3.617 \(\int \frac{x^{3/2}}{(2+b x)^{3/2}} \, dx\)

Optimal. Leaf size=63 \[ \frac{3 \sqrt{x} \sqrt{b x+2}}{b^2}-\frac{6 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{b \sqrt{b x+2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0138479, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 50, 54, 215} \[ \frac{3 \sqrt{x} \sqrt{b x+2}}{b^2}-\frac{6 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{b \sqrt{b x+2}} \]

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*ArcSinh[(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 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^{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 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.005861, size = 30, normalized size = 0.48 \[ \frac{x^{5/2} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};-\frac{b x}{2}\right )}{5 \sqrt{2}} \]

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])

________________________________________________________________________________________

Maple [B]  time = 0.019, size = 100, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{2}}\sqrt{x}\sqrt{bx+2}}+{ \left ( -3\,{\frac{1}{{b}^{5/2}}\ln \left ({\frac{bx+1}{\sqrt{b}}}+\sqrt{b{x}^{2}+2\,x} \right ) }+4\,{\frac{1}{{b}^{3}}\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^(3/2)/(b*x+2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.74284, size = 333, normalized size = 5.29 \begin{align*} \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}}{b \sqrt{x}}\right ) +{\left (b^{2} x + 6 \, b\right )} \sqrt{b x + 2} \sqrt{x}}{b^{4} x + 2 \, b^{3}}\right ] \end{align*}

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)/(b*sqrt(x))) + (b^2*x + 6*b)*sqrt(b*x + 2)*sq
rt(x))/(b^4*x + 2*b^3)]

________________________________________________________________________________________

Sympy [A]  time = 3.78501, size = 58, normalized size = 0.92 \begin{align*} \frac{x^{\frac{3}{2}}}{b \sqrt{b x + 2}} + \frac{6 \sqrt{x}}{b^{2} \sqrt{b x + 2}} - \frac{6 \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out