3.1702 \(\int (a+\frac{b}{x})^{3/2} x^2 \, dx\)

Optimal. Leaf size=90 \[ -\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{8 a^{3/2}}+\frac{b^2 x \sqrt{a+\frac{b}{x}}}{8 a}+\frac{1}{4} b x^2 \sqrt{a+\frac{b}{x}}+\frac{1}{3} x^3 \left (a+\frac{b}{x}\right )^{3/2} \]

[Out]

(b^2*Sqrt[a + b/x]*x)/(8*a) + (b*Sqrt[a + b/x]*x^2)/4 + ((a + b/x)^(3/2)*x^3)/3 - (b^3*ArcTanh[Sqrt[a + b/x]/S
qrt[a]])/(8*a^(3/2))

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Rubi [A]  time = 0.0391332, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 51, 63, 208} \[ -\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{8 a^{3/2}}+\frac{b^2 x \sqrt{a+\frac{b}{x}}}{8 a}+\frac{1}{4} b x^2 \sqrt{a+\frac{b}{x}}+\frac{1}{3} x^3 \left (a+\frac{b}{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)*x^2,x]

[Out]

(b^2*Sqrt[a + b/x]*x)/(8*a) + (b*Sqrt[a + b/x]*x^2)/4 + ((a + b/x)^(3/2)*x^3)/3 - (b^3*ArcTanh[Sqrt[a + b/x]/S
qrt[a]])/(8*a^(3/2))

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 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 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 \left (a+\frac{b}{x}\right )^{3/2} x^2 \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} x^3-\frac{1}{2} b \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{4} b \sqrt{a+\frac{b}{x}} x^2+\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} x^3-\frac{1}{8} b^2 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b^2 \sqrt{a+\frac{b}{x}} x}{8 a}+\frac{1}{4} b \sqrt{a+\frac{b}{x}} x^2+\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} x^3+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{16 a}\\ &=\frac{b^2 \sqrt{a+\frac{b}{x}} x}{8 a}+\frac{1}{4} b \sqrt{a+\frac{b}{x}} x^2+\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} x^3+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{8 a}\\ &=\frac{b^2 \sqrt{a+\frac{b}{x}} x}{8 a}+\frac{1}{4} b \sqrt{a+\frac{b}{x}} x^2+\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} x^3-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{8 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.013115, size = 39, normalized size = 0.43 \[ -\frac{2 b^3 \left (a+\frac{b}{x}\right )^{5/2} \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{b}{a x}+1\right )}{5 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)*x^2,x]

[Out]

(-2*b^3*(a + b/x)^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, 1 + b/(a*x)])/(5*a^4)

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Maple [A]  time = 0.007, size = 115, normalized size = 1.3 \begin{align*}{\frac{x}{48}\sqrt{{\frac{ax+b}{x}}} \left ( 16\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{5/2}+12\,\sqrt{a{x}^{2}+bx}{a}^{5/2}xb+6\,\sqrt{a{x}^{2}+bx}{a}^{3/2}{b}^{2}-3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) a{b}^{3} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((a*x+b)/x)^(1/2)*x/a^(5/2)*(16*(a*x^2+b*x)^(3/2)*a^(5/2)+12*(a*x^2+b*x)^(1/2)*a^(5/2)*x*b+6*(a*x^2+b*x)^
(1/2)*a^(3/2)*b^2-3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*b^3)/((a*x+b)*x)^(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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88656, size = 352, normalized size = 3.91 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{3} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (8 \, a^{3} x^{3} + 14 \, a^{2} b x^{2} + 3 \, a b^{2} x\right )} \sqrt{\frac{a x + b}{x}}}{48 \, a^{2}}, \frac{3 \, \sqrt{-a} b^{3} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (8 \, a^{3} x^{3} + 14 \, a^{2} b x^{2} + 3 \, a b^{2} x\right )} \sqrt{\frac{a x + b}{x}}}{24 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.16634, size = 124, normalized size = 1.38 \begin{align*} \frac{a^{2} x^{\frac{7}{2}}}{3 \sqrt{b} \sqrt{\frac{a x}{b} + 1}} + \frac{11 a \sqrt{b} x^{\frac{5}{2}}}{12 \sqrt{\frac{a x}{b} + 1}} + \frac{17 b^{\frac{3}{2}} x^{\frac{3}{2}}}{24 \sqrt{\frac{a x}{b} + 1}} + \frac{b^{\frac{5}{2}} \sqrt{x}}{8 a \sqrt{\frac{a x}{b} + 1}} - \frac{b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )}}{8 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1448, size = 126, normalized size = 1.4 \begin{align*} \frac{b^{3} \log \left ({\left | -2 \,{\left (\sqrt{a} x - \sqrt{a x^{2} + b x}\right )} \sqrt{a} - b \right |}\right ) \mathrm{sgn}\left (x\right )}{16 \, a^{\frac{3}{2}}} - \frac{b^{3} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{16 \, a^{\frac{3}{2}}} + \frac{1}{24} \, \sqrt{a x^{2} + b x}{\left (2 \,{\left (4 \, a x \mathrm{sgn}\left (x\right ) + 7 \, b \mathrm{sgn}\left (x\right )\right )} x + \frac{3 \, b^{2} \mathrm{sgn}\left (x\right )}{a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^3*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))*sgn(x)/a^(3/2) - 1/16*b^3*log(abs(b))*sgn(x)
/a^(3/2) + 1/24*sqrt(a*x^2 + b*x)*(2*(4*a*x*sgn(x) + 7*b*sgn(x))*x + 3*b^2*sgn(x)/a)