3.1898 \(\int \frac {\sqrt {a+\frac {b}{x^2}}}{x^4} \, dx\)

Optimal. Leaf size=74 \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{8 b^{3/2}}-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 b x}-\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3} \]

[Out]

1/8*a^2*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))/b^(3/2)-1/4*(a+b/x^2)^(1/2)/x^3-1/8*a*(a+b/x^2)^(1/2)/b/x

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {335, 279, 321, 217, 206} \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{8 b^{3/2}}-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 b x}-\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^2]/x^4,x]

[Out]

-Sqrt[a + b/x^2]/(4*x^3) - (a*Sqrt[a + b/x^2])/(8*b*x) + (a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(8*b^(3/2)
)

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 217

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 279

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

Rule 321

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 335

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*} \int \frac {\sqrt {a+\frac {b}{x^2}}}{x^4} \, dx &=-\operatorname {Subst}\left (\int x^2 \sqrt {a+b x^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}-\frac {1}{4} a \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 b x}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{8 b}\\ &=-\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 b x}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b}\\ &=-\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 b x}+\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.64 \[ -\frac {a^2 x \sqrt {a+\frac {b}{x^2}} \left (a x^2+b\right ) \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {a x^2}{b}+1\right )}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^2]/x^4,x]

[Out]

-1/3*(a^2*Sqrt[a + b/x^2]*x*(b + a*x^2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (a*x^2)/b])/b^3

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fricas [A]  time = 0.97, size = 158, normalized size = 2.14 \[ \left [\frac {a^{2} \sqrt {b} x^{3} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (a b x^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, b^{2} x^{3}}, -\frac {a^{2} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (a b x^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, b^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(a^2*sqrt(b)*x^3*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(a*b*x^2 + 2*b^2)*sqrt(
(a*x^2 + b)/x^2))/(b^2*x^3), -1/8*(a^2*sqrt(-b)*x^3*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (a*
b*x^2 + 2*b^2)*sqrt((a*x^2 + b)/x^2))/(b^2*x^3)]

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giac [A]  time = 0.23, size = 78, normalized size = 1.05 \[ -\frac {\frac {a^{3} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b} b} + \frac {{\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{3} \mathrm {sgn}\relax (x) + \sqrt {a x^{2} + b} a^{3} b \mathrm {sgn}\relax (x)}{a^{2} b x^{4}}}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(a^3*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b) + ((a*x^2 + b)^(3/2)*a^3*sgn(x) + sqrt(a*x^2 +
b)*a^3*b*sgn(x))/(a^2*b*x^4))/a

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maple [A]  time = 0.01, size = 106, normalized size = 1.43 \[ \frac {\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, \left (a^{2} \sqrt {b}\, x^{4} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-\sqrt {a \,x^{2}+b}\, a^{2} x^{4}+\left (a \,x^{2}+b \right )^{\frac {3}{2}} a \,x^{2}-2 \left (a \,x^{2}+b \right )^{\frac {3}{2}} b \right )}{8 \sqrt {a \,x^{2}+b}\, b^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*x^2+b)/x^2)^(1/2)/x^3*(b^(1/2)*ln(2*(b+(a*x^2+b)^(1/2)*b^(1/2))/x)*x^4*a^2-(a*x^2+b)^(1/2)*x^4*a^2+(a*
x^2+b)^(3/2)*x^2*a-2*(a*x^2+b)^(3/2)*b)/(a*x^2+b)^(1/2)/b^2

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maxima [A]  time = 1.98, size = 114, normalized size = 1.54 \[ -\frac {a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{16 \, b^{\frac {3}{2}}} - \frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2} x^{3} + \sqrt {a + \frac {b}{x^{2}}} a^{2} b x}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} b x^{4} - 2 \, {\left (a + \frac {b}{x^{2}}\right )} b^{2} x^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*a^2*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(3/2) - 1/8*((a + b/x^2)^(3/2)*a^
2*x^3 + sqrt(a + b/x^2)*a^2*b*x)/((a + b/x^2)^2*b*x^4 - 2*(a + b/x^2)*b^2*x^2 + b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^(1/2)/x^4,x)

[Out]

int((a + b/x^2)^(1/2)/x^4, x)

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sympy [A]  time = 4.08, size = 92, normalized size = 1.24 \[ - \frac {a^{\frac {3}{2}}}{8 b x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {3 \sqrt {a}}{8 x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{8 b^{\frac {3}{2}}} - \frac {b}{4 \sqrt {a} x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(3/2)/(8*b*x*sqrt(1 + b/(a*x**2))) - 3*sqrt(a)/(8*x**3*sqrt(1 + b/(a*x**2))) + a**2*asinh(sqrt(b)/(sqrt(a)
*x))/(8*b**(3/2)) - b/(4*sqrt(a)*x**5*sqrt(1 + b/(a*x**2)))

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