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

Optimal. Leaf size=89 \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{2-b x}}-\frac{5 \sqrt{x} \sqrt{2-b x}}{b^3}+\frac{10 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}+\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}} \]

[Out]

(2*x^(5/2))/(3*b*(2 - b*x)^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[2 - b*x]) - (5*Sqrt[x]*Sqrt[2 - b*x])/b^3 + (10*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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Rubi [A]  time = 0.0216252, 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} \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{2-b x}}-\frac{5 \sqrt{x} \sqrt{2-b x}}{b^3}+\frac{10 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}+\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(2 - b*x)^(5/2),x]

[Out]

(2*x^(5/2))/(3*b*(2 - b*x)^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[2 - b*x]) - (5*Sqrt[x]*Sqrt[2 - b*x])/b^3 + (10*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)^{5/2}} \, dx &=\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac{5 \int \frac{x^{3/2}}{(2-b x)^{3/2}} \, dx}{3 b}\\ &=\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{2-b x}}+\frac{5 \int \frac{\sqrt{x}}{\sqrt{2-b x}} \, dx}{b^2}\\ &=\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{2-b x}}-\frac{5 \sqrt{x} \sqrt{2-b x}}{b^3}+\frac{5 \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx}{b^3}\\ &=\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{2-b x}}-\frac{5 \sqrt{x} \sqrt{2-b x}}{b^3}+\frac{10 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )}{b^3}\\ &=\frac{2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{2-b x}}-\frac{5 \sqrt{x} \sqrt{2-b x}}{b^3}+\frac{10 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0062723, size = 30, normalized size = 0.34 \[ \frac{x^{7/2} \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};\frac{b x}{2}\right )}{14 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(2 - b*x)^(5/2),x]

[Out]

(x^(7/2)*Hypergeometric2F1[5/2, 7/2, 9/2, (b*x)/2])/(14*Sqrt[2])

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Maple [B]  time = 0.029, size = 168, normalized size = 1.9 \begin{align*}{\frac{bx-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 ( 5\,{\frac{1}{{b}^{7/2}}\arctan \left ({\frac{\sqrt{b}}{\sqrt{-b{x}^{2}+2\,x}} \left ( x-{b}^{-1} \right ) } \right ) }+{\frac{28}{3\,{b}^{4}}\sqrt{-b \left ( x-2\,{b}^{-1} \right ) ^{2}-2\,x+4\,{b}^{-1}} \left ( x-2\,{b}^{-1} \right ) ^{-1}}+{\frac{8}{3\,{b}^{5}}\sqrt{-b \left ( x-2\,{b}^{-1} \right ) ^{2}-2\,x+4\,{b}^{-1}} \left ( x-2\,{b}^{-1} \right ) ^{-2}} \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)^(5/2),x)

[Out]

1/b^3*x^(1/2)*(b*x-2)/(-x*(b*x-2))^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)+(5/b^(7/2)*arctan(b^(1/2)*(x-1/b)/(
-b*x^2+2*x)^(1/2))+28/3/b^4/(x-2/b)*(-b*(x-2/b)^2-2*x+4/b)^(1/2)+8/3/b^5/(x-2/b)^2*(-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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68149, size = 462, normalized size = 5.19 \begin{align*} \left [-\frac{15 \,{\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt{-b} \log \left (-b x + \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right ) +{\left (3 \, b^{3} x^{2} - 40 \, b^{2} x + 60 \, b\right )} \sqrt{-b x + 2} \sqrt{x}}{3 \,{\left (b^{6} x^{2} - 4 \, b^{5} x + 4 \, b^{4}\right )}}, -\frac{30 \,{\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt{b} \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right ) +{\left (3 \, b^{3} x^{2} - 40 \, b^{2} x + 60 \, b\right )} \sqrt{-b x + 2} \sqrt{x}}{3 \,{\left (b^{6} x^{2} - 4 \, b^{5} x + 4 \, b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(15*(b^2*x^2 - 4*b*x + 4)*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + (3*b^3*x^2 - 40*b^2
*x + 60*b)*sqrt(-b*x + 2)*sqrt(x))/(b^6*x^2 - 4*b^5*x + 4*b^4), -1/3*(30*(b^2*x^2 - 4*b*x + 4)*sqrt(b)*arctan(
sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (3*b^3*x^2 - 40*b^2*x + 60*b)*sqrt(-b*x + 2)*sqrt(x))/(b^6*x^2 - 4*b^5*x +
4*b^4)]

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Sympy [B]  time = 12.4095, size = 753, normalized size = 8.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*I*b**(23/2)*x**15/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) +
40*I*b**(21/2)*x**14/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) - 60*I*b**(19
/2)*x**13/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) - 30*I*b**10*x**(27/2)*s
qrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqr
t(b*x - 2)) + 15*pi*b**10*x**(27/2)*sqrt(b*x - 2)/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)
*sqrt(b*x - 2)) + 60*I*b**9*x**(25/2)*sqrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sq
rt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) - 30*pi*b**9*x**(25/2)*sqrt(b*x - 2)/(3*b**(27/2)*x**(27/2)
*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)), Abs(b*x)/2 > 1), (3*b**(23/2)*x**15/(3*b**(27/2)*x**(27
/2)*sqrt(-b*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) - 40*b**(21/2)*x**14/(3*b**(27/2)*x**(27/2)*sqrt(-b
*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) + 60*b**(19/2)*x**13/(3*b**(27/2)*x**(27/2)*sqrt(-b*x + 2) - 6
*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) + 30*b**10*x**(27/2)*sqrt(-b*x + 2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b*
*(27/2)*x**(27/2)*sqrt(-b*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) - 60*b**9*x**(25/2)*sqrt(-b*x + 2)*as
in(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sqrt(-b*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)), T
rue))

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Giac [B]  time = 18.1029, size = 270, normalized size = 3.03 \begin{align*} \frac{{\left (\frac{15 \, \log \left ({\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt{-b} b^{2}} - \frac{3 \, \sqrt{{\left (b x - 2\right )} b + 2 \, b} \sqrt{-b x + 2}}{b^{3}} - \frac{16 \,{\left (9 \,{\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{4} - 24 \,{\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} b + 28 \, b^{2}\right )}}{{\left ({\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )}^{3} \sqrt{-b} b}\right )}{\left | b \right |}}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(15*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/(sqrt(-b)*b^2) - 3*sqrt((b*x - 2)*b + 2*b)*
sqrt(-b*x + 2)/b^3 - 16*(9*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^4 - 24*(sqrt(-b*x + 2)*sqrt(-b)
- sqrt((b*x - 2)*b + 2*b))^2*b + 28*b^2)/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)^3*sqr
t(-b)*b))*abs(b)/b^2