\(\int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx\) [89]

Optimal result

Integrand size = 24, antiderivative size = 125 \[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=-\frac {\left (\sqrt {517}+\left (3+\frac {4}{x}\right )^2\right ) \sqrt {\frac {517-38 \left (3+\frac {4}{x}\right )^2+\left (3+\frac {4}{x}\right )^4}{\left (\sqrt {517}+\left (3+\frac {4}{x}\right )^2\right )^2}} x^2 \operatorname {EllipticF}\left (2 \arctan \left (\frac {3+\frac {4}{x}}{\sqrt [4]{517}}\right ),\frac {517+19 \sqrt {517}}{1034}\right )}{8 \sqrt [4]{517} \sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \] Output:

-1/4136*(517^(1/2)+(3+4/x)^2)*((517-38*(3+4/x)^2+(3+4/x)^4)/(517^(1/2)+(3+ 
4/x)^2)^2)^(1/2)*x^2*InverseJacobiAM(2*arctan(1/517*(3+4/x)*517^(3/4)),1/1 
034*(534578+19646*517^(1/2))^(1/2))*517^(3/4)/(8*x^4-15*x^3+8*x^2+24*x+8)^ 
(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4 in optimal.

Time = 11.10 (sec) , antiderivative size = 1148, normalized size of antiderivative = 9.18 \[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx =\text {Too large to display} \] Input:

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

Output:

(-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 + 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 & , 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*Sqrt[((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...
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2504, 27, 7270, 1416}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2504

\(\displaystyle -1024 \int \frac {1}{128 \sqrt {2} \left (3-4 \left (\frac {3}{4}+\frac {1}{x}\right )\right )^2 \sqrt {\frac {256 \left (\frac {3}{4}+\frac {1}{x}\right )^4-608 \left (\frac {3}{4}+\frac {1}{x}\right )^2+517}{\left (3-4 \left (\frac {3}{4}+\frac {1}{x}\right )\right )^4}}}d\left (\frac {3}{4}+\frac {1}{x}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -4 \sqrt {2} \int \frac {1}{\left (3-4 \left (\frac {3}{4}+\frac {1}{x}\right )\right )^2 \sqrt {\frac {256 \left (\frac {3}{4}+\frac {1}{x}\right )^4-608 \left (\frac {3}{4}+\frac {1}{x}\right )^2+517}{\left (3-4 \left (\frac {3}{4}+\frac {1}{x}\right )\right )^4}}}d\left (\frac {3}{4}+\frac {1}{x}\right )\)

\(\Big \downarrow \) 7270

\(\displaystyle -\frac {4 \sqrt {2} \sqrt {256 \left (\frac {1}{x}+\frac {3}{4}\right )^4-608 \left (\frac {1}{x}+\frac {3}{4}\right )^2+517} \int \frac {1}{\sqrt {256 \left (\frac {3}{4}+\frac {1}{x}\right )^4-608 \left (\frac {3}{4}+\frac {1}{x}\right )^2+517}}d\left (\frac {3}{4}+\frac {1}{x}\right )}{\left (3-4 \left (\frac {1}{x}+\frac {3}{4}\right )\right )^2 \sqrt {\frac {256 \left (\frac {1}{x}+\frac {3}{4}\right )^4-608 \left (\frac {1}{x}+\frac {3}{4}\right )^2+517}{\left (3-4 \left (\frac {1}{x}+\frac {3}{4}\right )\right )^4}}}\)

\(\Big \downarrow \) 1416

\(\displaystyle -\frac {\left (16 \left (\frac {1}{x}+\frac {3}{4}\right )^2+\sqrt {517}\right ) \sqrt {\frac {256 \left (\frac {1}{x}+\frac {3}{4}\right )^4-608 \left (\frac {1}{x}+\frac {3}{4}\right )^2+517}{\left (16 \left (\frac {1}{x}+\frac {3}{4}\right )^2+\sqrt {517}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {4 \left (\frac {3}{4}+\frac {1}{x}\right )}{\sqrt [4]{517}}\right ),\frac {517+19 \sqrt {517}}{1034}\right )}{\sqrt {2} \sqrt [4]{517} \left (3-4 \left (\frac {1}{x}+\frac {3}{4}\right )\right )^2 \sqrt {\frac {256 \left (\frac {1}{x}+\frac {3}{4}\right )^4-608 \left (\frac {1}{x}+\frac {3}{4}\right )^2+517}{\left (3-4 \left (\frac {1}{x}+\frac {3}{4}\right )\right )^4}}}\)

Input:

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

Output:

-(((Sqrt[517] + 16*(3/4 + x^(-1))^2)*Sqrt[(517 - 608*(3/4 + x^(-1))^2 + 25 
6*(3/4 + x^(-1))^4)/(Sqrt[517] + 16*(3/4 + x^(-1))^2)^2]*EllipticF[2*ArcTa 
n[(4*(3/4 + x^(-1)))/517^(1/4)], (517 + 19*Sqrt[517])/1034])/(Sqrt[2]*517^ 
(1/4)*(3 - 4*(3/4 + x^(-1)))^2*Sqrt[(517 - 608*(3/4 + x^(-1))^2 + 256*(3/4 
 + x^(-1))^4)/(3 - 4*(3/4 + x^(-1)))^4]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1416

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)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 2504

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]}, Simp[-16* 
a^2   Subst[Int[(1/(b - 4*a*x)^2)*(a*((-3*b^4 + 16*a*b^2*c - 64*a^2*b*d + 2 
56*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, 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 7270

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1179\) vs. \(2(110)=220\).

Time = 1.02 (sec) , antiderivative size = 1180, normalized size of antiderivative = 9.44

method	result	size
default	\(\text {Expression too large to display}\)	\(1180\)
elliptic	\(\text {Expression too large to display}\)	\(1180\)

Input:

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

Output:

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)-Ro 
otOf(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,index=1))/(x-RootOf(8*_Z^4-15*_Z^3+8*_Z^2+2 
4*_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,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=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^4-15*_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)-RootO 
f(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...
 

Fricas [F]

\[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=\int \frac {1}{\sqrt {8 x^{4} - 15 x^{3} + 8 x^{2} + 24 x + 8}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, x^{4} - 15 \, x^{3} + 8 \, x^{2} + 24 \, x + 8}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=\int \frac {1}{\sqrt {8\,x^4-15\,x^3+8\,x^2+24\,x+8}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {8+24 x+8 x^2-15 x^3+8 x^4}} \, dx=\int \frac {1}{\sqrt {8 x^{4}-15 x^{3}+8 x^{2}+24 x +8}}d x \] Input:

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

Output:

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