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

Optimal. Leaf size=50 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{3/2}}-\frac {\sqrt {x} \sqrt {a-b x}}{b} \]

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {50, 63, 217, 203} \begin {gather*} \frac {a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{3/2}}-\frac {\sqrt {x} \sqrt {a-b x}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/Sqrt[a - b*x],x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 71, normalized size = 1.42 \begin {gather*} \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} (b x-a)}{b^{3/2} \sqrt {a-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/Sqrt[a - b*x],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 59, normalized size = 1.18 \begin {gather*} \frac {a \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )}{b^2}-\frac {\sqrt {x} \sqrt {a-b x}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]/Sqrt[a - b*x],x]

[Out]

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

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fricas [A]  time = 1.21, size = 93, normalized size = 1.86 \begin {gather*} \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^{2}}, -\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + \sqrt {-b x + a} b \sqrt {x}}{b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(-b*x+a)^(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^2, -(a*s
qrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + sqrt(-b*x + a)*b*sqrt(x))/b^2]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 70, normalized size = 1.40 \begin {gather*} \frac {\sqrt {\left (-b x +a \right ) x}\, a \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{2 \sqrt {-b x +a}\, b^{\frac {3}{2}} \sqrt {x}}-\frac {\sqrt {-b x +a}\, \sqrt {x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 56, normalized size = 1.12 \begin {gather*} -\frac {a \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {3}{2}}} - \frac {\sqrt {-b x + a} a}{{\left (b^{2} - \frac {{\left (b x - a\right )} b}{x}\right )} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(3/2) - sqrt(-b*x + a)*a/((b^2 - (b*x - a)*b/x)*sqrt(x))

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mupad [B]  time = 0.52, size = 47, normalized size = 0.94 \begin {gather*} \frac {2\,a\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a-b\,x}-\sqrt {a}}\right )}{b^{3/2}}-\frac {\sqrt {x}\,\sqrt {a-b\,x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a - b*x)^(1/2),x)

[Out]

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

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sympy [A]  time = 2.28, size = 121, normalized size = 2.42 \begin {gather*} \begin {cases} - \frac {i \sqrt {a} \sqrt {x} \sqrt {-1 + \frac {b x}{a}}}{b} - \frac {i a \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {\sqrt {a} \sqrt {x}}{b \sqrt {1 - \frac {b x}{a}}} + \frac {a \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} \sqrt {1 - \frac {b x}{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)**(1/2),x)

[Out]

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

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