3.4.10 \(\int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx\) [310]

3.4.10.1 Optimal result

Integrand size = 18, antiderivative size = 71 \[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=-\frac {\sqrt {4+2 x^2+x^4}}{16 x^4}+\frac {3 \sqrt {4+2 x^2+x^4}}{64 x^2}+\frac {1}{128} \text {arctanh}\left (\frac {4+x^2}{2 \sqrt {4+2 x^2+x^4}}\right ) \]

output

1/128*arctanh(1/2*(x^2+4)/(x^4+2*x^2+4)^(1/2))-1/16*(x^4+2*x^2+4)^(1/2)/x^ 
4+3/64*(x^4+2*x^2+4)^(1/2)/x^2
 

3.4.10.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\frac {1}{64} \left (\frac {\left (-4+3 x^2\right ) \sqrt {4+2 x^2+x^4}}{x^4}-\text {arctanh}\left (\frac {1}{2} \left (x^2-\sqrt {4+2 x^2+x^4}\right )\right )\right ) \]

input

Integrate[1/(x^5*Sqrt[4 + 2*x^2 + x^4]),x]
 

output

(((-4 + 3*x^2)*Sqrt[4 + 2*x^2 + x^4])/x^4 - ArcTanh[(x^2 - Sqrt[4 + 2*x^2 
+ x^4])/2])/64
 

3.4.10.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1434, 1167, 1228, 1154, 219}

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.

\(\displaystyle \int \frac {1}{x^5 \sqrt {x^4+2 x^2+4}} \, dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \sqrt {x^4+2 x^2+4}}dx^2\)

\(\Big \downarrow \) 1167

\(\displaystyle \frac {1}{2} \left (-\frac {1}{8} \int \frac {x^2+3}{x^4 \sqrt {x^4+2 x^2+4}}dx^2-\frac {\sqrt {x^4+2 x^2+4}}{8 x^4}\right )\)

\(\Big \downarrow \) 1228

\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (\frac {3 \sqrt {x^4+2 x^2+4}}{4 x^2}-\frac {1}{4} \int \frac {1}{x^2 \sqrt {x^4+2 x^2+4}}dx^2\right )-\frac {\sqrt {x^4+2 x^2+4}}{8 x^4}\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{2} \int \frac {1}{16-x^4}d\frac {2 \left (x^2+4\right )}{\sqrt {x^4+2 x^2+4}}+\frac {3 \sqrt {x^4+2 x^2+4}}{4 x^2}\right )-\frac {\sqrt {x^4+2 x^2+4}}{8 x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (\frac {1}{8} \text {arctanh}\left (\frac {x^2+4}{2 \sqrt {x^4+2 x^2+4}}\right )+\frac {3 \sqrt {x^4+2 x^2+4}}{4 x^2}\right )-\frac {\sqrt {x^4+2 x^2+4}}{8 x^4}\right )\)

input

Int[1/(x^5*Sqrt[4 + 2*x^2 + x^4]),x]
 

output

(-1/8*Sqrt[4 + 2*x^2 + x^4]/x^4 + ((3*Sqrt[4 + 2*x^2 + x^4])/(4*x^2) + Arc 
Tanh[(4 + x^2)/(2*Sqrt[4 + 2*x^2 + x^4])]/8)/8)/2
 

3.4.10.3.1 Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1167

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2))   Int[ 
(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, 
 x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m 
, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp 
lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
 

rule 1228

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^ 
(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

3.4.10.4 Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73

method	result	size
trager	\(\frac {\left (3 x^{2}-4\right ) \sqrt {x^{4}+2 x^{2}+4}}{64 x^{4}}+\frac {\ln \left (\frac {x^{2}+2 \sqrt {x^{4}+2 x^{2}+4}+4}{x^{2}}\right )}{128}\)	\(52\)
default	\(-\frac {\sqrt {x^{4}+2 x^{2}+4}}{16 x^{4}}+\frac {3 \sqrt {x^{4}+2 x^{2}+4}}{64 x^{2}}+\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+8}{4 \sqrt {x^{4}+2 x^{2}+4}}\right )}{128}\)	\(60\)
risch	\(\frac {3 x^{6}+2 x^{4}+4 x^{2}-16}{64 x^{4} \sqrt {x^{4}+2 x^{2}+4}}+\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+8}{4 \sqrt {x^{4}+2 x^{2}+4}}\right )}{128}\)	\(60\)
elliptic	\(-\frac {\sqrt {x^{4}+2 x^{2}+4}}{16 x^{4}}+\frac {3 \sqrt {x^{4}+2 x^{2}+4}}{64 x^{2}}+\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+8}{4 \sqrt {x^{4}+2 x^{2}+4}}\right )}{128}\)	\(60\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {x^{2}+4}{2 \sqrt {x^{4}+2 x^{2}+4}}\right ) x^{4}+6 x^{2} \sqrt {x^{4}+2 x^{2}+4}-8 \sqrt {x^{4}+2 x^{2}+4}}{128 x^{4}}\)	\(62\)

input

int(1/x^5/(x^4+2*x^2+4)^(1/2),x,method=_RETURNVERBOSE)
 

output

1/64*(3*x^2-4)/x^4*(x^4+2*x^2+4)^(1/2)+1/128*ln((x^2+2*(x^4+2*x^2+4)^(1/2) 
+4)/x^2)
 

3.4.10.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\frac {x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 2 \, x^{2} + 4} + 2\right ) - x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 2 \, x^{2} + 4} - 2\right ) + 6 \, x^{4} + 2 \, \sqrt {x^{4} + 2 \, x^{2} + 4} {\left (3 \, x^{2} - 4\right )}}{128 \, x^{4}} \]

input

integrate(1/x^5/(x^4+2*x^2+4)^(1/2),x, algorithm="fricas")
 

output

1/128*(x^4*log(-x^2 + sqrt(x^4 + 2*x^2 + 4) + 2) - x^4*log(-x^2 + sqrt(x^4 
 + 2*x^2 + 4) - 2) + 6*x^4 + 2*sqrt(x^4 + 2*x^2 + 4)*(3*x^2 - 4))/x^4
 

3.4.10.6 Sympy [F]

\[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\int \frac {1}{x^{5} \sqrt {x^{4} + 2 x^{2} + 4}}\, dx \]

input

integrate(1/x**5/(x**4+2*x**2+4)**(1/2),x)
 

output

Integral(1/(x**5*sqrt(x**4 + 2*x**2 + 4)), x)
 

3.4.10.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\frac {3 \, \sqrt {x^{4} + 2 \, x^{2} + 4}}{64 \, x^{2}} - \frac {\sqrt {x^{4} + 2 \, x^{2} + 4}}{16 \, x^{4}} + \frac {1}{128} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} + \frac {4 \, \sqrt {3}}{3 \, x^{2}}\right ) \]

input

integrate(1/x^5/(x^4+2*x^2+4)^(1/2),x, algorithm="maxima")
 

output

3/64*sqrt(x^4 + 2*x^2 + 4)/x^2 - 1/16*sqrt(x^4 + 2*x^2 + 4)/x^4 + 1/128*ar 
csinh(1/3*sqrt(3) + 4/3*sqrt(3)/x^2)
 

3.4.10.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\frac {{\left (x^{2} - \sqrt {x^{4} + 2 \, x^{2} + 4}\right )}^{3} + 36 \, x^{2} - 36 \, \sqrt {x^{4} + 2 \, x^{2} + 4} + 64}{32 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 2 \, x^{2} + 4}\right )}^{2} - 4\right )}^{2}} - \frac {1}{128} \, \log \left (x^{2} - \sqrt {x^{4} + 2 \, x^{2} + 4} + 2\right ) + \frac {1}{128} \, \log \left (-x^{2} + \sqrt {x^{4} + 2 \, x^{2} + 4} + 2\right ) \]

input

integrate(1/x^5/(x^4+2*x^2+4)^(1/2),x, algorithm="giac")
 

output

1/32*((x^2 - sqrt(x^4 + 2*x^2 + 4))^3 + 36*x^2 - 36*sqrt(x^4 + 2*x^2 + 4) 
+ 64)/((x^2 - sqrt(x^4 + 2*x^2 + 4))^2 - 4)^2 - 1/128*log(x^2 - sqrt(x^4 + 
 2*x^2 + 4) + 2) + 1/128*log(-x^2 + sqrt(x^4 + 2*x^2 + 4) + 2)
 

3.4.10.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\int \frac {1}{x^5\,\sqrt {x^4+2\,x^2+4}} \,d x \]

input

int(1/(x^5*(2*x^2 + x^4 + 4)^(1/2)),x)
 

output

int(1/(x^5*(2*x^2 + x^4 + 4)^(1/2)), x)
 

3.4.10.10 Reduce [F]

\[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\int \frac {1}{\sqrt {x^{4}+2 x^{2}+4}\, x^{5}}d x \]

input

int(1/(sqrt(x**4 + 2*x**2 + 4)*x**5),x)
 

output

int(1/(sqrt(x**4 + 2*x**2 + 4)*x**5),x)