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

Optimal. Leaf size=75 \[ -\frac{2 \sqrt{a^2-b^2 x} \tan ^{-1}\left (\frac{\sqrt{a^2-b^2 x}}{\sqrt{a^2+b^2 x}}\right )}{b^2 \sqrt{a-b \sqrt{x}} \sqrt{a+b \sqrt{x}}} \]

[Out]

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

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Rubi [A]  time = 0.0502034, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {519, 63, 217, 203} \[ -\frac{2 \sqrt{a^2-b^2 x} \tan ^{-1}\left (\frac{\sqrt{a^2-b^2 x}}{\sqrt{a^2+b^2 x}}\right )}{b^2 \sqrt{a-b \sqrt{x}} \sqrt{a+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 519

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_)*((a2_) + (b2_.)*(x_)^(non2_.))^(
p_), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1*a2 + b1*b2*x^n)^FracP
art[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x] && EqQ[
non2, n/2] && EqQ[a2*b1 + a1*b2, 0] &&  !(EqQ[n, 2] && IGtQ[q, 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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0332435, size = 75, normalized size = 1. \[ -\frac{2 \sqrt{a^2-b^2 x} \tan ^{-1}\left (\frac{\sqrt{a^2-b^2 x}}{\sqrt{a^2+b^2 x}}\right )}{b^2 \sqrt{a-b \sqrt{x}} \sqrt{a+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.663, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{{b}^{2}x+{a}^{2}}}}{\frac{1}{\sqrt{a-b\sqrt{x}}}}{\frac{1}{\sqrt{a+b\sqrt{x}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.8122, size = 124, normalized size = 1.65 \begin{align*} -\frac{2 \, \arctan \left (-\frac{a^{2} - \sqrt{b^{2} x + a^{2}} \sqrt{b \sqrt{x} + a} \sqrt{-b \sqrt{x} + a}}{b^{2} x}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out