\(\int \frac {1}{\sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}} \, dx\) [208]

Optimal result

Integrand size = 29, antiderivative size = 77 \[ \int \frac {1}{\sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1166, 385, 214} \[ \int \frac {1}{\sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[In]

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

[Out]

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^FracPart[p]/((d + e*x
^2)^FracPart[p]*(a/d + c*(x^2/e))^FracPart[p]), Int[(d + e*x^2)^(p + q)*(a/d + (c/e)*x^2)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {1}{\left (a-b x^2\right ) \sqrt {a+b x^2}} \, dx}{\sqrt {a^2-b^2 x^4}} \\ & = \frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{a-2 a b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{\sqrt {a^2-b^2 x^4}} \\ & = \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a^2-b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a^2-b^2 x^4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a-b x^2} \sqrt {a+b x^2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(255\) vs. \(2(62)=124\).

Time = 0.28 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.32

method	result	size
default	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \sqrt {b}\, \left (\sqrt {a}\, \sqrt {2}\, \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {b \,x^{2}+a}-x \sqrt {a b}+a \right )}{b x +\sqrt {a b}}\right ) \sqrt {b}-\sqrt {a}\, \sqrt {2}\, \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {b \,x^{2}+a}+x \sqrt {a b}+a \right )}{b x -\sqrt {a b}}\right ) \sqrt {b}+2 \sqrt {a b}\, \ln \left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+a}+b x}{\sqrt {b}}\right )-2 \ln \left (\frac {\sqrt {b}\, \sqrt {-\frac {\left (-b x +\sqrt {-a b}\right ) \left (b x +\sqrt {-a b}\right )}{b}}+b x}{\sqrt {b}}\right ) \sqrt {a b}\right )}{2 \sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}\, \sqrt {a b}\, \left (\sqrt {-a b}-\sqrt {a b}\right ) \left (\sqrt {-a b}+\sqrt {a b}\right )}\)	\(256\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.01 \[ \int \frac {1}{\sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\left [\frac {\sqrt {2} \log \left (-\frac {3 \, b^{2} x^{4} - 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {b} x - a^{2}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}}\right )}{4 \, a \sqrt {b}}, \frac {\sqrt {2} \sqrt {-b} \arctan \left (\frac {\sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {-b}}{2 \, {\left (b^{2} x^{3} - a b x\right )}}\right )}{2 \, a b}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(1/(sqrt(-(-a + b*x**2)*(a + b*x**2))*sqrt(a - b*x**2)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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