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]
-((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])
________________________________________________________________________________________
Rubi [A] time = 0.325988, antiderivative size = 126, normalized size of antiderivative = 1.,
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 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+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*}
Mathematica [C] time = 0.355084, 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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Maple [C] time = 0.76, size = 1180, normalized size = 9.4 \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^4-15*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=4)+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=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,in
dex=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-RootO
f(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,in
dex=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-RootO
f(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-
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=2)-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+2
4*_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,inde
x=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,in
dex=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)-RootO
f(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=4)+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,ind
ex=1)-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=2)-RootOf(8*_
Z^4-15*_Z^3+8*_Z^2+24*_Z+8,index=4)))^(1/2))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}\,{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)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/sqrt(8*x^4 - 15*x^3 + 8*x^2 + 24*x + 8), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}, 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)^(1/2),x, algorithm="fricas")
[Out]
integral(1/sqrt(8*x^4 - 15*x^3 + 8*x^2 + 24*x + 8), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 x^{4} - 15 x^{3} + 8 x^{2} + 24 x + 8}}\, 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)**(1/2),x)
[Out]
Integral(1/sqrt(8*x**4 - 15*x**3 + 8*x**2 + 24*x + 8), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}}\,{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)^(1/2),x, algorithm="giac")
[Out]
integrate(1/sqrt(8*x^4 - 15*x^3 + 8*x^2 + 24*x + 8), x)