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

Optimal. Leaf size=79 \[ -\frac{2 (b x+2)^{5/2}}{\sqrt{x}}+\frac{5}{2} b \sqrt{x} (b x+2)^{3/2}+\frac{15}{2} b \sqrt{x} \sqrt{b x+2}+15 \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

[Out]

(15*b*Sqrt[x]*Sqrt[2 + b*x])/2 + (5*b*Sqrt[x]*(2 + b*x)^(3/2))/2 - (2*(2 + b*x)^(5/2))/Sqrt[x] + 15*Sqrt[b]*Ar
cSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]]

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Rubi [A]  time = 0.0162733, antiderivative size = 79, 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} \[ -\frac{2 (b x+2)^{5/2}}{\sqrt{x}}+\frac{5}{2} b \sqrt{x} (b x+2)^{3/2}+\frac{15}{2} b \sqrt{x} \sqrt{b x+2}+15 \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(5/2)/x^(3/2),x]

[Out]

(15*b*Sqrt[x]*Sqrt[2 + b*x])/2 + (5*b*Sqrt[x]*(2 + b*x)^(3/2))/2 - (2*(2 + b*x)^(5/2))/Sqrt[x] + 15*Sqrt[b]*Ar
cSinh[(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)^{5/2}}{x^{3/2}} \, dx &=-\frac{2 (2+b x)^{5/2}}{\sqrt{x}}+(5 b) \int \frac{(2+b x)^{3/2}}{\sqrt{x}} \, dx\\ &=\frac{5}{2} b \sqrt{x} (2+b x)^{3/2}-\frac{2 (2+b x)^{5/2}}{\sqrt{x}}+\frac{1}{2} (15 b) \int \frac{\sqrt{2+b x}}{\sqrt{x}} \, dx\\ &=\frac{15}{2} b \sqrt{x} \sqrt{2+b x}+\frac{5}{2} b \sqrt{x} (2+b x)^{3/2}-\frac{2 (2+b x)^{5/2}}{\sqrt{x}}+\frac{1}{2} (15 b) \int \frac{1}{\sqrt{x} \sqrt{2+b x}} \, dx\\ &=\frac{15}{2} b \sqrt{x} \sqrt{2+b x}+\frac{5}{2} b \sqrt{x} (2+b x)^{3/2}-\frac{2 (2+b x)^{5/2}}{\sqrt{x}}+(15 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{15}{2} b \sqrt{x} \sqrt{2+b x}+\frac{5}{2} b \sqrt{x} (2+b x)^{3/2}-\frac{2 (2+b x)^{5/2}}{\sqrt{x}}+15 \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0050738, size = 28, normalized size = 0.35 \[ -\frac{8 \sqrt{2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x}{2}\right )}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(5/2)/x^(3/2),x]

[Out]

(-8*Sqrt[2]*Hypergeometric2F1[-5/2, -1/2, 1/2, -(b*x)/2])/Sqrt[x]

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Maple [A]  time = 0.015, size = 81, normalized size = 1. \begin{align*}{\frac{{b}^{3}{x}^{3}+11\,{b}^{2}{x}^{2}+2\,bx-32}{2}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+2}}}}+{\frac{15}{2}\sqrt{b}\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)^(5/2)/x^(3/2),x)

[Out]

1/2*(b^3*x^3+11*b^2*x^2+2*b*x-32)/x^(1/2)/(b*x+2)^(1/2)+15/2*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)^(5/2)/x^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86295, size = 313, normalized size = 3.96 \begin{align*} \left [\frac{15 \, \sqrt{b} x \log \left (b x + \sqrt{b x + 2} \sqrt{b} \sqrt{x} + 1\right ) +{\left (b^{2} x^{2} + 9 \, b x - 16\right )} \sqrt{b x + 2} \sqrt{x}}{2 \, x}, -\frac{30 \, \sqrt{-b} x \arctan \left (\frac{\sqrt{b x + 2} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (b^{2} x^{2} + 9 \, b x - 16\right )} \sqrt{b x + 2} \sqrt{x}}{2 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(15*sqrt(b)*x*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (b^2*x^2 + 9*b*x - 16)*sqrt(b*x + 2)*sqrt(x)
)/x, -1/2*(30*sqrt(-b)*x*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - (b^2*x^2 + 9*b*x - 16)*sqrt(b*x + 2)*sqr
t(x))/x]

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Sympy [A]  time = 11.0624, size = 94, normalized size = 1.19 \begin{align*} 15 \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} + \frac{b^{3} x^{\frac{5}{2}}}{2 \sqrt{b x + 2}} + \frac{11 b^{2} x^{\frac{3}{2}}}{2 \sqrt{b x + 2}} + \frac{b \sqrt{x}}{\sqrt{b x + 2}} - \frac{16}{\sqrt{x} \sqrt{b x + 2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(b)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2) + b**3*x**(5/2)/(2*sqrt(b*x + 2)) + 11*b**2*x**(3/2)/(2*sqrt(b*x +
2)) + b*sqrt(x)/sqrt(b*x + 2) - 16/(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)^(5/2)/x^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError