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

Optimal. Leaf size=126 \[ -\frac {\left (\left (\frac {4}{x}+3\right )^2+\sqrt {517}\right ) \sqrt {\frac {\left (\frac {4}{x}+3\right )^4-38 \left (\frac {4}{x}+3\right )^2+517}{\left (\left (\frac {4}{x}+3\right )^2+\sqrt {517}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac {3 x+4}{\sqrt [4]{517} x}\right )|\frac {517+19 \sqrt {517}}{1034}\right )}{8 \sqrt [4]{517} \sqrt {8 x^4-15 x^3+8 x^2+24 x+8}} \]

[Out]

-1/4136*x^2*(cos(2*arctan(1/517*(4+3*x)*517^(3/4)/x))^2)^(1/2)/cos(2*arctan(1/517*(4+3*x)*517^(3/4)/x))*Ellipt
icF(sin(2*arctan(1/517*(4+3*x)*517^(3/4)/x)),1/1034*(534578+19646*517^(1/2))^(1/2))*((3+4/x)^2+517^(1/2))*((51
7-38*(3+4/x)^2+(3+4/x)^4)/((3+4/x)^2+517^(1/2))^2)^(1/2)*517^(3/4)/(8*x^4-15*x^3+8*x^2+24*x+8)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2069, 12, 6719, 1103} \[ -\frac {\left (\left (\frac {4}{x}+3\right )^2+\sqrt {517}\right ) \sqrt {\frac {\left (\frac {4}{x}+3\right )^4-38 \left (\frac {4}{x}+3\right )^2+517}{\left (\left (\frac {4}{x}+3\right )^2+\sqrt {517}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac {3 x+4}{\sqrt [4]{517} x}\right )|\frac {517+19 \sqrt {517}}{1034}\right )}{8 \sqrt [4]{517} \sqrt {8 x^4-15 x^3+8 x^2+24 x+8}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4],x]

[Out]

-((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*Ellipti
cF[2*ArcTan[(4 + 3*x)/(517^(1/4)*x)], (517 + 19*Sqrt[517])/1034])/(8*517^(1/4)*Sqrt[8 + 24*x + 8*x^2 - 15*x^3
+ 8*x^4])

Rule 12

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

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx &=-\left (1024 \operatorname {Subst}\left (\int \frac {1}{2 \sqrt {2} (24-32 x)^2 \sqrt {\frac {2117632-2490368 x^2+1048576 x^4}{(24-32 x)^4}}} \, dx,x,\frac {3}{4}+\frac {1}{x}\right )\right )\\ &=-\left (\left (256 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{(24-32 x)^2 \sqrt {\frac {2117632-2490368 x^2+1048576 x^4}{(24-32 x)^4}}} \, 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 {1}{\sqrt {2117632-2490368 x^2+1048576 x^4}} \, dx,x,\frac {3}{4}+\frac {1}{x}\right )}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}}\\ &=-\frac {\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 )}{8 \sqrt [4]{517} \sqrt {8+24 x+8 x^2-15 x^3+8 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.98, size = 1148, normalized size = 9.11 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[8 + 24*x + 8*x^2 - 15*x^3 + 8*x^4],x]

[Out]

(-2*EllipticF[ArcSin[Sqrt[((x - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0])*(Root[8 + 24*#1 + 8*#1^2
- 15*#1^3 + 8*#1^4 & , 2, 0] - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 4, 0]))/((x - Root[8 + 24*#1 +
8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*#1 +
8*#1^2 - 15*#1^3 + 8*#1^4 & , 4, 0]))]], ((Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0] - Root[8 + 24*
#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 3, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*
#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 4, 0]))/((Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 2
4*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 3, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0] - Root[8 + 2
4*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 4, 0]))]*(x - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0])^2*Sqr
t[((Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0
])*(x - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 3, 0]))/((x - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^
4 & , 2, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^
4 & , 3, 0]))]*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#
1^4 & , 4, 0])*Sqrt[((x - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*
#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0])*(x - Root[8 + 24*#1 + 8*#1^2 -
15*#1^3 + 8*#1^4 & , 4, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0] - Root[8 + 24*#1 + 8*#1^2 -
15*#1^3 + 8*#1^4 & , 4, 0]))/((x - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0])^2*(Root[8 + 24*#1 +
8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] - Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 4, 0])^2)])/(Sqrt[8 + 24*
x + 8*x^2 - 15*x^3 + 8*x^4]*(-Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 1, 0] + Root[8 + 24*#1 + 8*#1^2 -
15*#1^3 + 8*#1^4 & , 2, 0])*(Root[8 + 24*#1 + 8*#1^2 - 15*#1^3 + 8*#1^4 & , 2, 0] - Root[8 + 24*#1 + 8*#1^2 -
15*#1^3 + 8*#1^4 & , 4, 0]))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}\,{d x} \]

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.68, size = 1180, normalized size = 9.37 \[ \text {result too large to display} \]

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)^(1/2),x)

[Out]

1/2*(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4))*((RootOf(8*_
Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2))*(x-RootOf(8*_Z^4-15*_Z^3+8*
_Z^2+24*_Z+8,index=1))/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,ind
ex=1))/(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)))^(1/2)*(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=
2))^2*((RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))*(x-RootOf
(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=3))/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=3)-RootOf(8*_Z^4-15*_Z^3+
8*_Z^2+24*_Z+8,index=1))/(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)))^(1/2)*((RootOf(8*_Z^4-15*_Z^3+8*_Z
^2+24*_Z+8,index=2)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))*(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,ind
ex=4))/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))/(x-RootOf
(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)))^(1/2)/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)-RootOf(8*_Z^4-1
5*_Z^3+8*_Z^2+24*_Z+8,index=2))/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24
*_Z+8,index=1))*2^(1/2)/((x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))*(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*
_Z+8,index=2))*(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=3))*(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index
=4)))^(1/2)*EllipticF(((RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,ind
ex=2))*(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)-RootOf
(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))/(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)))^(1/2),((RootOf(8*_
Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=3))*(RootOf(8*_Z^4-15*_Z^3+8*_Z
^2+24*_Z+8,index=1)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4))/(-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,inde
x=3)+RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=1))/(RootOf(8*_Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=2)-RootOf(8*_Z
^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}\,{d x} \]

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)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {8\,x^4-15\,x^3+8\,x^2+24\,x+8}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {8 x^{4} - 15 x^{3} + 8 x^{2} + 24 x + 8}}\, dx \]

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)**(1/2),x)

[Out]

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

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