3.623 \(\int \frac{x^{5/2}}{(2+b x)^{5/2}} \, dx\)

Optimal. Leaf size=86 \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{b x+2}}+\frac{5 \sqrt{x} \sqrt{b x+2}}{b^3}-\frac{10 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b (b x+2)^{3/2}} \]

[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*
ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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Rubi [A]  time = 0.0201415, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 50, 54, 215} \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{b x+2}}+\frac{5 \sqrt{x} \sqrt{b x+2}}{b^3}-\frac{10 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b (b x+2)^{3/2}} \]

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*
ArcSinh[(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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[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 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.006098, size = 30, normalized size = 0.35 \[ \frac{x^{7/2} \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};-\frac{b x}{2}\right )}{14 \sqrt{2}} \]

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 = 0.028, size = 136, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{3}}\sqrt{x}\sqrt{bx+2}}+{ \left ( -5\,{\frac{1}{{b}^{7/2}}\ln \left ({\frac{bx+1}{\sqrt{b}}}+\sqrt{b{x}^{2}+2\,x} \right ) }-{\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}}+{\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}} \right ) \sqrt{x \left ( bx+2 \right ) }{\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]

x^(1/2)*(b*x+2)^(1/2)/b^3+(-5/b^(7/2)*ln((b*x+1)/b^(1/2)+(b*x^2+2*x)^(1/2))-8/3/b^5/(x+2/b)^2*(b*(x+2/b)^2-2*x
-4/b)^(1/2)+28/3/b^4/(x+2/b)*(b*(x+2/b)^2-2*x-4/b)^(1/2))*(x*(b*x+2))^(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 = 1.55062, size = 455, normalized size = 5.29 \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}}{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)/(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 = 12.9109, size = 308, normalized size = 3.58 \begin{align*} \frac{3 b^{\frac{23}{2}} x^{15}}{3 b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{b x + 2} + 6 b^{\frac{25}{2}} x^{\frac{25}{2}} \sqrt{b x + 2}} + \frac{40 b^{\frac{21}{2}} x^{14}}{3 b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{b x + 2} + 6 b^{\frac{25}{2}} x^{\frac{25}{2}} \sqrt{b x + 2}} + \frac{60 b^{\frac{19}{2}} x^{13}}{3 b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{b x + 2} + 6 b^{\frac{25}{2}} x^{\frac{25}{2}} \sqrt{b x + 2}} - \frac{30 b^{10} x^{\frac{27}{2}} \sqrt{b x + 2} \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{3 b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{b x + 2} + 6 b^{\frac{25}{2}} x^{\frac{25}{2}} \sqrt{b x + 2}} - \frac{60 b^{9} x^{\frac{25}{2}} \sqrt{b x + 2} \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{3 b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{b x + 2} + 6 b^{\frac{25}{2}} x^{\frac{25}{2}} \sqrt{b x + 2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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**(2
7/2)*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)*asinh(s
qrt(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)*asinh(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))

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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)^(5/2),x, algorithm="giac")

[Out]

Timed out