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

Optimal. Leaf size=40 \[ \frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 40, 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 = {50, 63, 208} \begin {gather*} \frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(-a + b*x),x]

[Out]

(2*Sqrt[x])/b - (2*Sqrt[a]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {x}}{-a+b x} \, dx &=\frac {2 \sqrt {x}}{b}+\frac {a \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{b}\\ &=\frac {2 \sqrt {x}}{b}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{b}\\ &=\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(-a + b*x),x]

[Out]

(2*Sqrt[x])/b - (2*Sqrt[a]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)

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IntegrateAlgebraic [A]  time = 0.03, size = 40, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x])/b - (2*Sqrt[a]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)

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fricas [A]  time = 0.76, size = 83, normalized size = 2.08 \begin {gather*} \left [\frac {\sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, \sqrt {x}}{b}, \frac {2 \, {\left (\sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + \sqrt {x}\right )}}{b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.04, size = 33, normalized size = 0.82 \begin {gather*} \frac {2 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} b} + \frac {2 \, \sqrt {x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 32, normalized size = 0.80 \begin {gather*} -\frac {2 a \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}+\frac {2 \sqrt {x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.08, size = 47, normalized size = 1.18 \begin {gather*} \frac {a \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {2 \, \sqrt {x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*b) + 2*sqrt(x)/b

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mupad [B]  time = 0.11, size = 28, normalized size = 0.70 \begin {gather*} \frac {2\,\sqrt {x}}{b}-\frac {2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^(1/2)/(a - b*x),x)

[Out]

(2*x^(1/2))/b - (2*a^(1/2)*atanh((b^(1/2)*x^(1/2))/a^(1/2)))/b^(3/2)

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sympy [A]  time = 0.71, size = 87, normalized size = 2.18 \begin {gather*} \begin {cases} \frac {\sqrt {a} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{2} \sqrt {\frac {1}{b}}} - \frac {\sqrt {a} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{2} \sqrt {\frac {1}{b}}} + \frac {2 \sqrt {x}}{b} & \text {for}\: b \neq 0 \\- \frac {2 x^{\frac {3}{2}}}{3 a} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(a)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(b**2*sqrt(1/b)) - sqrt(a)*log(sqrt(a)*sqrt(1/b) + sqrt(x
))/(b**2*sqrt(1/b)) + 2*sqrt(x)/b, Ne(b, 0)), (-2*x**(3/2)/(3*a), True))

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