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

Optimal. Leaf size=131 \[ \frac {2 \sqrt {x^2+1} \sqrt {a x}}{x+1}+\frac {\sqrt {a} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {a x}}{\sqrt {a}}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}}-\frac {2 \sqrt {a} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {a x}}{\sqrt {a}}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}} \]

[Out]

2*(a*x)^(1/2)*(x^2+1)^(1/2)/(1+x)-2*(1+x)*(cos(2*arctan((a*x)^(1/2)/a^(1/2)))^2)^(1/2)/cos(2*arctan((a*x)^(1/2
)/a^(1/2)))*EllipticE(sin(2*arctan((a*x)^(1/2)/a^(1/2))),1/2*2^(1/2))*a^(1/2)*((x^2+1)/(1+x)^2)^(1/2)/(x^2+1)^
(1/2)+(1+x)*(cos(2*arctan((a*x)^(1/2)/a^(1/2)))^2)^(1/2)/cos(2*arctan((a*x)^(1/2)/a^(1/2)))*EllipticF(sin(2*ar
ctan((a*x)^(1/2)/a^(1/2))),1/2*2^(1/2))*a^(1/2)*((x^2+1)/(1+x)^2)^(1/2)/(x^2+1)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {329, 305, 220, 1196} \[ \frac {2 \sqrt {x^2+1} \sqrt {a x}}{x+1}+\frac {\sqrt {a} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {a x}}{\sqrt {a}}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}}-\frac {2 \sqrt {a} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {a x}}{\sqrt {a}}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x]/Sqrt[1 + x^2],x]

[Out]

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

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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.21 \[ \frac {2}{3} x \sqrt {a x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x]/Sqrt[1 + x^2],x]

[Out]

(2*x*Sqrt[a*x]*Hypergeometric2F1[1/2, 3/4, 7/4, -x^2])/3

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a x}}{\sqrt {x^{2} + 1}}, x\right ) \]

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 \[ \int \frac {\sqrt {a x}}{\sqrt {x^{2} + 1}}\,{d x} \]

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 = 0.03, size = 81, normalized size = 0.62 \[ \frac {\sqrt {a x}\, \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \left (2 \EllipticE \left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-\EllipticF \left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {x^{2}+1}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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 = -0.01 \[ \int \frac {\sqrt {a\,x}}{\sqrt {x^2+1}} \,d x \]

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 [C]  time = 1.02, size = 36, normalized size = 0.27 \[ \frac {\sqrt {a} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {x^{2} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*x**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), x**2*exp_polar(I*pi))/(2*gamma(7/4))

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