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

Optimal. Leaf size=162 \[ \frac {\sqrt {\sqrt {a x^2+b^2}+b} \left (\sqrt {a} \left (2 b \sqrt {a x^2+b^2}-10 b^2\right )-3 a^{3/2} x^2\right )}{24 \sqrt {a} b^2 x^3}-\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {\sqrt {a x^2+b^2}+b}}-\frac {\sqrt {\sqrt {a x^2+b^2}+b}}{\sqrt {2} \sqrt {b}}\right )}{4 \sqrt {2} b^{5/2}} \]

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Rubi [F]  time = 0.14, 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 {\sqrt {b+\sqrt {b^2+a x^2}}}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[Sqrt[b + Sqrt[b^2 + a*x^2]]/x^4,x]

[Out]

Defer[Int][Sqrt[b + Sqrt[b^2 + a*x^2]]/x^4, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{x^4} \, dx &=\int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{x^4} \, dx\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 81, normalized size = 0.50 \begin {gather*} \frac {\sqrt {\sqrt {a x^2+b^2}+b} \left (\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 )-6 b\right )}{12 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b + Sqrt[b^2 + a*x^2]]/x^4,x]

[Out]

(Sqrt[b + Sqrt[b^2 + a*x^2]]*(-6*b + (b + Sqrt[b^2 + a*x^2])*Hypergeometric2F1[-3/2, 1, -1/2, (b - Sqrt[b^2 +
a*x^2])/(2*b)]))/(12*b*x^3)

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IntegrateAlgebraic [A]  time = 0.28, size = 130, normalized size = 0.80 \begin {gather*} \frac {\sqrt {b+\sqrt {b^2+a x^2}} \left (-3 a^{3/2} x^2+\sqrt {a} \left (-10 b^2+2 b \sqrt {b^2+a x^2}\right )\right )}{24 \sqrt {a} b^2 x^3}-\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}\right )}{8 \sqrt {2} b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[b + Sqrt[b^2 + a*x^2]]/x^4,x]

[Out]

(Sqrt[b + Sqrt[b^2 + a*x^2]]*(-3*a^(3/2)*x^2 + Sqrt[a]*(-10*b^2 + 2*b*Sqrt[b^2 + a*x^2])))/(24*Sqrt[a]*b^2*x^3
) - (a^(3/2)*ArcTan[(Sqrt[a]*x)/(Sqrt[2]*Sqrt[b]*Sqrt[b + Sqrt[b^2 + a*x^2]])])/(8*Sqrt[2]*b^(5/2))

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fricas [A]  time = 99.87, size = 280, normalized size = 1.73 \begin {gather*} \left [\frac {3 \, \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 (3 \, a x^{2} + 10 \, b^{2} - 2 \, \sqrt {a x^{2} + b^{2}} b\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{48 \, b^{2} x^{3}}, \frac {3 \, \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 (3 \, a x^{2} + 10 \, b^{2} - 2 \, \sqrt {a x^{2} + b^{2}} b\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{24 \, b^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*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)*sqr
t(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*
(3*a*x^2 + 10*b^2 - 2*sqrt(a*x^2 + b^2)*b)*sqrt(b + sqrt(a*x^2 + b^2)))/(b^2*x^3), 1/24*(3*sqrt(1/2)*a*x^3*sqr
t(a/b)*arctan(2*sqrt(1/2)*sqrt(b + sqrt(a*x^2 + b^2))*b*sqrt(a/b)/(a*x)) - (3*a*x^2 + 10*b^2 - 2*sqrt(a*x^2 +
b^2)*b)*sqrt(b + sqrt(a*x^2 + b^2)))/(b^2*x^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {b + \sqrt {a x^{2} + b^{2}}}}{x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.04, size = 31, normalized size = 0.19

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {b + \sqrt {a x^{2} + b^{2}}}}{x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 1.12, size = 51, normalized size = 0.31 \begin {gather*} \frac {\sqrt {b} \Gamma \left (- \frac {1}{4}\right ) \Gamma \left (\frac {1}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4}, \frac {1}{4} \\ - \frac {1}{2}, \frac {1}{2} \end {matrix}\middle | {\frac {a x^{2} e^{i \pi }}{b^{2}}} \right )}}{12 \pi x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b)*gamma(-1/4)*gamma(1/4)*hyper((-3/2, -1/4, 1/4), (-1/2, 1/2), a*x**2*exp_polar(I*pi)/b**2)/(12*pi*x**3)

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