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

Optimal. Leaf size=54 \[ x \left (a+\frac{b}{x}\right )^{3/2}-3 b \sqrt{a+\frac{b}{x}}+3 \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0253246, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {242, 47, 50, 63, 208} \[ x \left (a+\frac{b}{x}\right )^{3/2}-3 b \sqrt{a+\frac{b}{x}}+3 \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2),x]

[Out]

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

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

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

Mathematica [A]  time = 0.0191561, size = 46, normalized size = 0.85 \[ \sqrt{a+\frac{b}{x}} (a x-2 b)+3 \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2),x]

[Out]

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

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Maple [B]  time = 0.007, size = 100, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,x}\sqrt{{\frac{ax+b}{x}}} \left ( -6\,\sqrt{a{x}^{2}+bx}{a}^{3/2}{x}^{2}-3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}ab+4\, \left ( a{x}^{2}+bx \right ) ^{3/2}\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)

[Out]

-1/2*((a*x+b)/x)^(1/2)/x*(-6*(a*x^2+b*x)^(1/2)*a^(3/2)*x^2-3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1
/2))*x^2*a*b+4*(a*x^2+b*x)^(3/2)*a^(1/2))/((a*x+b)*x)^(1/2)/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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52256, size = 246, normalized size = 4.56 \begin{align*} \left [\frac{3}{2} \, \sqrt{a} b \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) +{\left (a x - 2 \, b\right )} \sqrt{\frac{a x + b}{x}}, -3 \, \sqrt{-a} b \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (a x - 2 \, b\right )} \sqrt{\frac{a x + b}{x}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[3/2*sqrt(a)*b*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + (a*x - 2*b)*sqrt((a*x + b)/x), -3*sqrt(-a)*b*a
rctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (a*x - 2*b)*sqrt((a*x + b)/x)]

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Sympy [B]  time = 2.37587, size = 92, normalized size = 1.7 \begin{align*} 3 \sqrt{a} b \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )} + \frac{a^{2} x^{\frac{3}{2}}}{\sqrt{b} \sqrt{\frac{a x}{b} + 1}} - \frac{a \sqrt{b} \sqrt{x}}{\sqrt{\frac{a x}{b} + 1}} - \frac{2 b^{\frac{3}{2}}}{\sqrt{x} \sqrt{\frac{a x}{b} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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, algorithm="giac")

[Out]

Exception raised: TypeError