-2455 \left (\frac{4}{x}+3\right )^2\right ) \left (\frac{4}{x}+3\right ) x^2}{322608 \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}+\frac{2455 \left (\left (\frac{4}{x}+3\right )^4-38 \left (\frac{4}{x}+3\right )^2+517\right ) \left (\frac{4}{x}+3\right ) x^2}{322608 \left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right ) \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}+\frac{\left (4910-203 \sqrt{517}\right ) \left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right ) \sqrt{\frac{\left (\frac{4}{x}+3\right )^4-38 \left (\frac{4}{x}+3\right )^2+517}{\left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac{3 x+4}{\sqrt [4]{517} x}\right )|\frac{517+19 \sqrt{517}}{1034}\right )}{2496\ 517^{3/4} \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}}-\frac{2455 \left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right ) \sqrt{\frac{\left (\frac{4}{x}+3\right )^4-38 \left (\frac{4}{x}+3\right )^2+517}{\left (\left (\frac{4}{x}+3\right )^2+\sqrt{517}\right )^2}} x^2 E\left (2 \tan ^{-1}\left (\frac{3 x+4}{\sqrt [4]{517} x}\right )|\frac{517+19 \sqrt{517}}{1034}\right )}{624\ 517^{3/4} \sqrt{8 x^4-15 x^3+8 x^2+24 x+8}} \]

Antiderivative was successfully verified.

[In]

Int[(8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4)^(-3/2),x]

[Out]

-((172 - 7*(3 + 4/x)^2)*x^2)/(208*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) + ((50896 - 2455*(3 + 4/x)^2)*(3 +
4/x)*x^2)/(322608*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) + (2455*(517 - 38*(3 + 4/x)^2 + (3 + 4/x)^4)*(3 + 4
/x)*x^2)/(322608*(Sqrt[517] + (3 + 4/x)^2)*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) - (2455*(Sqrt[517] + (3 +
4/x)^2)*Sqrt[(517 - 38*(3 + 4/x)^2 + (3 + 4/x)^4)/(Sqrt[517] + (3 + 4/x)^2)^2]*x^2*EllipticE[2*ArcTan[(4 + 3*x
)/(517^(1/4)*x)], (517 + 19*Sqrt[517])/1034])/(624*517^(3/4)*Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4]) + ((4910
- 203*Sqrt[517])*(Sqrt[517] + (3 + 4/x)^2)*Sqrt[(517 - 38*(3 + 4/x)^2 + (3 + 4/x)^4)/(Sqrt[517] + (3 + 4/x)^2
)^2]*x^2*EllipticF[2*ArcTan[(4 + 3*x)/(517^(1/4)*x)], (517 + 19*Sqrt[517])/1034])/(2496*517^(3/4)*Sqrt[8 + 24*
x + 8*x^2 - 15*x^3 + 8*x^4])

Rule 2069

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Dist[-16*a^2, Subst[Int[(1*((a*(-3*b^4 + 16*a*b^2*c - 64*a^2*b*d + 256*a^3*e -
32*a^2*(3*b^2 - 8*a*c)*x^2 + 256*a^4*x^4))/(b - 4*a*x)^4)^p)/(b - 4*a*x)^2, x], x, b/(4*a) + 1/x], x] /; NeQ[a
, 0] && NeQ[b, 0] && EqQ[b^3 - 4*a*b*c + 8*a^2*d, 0]] /; FreeQ[p, x] && PolyQ[P4, x, 4] && IntegerQ[2*p] &&  !
IGtQ[p, 0]

Rule 12

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

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2
+ c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*
a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (8+24 x+8 x^2-15 x^3+8 x^4\right )^{3/2}} \, dx &=-\left (1024 \operatorname{Subst}\left (\int \frac{1}{16 \sqrt{2} (24-32 x)^2 \left (\frac{2117632-2490368 x^2+1048576 x^4}{(24-32 x)^4}\right )^{3/2}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )\right )\\ &=-\left (\left (32 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{(24-32 x)^2 \left (\frac{2117632-2490368 x^2+1048576 x^4}{(24-32 x)^4}\right )^{3/2}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )\right )\\ &=-\frac{\left (\sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{(24-32 x)^4}{\left (2117632-2490368 x^2+1048576 x^4\right )^{3/2}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )}{8 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}\\ &=-\frac{\left (\sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{x \left (-1769472-3145728 x^2\right )}{\left (2117632-2490368 x^2+1048576 x^4\right )^{3/2}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )}{8 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}-\frac{\left (\sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{331776+3538944 x^2+1048576 x^4}{\left (2117632-2490368 x^2+1048576 x^4\right )^{3/2}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )}{8 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}\\ &=\frac{\left (50896-2455 \left (3+\frac{4}{x}\right )^2\right ) \left (3+\frac{4}{x}\right ) x^2}{322608 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}-\frac{\left (\sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{29541080280760057856-22112674170389135360 x^2}{\sqrt{2117632-2490368 x^2+1048576 x^4}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )}{45403039643335655424 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}-\frac{\left (\sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{-1769472-3145728 x}{\left (2117632-2490368 x+1048576 x^2\right )^{3/2}} \, dx,x,\left (\frac{3}{4}+\frac{1}{x}\right )^2\right )}{16 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}\\ &=-\frac{\left (172-7 \left (3+\frac{4}{x}\right )^2\right ) x^2}{208 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}+\frac{\left (50896-2455 \left (3+\frac{4}{x}\right )^2\right ) \left (3+\frac{4}{x}\right ) x^2}{322608 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}-\frac{\left (2455 \sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{1-\frac{16 x^2}{\sqrt{517}}}{\sqrt{2117632-2490368 x^2+1048576 x^4}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )}{156 \sqrt{517} \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}-\frac{\left (\left (104951-4910 \sqrt{517}\right ) \sqrt{2117632-2490368 \left (\frac{3}{4}+\frac{1}{x}\right )^2+1048576 \left (\frac{3}{4}+\frac{1}{x}\right )^4} x^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2117632-2490368 x^2+1048576 x^4}} \, dx,x,\frac{3}{4}+\frac{1}{x}\right )}{161304 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}\\ &=-\frac{\left (172-7 \left (3+\frac{4}{x}\right )^2\right ) x^2}{208 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}+\frac{\left (50896-2455 \left (3+\frac{4}{x}\right )^2\right ) \left (3+\frac{4}{x}\right ) x^2}{322608 \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}+\frac{2455 \left (517-38 \left (3+\frac{4}{x}\right )^2+\left (3+\frac{4}{x}\right )^4\right ) \left (3+\frac{4}{x}\right ) x^2}{322608 \left (\sqrt{517}+\left (3+\frac{4}{x}\right )^2\right ) \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}-\frac{2455 \left (\sqrt{517}+\left (3+\frac{4}{x}\right )^2\right ) \sqrt{\frac{517-38 \left (3+\frac{4}{x}\right )^2+\left (3+\frac{4}{x}\right )^4}{\left (\sqrt{517}+\left (3+\frac{4}{x}\right )^2\right )^2}} x^2 E\left (2 \tan ^{-1}\left (\frac{4+3 x}{\sqrt [4]{517} x}\right )|\frac{517+19 \sqrt{517}}{1034}\right )}{624\ 517^{3/4} \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}+\frac{\left (4910-203 \sqrt{517}\right ) \left (\sqrt{517}+\left (3+\frac{4}{x}\right )^2\right ) \sqrt{\frac{517-38 \left (3+\frac{4}{x}\right )^2+\left (3+\frac{4}{x}\right )^4}{\left (\sqrt{517}+\left (3+\frac{4}{x}\right )^2\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac{4+3 x}{\sqrt [4]{517} x}\right )|\frac{517+19 \sqrt{517}}{1034}\right )}{2496\ 517^{3/4} \sqrt{8+24 x+8 x^2-15 x^3+8 x^4}}\\ \end{align*}

Mathematica [C]  time = 6.04897, size = 6019, normalized size = 13.87 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4)^(-3/2),x]

[Out]

Result too large to show

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Maple [C]  time = 0.017, size = 5421, normalized size = 12.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(8*x^4-15*x^3+8*x^2+24*x+8)^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*x^4-15*x^3+8*x^2+24*x+8)^(3/2),x, algorithm="maxima")

[Out]

integrate((8*x^4 - 15*x^3 + 8*x^2 + 24*x + 8)^(-3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}{64 \, x^{8} - 240 \, x^{7} + 353 \, x^{6} + 144 \, x^{5} - 528 \, x^{4} + 144 \, x^{3} + 704 \, x^{2} + 384 \, x + 64}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*x^4-15*x^3+8*x^2+24*x+8)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(8*x^4 - 15*x^3 + 8*x^2 + 24*x + 8)/(64*x^8 - 240*x^7 + 353*x^6 + 144*x^5 - 528*x^4 + 144*x^3 + 7
04*x^2 + 384*x + 64), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (8 x^{4} - 15 x^{3} + 8 x^{2} + 24 x + 8\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*x**4-15*x**3+8*x**2+24*x+8)**(3/2),x)

[Out]

Integral((8*x**4 - 15*x**3 + 8*x**2 + 24*x + 8)**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*x^4-15*x^3+8*x^2+24*x+8)^(3/2),x, algorithm="giac")

[Out]

integrate((8*x^4 - 15*x^3 + 8*x^2 + 24*x + 8)^(-3/2), x)