3.891 \(\int \frac{1}{\sqrt{a+\frac{b}{x^2}} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=70 \[ \frac{\sqrt{a x^2+b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a x^2+b}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} \sqrt{d} x \sqrt{a+\frac{b}{x^2}}} \]

[Out]

(Sqrt[b + a*x^2]*ArcTanh[(Sqrt[d]*Sqrt[b + a*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*Sqrt[d]*Sqrt[a + b/x^2
]*x)

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Rubi [A]  time = 0.0681262, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {435, 444, 63, 217, 206} \[ \frac{\sqrt{a x^2+b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a x^2+b}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} \sqrt{d} x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b + a*x^2]*ArcTanh[(Sqrt[d]*Sqrt[b + a*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*Sqrt[d]*Sqrt[a + b/x^2
]*x)

Rule 435

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(x^(n*FracPart[q])*(c +
d/x^n)^FracPart[q])/(d + c*x^n)^FracPart[q], Int[((a + b*x^n)^p*(d + c*x^n)^q)/x^(n*q), x], x] /; FreeQ[{a, b,
c, d, n, p, q}, x] && EqQ[mn, -n] &&  !IntegerQ[q] &&  !IntegerQ[p]

Rule 444

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

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 206

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+\frac{b}{x^2}} \sqrt{c+d x^2}} \, dx &=\frac{\sqrt{b+a x^2} \int \frac{x}{\sqrt{b+a x^2} \sqrt{c+d x^2}} \, dx}{\sqrt{a+\frac{b}{x^2}} x}\\ &=\frac{\sqrt{b+a x^2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x} \sqrt{c+d x}} \, dx,x,x^2\right )}{2 \sqrt{a+\frac{b}{x^2}} x}\\ &=\frac{\sqrt{b+a x^2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{b d}{a}+\frac{d x^2}{a}}} \, dx,x,\sqrt{b+a x^2}\right )}{a \sqrt{a+\frac{b}{x^2}} x}\\ &=\frac{\sqrt{b+a x^2} \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{a}} \, dx,x,\frac{\sqrt{b+a x^2}}{\sqrt{c+d x^2}}\right )}{a \sqrt{a+\frac{b}{x^2}} x}\\ &=\frac{\sqrt{b+a x^2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{b+a x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} \sqrt{d} \sqrt{a+\frac{b}{x^2}} x}\\ \end{align*}

Mathematica [A]  time = 0.125109, size = 105, normalized size = 1.5 \[ \frac{x \sqrt{a+\frac{b}{x^2}} \sqrt{c+d x^2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a x^2+b}}{\sqrt{a c-b d}}\right )}{\sqrt{d} \sqrt{a x^2+b} \sqrt{a c-b d} \sqrt{\frac{a \left (c+d x^2\right )}{a c-b d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b/x^2]*x*Sqrt[c + d*x^2]*ArcSinh[(Sqrt[d]*Sqrt[b + a*x^2])/Sqrt[a*c - b*d]])/(Sqrt[d]*Sqrt[a*c - b*d
]*Sqrt[b + a*x^2]*Sqrt[(a*(c + d*x^2))/(a*c - b*d)])

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Maple [B]  time = 0.042, size = 117, normalized size = 1.7 \begin{align*}{\frac{a{x}^{2}+b}{2\,x}\ln \left ({\frac{1}{2} \left ( 2\,ad{x}^{2}+2\,\sqrt{ad{x}^{4}+ac{x}^{2}+bd{x}^{2}+bc}\sqrt{ad}+ac+bd \right ){\frac{1}{\sqrt{ad}}}} \right ) \sqrt{d{x}^{2}+c}{\frac{1}{\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}{\frac{1}{\sqrt{ad}}}{\frac{1}{\sqrt{ad{x}^{4}+ac{x}^{2}+bd{x}^{2}+bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(d*x^2 + c)*sqrt(a + b/x^2)), x)

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Fricas [A]  time = 1.61428, size = 467, normalized size = 6.67 \begin{align*} \left [\frac{\sqrt{a d} \log \left (8 \, a^{2} d^{2} x^{4} + a^{2} c^{2} + 6 \, a b c d + b^{2} d^{2} + 8 \,{\left (a^{2} c d + a b d^{2}\right )} x^{2} + 4 \,{\left (2 \, a d x^{3} +{\left (a c + b d\right )} x\right )} \sqrt{d x^{2} + c} \sqrt{a d} \sqrt{\frac{a x^{2} + b}{x^{2}}}\right )}{4 \, a d}, -\frac{\sqrt{-a d} \arctan \left (\frac{{\left (2 \, a d x^{3} +{\left (a c + b d\right )} x\right )} \sqrt{d x^{2} + c} \sqrt{-a d} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{2 \,{\left (a^{2} d^{2} x^{4} + a b c d +{\left (a^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right )}{2 \, a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(a*d)*log(8*a^2*d^2*x^4 + a^2*c^2 + 6*a*b*c*d + b^2*d^2 + 8*(a^2*c*d + a*b*d^2)*x^2 + 4*(2*a*d*x^3 +
(a*c + b*d)*x)*sqrt(d*x^2 + c)*sqrt(a*d)*sqrt((a*x^2 + b)/x^2))/(a*d), -1/2*sqrt(-a*d)*arctan(1/2*(2*a*d*x^3 +
(a*c + b*d)*x)*sqrt(d*x^2 + c)*sqrt(-a*d)*sqrt((a*x^2 + b)/x^2)/(a^2*d^2*x^4 + a*b*c*d + (a^2*c*d + a*b*d^2)*
x^2))/(a*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(d*x^2 + c)*sqrt(a + b/x^2)), x)