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

Optimal. Leaf size=42 \[ \frac {b \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {b x-a}}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {47, 63, 205} \begin {gather*} \frac {b \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {b x-a}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-a + b*x]/x^2,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 205

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 52, normalized size = 1.24 \begin {gather*} \frac {-b x \sqrt {1-\frac {b x}{a}} \tanh ^{-1}\left (\sqrt {1-\frac {b x}{a}}\right )+a-b x}{x \sqrt {b x-a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-a + b*x]/x^2,x]

[Out]

(a - b*x - b*x*Sqrt[1 - (b*x)/a]*ArcTanh[Sqrt[1 - (b*x)/a]])/(x*Sqrt[-a + b*x])

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IntegrateAlgebraic [A]  time = 0.04, size = 42, normalized size = 1.00 \begin {gather*} \frac {b \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {b x-a}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.00, size = 98, normalized size = 2.33 \begin {gather*} \left [-\frac {\sqrt {-a} b x \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, \sqrt {b x - a} a}{2 \, a x}, \frac {\sqrt {a} b x \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - \sqrt {b x - a} a}{a x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a)*b*x*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2*sqrt(b*x - a)*a)/(a*x), (sqrt(a)*b*x*arc
tan(sqrt(b*x - a)/sqrt(a)) - sqrt(b*x - a)*a)/(a*x)]

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giac [A]  time = 1.08, size = 41, normalized size = 0.98 \begin {gather*} \frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {\sqrt {b x - a} b}{x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 35, normalized size = 0.83 \begin {gather*} \frac {b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {b x -a}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 34, normalized size = 0.81 \begin {gather*} \frac {b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {\sqrt {b x - a}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(sqrt(b*x - a)/sqrt(a))/sqrt(a) - sqrt(b*x - a)/x

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mupad [B]  time = 0.10, size = 34, normalized size = 0.81 \begin {gather*} \frac {b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {b\,x-a}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(1/2)/x^2,x)

[Out]

(b*atan((b*x - a)^(1/2)/a^(1/2)))/a^(1/2) - (b*x - a)^(1/2)/x

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sympy [A]  time = 2.14, size = 121, normalized size = 2.88 \begin {gather*} \begin {cases} - \frac {i a}{\sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {i \sqrt {b}}{\sqrt {x} \sqrt {\frac {a}{b x} - 1}} + \frac {i b \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {\sqrt {b} \sqrt {- \frac {a}{b x} + 1}}{\sqrt {x}} - \frac {b \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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