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

Optimal. Leaf size=37 \[ \sqrt {a+b x^2}-\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ \sqrt {a+b x^2}-\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/x,x]

[Out]

Sqrt[a + b*x^2] - Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 37, normalized size = 1.00 \[ \sqrt {a+b x^2}-\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2]/x,x]

[Out]

Sqrt[a + b*x^2] - Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

________________________________________________________________________________________

fricas [A]  time = 0.99, size = 77, normalized size = 2.08 \[ \left [\frac {1}{2} \, \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + \sqrt {b x^{2} + a}, \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + \sqrt {b x^{2} + a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + sqrt(b*x^2 + a), sqrt(-a)*arctan(sqrt(-a)/s
qrt(b*x^2 + a)) + sqrt(b*x^2 + a)]

________________________________________________________________________________________

giac [A]  time = 0.59, size = 33, normalized size = 0.89 \[ \frac {a \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \sqrt {b x^{2} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 39, normalized size = 1.05 \[ -\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\sqrt {b \,x^{2}+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.36, size = 27, normalized size = 0.73 \[ -\sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \sqrt {b x^{2} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.66, size = 29, normalized size = 0.78 \[ \sqrt {b\,x^2+a}-\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)/x,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.41, size = 56, normalized size = 1.51 \[ - \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {a}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {\sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________