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
Time = 0.08 (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
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
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])
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]
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])
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]
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]
Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
trager | \(\frac {\left (3 x^{2}-4\right ) \sqrt {x^{4}+2 x^{2}+4}}{64 x^{4}}-\frac {\ln \left (\frac {-x^{2}+2 \sqrt {x^{4}+2 x^{2}+4}-4}{x^{2}}\right )}{128}\) | \(54\) |
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)
Time = 0.06 (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
\[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\int \frac {1}{x^{5} \sqrt {x^{4} + 2 x^{2} + 4}}\, dx \] 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)
Time = 0.11 (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)
Time = 0.13 (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)
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)
\[ \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx=\int \frac {1}{\sqrt {x^{4}+2 x^{2}+4}\, x^{5}}d x \] Input:
int(1/x^5/(x^4+2*x^2+4)^(1/2),x)
Output:
int(1/(sqrt(x**4 + 2*x**2 + 4)*x**5),x)