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

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Rubi [A]  time = 0.05, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 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 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])

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 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])

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*}

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Mathematica [A]  time = 0.06, size = 75, normalized size = 1.00 \[ -\frac {2 \sqrt {a^2-b^2 x} \tan ^{-1}\left (\frac {\sqrt {a^2-b^2 x}}{\sqrt {a^2+b^2 x}}\right )}{b^2 \sqrt {a-b \sqrt {x}} \sqrt {a+b \sqrt {x}}} \]

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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fricas [A]  time = 1.32, size = 50, normalized size = 0.67 \[ -\frac {2 \, \arctan \left (-\frac {a^{2} - \sqrt {b^{2} x + a^{2}} \sqrt {b \sqrt {x} + a} \sqrt {-b \sqrt {x} + a}}{b^{2} x}\right )}{b^{2}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b^{2} x + a^{2}} \sqrt {b \sqrt {x} + a} \sqrt {-b \sqrt {x} + a}}\,{d x} \]

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]

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

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maple [F]  time = 0.88, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b^{2} x +a^{2}}\, \sqrt {-b \sqrt {x}+a}\, \sqrt {b \sqrt {x}+a}}\, dx \]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b^{2} x + a^{2}} \sqrt {b \sqrt {x} + a} \sqrt {-b \sqrt {x} + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a+b\,\sqrt {x}}\,\sqrt {a-b\,\sqrt {x}}\,\sqrt {a^2+x\,b^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a - b \sqrt {x}} \sqrt {a + b \sqrt {x}} \sqrt {a^{2} + b^{2} x}}\, dx \]

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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