3.796 \(\int \frac {1}{\sqrt {8+8 x-x^3+8 x^4}} \, dx\)
Optimal. Leaf size=129 \[ -\frac {x^2 \sqrt {\frac {\left (\frac {4}{x}+1\right )^4-6 \left (\frac {4}{x}+1\right )^2+261}{\left (\frac {\sqrt {29} (x+4)^2}{x^2}+87\right )^2}} \left (\frac {\sqrt {29} (x+4)^2}{x^2}+87\right ) F\left (2 \tan ^{-1}\left (\frac {x+4}{\sqrt {3} \sqrt [4]{29} x}\right )|\frac {1}{58} \left (29+\sqrt {29}\right )\right )}{8 \sqrt {3} \sqrt [4]{29} \sqrt {8 x^4-x^3+8 x+8}} \]
[Out]
-1/696*x^2*(cos(2*arctan(1/87*(4+x)*29^(3/4)/x*3^(1/2)))^2)^(1/2)/cos(2*arctan(1/87*(4+x)*29^(3/4)/x*3^(1/2)))
*EllipticF(sin(2*arctan(1/87*(4+x)*29^(3/4)/x*3^(1/2))),1/58*(1682+58*29^(1/2))^(1/2))*(87+(4+x)^2*29^(1/2)/x^
2)*((261-6*(1+4/x)^2+(1+4/x)^4)/(87+(4+x)^2*29^(1/2)/x^2)^2)^(1/2)*29^(3/4)*3^(1/2)/(8*x^4-x^3+8*x+8)^(1/2)
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Rubi [A] time = 0.30, antiderivative size = 129, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used =
{2069, 12, 6719, 1103} \[ -\frac {x^2 \sqrt {\frac {\left (\frac {4}{x}+1\right )^4-6 \left (\frac {4}{x}+1\right )^2+261}{\left (\frac {\sqrt {29} (x+4)^2}{x^2}+87\right )^2}} \left (\frac {\sqrt {29} (x+4)^2}{x^2}+87\right ) F\left (2 \tan ^{-1}\left (\frac {x+4}{\sqrt {3} \sqrt [4]{29} x}\right )|\frac {1}{58} \left (29+\sqrt {29}\right )\right )}{8 \sqrt {3} \sqrt [4]{29} \sqrt {8 x^4-x^3+8 x+8}} \]
Antiderivative was successfully verified.
[In]
Int[1/Sqrt[8 + 8*x - x^3 + 8*x^4],x]
[Out]
-(x^2*Sqrt[(261 - 6*(1 + 4/x)^2 + (1 + 4/x)^4)/(87 + (Sqrt[29]*(4 + x)^2)/x^2)^2]*(87 + (Sqrt[29]*(4 + x)^2)/x
^2)*EllipticF[2*ArcTan[(4 + x)/(Sqrt[3]*29^(1/4)*x)], (29 + Sqrt[29])/58])/(8*Sqrt[3]*29^(1/4)*Sqrt[8 + 8*x -
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+8 x-x^3+8 x^4}} \, dx &=-\left (1024 \operatorname {Subst}\left (\int \frac {1}{2 \sqrt {2} (8-32 x)^2 \sqrt {\frac {1069056-393216 x^2+1048576 x^4}{(8-32 x)^4}}} \, dx,x,\frac {1}{4}+\frac {1}{x}\right )\right )\\ &=-\left (\left (256 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{(8-32 x)^2 \sqrt {\frac {1069056-393216 x^2+1048576 x^4}{(8-32 x)^4}}} \, dx,x,\frac {1}{4}+\frac {1}{x}\right )\right )\\ &=-\frac {\left (\sqrt {1069056-393216 \left (\frac {1}{4}+\frac {1}{x}\right )^2+1048576 \left (\frac {1}{4}+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1069056-393216 x^2+1048576 x^4}} \, dx,x,\frac {1}{4}+\frac {1}{x}\right )}{\sqrt {8+8 x-x^3+8 x^4}}\\ &=-\frac {x^2 \sqrt {\frac {261-6 \left (1+\frac {4}{x}\right )^2+\left (1+\frac {4}{x}\right )^4}{\left (87+\frac {\sqrt {29} (4+x)^2}{x^2}\right )^2}} \left (87+\frac {\sqrt {29} (4+x)^2}{x^2}\right ) F\left (2 \tan ^{-1}\left (\frac {4+x}{\sqrt {3} \sqrt [4]{29} x}\right )|\frac {1}{58} \left (29+\sqrt {29}\right )\right )}{8 \sqrt {3} \sqrt [4]{29} \sqrt {8+8 x-x^3+8 x^4}}\\ \end {align*}
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Mathematica [C] time = 0.84, size = 927, normalized size = 7.19 \[ \text {result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[1/Sqrt[8 + 8*x - x^3 + 8*x^4],x]
[Out]
(-2*EllipticF[ArcSin[Sqrt[((x - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0])*(Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2,
0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4, 0]))/((x - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0])*(Root[8 + 8*#1
- #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4, 0]))]], ((Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2
, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 3, 0])*(Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^
3 + 8*#1^4 & , 4, 0]))/((Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 3, 0])*(R
oot[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4, 0]))]*(x - Root[8 + 8*#1 - #1^3
+ 8*#1^4 & , 2, 0])^2*Sqrt[((Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0]
)*(x - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 3, 0]))/((x - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0])*(Root[8 + 8*#1
- #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 3, 0]))]*(Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1,
0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4, 0])*Sqrt[((x - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0])*(Root[8 + 8*
#1 - #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0])*(x - Root[8 + 8*#1 - #1^3 + 8*#1^4 & ,
4, 0])*(Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4, 0]))/((x - Root[8 + 8*
#1 - #1^3 + 8*#1^4 & , 2, 0])^2*(Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4
, 0])^2)])/(Sqrt[8 + 8*x - x^3 + 8*x^4]*(-Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 1, 0] + Root[8 + 8*#1 - #1^3 + 8*#
1^4 & , 2, 0])*(Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 2, 0] - Root[8 + 8*#1 - #1^3 + 8*#1^4 & , 4, 0]))
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {8 \, x^{4} - x^{3} + 8 \, x + 8}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(8*x^4-x^3+8*x+8)^(1/2),x, algorithm="fricas")
[Out]
integral(1/sqrt(8*x^4 - x^3 + 8*x + 8), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {8 \, x^{4} - x^{3} + 8 \, x + 8}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(8*x^4-x^3+8*x+8)^(1/2),x, algorithm="giac")
[Out]
integrate(1/sqrt(8*x^4 - x^3 + 8*x + 8), x)
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maple [C] time = 1.13, size = 965, normalized size = 7.48 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(8*x^4-x^3+8*x+8)^(1/2),x)
[Out]
1/2*(-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=4)+RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))*((RootOf(8*_Z^4-_Z^3+8*_Z+8,index
=4)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2))*(x-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(RootOf(8*_Z^4-_Z^3+8*_Z+8,inde
x=4)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(x-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)))^(1/2)*(x-RootOf(8*_Z^4-_Z^3+8
*_Z+8,index=2))^2*((RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))*(x-RootOf(8*_Z^4-_Z
^3+8*_Z+8,index=3))/(RootOf(8*_Z^4-_Z^3+8*_Z+8,index=3)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(x-RootOf(8*_Z^4-_
Z^3+8*_Z+8,index=2)))^(1/2)*((RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))*(x-RootOf
(8*_Z^4-_Z^3+8*_Z+8,index=4))/(RootOf(8*_Z^4-_Z^3+8*_Z+8,index=4)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(x-RootO
f(8*_Z^4-_Z^3+8*_Z+8,index=2)))^(1/2)/(RootOf(8*_Z^4-_Z^3+8*_Z+8,index=4)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2))/
(RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))*2^(1/2)/((x-RootOf(8*_Z^4-_Z^3+8*_Z+8,
index=1))*(x-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2))*(x-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=3))*(x-RootOf(8*_Z^4-_Z^3+
8*_Z+8,index=4)))^(1/2)*EllipticF(((RootOf(8*_Z^4-_Z^3+8*_Z+8,index=4)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2))*(x-
RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(RootOf(8*_Z^4-_Z^3+8*_Z+8,index=4)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(x
-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)))^(1/2),((RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)-RootOf(8*_Z^4-_Z^3+8*_Z+8,ind
ex=3))*(-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=4)+RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(-RootOf(8*_Z^4-_Z^3+8*_Z+8,in
dex=3)+RootOf(8*_Z^4-_Z^3+8*_Z+8,index=1))/(RootOf(8*_Z^4-_Z^3+8*_Z+8,index=2)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index
=4)))^(1/2))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {8 \, x^{4} - x^{3} + 8 \, x + 8}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(8*x^4-x^3+8*x+8)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/sqrt(8*x^4 - x^3 + 8*x + 8), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {8\,x^4-x^3+8\,x+8}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(8*x - x^3 + 8*x^4 + 8)^(1/2),x)
[Out]
int(1/(8*x - x^3 + 8*x^4 + 8)^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {8 x^{4} - x^{3} + 8 x + 8}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(8*x**4-x**3+8*x+8)**(1/2),x)
[Out]
Integral(1/sqrt(8*x**4 - x**3 + 8*x + 8), x)
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