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

Optimal. Leaf size=51 \[ 2 x \sqrt{\frac{a}{x^2}+\frac{b}{x}}-2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+\frac{b}{x}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0765409, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {1979, 2007, 2013, 620, 206} \[ 2 x \sqrt{\frac{a}{x^2}+\frac{b}{x}}-2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+\frac{b}{x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(a + b*x)/x^2],x]

[Out]

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

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2007

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(p*(n - j)), x] + Dist
[a, Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, j, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[
Simplify[j*p + 1], 0]

Rule 2013

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(a*x^Simplify[j/n]
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && IntegerQ[Simplify[j
/n]] && EqQ[Simplify[m - n + 1], 0]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0256985, size = 58, normalized size = 1.14 \[ \frac{2 x \sqrt{\frac{a+b x}{x^2}} \left (\sqrt{a+b x}-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\right )}{\sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(a + b*x)/x^2],x]

[Out]

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

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Maple [A]  time = 0.007, size = 47, normalized size = 0.9 \begin{align*} 2\,{\frac{x}{\sqrt{bx+a}}\sqrt{{\frac{bx+a}{{x}^{2}}}} \left ( -\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) +\sqrt{bx+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x + a)/x^2), x)

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Fricas [A]  time = 0.916763, size = 231, normalized size = 4.53 \begin{align*} \left [2 \, x \sqrt{\frac{b x + a}{x^{2}}} + \sqrt{a} \log \left (\frac{b x - 2 \, \sqrt{a} x \sqrt{\frac{b x + a}{x^{2}}} + 2 \, a}{x}\right ), 2 \, x \sqrt{\frac{b x + a}{x^{2}}} + 2 \, \sqrt{-a} \arctan \left (\frac{\sqrt{-a} x \sqrt{\frac{b x + a}{x^{2}}}}{a}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((a + b*x)/x**2), x)

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Giac [A]  time = 1.12506, size = 88, normalized size = 1.73 \begin{align*} 2 \,{\left (\frac{a \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \sqrt{b x + a}\right )} \mathrm{sgn}\left (x\right ) - \frac{2 \,{\left (a \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + \sqrt{-a} \sqrt{a}\right )} \mathrm{sgn}\left (x\right )}{\sqrt{-a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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