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]
-(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])
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Rubi [A] time = 0.302434, antiderivative size = 129, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, 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 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 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]
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*}
Mathematica [C] time = 0.344436, 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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Maple [C] time = 1.415, size = 965, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}
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=1)-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=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)*(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-RootOf
(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,i
ndex=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-R
ootOf(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,inde
x=3))*(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)-RootOf(8*_Z^4-_Z^3+8*_Z+8,index=3))/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 \, x^{4} - x^{3} + 8 \, x + 8}}\,{d x} \end{align*}
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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{8 \, x^{4} - x^{3} + 8 \, x + 8}}, x\right ) \end{align*}
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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 x^{4} - x^{3} + 8 x + 8}}\, dx \end{align*}
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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 \, x^{4} - x^{3} + 8 \, x + 8}}\,{d x} \end{align*}
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)