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

Optimal. Leaf size=78 \[ \frac{\sqrt{a-b x^2} \sqrt{a+b x^2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{2} a \sqrt{b} \sqrt{a^2-b^2 x^4}} \]

[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])

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Rubi [A]  time = 0.035809, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {1152, 377, 205} \[ \frac{\sqrt{a-b x^2} \sqrt{a+b x^2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{2} a \sqrt{b} \sqrt{a^2-b^2 x^4}} \]

Antiderivative was successfully verified.

[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 1152

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*x^2)/e)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rule 377

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 205

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}} \, dx &=\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 ) \operatorname{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]  time = 0.0542164, size = 78, normalized size = 1. \[ \frac{\sqrt{a^2-b^2 x^4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{2} a \sqrt{b} \sqrt{a-b x^2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[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])

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Maple [B]  time = 0.043, size = 249, normalized size = 3.2 \begin{align*} -{\frac{1}{2}\sqrt{-{b}^{2}{x}^{4}+{a}^{2}}\sqrt{b} \left ( \sqrt{a}\sqrt{2}\ln \left ( 2\,{\frac{b \left ( \sqrt{2}\sqrt{a}\sqrt{-b{x}^{2}+a}-\sqrt{-ab}x+a \right ) }{bx-\sqrt{-ab}}} \right ) \sqrt{b}-\sqrt{a}\sqrt{2}\ln \left ( 2\,{\frac{b \left ( \sqrt{2}\sqrt{a}\sqrt{-b{x}^{2}+a}+\sqrt{-ab}x+a \right ) }{bx+\sqrt{-ab}}} \right ) \sqrt{b}+2\,\sqrt{-ab}\arctan \left ({\frac{x\sqrt{b}}{\sqrt{-b{x}^{2}+a}}} \right ) -2\,\arctan \left ({x\sqrt{b}{\frac{1}{\sqrt{{\frac{ \left ( -bx+\sqrt{ab} \right ) \left ( bx+\sqrt{ab} \right ) }{b}}}}}} \right ) \sqrt{-ab} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{-b{x}^{2}+a}}} \left ( \sqrt{-ab}+\sqrt{ab} \right ) ^{-1} \left ( -\sqrt{-ab}+\sqrt{ab} \right ) ^{-1}{\frac{1}{\sqrt{-ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{4} + a^{2}} \sqrt{b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.16544, size = 359, normalized size = 4.6 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- \left (- a + b x^{2}\right ) \left (a + b x^{2}\right )} \sqrt{a + b x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{4} + a^{2}} \sqrt{b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)