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

Optimal. Leaf size=65 \[ -\frac {6 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}+\frac {3 \sqrt {x} \sqrt {2-b x}}{b^2}+\frac {2 x^{3/2}}{b \sqrt {2-b x}} \]

[Out]

-6*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)+2*x^(3/2)/b/(-b*x+2)^(1/2)+3*x^(1/2)*(-b*x+2)^(1/2)/b^2

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Rubi [A]  time = 0.01, antiderivative size = 65, 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, 50, 54, 216} \[ \frac {3 \sqrt {x} \sqrt {2-b x}}{b^2}-\frac {6 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}+\frac {2 x^{3/2}}{b \sqrt {2-b x}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(2 - b*x)^(3/2),x]

[Out]

(2*x^(3/2))/(b*Sqrt[2 - b*x]) + (3*Sqrt[x]*Sqrt[2 - b*x])/b^2 - (6*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(5/2)

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 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 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 {x^{3/2}}{(2-b x)^{3/2}} \, dx &=\frac {2 x^{3/2}}{b \sqrt {2-b x}}-\frac {3 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{b}\\ &=\frac {2 x^{3/2}}{b \sqrt {2-b x}}+\frac {3 \sqrt {x} \sqrt {2-b x}}{b^2}-\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{b^2}\\ &=\frac {2 x^{3/2}}{b \sqrt {2-b x}}+\frac {3 \sqrt {x} \sqrt {2-b x}}{b^2}-\frac {6 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=\frac {2 x^{3/2}}{b \sqrt {2-b x}}+\frac {3 \sqrt {x} \sqrt {2-b x}}{b^2}-\frac {6 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.46 \[ \frac {x^{5/2} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {b x}{2}\right )}{5 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(2 - b*x)^(3/2),x]

[Out]

(x^(5/2)*Hypergeometric2F1[3/2, 5/2, 7/2, (b*x)/2])/(5*Sqrt[2])

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fricas [A]  time = 0.46, size = 138, normalized size = 2.12 \[ \left [-\frac {3 \, {\left (b x - 2\right )} \sqrt {-b} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) - {\left (b^{2} x - 6 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{b^{4} x - 2 \, b^{3}}, \frac {6 \, {\left (b x - 2\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + {\left (b^{2} x - 6 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{b^{4} x - 2 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(3*(b*x - 2)*sqrt(-b)*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) - (b^2*x - 6*b)*sqrt(-b*x + 2)*sqrt(x)
)/(b^4*x - 2*b^3), (6*(b*x - 2)*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (b^2*x - 6*b)*sqrt(-b*x + 2
)*sqrt(x))/(b^4*x - 2*b^3)]

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giac [B]  time = 10.41, size = 119, normalized size = 1.83 \[ -\frac {{\left (\frac {3 \, \log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b}} - \frac {\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}}{b} + \frac {16 \, \sqrt {-b}}{{\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b}\right )} {\left | b \right |}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/sqrt(-b) - sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x +
2)/b + 16*sqrt(-b)/((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b))*abs(b)/b^3

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maple [B]  time = 0.03, size = 133, normalized size = 2.05 \[ -\frac {\left (\frac {3 \arctan \left (\frac {\left (x -\frac {1}{b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+2 x}}\right )}{b^{\frac {5}{2}}}+\frac {4 \sqrt {-\left (x -\frac {2}{b}\right )^{2} b -2 x +\frac {4}{b}}}{\left (x -\frac {2}{b}\right ) b^{3}}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {-b x +2}\, \sqrt {x}}-\frac {\left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}\, \sqrt {x}}{\sqrt {-\left (b x -2\right ) x}\, \sqrt {-b x +2}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^2*(b*x-2)*x^(1/2)/(-(b*x-2)*x)^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)-(3/b^(5/2)*arctan((x-1/b)/(-b*x^2+
2*x)^(1/2)*b^(1/2))+4/b^3/(x-2/b)*(-(x-2/b)^2*b-2*x+4/b)^(1/2))*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)/x^(1/2)

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maxima [A]  time = 3.00, size = 71, normalized size = 1.09 \[ \frac {2 \, {\left (2 \, b - \frac {3 \, {\left (b x - 2\right )}}{x}\right )}}{\frac {\sqrt {-b x + 2} b^{3}}{\sqrt {x}} + \frac {{\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}}} + \frac {6 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*b - 3*(b*x - 2)/x)/(sqrt(-b*x + 2)*b^3/sqrt(x) + (-b*x + 2)^(3/2)*b^2/x^(3/2)) + 6*arctan(sqrt(-b*x + 2)/
(sqrt(b)*sqrt(x)))/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(2 - b*x)^(3/2),x)

[Out]

int(x^(3/2)/(2 - b*x)^(3/2), x)

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sympy [A]  time = 3.20, size = 128, normalized size = 1.97 \[ \begin {cases} \frac {i x^{\frac {3}{2}}}{b \sqrt {b x - 2}} - \frac {6 i \sqrt {x}}{b^{2} \sqrt {b x - 2}} + \frac {6 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} & \text {for}\: \frac {\left |{b x}\right |}{2} > 1 \\- \frac {x^{\frac {3}{2}}}{b \sqrt {- b x + 2}} + \frac {6 \sqrt {x}}{b^{2} \sqrt {- b x + 2}} - \frac {6 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*x**(3/2)/(b*sqrt(b*x - 2)) - 6*I*sqrt(x)/(b**2*sqrt(b*x - 2)) + 6*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)
/2)/b**(5/2), Abs(b*x)/2 > 1), (-x**(3/2)/(b*sqrt(-b*x + 2)) + 6*sqrt(x)/(b**2*sqrt(-b*x + 2)) - 6*asin(sqrt(2
)*sqrt(b)*sqrt(x)/2)/b**(5/2), True))

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