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

Optimal. Leaf size=71 \[ \frac {3 \sqrt {x^4+2 x^2+4}}{64 x^2}-\frac {\sqrt {x^4+2 x^2+4}}{16 x^4}+\frac {1}{128} \tanh ^{-1}\left (\frac {x^2+4}{2 \sqrt {x^4+2 x^2+4}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1114, 744, 806, 724, 206} \[ \frac {3 \sqrt {x^4+2 x^2+4}}{64 x^2}-\frac {\sqrt {x^4+2 x^2+4}}{16 x^4}+\frac {1}{128} \tanh ^{-1}\left (\frac {x^2+4}{2 \sqrt {x^4+2 x^2+4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[4 + 2*x^2 + x^4]),x]

[Out]

-Sqrt[4 + 2*x^2 + x^4]/(16*x^4) + (3*Sqrt[4 + 2*x^2 + x^4])/(64*x^2) + ArcTanh[(4 + x^2)/(2*Sqrt[4 + 2*x^2 + x
^4])]/128

Rule 206

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> 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] + Dist[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[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e
, 0] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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] - Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 1114

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

Rubi steps

\begin {align*} \int \frac {1}{x^5 \sqrt {4+2 x^2+x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {4+2 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {4+2 x^2+x^4}}{16 x^4}-\frac {1}{16} \operatorname {Subst}\left (\int \frac {3+x}{x^2 \sqrt {4+2 x+x^2}} \, dx,x,x^2\right )\\ &=-\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}{64} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {4+2 x+x^2}} \, dx,x,x^2\right )\\ &=-\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}{32} \operatorname {Subst}\left (\int \frac {1}{16-x^2} \, dx,x,\frac {2 \left (4+x^2\right )}{\sqrt {4+2 x^2+x^4}}\right )\\ &=-\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} \tanh ^{-1}\left (\frac {4+x^2}{2 \sqrt {4+2 x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 55, normalized size = 0.77 \[ \frac {1}{128} \left (\frac {2 \sqrt {x^4+2 x^2+4} \left (3 x^2-4\right )}{x^4}+\tanh ^{-1}\left (\frac {x^2+4}{2 \sqrt {x^4+2 x^2+4}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(-4 + 3*x^2)*Sqrt[4 + 2*x^2 + x^4])/x^4 + ArcTanh[(4 + x^2)/(2*Sqrt[4 + 2*x^2 + x^4])])/128

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IntegrateAlgebraic [A]  time = 0.21, size = 60, normalized size = 0.85 \[ \frac {\left (3 x^2-4\right ) \sqrt {x^4+2 x^2+4}}{64 x^4}-\frac {1}{64} \tanh ^{-1}\left (\frac {x^2}{2}-\frac {1}{2} \sqrt {x^4+2 x^2+4}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.76, size = 81, normalized size = 1.14 \[ \frac {x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 2 \, x^{2} + 4} + 2\right ) - x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 2 \, x^{2} + 4} - 2\right ) + 6 \, x^{4} + 2 \, \sqrt {x^{4} + 2 \, x^{2} + 4} {\left (3 \, x^{2} - 4\right )}}{128 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.67, size = 112, normalized size = 1.58 \[ \frac {{\left (x^{2} - \sqrt {x^{4} + 2 \, x^{2} + 4}\right )}^{3} + 36 \, x^{2} - 36 \, \sqrt {x^{4} + 2 \, x^{2} + 4} + 64}{32 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 2 \, x^{2} + 4}\right )}^{2} - 4\right )}^{2}} - \frac {1}{128} \, \log \left (x^{2} - \sqrt {x^{4} + 2 \, x^{2} + 4} + 2\right ) + \frac {1}{128} \, \log \left (-x^{2} + \sqrt {x^{4} + 2 \, x^{2} + 4} + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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maple [A]  time = 0.22, size = 52, normalized size = 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	\(-\frac {\sqrt {x^{4}+2 x^{2}+4}}{16 x^{4}}+\frac {3 \sqrt {x^{4}+2 x^{2}+4}}{64 x^{2}}+\frac {\arctanh \left (\frac {2 x^{2}+8}{4 \sqrt {x^{4}+2 x^{2}+4}}\right )}{128}\)	\(60\)
risch	\(\frac {3 x^{6}+2 x^{4}+4 x^{2}-16}{64 x^{4} \sqrt {x^{4}+2 x^{2}+4}}+\frac {\arctanh \left (\frac {2 x^{2}+8}{4 \sqrt {x^{4}+2 x^{2}+4}}\right )}{128}\)	\(60\)
elliptic	\(-\frac {\sqrt {x^{4}+2 x^{2}+4}}{16 x^{4}}+\frac {3 \sqrt {x^{4}+2 x^{2}+4}}{64 x^{2}}+\frac {\arctanh \left (\frac {2 x^{2}+8}{4 \sqrt {x^{4}+2 x^{2}+4}}\right )}{128}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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maxima [A]  time = 1.21, size = 52, normalized size = 0.73 \[ \frac {3 \, \sqrt {x^{4} + 2 \, x^{2} + 4}}{64 \, x^{2}} - \frac {\sqrt {x^{4} + 2 \, x^{2} + 4}}{16 \, x^{4}} + \frac {1}{128} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} + \frac {4 \, \sqrt {3}}{3 \, x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^5\,\sqrt {x^4+2\,x^2+4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{5} \sqrt {x^{4} + 2 x^{2} + 4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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