3.6.1 \(\int \frac {\sqrt {a-b x}}{x^{3/2}} \, dx\)

Optimal. Leaf size=47 \[ -\frac {2 \sqrt {a-b x}}{\sqrt {x}}-2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 47, 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 = {47, 63, 217, 203} \begin {gather*} -\frac {2 \sqrt {a-b x}}{\sqrt {x}}-2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - b*x]/x^(3/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 69, normalized size = 1.47 \begin {gather*} -\frac {2 \left (\sqrt {a} \sqrt {b} \sqrt {x} \sqrt {1-\frac {b x}{a}} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+a-b x\right )}{\sqrt {x} \sqrt {a-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - b*x]/x^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 53, normalized size = 1.13 \begin {gather*} -\frac {2 \sqrt {a-b x}}{\sqrt {x}}-2 \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 91, normalized size = 1.94 \begin {gather*} \left [\frac {\sqrt {-b} x \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, \sqrt {-b x + a} \sqrt {x}}{x}, \frac {2 \, {\left (\sqrt {b} x \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - \sqrt {-b x + a} \sqrt {x}\right )}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(1/2)/x^(3/2),x, algorithm="fricas")

[Out]

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

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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((-b*x+a)^(1/2)/x^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-b x +a}}{x^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.93, size = 35, normalized size = 0.74 \begin {gather*} 2 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - \frac {2 \, \sqrt {-b x + a}}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(1/2)/x^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {a-b\,x}}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x)^(1/2)/x^(3/2),x)

[Out]

int((a - b*x)^(1/2)/x^(3/2), x)

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sympy [A]  time = 1.70, size = 148, normalized size = 3.15 \begin {gather*} \begin {cases} \frac {2 i \sqrt {a}}{\sqrt {x} \sqrt {-1 + \frac {b x}{a}}} + 2 i \sqrt {b} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 i b \sqrt {x}}{\sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {2 \sqrt {a}}{\sqrt {x} \sqrt {1 - \frac {b x}{a}}} - 2 \sqrt {b} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + \frac {2 b \sqrt {x}}{\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((-b*x+a)**(1/2)/x**(3/2),x)

[Out]

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

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