\(\int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx\) [2691]

Optimal result

Integrand size = 23, antiderivative size = 243 \[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=\frac {a^{3/2} \left (-56 b^3-8 b^2 \sqrt {b^2+a x^2}\right )+a^{5/2} \left (5 b x^2+15 x^2 \sqrt {b^2+a x^2}\right )}{\frac {384 a^{3/2} b^4 x^3 \sqrt {b^2+a x^2}}{\left (b+\sqrt {b^2+a x^2}\right )^{3/2}}+\frac {192 a^{3/2} b^3 x^3 \left (2 b^2+a x^2\right )}{\left (b+\sqrt {b^2+a x^2}\right )^{3/2}}}+\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}-\frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\sqrt {2} \sqrt {b}}\right )}{32 \sqrt {2} b^{7/2}} \]

[Out]

(a^(3/2)*(-56*b^3-8*b^2*(a*x^2+b^2)^(1/2))+a^(5/2)*(5*b*x^2+15*x^2*(a*x^2+b^2)^(1/2)))/(384*a^(3/2)*b^4*x^3*(a
*x^2+b^2)^(1/2)/(b+(a*x^2+b^2)^(1/2))^(3/2)+192*a^(3/2)*b^3*x^3*(a*x^2+2*b^2)/(b+(a*x^2+b^2)^(1/2))^(3/2))+5/6
4*a^(3/2)*arctan(1/2*a^(1/2)*x*2^(1/2)/b^(1/2)/(b+(a*x^2+b^2)^(1/2))^(1/2)-1/2*(b+(a*x^2+b^2)^(1/2))^(1/2)*2^(
1/2)/b^(1/2))*2^(1/2)/b^(7/2)

Rubi [F]

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

[In]

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

[Out]

Defer[Int][1/(x^4*Sqrt[b + Sqrt[b^2 + a*x^2]]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=\frac {\sqrt {b+\sqrt {b^2+a x^2}} \left (-64 b^4+12 a b^2 x^2+15 a^2 x^4-64 b^3 \sqrt {b^2+a x^2}+20 a b x^2 \sqrt {b^2+a x^2}\right )}{192 b^3 x^3 \left (2 b^2+a x^2+2 b \sqrt {b^2+a x^2}\right )}+\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}\right )}{64 \sqrt {2} b^{7/2}} \]

[In]

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

[Out]

(Sqrt[b + Sqrt[b^2 + a*x^2]]*(-64*b^4 + 12*a*b^2*x^2 + 15*a^2*x^4 - 64*b^3*Sqrt[b^2 + a*x^2] + 20*a*b*x^2*Sqrt
[b^2 + a*x^2]))/(192*b^3*x^3*(2*b^2 + a*x^2 + 2*b*Sqrt[b^2 + a*x^2])) + (5*a^(3/2)*ArcTan[(Sqrt[a]*x)/(Sqrt[2]
*Sqrt[b]*Sqrt[b + Sqrt[b^2 + a*x^2]])])/(64*Sqrt[2]*b^(7/2))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.13

method	result	size
meijerg	\(-\frac {\sqrt {2}\, \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {1}{4}, \frac {3}{4}\right ], \left [-\frac {1}{2}, \frac {3}{2}\right ], -\frac {x^{2} a}{b^{2}}\right )}{6 \left (b^{2}\right )^{\frac {1}{4}} x^{3}}\)	\(31\)

[In]

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

[Out]

-1/6/(b^2)^(1/4)*2^(1/2)/x^3*hypergeom([-3/2,1/4,3/4],[-1/2,3/2],-x^2*a/b^2)

Fricas [A] (verification not implemented)

none

Time = 93.78 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.38 \[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=\left [\frac {15 \, \sqrt {\frac {1}{2}} a^{2} x^{5} \sqrt {-\frac {a}{b}} \log \left (-\frac {a^{2} x^{3} + 4 \, a b^{2} x - 4 \, \sqrt {a x^{2} + b^{2}} a b x - 4 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {a x^{2} + b^{2}} b^{2} \sqrt {-\frac {a}{b}} - \sqrt {\frac {1}{2}} {\left (a b x^{2} + 2 \, b^{3}\right )} \sqrt {-\frac {a}{b}}\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{x^{3}}\right ) + 2 \, {\left (15 \, a^{2} x^{4} + 2 \, a b^{2} x^{2} + 48 \, b^{4} - 2 \, {\left (5 \, a b x^{2} + 24 \, b^{3}\right )} \sqrt {a x^{2} + b^{2}}\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{384 \, a b^{3} x^{5}}, -\frac {15 \, \sqrt {\frac {1}{2}} a^{2} x^{5} \sqrt {\frac {a}{b}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {b + \sqrt {a x^{2} + b^{2}}} b \sqrt {\frac {a}{b}}}{a x}\right ) - {\left (15 \, a^{2} x^{4} + 2 \, a b^{2} x^{2} + 48 \, b^{4} - 2 \, {\left (5 \, a b x^{2} + 24 \, b^{3}\right )} \sqrt {a x^{2} + b^{2}}\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{192 \, a b^{3} x^{5}}\right ] \]

[In]

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

[Out]

[1/384*(15*sqrt(1/2)*a^2*x^5*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*(15*a^2*x^4 + 2*a*b^2*x^2 + 48*b^4 - 2*(5*a*b*x^2 + 24*b^3)*sqrt(a*x^2 + b^2))*sqrt(b + sqrt(a*x^2 + b^2))
)/(a*b^3*x^5), -1/192*(15*sqrt(1/2)*a^2*x^5*sqrt(a/b)*arctan(2*sqrt(1/2)*sqrt(b + sqrt(a*x^2 + b^2))*b*sqrt(a/
b)/(a*x)) - (15*a^2*x^4 + 2*a*b^2*x^2 + 48*b^4 - 2*(5*a*b*x^2 + 24*b^3)*sqrt(a*x^2 + b^2))*sqrt(b + sqrt(a*x^2
+ b^2)))/(a*b^3*x^5)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=- \frac {\Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4}, \frac {3}{4} \\ - \frac {1}{2}, \frac {3}{2} \end {matrix}\middle | {\frac {a x^{2} e^{i \pi }}{b^{2}}} \right )}}{6 \pi \sqrt {b} x^{3}} \]

[In]

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

[Out]

-gamma(1/4)*gamma(3/4)*hyper((-3/2, 1/4, 3/4), (-1/2, 3/2), a*x**2*exp_polar(I*pi)/b**2)/(6*pi*sqrt(b)*x**3)

Maxima [F]

\[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=\int { \frac {1}{\sqrt {b + \sqrt {a x^{2} + b^{2}}} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(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^4), x)

Giac [F]

\[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=\int { \frac {1}{\sqrt {b + \sqrt {a x^{2} + b^{2}}} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(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^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \sqrt {b+\sqrt {b^2+a x^2}}} \, dx=\int \frac {1}{x^4\,\sqrt {b+\sqrt {b^2+a\,x^2}}} \,d x \]

[In]

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

[Out]

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