3.635 \(\int \frac{x^{5/2}}{(2-b x)^{3/2}} \, dx\)

Optimal. Leaf size=89 \[ \frac{5 x^{3/2} \sqrt{2-b x}}{2 b^2}+\frac{15 \sqrt{x} \sqrt{2-b x}}{2 b^3}-\frac{15 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}+\frac{2 x^{5/2}}{b \sqrt{2-b x}} \]

[Out]

(2*x^(5/2))/(b*Sqrt[2 - b*x]) + (15*Sqrt[x]*Sqrt[2 - b*x])/(2*b^3) + (5*x^(3/2)*Sqrt[2 - b*x])/(2*b^2) - (15*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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Rubi [A]  time = 0.0207852, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {47, 50, 54, 216} \[ \frac{5 x^{3/2} \sqrt{2-b x}}{2 b^2}+\frac{15 \sqrt{x} \sqrt{2-b x}}{2 b^3}-\frac{15 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}+\frac{2 x^{5/2}}{b \sqrt{2-b x}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

(2*x^(5/2))/(b*Sqrt[2 - b*x]) + (15*Sqrt[x]*Sqrt[2 - b*x])/(2*b^3) + (5*x^(3/2)*Sqrt[2 - b*x])/(2*b^2) - (15*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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^{5/2}}{(2-b x)^{3/2}} \, dx &=\frac{2 x^{5/2}}{b \sqrt{2-b x}}-\frac{5 \int \frac{x^{3/2}}{\sqrt{2-b x}} \, dx}{b}\\ &=\frac{2 x^{5/2}}{b \sqrt{2-b x}}+\frac{5 x^{3/2} \sqrt{2-b x}}{2 b^2}-\frac{15 \int \frac{\sqrt{x}}{\sqrt{2-b x}} \, dx}{2 b^2}\\ &=\frac{2 x^{5/2}}{b \sqrt{2-b x}}+\frac{15 \sqrt{x} \sqrt{2-b x}}{2 b^3}+\frac{5 x^{3/2} \sqrt{2-b x}}{2 b^2}-\frac{15 \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx}{2 b^3}\\ &=\frac{2 x^{5/2}}{b \sqrt{2-b x}}+\frac{15 \sqrt{x} \sqrt{2-b x}}{2 b^3}+\frac{5 x^{3/2} \sqrt{2-b x}}{2 b^2}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )}{b^3}\\ &=\frac{2 x^{5/2}}{b \sqrt{2-b x}}+\frac{15 \sqrt{x} \sqrt{2-b x}}{2 b^3}+\frac{5 x^{3/2} \sqrt{2-b x}}{2 b^2}-\frac{15 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0061238, size = 30, normalized size = 0.34 \[ \frac{x^{7/2} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};\frac{b x}{2}\right )}{7 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

(x^(7/2)*Hypergeometric2F1[3/2, 7/2, 9/2, (b*x)/2])/(7*Sqrt[2])

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Maple [B]  time = 0.026, size = 138, normalized size = 1.6 \begin{align*} -{\frac{ \left ( bx+7 \right ) \left ( bx-2 \right ) }{2\,{b}^{3}}\sqrt{x}\sqrt{ \left ( -bx+2 \right ) x}{\frac{1}{\sqrt{-x \left ( bx-2 \right ) }}}{\frac{1}{\sqrt{-bx+2}}}}-{ \left ({\frac{15}{2}\arctan \left ({\sqrt{b} \left ( x-{b}^{-1} \right ){\frac{1}{\sqrt{-b{x}^{2}+2\,x}}}} \right ){b}^{-{\frac{7}{2}}}}+8\,{\frac{1}{{b}^{4}}\sqrt{-b \left ( x-2\,{b}^{-1} \right ) ^{2}-2\,x+4\,{b}^{-1}} \left ( x-2\,{b}^{-1} \right ) ^{-1}} \right ) \sqrt{ \left ( -bx+2 \right ) x}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(-b*x+2)^(3/2),x)

[Out]

-1/2*(b*x+7)*x^(1/2)*(b*x-2)/b^3/(-x*(b*x-2))^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)-(15/2/b^(7/2)*arctan(b^(
1/2)*(x-1/b)/(-b*x^2+2*x)^(1/2))+8/b^4/(x-2/b)*(-b*(x-2/b)^2-2*x+4/b)^(1/2))*((-b*x+2)*x)^(1/2)/x^(1/2)/(-b*x+
2)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(-b*x+2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53743, size = 387, normalized size = 4.35 \begin{align*} \left [-\frac{15 \,{\left (b x - 2\right )} \sqrt{-b} \log \left (-b x - \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right ) -{\left (b^{3} x^{2} + 5 \, b^{2} x - 30 \, b\right )} \sqrt{-b x + 2} \sqrt{x}}{2 \,{\left (b^{5} x - 2 \, b^{4}\right )}}, \frac{30 \,{\left (b x - 2\right )} \sqrt{b} \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right ) +{\left (b^{3} x^{2} + 5 \, b^{2} x - 30 \, b\right )} \sqrt{-b x + 2} \sqrt{x}}{2 \,{\left (b^{5} x - 2 \, b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(-b*x+2)^(3/2),x, algorithm="fricas")

[Out]

[-1/2*(15*(b*x - 2)*sqrt(-b)*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) - (b^3*x^2 + 5*b^2*x - 30*b)*sqrt
(-b*x + 2)*sqrt(x))/(b^5*x - 2*b^4), 1/2*(30*(b*x - 2)*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (b^3
*x^2 + 5*b^2*x - 30*b)*sqrt(-b*x + 2)*sqrt(x))/(b^5*x - 2*b^4)]

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Sympy [A]  time = 12.8081, size = 173, normalized size = 1.94 \begin{align*} \begin{cases} \frac{i x^{\frac{5}{2}}}{2 b \sqrt{b x - 2}} + \frac{5 i x^{\frac{3}{2}}}{2 b^{2} \sqrt{b x - 2}} - \frac{15 i \sqrt{x}}{b^{3} \sqrt{b x - 2}} + \frac{15 i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{b^{\frac{7}{2}}} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\- \frac{x^{\frac{5}{2}}}{2 b \sqrt{- b x + 2}} - \frac{5 x^{\frac{3}{2}}}{2 b^{2} \sqrt{- b x + 2}} + \frac{15 \sqrt{x}}{b^{3} \sqrt{- b x + 2}} - \frac{15 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{b^{\frac{7}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(-b*x+2)**(3/2),x)

[Out]

Piecewise((I*x**(5/2)/(2*b*sqrt(b*x - 2)) + 5*I*x**(3/2)/(2*b**2*sqrt(b*x - 2)) - 15*I*sqrt(x)/(b**3*sqrt(b*x
- 2)) + 15*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), Abs(b*x)/2 > 1), (-x**(5/2)/(2*b*sqrt(-b*x + 2)) - 5*x
**(3/2)/(2*b**2*sqrt(-b*x + 2)) + 15*sqrt(x)/(b**3*sqrt(-b*x + 2)) - 15*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/
2), True))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(-b*x+2)^(3/2),x, algorithm="giac")

[Out]

Timed out