\(\int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx\) [1926]

Optimal result

Integrand size = 15, antiderivative size = 53 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=-\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}+\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b^{3/2}} \]

[Out]

1/2*a*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))/b^(3/2)-1/2*(a+b/x^2)^(1/2)/b/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {342, 327, 223, 212} \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 b^{3/2}}-\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x} \]

[In]

Int[1/(Sqrt[a + b/x^2]*x^4),x]

[Out]

-1/2*Sqrt[a + b/x^2]/(b*x) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(3/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{2 b} \\ & = -\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}+\frac {a \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b} \\ & = -\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=\frac {-\sqrt {b} \left (b+a x^2\right )+a x^2 \sqrt {b+a x^2} \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{2 b^{3/2} \sqrt {a+\frac {b}{x^2}} x^3} \]

[In]

Integrate[1/(Sqrt[a + b/x^2]*x^4),x]

[Out]

(-(Sqrt[b]*(b + a*x^2)) + a*x^2*Sqrt[b + a*x^2]*ArcTanh[Sqrt[b + a*x^2]/Sqrt[b]])/(2*b^(3/2)*Sqrt[a + b/x^2]*x
^3)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38

method	result	size
default	\(-\frac {\sqrt {a \,x^{2}+b}\, \left (-a \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) b \,x^{2}+\sqrt {a \,x^{2}+b}\, b^{\frac {3}{2}}\right )}{2 \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x^{3} b^{\frac {5}{2}}}\)	\(73\)
risch	\(-\frac {a \,x^{2}+b}{2 b \,x^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {a \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \sqrt {a \,x^{2}+b}}{2 b^{\frac {3}{2}} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\)	\(84\)

[In]

int(1/(a+b/x^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/2*(a*x^2+b)^(1/2)*(-a*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*b*x^2+(a*x^2+b)^(1/2)*b^(3/2))/((a*x^2+b)/x^2)^(1
/2)/x^3/b^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=\left [\frac {a \sqrt {b} x \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, b \sqrt {\frac {a x^{2} + b}{x^{2}}}}{4 \, b^{2} x}, -\frac {a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + b \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, b^{2} x}\right ] \]

[In]

integrate(1/(a+b/x^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/4*(a*sqrt(b)*x*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*b*sqrt((a*x^2 + b)/x^2))/(b^
2*x), -1/2*(a*sqrt(-b)*x*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + b*sqrt((a*x^2 + b)/x^2))/(b^2*
x)]

Sympy [A] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=- \frac {\sqrt {a} \sqrt {1 + \frac {b}{a x^{2}}}}{2 b x} + \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{2 b^{\frac {3}{2}}} \]

[In]

integrate(1/(a+b/x**2)**(1/2)/x**4,x)

[Out]

-sqrt(a)*sqrt(1 + b/(a*x**2))/(2*b*x) + a*asinh(sqrt(b)/(sqrt(a)*x))/(2*b**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=-\frac {\sqrt {a + \frac {b}{x^{2}}} a x}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )} b x^{2} - b^{2}\right )}} - \frac {a \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{4 \, b^{\frac {3}{2}}} \]

[In]

integrate(1/(a+b/x^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

-1/2*sqrt(a + b/x^2)*a*x/((a + b/x^2)*b*x^2 - b^2) - 1/4*a*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*
x + sqrt(b)))/b^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=-\frac {\frac {a^{2} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {a x^{2} + b} a}{b x^{2}}}{2 \, a \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/(a+b/x^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/2*(a^2*arctan(sqrt(a*x^2 + b)/sqrt(-b))/(sqrt(-b)*b) + sqrt(a*x^2 + b)*a/(b*x^2))/(a*sgn(x))

Mupad [B] (verification not implemented)

Time = 6.71 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4} \, dx=\left \{\begin {array}{cl} -\frac {1}{3\,\sqrt {a}\,x^3} & \text {\ if\ \ }b=0\\ \frac {a\,\ln \left (2\,\sqrt {a+\frac {b}{x^2}}+\frac {2\,\sqrt {b}}{x}\right )}{2\,b^{3/2}}-\frac {\sqrt {a+\frac {b}{x^2}}}{2\,b\,x} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

[In]

int(1/(x^4*(a + b/x^2)^(1/2)),x)

[Out]

piecewise(b == 0, -1/(3*a^(1/2)*x^3), b ~= 0, (a*log(2*(a + b/x^2)^(1/2) + (2*b^(1/2))/x))/(2*b^(3/2)) - (a +
b/x^2)^(1/2)/(2*b*x))