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

Optimal. Leaf size=58 \[ -\frac{2 (b x+2)^{3/2}}{\sqrt{x}}+3 b \sqrt{x} \sqrt{b x+2}+6 \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0117686, antiderivative size = 58, 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} \[ -\frac{2 (b x+2)^{3/2}}{\sqrt{x}}+3 b \sqrt{x} \sqrt{b x+2}+6 \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(3/2)/x^(3/2),x]

[Out]

3*b*Sqrt[x]*Sqrt[2 + b*x] - (2*(2 + b*x)^(3/2))/Sqrt[x] + 6*Sqrt[b]*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 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{(2+b x)^{3/2}}{x^{3/2}} \, dx &=-\frac{2 (2+b x)^{3/2}}{\sqrt{x}}+(3 b) \int \frac{\sqrt{2+b x}}{\sqrt{x}} \, dx\\ &=3 b \sqrt{x} \sqrt{2+b x}-\frac{2 (2+b x)^{3/2}}{\sqrt{x}}+(3 b) \int \frac{1}{\sqrt{x} \sqrt{2+b x}} \, dx\\ &=3 b \sqrt{x} \sqrt{2+b x}-\frac{2 (2+b x)^{3/2}}{\sqrt{x}}+(6 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=3 b \sqrt{x} \sqrt{2+b x}-\frac{2 (2+b x)^{3/2}}{\sqrt{x}}+6 \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0043338, size = 28, normalized size = 0.48 \[ -\frac{4 \sqrt{2} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x}{2}\right )}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(3/2)/x^(3/2),x]

[Out]

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

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Maple [A]  time = 0.012, size = 72, normalized size = 1.2 \begin{align*}{({b}^{2}{x}^{2}-2\,bx-8){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+2}}}}+3\,{\frac{\sqrt{b}\sqrt{x \left ( bx+2 \right ) }}{\sqrt{x}\sqrt{bx+2}}\ln \left ({\frac{bx+1}{\sqrt{b}}}+\sqrt{b{x}^{2}+2\,x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*x^2-2*b*x-8)/x^(1/2)/(b*x+2)^(1/2)+3*b^(1/2)*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^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88847, size = 265, normalized size = 4.57 \begin{align*} \left [\frac{3 \, \sqrt{b} x \log \left (b x + \sqrt{b x + 2} \sqrt{b} \sqrt{x} + 1\right ) + \sqrt{b x + 2}{\left (b x - 4\right )} \sqrt{x}}{x}, -\frac{6 \, \sqrt{-b} x \arctan \left (\frac{\sqrt{b x + 2} \sqrt{-b}}{b \sqrt{x}}\right ) - \sqrt{b x + 2}{\left (b x - 4\right )} \sqrt{x}}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(3*sqrt(b)*x*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + sqrt(b*x + 2)*(b*x - 4)*sqrt(x))/x, -(6*sqrt(-b)*
x*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - sqrt(b*x + 2)*(b*x - 4)*sqrt(x))/x]

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Sympy [A]  time = 2.84553, size = 73, normalized size = 1.26 \begin{align*} 6 \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} + \frac{b^{2} x^{\frac{3}{2}}}{\sqrt{b x + 2}} - \frac{2 b \sqrt{x}}{\sqrt{b x + 2}} - \frac{8}{\sqrt{x} \sqrt{b x + 2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*sqrt(b)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2) + b**2*x**(3/2)/sqrt(b*x + 2) - 2*b*sqrt(x)/sqrt(b*x + 2) - 8/(sqrt
(x)*sqrt(b*x + 2))

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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^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError