3.1767 \(\int \frac{(a+\frac{b}{x})^{3/2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=103 \[ \frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{8 b^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x}}}{8 b \sqrt{x}}-\frac{a \sqrt{a+\frac{b}{x}}}{4 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}} \]

[Out]

-(a*Sqrt[a + b/x])/(4*x^(3/2)) - (a + b/x)^(3/2)/(3*x^(3/2)) - (a^2*Sqrt[a + b/x])/(8*b*Sqrt[x]) + (a^3*ArcTan
h[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(8*b^(3/2))

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Rubi [A]  time = 0.053141, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {337, 279, 321, 217, 206} \[ \frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{8 b^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x}}}{8 b \sqrt{x}}-\frac{a \sqrt{a+\frac{b}{x}}}{4 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)/x^(5/2),x]

[Out]

-(a*Sqrt[a + b/x])/(4*x^(3/2)) - (a + b/x)^(3/2)/(3*x^(3/2)) - (a^2*Sqrt[a + b/x])/(8*b*Sqrt[x]) + (a^3*ArcTan
h[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(8*b^(3/2))

Rule 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x}\right )^{3/2}}{x^{5/2}} \, dx &=-\left (2 \operatorname{Subst}\left (\int x^2 \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}}-a \operatorname{Subst}\left (\int x^2 \sqrt{a+b x^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{a \sqrt{a+\frac{b}{x}}}{4 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}}-\frac{1}{4} a^2 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{a \sqrt{a+\frac{b}{x}}}{4 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x}}}{8 b \sqrt{x}}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{8 b}\\ &=-\frac{a \sqrt{a+\frac{b}{x}}}{4 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x}}}{8 b \sqrt{x}}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{8 b}\\ &=-\frac{a \sqrt{a+\frac{b}{x}}}{4 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{3/2}}{3 x^{3/2}}-\frac{a^2 \sqrt{a+\frac{b}{x}}}{8 b \sqrt{x}}+\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{8 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.122791, size = 90, normalized size = 0.87 \[ \frac{\sqrt{a+\frac{b}{x}} \left (\frac{3 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}}\right )}{\sqrt{\frac{b}{a x}+1}}-\frac{\sqrt{b} \left (3 a^2 x^2+14 a b x+8 b^2\right )}{x^{5/2}}\right )}{24 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)/x^(5/2),x]

[Out]

(Sqrt[a + b/x]*(-((Sqrt[b]*(8*b^2 + 14*a*b*x + 3*a^2*x^2))/x^(5/2)) + (3*a^(5/2)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt
[x])])/Sqrt[1 + b/(a*x)]))/(24*b^(3/2))

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Maple [A]  time = 0.016, size = 92, normalized size = 0.9 \begin{align*} -{\frac{1}{24}\sqrt{{\frac{ax+b}{x}}} \left ( -3\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ){a}^{3}{x}^{3}+8\,{b}^{5/2}\sqrt{ax+b}+14\,xa{b}^{3/2}\sqrt{ax+b}+3\,{x}^{2}{a}^{2}\sqrt{b}\sqrt{ax+b} \right ){x}^{-{\frac{5}{2}}}{b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ax+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(3/2)/x^(5/2),x)

[Out]

-1/24*((a*x+b)/x)^(1/2)*(-3*arctanh((a*x+b)^(1/2)/b^(1/2))*a^3*x^3+8*b^(5/2)*(a*x+b)^(1/2)+14*x*a*b^(3/2)*(a*x
+b)^(1/2)+3*x^2*a^2*b^(1/2)*(a*x+b)^(1/2))/x^(5/2)/b^(3/2)/(a*x+b)^(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((a+b/x)^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52383, size = 416, normalized size = 4.04 \begin{align*} \left [\frac{3 \, a^{3} \sqrt{b} x^{3} \log \left (\frac{a x + 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) - 2 \,{\left (3 \, a^{2} b x^{2} + 14 \, a b^{2} x + 8 \, b^{3}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{48 \, b^{2} x^{3}}, -\frac{3 \, a^{3} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (3 \, a^{2} b x^{2} + 14 \, a b^{2} x + 8 \, b^{3}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{24 \, b^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*a^3*sqrt(b)*x^3*log((a*x + 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(3*a^2*b*x^2 + 14*a*b^2*
x + 8*b^3)*sqrt(x)*sqrt((a*x + b)/x))/(b^2*x^3), -1/24*(3*a^3*sqrt(-b)*x^3*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x +
b)/x)/b) + (3*a^2*b*x^2 + 14*a*b^2*x + 8*b^3)*sqrt(x)*sqrt((a*x + b)/x))/(b^2*x^3)]

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Sympy [A]  time = 27.4767, size = 124, normalized size = 1.2 \begin{align*} - \frac{a^{\frac{5}{2}}}{8 b \sqrt{x} \sqrt{1 + \frac{b}{a x}}} - \frac{17 a^{\frac{3}{2}}}{24 x^{\frac{3}{2}} \sqrt{1 + \frac{b}{a x}}} - \frac{11 \sqrt{a} b}{12 x^{\frac{5}{2}} \sqrt{1 + \frac{b}{a x}}} + \frac{a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}} \right )}}{8 b^{\frac{3}{2}}} - \frac{b^{2}}{3 \sqrt{a} x^{\frac{7}{2}} \sqrt{1 + \frac{b}{a x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)/(8*b*sqrt(x)*sqrt(1 + b/(a*x))) - 17*a**(3/2)/(24*x**(3/2)*sqrt(1 + b/(a*x))) - 11*sqrt(a)*b/(12*x**
(5/2)*sqrt(1 + b/(a*x))) + a**3*asinh(sqrt(b)/(sqrt(a)*sqrt(x)))/(8*b**(3/2)) - b**2/(3*sqrt(a)*x**(7/2)*sqrt(
1 + b/(a*x)))

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Giac [A]  time = 1.28967, size = 97, normalized size = 0.94 \begin{align*} -\frac{1}{24} \, a^{3}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{3 \,{\left (a x + b\right )}^{\frac{5}{2}} + 8 \,{\left (a x + b\right )}^{\frac{3}{2}} b - 3 \, \sqrt{a x + b} b^{2}}{a^{3} b x^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*a^3*(3*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b) + (3*(a*x + b)^(5/2) + 8*(a*x + b)^(3/2)*b - 3*sqrt(a
*x + b)*b^2)/(a^3*b*x^3))