3.1705 \(\int \frac{(a+\frac{b}{x})^{3/2}}{x} \, dx\)

Optimal. Leaf size=55 \[ 2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )-2 a \sqrt{a+\frac{b}{x}}-\frac{2}{3} \left (a+\frac{b}{x}\right )^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.0269337, antiderivative size = 55, 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 = {266, 50, 63, 208} \[ 2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )-2 a \sqrt{a+\frac{b}{x}}-\frac{2}{3} \left (a+\frac{b}{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)/x,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0357304, size = 50, normalized size = 0.91 \[ 2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )-\frac{2 \sqrt{a+\frac{b}{x}} (4 a x+b)}{3 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)/x,x]

[Out]

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

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Maple [B]  time = 0.01, size = 123, normalized size = 2.2 \begin{align*} -{\frac{1}{3\,b{x}^{2}}\sqrt{{\frac{ax+b}{x}}} \left ( -6\,\sqrt{a{x}^{2}+bx}{a}^{5/2}{x}^{3}-3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}{a}^{2}b+6\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}x+2\, \left ( a{x}^{2}+bx \right ) ^{3/2}b\sqrt{a} \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((a*x+b)/x)^(1/2)/x^2*(-6*(a*x^2+b*x)^(1/2)*a^(5/2)*x^3-3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^
(1/2))*x^3*a^2*b+6*(a*x^2+b*x)^(3/2)*a^(3/2)*x+2*(a*x^2+b*x)^(3/2)*b*a^(1/2))/((a*x+b)*x)^(1/2)/b/a^(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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81818, size = 270, normalized size = 4.91 \begin{align*} \left [\frac{3 \, a^{\frac{3}{2}} x \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) - 2 \,{\left (4 \, a x + b\right )} \sqrt{\frac{a x + b}{x}}}{3 \, x}, -\frac{2 \,{\left (3 \, \sqrt{-a} a x \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (4 \, a x + b\right )} \sqrt{\frac{a x + b}{x}}\right )}}{3 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.96372, size = 71, normalized size = 1.29 \begin{align*} - \frac{8 a^{\frac{3}{2}} \sqrt{1 + \frac{b}{a x}}}{3} - a^{\frac{3}{2}} \log{\left (\frac{b}{a x} \right )} + 2 a^{\frac{3}{2}} \log{\left (\sqrt{1 + \frac{b}{a x}} + 1 \right )} - \frac{2 \sqrt{a} b \sqrt{1 + \frac{b}{a x}}}{3 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a**(3/2)*sqrt(1 + b/(a*x))/3 - a**(3/2)*log(b/(a*x)) + 2*a**(3/2)*log(sqrt(1 + b/(a*x)) + 1) - 2*sqrt(a)*b*
sqrt(1 + b/(a*x))/(3*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError