3.500 \(\int \frac{\sqrt{a-b x}}{\sqrt{x}} \, dx\)

Optimal. Leaf size=46 \[ \sqrt{x} \sqrt{a-b x}+\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{\sqrt{b}} \]

[Out]

Sqrt[x]*Sqrt[a - b*x] + (a*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/Sqrt[b]

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Rubi [A]  time = 0.0169224, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {50, 63, 217, 203} \[ \sqrt{x} \sqrt{a-b x}+\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - b*x]/Sqrt[x],x]

[Out]

Sqrt[x]*Sqrt[a - b*x] + (a*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/Sqrt[b]

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0902888, size = 65, normalized size = 1.41 \[ \frac{\frac{a^{3/2} \sqrt{1-\frac{b x}{a}} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b}}+\sqrt{x} (a-b x)}{\sqrt{a-b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - b*x]/Sqrt[x],x]

[Out]

(Sqrt[x]*(a - b*x) + (a^(3/2)*Sqrt[1 - (b*x)/a]*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[b])/Sqrt[a - b*x]

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Maple [A]  time = 0.004, size = 66, normalized size = 1.4 \begin{align*} \sqrt{x}\sqrt{-bx+a}+{\frac{a}{2}\sqrt{x \left ( -bx+a \right ) }\arctan \left ({\sqrt{b} \left ( x-{\frac{a}{2\,b}} \right ){\frac{1}{\sqrt{-b{x}^{2}+ax}}}} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+a}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58569, size = 257, normalized size = 5.59 \begin{align*} \left [-\frac{a \sqrt{-b} \log \left (-2 \, b x + 2 \, \sqrt{-b x + a} \sqrt{-b} \sqrt{x} + a\right ) - 2 \, \sqrt{-b x + a} b \sqrt{x}}{2 \, b}, -\frac{a \sqrt{b} \arctan \left (\frac{\sqrt{-b x + a}}{\sqrt{b} \sqrt{x}}\right ) - \sqrt{-b x + a} b \sqrt{x}}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*sqrt(-b*x + a)*b*sqrt(x))/b, -(a*sqr
t(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) - sqrt(-b*x + a)*b*sqrt(x))/b]

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Sympy [A]  time = 2.00933, size = 121, normalized size = 2.63 \begin{align*} \begin{cases} - \frac{i \sqrt{a} \sqrt{x}}{\sqrt{-1 + \frac{b x}{a}}} - \frac{i a \operatorname{acosh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{\sqrt{b}} + \frac{i b x^{\frac{3}{2}}}{\sqrt{a} \sqrt{-1 + \frac{b x}{a}}} & \text{for}\: \frac{\left |{b x}\right |}{\left |{a}\right |} > 1 \\\sqrt{a} \sqrt{x} \sqrt{1 - \frac{b x}{a}} + \frac{a \operatorname{asin}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{\sqrt{b}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*sqrt(a)*sqrt(x)/sqrt(-1 + b*x/a) - I*a*acosh(sqrt(b)*sqrt(x)/sqrt(a))/sqrt(b) + I*b*x**(3/2)/(sq
rt(a)*sqrt(-1 + b*x/a)), Abs(b*x)/Abs(a) > 1), (sqrt(a)*sqrt(x)*sqrt(1 - b*x/a) + a*asin(sqrt(b)*sqrt(x)/sqrt(
a))/sqrt(b), True))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out