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

Optimal. Leaf size=109 \[ \frac {2 x}{\sqrt {\sqrt {a x^2+b^2}+b}}-\frac {2 \sqrt {2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {\sqrt {a x^2+b^2}+b}}-\frac {\sqrt {\sqrt {a x^2+b^2}+b}}{\sqrt {2} \sqrt {b}}\right )}{\sqrt {a}} \]

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Rubi [F]  time = 0.05, 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}{\sqrt {b+\sqrt {b^2+a x^2}}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[1/Sqrt[b + Sqrt[b^2 + a*x^2]],x]

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 0.39, size = 149, normalized size = 1.37 \begin {gather*} \frac {\sqrt {\sqrt {a x^2+b^2}+b} \left (-2 b \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b-\sqrt {b^2+a x^2}}{2 b}\right )+4 \sqrt {a x^2+b^2}-\sqrt {2} \sqrt {b} \sqrt {\sqrt {a x^2+b^2}-b} \tan ^{-1}\left (\frac {\sqrt {\sqrt {a x^2+b^2}-b}}{\sqrt {2} \sqrt {b}}\right )-2 b\right )}{2 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b + Sqrt[b^2 + a*x^2]]*(-2*b + 4*Sqrt[b^2 + a*x^2] - Sqrt[2]*Sqrt[b]*Sqrt[-b + Sqrt[b^2 + a*x^2]]*ArcTan
[Sqrt[-b + Sqrt[b^2 + a*x^2]]/(Sqrt[2]*Sqrt[b])] - 2*b*Hypergeometric2F1[-1/2, 1, 1/2, (b - Sqrt[b^2 + a*x^2])
/(2*b)]))/(2*a*x)

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IntegrateAlgebraic [A]  time = 0.15, size = 77, normalized size = 0.71 \begin {gather*} \frac {2 x}{\sqrt {b+\sqrt {b^2+a x^2}}}-\frac {\sqrt {2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 45.69, size = 239, normalized size = 2.19 \begin {gather*} \left [\frac {\sqrt {2} a x \sqrt {-\frac {b}{a}} \log \left (-\frac {a x^{3} + 4 \, b^{2} x - 4 \, \sqrt {a x^{2} + b^{2}} b x + 2 \, {\left (2 \, \sqrt {2} \sqrt {a x^{2} + b^{2}} b \sqrt {-\frac {b}{a}} - \sqrt {2} {\left (a x^{2} + 2 \, b^{2}\right )} \sqrt {-\frac {b}{a}}\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{x^{3}}\right ) - 4 \, \sqrt {b + \sqrt {a x^{2} + b^{2}}} {\left (b - \sqrt {a x^{2} + b^{2}}\right )}}{2 \, a x}, \frac {\sqrt {2} a x \sqrt {\frac {b}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {b + \sqrt {a x^{2} + b^{2}}} \sqrt {\frac {b}{a}}}{x}\right ) - 2 \, \sqrt {b + \sqrt {a x^{2} + b^{2}}} {\left (b - \sqrt {a x^{2} + b^{2}}\right )}}{a x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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maple [C]  time = 0.04, size = 29, normalized size = 0.27

method result size
meijerg \(\frac {\sqrt {2}\, x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}, \frac {3}{2}\right ], -\frac {x^{2} a}{b^{2}}\right )}{2 \left (b^{2}\right )^{\frac {1}{4}}}\) \(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(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([1/4,1/2,3/4],[3/2,3/2],-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}}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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

[Out]

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

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sympy [C]  time = 0.81, size = 42, normalized size = 0.39 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2}, \frac {3}{2} \end {matrix}\middle | {\frac {a x^{2} e^{i \pi }}{b^{2}}} \right )}}{2 \pi \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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