3.5.72 \(\int \frac {1}{\sqrt {x} (-a+b x)} \, dx\) [472]

Optimal. Leaf size=29 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {65, 214} \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(-a + b*x)),x]

[Out]

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

Rule 65

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 214

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(-a + b*x)),x]

[Out]

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.97, size = 88, normalized size = 3.03 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\text {DirectedInfinity}\left [\frac {1}{\sqrt {x}}\right ],a\text {==}0\text {\&\&}b\text {==}0\right \},\left \{\frac {-2}{b \sqrt {x}},a\text {==}0\right \},\left \{\frac {-2 \sqrt {x}}{a},b\text {==}0\right \}\right \},-\frac {\text {Log}\left [\sqrt {x}+\sqrt {\frac {a}{b}}\right ]}{b \sqrt {\frac {a}{b}}}+\frac {\text {Log}\left [\sqrt {x}-\sqrt {\frac {a}{b}}\right ]}{b \sqrt {\frac {a}{b}}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[1/(Sqrt[x]*(-a + b*x)),x]')

[Out]

Piecewise[{{DirectedInfinity[1 / Sqrt[x]], a == 0 && b == 0}, {-2 / (b Sqrt[x]), a == 0}, {-2 Sqrt[x] / a, b =
= 0}}, -Log[Sqrt[x] + Sqrt[a / b]] / (b Sqrt[a / b]) + Log[Sqrt[x] - Sqrt[a / b]] / (b Sqrt[a / b])]

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Maple [A]
time = 0.10, size = 19, normalized size = 0.66

method	result	size
derivativedivides	\(-\frac {2 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\)	\(19\)
default	\(-\frac {2 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.37, size = 34, normalized size = 1.17 \begin {gather*} \frac {\log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/sqrt(a*b)

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Fricas [A]
time = 0.31, size = 67, normalized size = 2.31 \begin {gather*} \left [\frac {\sqrt {a b} \log \left (\frac {b x + a - 2 \, \sqrt {a b} \sqrt {x}}{b x - a}\right )}{a b}, \frac {2 \, \sqrt {-a b} \arctan \left (\frac {\sqrt {-a b}}{b \sqrt {x}}\right )}{a b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.47, size = 68, normalized size = 2.34 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{b \sqrt {x}} & \text {for}\: a = 0 \\- \frac {2 \sqrt {x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{b \sqrt {\frac {a}{b}}} - \frac {\log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{b \sqrt {\frac {a}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (-2/(b*sqrt(x)), Eq(a, 0)), (-2*sqrt(x)/a, Eq(b, 0)), (log(sqrt(
x) - sqrt(a/b))/(b*sqrt(a/b)) - log(sqrt(x) + sqrt(a/b))/(b*sqrt(a/b)), True))

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Giac [A]
time = 0.00, size = 28, normalized size = 0.97 \begin {gather*} \frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.13, size = 19, normalized size = 0.66 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^(1/2)*(a - b*x)),x)

[Out]

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

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