3.1703 \(\int (a+\frac {b}{x})^{3/2} x \, dx\)

Optimal. Leaf size=66 \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}+\frac {3}{4} b x \sqrt {a+\frac {b}{x}} \]

[Out]

1/2*(a+b/x)^(3/2)*x^2+3/4*b^2*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(1/2)+3/4*b*x*(a+b/x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 47, 63, 208} \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}+\frac {3}{4} b x \sqrt {a+\frac {b}{x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)*x,x]

[Out]

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

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 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]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 76, normalized size = 1.15 \[ \frac {x \sqrt {a+\frac {b}{x}} \left (2 a^2 x^2+3 b^2 \sqrt {\frac {b}{a x}+1} \tanh ^{-1}\left (\sqrt {\frac {b}{a x}+1}\right )+7 a b x+5 b^2\right )}{4 (a x+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)*x,x]

[Out]

(Sqrt[a + b/x]*x*(5*b^2 + 7*a*b*x + 2*a^2*x^2 + 3*b^2*Sqrt[1 + b/(a*x)]*ArcTanh[Sqrt[1 + b/(a*x)]]))/(4*(b + a
*x))

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fricas [A]  time = 1.04, size = 130, normalized size = 1.97 \[ \left [\frac {3 \, \sqrt {a} b^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (2 \, a^{2} x^{2} + 5 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{8 \, a}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (2 \, a^{2} x^{2} + 5 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{4 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 79, normalized size = 1.20 \[ -\frac {3 \, b^{2} \log \left ({\left | -2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} - b \right |}\right ) \mathrm {sgn}\relax (x)}{8 \, \sqrt {a}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{8 \, \sqrt {a}} + \frac {1}{4} \, \sqrt {a x^{2} + b x} {\left (2 \, a x \mathrm {sgn}\relax (x) + 5 \, b \mathrm {sgn}\relax (x)\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*b^2*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))*sgn(x)/sqrt(a) + 3/8*b^2*log(abs(b))*sgn(x)/
sqrt(a) + 1/4*sqrt(a*x^2 + b*x)*(2*a*x*sgn(x) + 5*b*sgn(x))

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maple [A]  time = 0.01, size = 96, normalized size = 1.45 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (3 a \,b^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} x +10 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b \right ) x}{8 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*x+b)/x)^(1/2)*x*(4*(a*x^2+b*x)^(1/2)*a^(5/2)*x+10*(a*x^2+b*x)^(1/2)*a^(3/2)*b+3*a*b^2*ln(1/2*(2*a*x+b+
2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2)))/((a*x+b)*x)^(1/2)/a^(3/2)

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maxima [A]  time = 2.23, size = 98, normalized size = 1.48 \[ -\frac {3 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{8 \, \sqrt {a}} + \frac {5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {a + \frac {b}{x}} a b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} - 2 \, {\left (a + \frac {b}{x}\right )} a + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a) + 1/4*(5*(a + b/x)^(3/2)*b^2 - 3*sqr
t(a + b/x)*a*b^2)/((a + b/x)^2 - 2*(a + b/x)*a + a^2)

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mupad [B]  time = 1.17, size = 52, normalized size = 0.79 \[ \frac {5\,x^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{4}+\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4\,\sqrt {a}}-\frac {3\,a\,x^2\,\sqrt {a+\frac {b}{x}}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x)^(3/2),x)

[Out]

(5*x^2*(a + b/x)^(3/2))/4 + (3*b^2*atanh((a + b/x)^(1/2)/a^(1/2)))/(4*a^(1/2)) - (3*a*x^2*(a + b/x)^(1/2))/4

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sympy [A]  time = 4.11, size = 75, normalized size = 1.14 \[ \frac {a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a x}{b} + 1}}{2} + \frac {5 b^{\frac {3}{2}} \sqrt {x} \sqrt {\frac {a x}{b} + 1}}{4} + \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*sqrt(b)*x**(3/2)*sqrt(a*x/b + 1)/2 + 5*b**(3/2)*sqrt(x)*sqrt(a*x/b + 1)/4 + 3*b**2*asinh(sqrt(a)*sqrt(x)/sqr
t(b))/(4*sqrt(a))

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