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

Optimal. Leaf size=54 \[ \frac {\sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \sqrt {\frac {a}{x}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}} \]

[Out]

(1+x)*(cos(2*arctan(x^(1/2)))^2)^(1/2)/cos(2*arctan(x^(1/2)))*EllipticF(sin(2*arctan(x^(1/2))),1/2*2^(1/2))*(a
/x)^(1/2)*x^(1/2)*((x^2+1)/(1+x)^2)^(1/2)/(x^2+1)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {15, 329, 220} \[ \frac {\sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \sqrt {\frac {a}{x}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a/x]/Sqrt[1 + x^2],x]

[Out]

(Sqrt[a/x]*Sqrt[x]*(1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTan[Sqrt[x]], 1/2])/Sqrt[1 + x^2]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^2}} \, dx &=\left (\sqrt {\frac {a}{x}} \sqrt {x}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^2}} \, dx\\ &=\left (2 \sqrt {\frac {a}{x}} \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {\sqrt {\frac {a}{x}} \sqrt {x} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {1+x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.50 \[ 2 x \sqrt {\frac {a}{x}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a/x]/Sqrt[1 + x^2],x]

[Out]

2*Sqrt[a/x]*x*Hypergeometric2F1[1/4, 1/2, 5/4, -x^2]

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {a}{x}}}{\sqrt {x^{2} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a/x)/sqrt(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a/x)/sqrt(x^2 + 1), x)

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maple [C]  time = 0.04, size = 62, normalized size = 1.15 \[ \frac {i \sqrt {\frac {a}{x}}\, \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \EllipticF \left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/x)^(1/2)/(x^2+1)^(1/2),x)

[Out]

I*(a/x)^(1/2)/(x^2+1)^(1/2)*(-I*(x+I))^(1/2)*2^(1/2)*(-I*(-x+I))^(1/2)*(I*x)^(1/2)*EllipticF((-I*(x+I))^(1/2),
1/2*2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a/x)/sqrt(x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/x)^(1/2)/(x^2 + 1)^(1/2),x)

[Out]

int((a/x)^(1/2)/(x^2 + 1)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a/x)/sqrt(x**2 + 1), x)

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