\(\int \frac {\sqrt {a+b x^2}}{\sqrt {a^2-b^2 x^4}} \, dx\) [201]

Optimal result

Integrand size = 28, antiderivative size = 65 \[ \int \frac {\sqrt {a+b x^2}}{\sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, 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, 223, 209} \[ \int \frac {\sqrt {a+b x^2}}{\sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[In]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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}} \, 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}{1+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 {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {b} \sqrt {a^2-b^2 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.96 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x^2}}{\sqrt {a^2-b^2 x^4}} \, dx=\frac {i \log \left (-2 i \sqrt {b} x+\frac {2 \sqrt {a^2-b^2 x^4}}{\sqrt {a+b x^2}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}\, \sqrt {b}}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.86 \[ \int \frac {\sqrt {a+b x^2}}{\sqrt {a^2-b^2 x^4}} \, dx=\left [-\frac {\sqrt {-b} \log \left (-\frac {2 \, b^{2} x^{4} + a b x^{2} - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {-b} x - a^{2}}{b x^{2} + a}\right )}{2 \, b}, -\frac {\arctan \left (\frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {b}}{b^{2} x^{3} + a b x}\right )}{\sqrt {b}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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