3.1733 \(\int \frac{x}{(a+\frac{b}{x})^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ -\frac{15 b^2}{4 a^3 \sqrt{a+\frac{b}{x}}}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{5 b x}{4 a^2 \sqrt{a+\frac{b}{x}}}+\frac{x^2}{2 a \sqrt{a+\frac{b}{x}}} \]

[Out]

(-15*b^2)/(4*a^3*Sqrt[a + b/x]) - (5*b*x)/(4*a^2*Sqrt[a + b/x]) + x^2/(2*a*Sqrt[a + b/x]) + (15*b^2*ArcTanh[Sq
rt[a + b/x]/Sqrt[a]])/(4*a^(7/2))

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Rubi [A]  time = 0.0378823, antiderivative size = 91, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 51, 63, 208} \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{5 x^2 \sqrt{a+\frac{b}{x}}}{2 a^2}-\frac{15 b x \sqrt{a+\frac{b}{x}}}{4 a^3}-\frac{2 x^2}{a \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x)^(3/2),x]

[Out]

(-15*b*Sqrt[a + b/x]*x)/(4*a^3) - (2*x^2)/(a*Sqrt[a + b/x]) + (5*Sqrt[a + b/x]*x^2)/(2*a^2) + (15*b^2*ArcTanh[
Sqrt[a + b/x]/Sqrt[a]])/(4*a^(7/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{x}{\left (a+\frac{b}{x}\right )^{3/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)^{3/2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 x^2}{a \sqrt{a+\frac{b}{x}}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{2 x^2}{a \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x^2}{2 a^2}+\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{4 a^2}\\ &=-\frac{15 b \sqrt{a+\frac{b}{x}} x}{4 a^3}-\frac{2 x^2}{a \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x^2}{2 a^2}-\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{8 a^3}\\ &=-\frac{15 b \sqrt{a+\frac{b}{x}} x}{4 a^3}-\frac{2 x^2}{a \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x^2}{2 a^2}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{4 a^3}\\ &=-\frac{15 b \sqrt{a+\frac{b}{x}} x}{4 a^3}-\frac{2 x^2}{a \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x^2}{2 a^2}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.010619, size = 37, normalized size = 0.4 \[ -\frac{2 b^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b}{a x}+1\right )}{a^3 \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x)^(3/2),x]

[Out]

(-2*b^2*Hypergeometric2F1[-1/2, 3, 1/2, 1 + b/(a*x)])/(a^3*Sqrt[a + b/x])

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Maple [B]  time = 0.01, size = 395, normalized size = 4.3 \begin{align*}{\frac{x}{8\, \left ( ax+b \right ) ^{2}}\sqrt{{\frac{ax+b}{x}}} \left ( 4\,\sqrt{a{x}^{2}+bx}{a}^{9/2}{x}^{3}-32\,\sqrt{ \left ( ax+b \right ) x}{a}^{7/2}{x}^{2}b+10\,\sqrt{a{x}^{2}+bx}{a}^{7/2}{x}^{2}b+16\,{a}^{3}\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}{b}^{2}+16\, \left ( \left ( ax+b \right ) x \right ) ^{3/2}{a}^{5/2}b-64\,\sqrt{ \left ( ax+b \right ) x}{a}^{5/2}x{b}^{2}+8\,{a}^{5/2}\sqrt{a{x}^{2}+bx}x{b}^{2}+32\,{a}^{2}\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) x{b}^{3}-\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ){x}^{2}{a}^{3}{b}^{2}-32\,\sqrt{ \left ( ax+b \right ) x}{a}^{3/2}{b}^{3}+2\,{a}^{3/2}\sqrt{a{x}^{2}+bx}{b}^{3}+16\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) a{b}^{4}-2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) x{a}^{2}{b}^{3}-\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ) a{b}^{4} \right ){a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*x+b)/x)^(1/2)*x/a^(9/2)*(4*(a*x^2+b*x)^(1/2)*a^(9/2)*x^3-32*((a*x+b)*x)^(1/2)*a^(7/2)*x^2*b+10*(a*x^2+
b*x)^(1/2)*a^(7/2)*x^2*b+16*a^3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*b^2+16*((a*x+b)*x)^(
3/2)*a^(5/2)*b-64*((a*x+b)*x)^(1/2)*a^(5/2)*x*b^2+8*a^(5/2)*(a*x^2+b*x)^(1/2)*x*b^2+32*a^2*ln(1/2*(2*((a*x+b)*
x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*b^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^3*b^2-32*
((a*x+b)*x)^(1/2)*a^(3/2)*b^3+2*a^(3/2)*(a*x^2+b*x)^(1/2)*b^3+16*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/
a^(1/2))*a*b^4-2*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^2*b^3-ln(1/2*(2*(a*x^2+b*x)^(1/2)*a
^(1/2)+2*a*x+b)/a^(1/2))*a*b^4)/((a*x+b)*x)^(1/2)/(a*x+b)^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(x/(a+b/x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52712, size = 419, normalized size = 4.51 \begin{align*} \left [\frac{15 \,{\left (a b^{2} x + b^{3}\right )} \sqrt{a} \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (2 \, a^{3} x^{3} - 5 \, a^{2} b x^{2} - 15 \, a b^{2} x\right )} \sqrt{\frac{a x + b}{x}}}{8 \,{\left (a^{5} x + a^{4} b\right )}}, -\frac{15 \,{\left (a b^{2} x + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (2 \, a^{3} x^{3} - 5 \, a^{2} b x^{2} - 15 \, a b^{2} x\right )} \sqrt{\frac{a x + b}{x}}}{4 \,{\left (a^{5} x + a^{4} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(a*b^2*x + b^3)*sqrt(a)*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(2*a^3*x^3 - 5*a^2*b*x^2 -
15*a*b^2*x)*sqrt((a*x + b)/x))/(a^5*x + a^4*b), -1/4*(15*(a*b^2*x + b^3)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x +
b)/x)/a) - (2*a^3*x^3 - 5*a^2*b*x^2 - 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^5*x + a^4*b)]

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Sympy [A]  time = 5.06594, size = 105, normalized size = 1.13 \begin{align*} \frac{x^{\frac{5}{2}}}{2 a \sqrt{b} \sqrt{\frac{a x}{b} + 1}} - \frac{5 \sqrt{b} x^{\frac{3}{2}}}{4 a^{2} \sqrt{\frac{a x}{b} + 1}} - \frac{15 b^{\frac{3}{2}} \sqrt{x}}{4 a^{3} \sqrt{\frac{a x}{b} + 1}} + \frac{15 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )}}{4 a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(5/2)/(2*a*sqrt(b)*sqrt(a*x/b + 1)) - 5*sqrt(b)*x**(3/2)/(4*a**2*sqrt(a*x/b + 1)) - 15*b**(3/2)*sqrt(x)/(4*
a**3*sqrt(a*x/b + 1)) + 15*b**2*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(4*a**(7/2))

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Giac [A]  time = 1.16664, size = 142, normalized size = 1.53 \begin{align*} -\frac{1}{4} \, b^{2}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{8}{a^{3} \sqrt{\frac{a x + b}{x}}} - \frac{9 \, a \sqrt{\frac{a x + b}{x}} - \frac{7 \,{\left (a x + b\right )} \sqrt{\frac{a x + b}{x}}}{x}}{{\left (a - \frac{a x + b}{x}\right )}^{2} a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b^2*(15*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^3) + 8/(a^3*sqrt((a*x + b)/x)) - (9*a*sqrt((a*x +
b)/x) - 7*(a*x + b)*sqrt((a*x + b)/x)/x)/((a - (a*x + b)/x)^2*a^3))