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

Optimal. Leaf size=45 \[ \frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {47, 54, 216} \begin {gather*} \frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(2 - b*x)^(3/2),x]

[Out]

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

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 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 45, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(2 - b*x)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 66, normalized size = 1.47 \begin {gather*} -\frac {2 \sqrt {-b} \log \left (\sqrt {2-b x}-\sqrt {-b} \sqrt {x}\right )}{b^2}-\frac {2 \sqrt {x} \sqrt {2-b x}}{b (b x-2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 122, normalized size = 2.71 \begin {gather*} \left [-\frac {{\left (b x - 2\right )} \sqrt {-b} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) + 2 \, \sqrt {-b x + 2} b \sqrt {x}}{b^{3} x - 2 \, b^{2}}, \frac {2 \, {\left ({\left (b x - 2\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - \sqrt {-b x + 2} b \sqrt {x}\right )}}{b^{3} x - 2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((b*x - 2)*sqrt(-b)*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + 2*sqrt(-b*x + 2)*b*sqrt(x))/(b^3*x - 2
*b^2), 2*((b*x - 2)*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - sqrt(-b*x + 2)*b*sqrt(x))/(b^3*x - 2*b^
2)]

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giac [B]  time = 10.06, size = 92, normalized size = 2.04 \begin {gather*} -\frac {{\left (\frac {\log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b}} + \frac {8 \, \sqrt {-b}}{{\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b}\right )} {\left | b \right |}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/sqrt(-b) + 8*sqrt(-b)/((sqrt(-b*x + 2)*sqrt(-b) -
sqrt((b*x - 2)*b + 2*b))^2 - 2*b))*abs(b)/b^2

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maple [A]  time = 0.05, size = 67, normalized size = 1.49 \begin {gather*} -\frac {2 \left (\frac {\sqrt {\pi }\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \sqrt {x}}{2 \sqrt {-\frac {b x}{2}+1}\, b}-\frac {\sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {2}\, \sqrt {b}\, \sqrt {x}}{2}\right )}{b^{\frac {3}{2}}}\right )}{\sqrt {-b}\, \sqrt {\pi }\, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(-b)^(1/2)/Pi^(1/2)/b*(1/2*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(3/2)/b/(-1/2*b*x+1)^(1/2)-Pi^(1/2)*(-b)^(3/2)/b^(
3/2)*arcsin(1/2*2^(1/2)*b^(1/2)*x^(1/2)))

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maxima [A]  time = 2.98, size = 38, normalized size = 0.84 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, \sqrt {x}}{\sqrt {-b x + 2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(2 - b*x)^(3/2),x)

[Out]

int(x^(1/2)/(2 - b*x)^(3/2), x)

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sympy [A]  time = 1.69, size = 92, normalized size = 2.04 \begin {gather*} \begin {cases} - \frac {2 i \sqrt {x}}{b \sqrt {b x - 2}} + \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {for}\: \frac {\left |{b x}\right |}{2} > 1 \\\frac {2 \sqrt {x}}{b \sqrt {- b x + 2}} - \frac {2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(-b*x+2)**(3/2),x)

[Out]

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

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