3.21.28 \(\int \frac {1}{x^2 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx\)

Optimal. Leaf size=144 \[ -\frac {\sqrt {\sqrt {a x^2+b^2}+b}}{4 b x}-\frac {1}{2 x \sqrt {\sqrt {a x^2+b^2}+b}}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {\sqrt {a x^2+b^2}+b}}-\frac {\sqrt {\sqrt {a x^2+b^2}+b}}{\sqrt {2} \sqrt {b}}\right )}{2 \sqrt {2} b^{3/2}} \]

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Rubi [F]  time = 0.16, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{x^2 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[1/(x^2*Sqrt[b + Sqrt[b^2 + a*x^2]]),x]

[Out]

Defer[Int][1/(x^2*Sqrt[b + Sqrt[b^2 + a*x^2]]), x]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx &=\int \frac {1}{x^2 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx\\ \end {align*}

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Mathematica [C]  time = 0.40, size = 143, normalized size = 0.99 \begin {gather*} -\frac {\sqrt {\sqrt {a x^2+b^2}+b} \left (9 a x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b-\sqrt {b^2+a x^2}}{2 b}\right )+2 b \left (\sqrt {a x^2+b^2}+b\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b-\sqrt {b^2+a x^2}}{2 b}\right )+2 b \left (5 \sqrt {a x^2+b^2}-7 b\right )\right )}{24 a b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[b + Sqrt[b^2 + a*x^2]]),x]

[Out]

-1/24*(Sqrt[b + Sqrt[b^2 + a*x^2]]*(2*b*(-7*b + 5*Sqrt[b^2 + a*x^2]) + 2*b*(b + Sqrt[b^2 + a*x^2])*Hypergeomet
ric2F1[-3/2, 1, -1/2, (b - Sqrt[b^2 + a*x^2])/(2*b)] + 9*a*x^2*Hypergeometric2F1[-1/2, 1, 1/2, (b - Sqrt[b^2 +
 a*x^2])/(2*b)]))/(a*b*x^3)

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IntegrateAlgebraic [A]  time = 0.23, size = 112, normalized size = 0.78 \begin {gather*} -\frac {1}{2 x \sqrt {b+\sqrt {b^2+a x^2}}}-\frac {\sqrt {b+\sqrt {b^2+a x^2}}}{4 b x}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}\right )}{4 \sqrt {2} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*Sqrt[b + Sqrt[b^2 + a*x^2]]),x]

[Out]

-1/2*1/(x*Sqrt[b + Sqrt[b^2 + a*x^2]]) - Sqrt[b + Sqrt[b^2 + a*x^2]]/(4*b*x) - (Sqrt[a]*ArcTan[(Sqrt[a]*x)/(Sq
rt[2]*Sqrt[b]*Sqrt[b + Sqrt[b^2 + a*x^2]])])/(4*Sqrt[2]*b^(3/2))

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fricas [A]  time = 45.26, size = 282, normalized size = 1.96 \begin {gather*} \left [\frac {\sqrt {\frac {1}{2}} a x^{3} \sqrt {-\frac {a}{b}} \log \left (-\frac {a^{2} x^{3} + 4 \, a b^{2} x - 4 \, \sqrt {a x^{2} + b^{2}} a b x + 4 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {a x^{2} + b^{2}} b^{2} \sqrt {-\frac {a}{b}} - \sqrt {\frac {1}{2}} {\left (a b x^{2} + 2 \, b^{3}\right )} \sqrt {-\frac {a}{b}}\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{x^{3}}\right ) - 2 \, {\left (a x^{2} - 2 \, b^{2} + 2 \, \sqrt {a x^{2} + b^{2}} b\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{8 \, a b x^{3}}, \frac {\sqrt {\frac {1}{2}} a x^{3} \sqrt {\frac {a}{b}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {b + \sqrt {a x^{2} + b^{2}}} b \sqrt {\frac {a}{b}}}{a x}\right ) - {\left (a x^{2} - 2 \, b^{2} + 2 \, \sqrt {a x^{2} + b^{2}} b\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{4 \, a b x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(1/2)*a*x^3*sqrt(-a/b)*log(-(a^2*x^3 + 4*a*b^2*x - 4*sqrt(a*x^2 + b^2)*a*b*x + 4*(2*sqrt(1/2)*sqrt(a
*x^2 + b^2)*b^2*sqrt(-a/b) - sqrt(1/2)*(a*b*x^2 + 2*b^3)*sqrt(-a/b))*sqrt(b + sqrt(a*x^2 + b^2)))/x^3) - 2*(a*
x^2 - 2*b^2 + 2*sqrt(a*x^2 + b^2)*b)*sqrt(b + sqrt(a*x^2 + b^2)))/(a*b*x^3), 1/4*(sqrt(1/2)*a*x^3*sqrt(a/b)*ar
ctan(2*sqrt(1/2)*sqrt(b + sqrt(a*x^2 + b^2))*b*sqrt(a/b)/(a*x)) - (a*x^2 - 2*b^2 + 2*sqrt(a*x^2 + b^2)*b)*sqrt
(b + sqrt(a*x^2 + b^2)))/(a*b*x^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b + \sqrt {a x^{2} + b^{2}}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.04, size = 31, normalized size = 0.22

method result size
meijerg \(-\frac {\sqrt {2}\, \hypergeom \left (\left [-\frac {1}{2}, \frac {1}{4}, \frac {3}{4}\right ], \left [\frac {1}{2}, \frac {3}{2}\right ], -\frac {x^{2} a}{b^{2}}\right )}{2 \left (b^{2}\right )^{\frac {1}{4}} x}\) \(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b^2)^(1/4)*2^(1/2)/x*hypergeom([-1/2,1/4,3/4],[1/2,3/2],-x^2*a/b^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b + \sqrt {a x^{2} + b^{2}}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 0.99, size = 46, normalized size = 0.32 \begin {gather*} - \frac {\Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4}, \frac {3}{4} \\ \frac {1}{2}, \frac {3}{2} \end {matrix}\middle | {\frac {a x^{2} e^{i \pi }}{b^{2}}} \right )}}{2 \pi \sqrt {b} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(1/4)*gamma(3/4)*hyper((-1/2, 1/4, 3/4), (1/2, 3/2), a*x**2*exp_polar(I*pi)/b**2)/(2*pi*sqrt(b)*x)

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