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

Optimal. Leaf size=60 \[ 2 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )-\frac{2 (b x+2)^{3/2}}{3 x^{3/2}}-\frac{2 b \sqrt{b x+2}}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.013916, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {47, 54, 215} \[ 2 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )-\frac{2 (b x+2)^{3/2}}{3 x^{3/2}}-\frac{2 b \sqrt{b x+2}}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(3/2)/x^(5/2),x]

[Out]

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

Mathematica [C]  time = 0.0052794, size = 30, normalized size = 0.5 \[ -\frac{4 \sqrt{2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x}{2}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(3/2)/x^(5/2),x]

[Out]

(-4*Sqrt[2]*Hypergeometric2F1[-3/2, -3/2, -1/2, -(b*x)/2])/(3*x^(3/2))

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Maple [A]  time = 0.016, size = 73, normalized size = 1.2 \begin{align*} -{\frac{8\,{b}^{2}{x}^{2}+20\,bx+8}{3}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{bx+2}}}}+{{b}^{{\frac{3}{2}}}\ln \left ({(bx+1){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+2\,x} \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((b*x+2)^(3/2)/x^(5/2),x)

[Out]

-4/3*(2*b^2*x^2+5*b*x+2)/x^(3/2)/(b*x+2)^(1/2)+b^(3/2)*ln((b*x+1)/b^(1/2)+(b*x^2+2*x)^(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((b*x+2)^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89723, size = 300, normalized size = 5. \begin{align*} \left [\frac{3 \, b^{\frac{3}{2}} x^{2} \log \left (b x + \sqrt{b x + 2} \sqrt{b} \sqrt{x} + 1\right ) - 4 \,{\left (2 \, b x + 1\right )} \sqrt{b x + 2} \sqrt{x}}{3 \, x^{2}}, -\frac{2 \,{\left (3 \, \sqrt{-b} b x^{2} \arctan \left (\frac{\sqrt{b x + 2} \sqrt{-b}}{b \sqrt{x}}\right ) + 2 \,{\left (2 \, b x + 1\right )} \sqrt{b x + 2} \sqrt{x}\right )}}{3 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.73204, size = 70, normalized size = 1.17 \begin{align*} - \frac{8 b^{\frac{3}{2}} \sqrt{1 + \frac{2}{b x}}}{3} - b^{\frac{3}{2}} \log{\left (\frac{1}{b x} \right )} + 2 b^{\frac{3}{2}} \log{\left (\sqrt{1 + \frac{2}{b x}} + 1 \right )} - \frac{4 \sqrt{b} \sqrt{1 + \frac{2}{b x}}}{3 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*b**(3/2)*sqrt(1 + 2/(b*x))/3 - b**(3/2)*log(1/(b*x)) + 2*b**(3/2)*log(sqrt(1 + 2/(b*x)) + 1) - 4*sqrt(b)*sq
rt(1 + 2/(b*x))/(3*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError