\(\int \frac {1}{\sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}} \, dx\) [202]

Optimal result

Integrand size = 28, antiderivative size = 78 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \arctan \left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[Out]

1/2*arctan(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 = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1166, 385, 211} \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \arctan \left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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}{\sqrt {a-b x^2} \left (a+b x^2\right )} \, 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} \tan ^{-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 = 1.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a^2-b^2 x^4} \arctan \left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {2} a \sqrt {b} \sqrt {a-b x^2} \sqrt {a+b x^2}} \]

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^4]*ArcTan[(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. \(248\) vs. \(2(63)=126\).

Time = 0.59 (sec) , antiderivative size = 249, normalized size of antiderivative = 3.19

method	result	size
default	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \sqrt {b}\, \left (\sqrt {a}\, \sqrt {2}\, \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {-b \,x^{2}+a}+\sqrt {-a b}\, x +a \right )}{b x +\sqrt {-a b}}\right ) \sqrt {b}-\sqrt {a}\, \sqrt {2}\, \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {-b \,x^{2}+a}-\sqrt {-a b}\, x +a \right )}{b x -\sqrt {-a b}}\right ) \sqrt {b}+2 \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right ) \sqrt {-a b}-2 \sqrt {-a b}\, \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )\right )}{2 \sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}\, \left (-\sqrt {-a b}+\sqrt {a b}\right ) \left (\sqrt {-a b}+\sqrt {a b}\right ) \sqrt {-a b}}\)	\(249\)

[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)+(-a*b)^(1/2)*x+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)-(-a*b)^(1/2)*x+a)/(b*x-(-a
*b)^(1/2)))*b^(1/2)+2*arctan(b^(1/2)*x/(1/b*(-b*x+(a*b)^(1/2))*(b*x+(a*b)^(1/2)))^(1/2))*(-a*b)^(1/2)-2*(-a*b)
^(1/2)*arctan(b^(1/2)*x/(-b*x^2+a)^(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 = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.95 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}} \, dx=\left [-\frac {\sqrt {2} \sqrt {-b} \log \left (-\frac {3 \, b^{2} x^{4} + 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {-b} x - a^{2}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, a b}, -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {b}}{2 \, {\left (b^{2} x^{3} + a b x\right )}}\right )}{2 \, a \sqrt {b}}\right ] \]

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

[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)