3.1704 \(\int (a+\frac {b}{x})^{3/2} \, dx\)

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]

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

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 1.07, size = 100, normalized size = 1.85 \[ \left [\frac {3}{2} \, \sqrt {a} b \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + {\left (a x - 2 \, b\right )} \sqrt {\frac {a x + b}{x}}, -3 \, \sqrt {-a} b \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (a x - 2 \, b\right )} \sqrt {\frac {a x + b}{x}}\right ] \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-2,[1,2,0]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[2,4,0]%%
%}+%%%{-2,[2,2,0]%%%}+%%%{1,[2,0,0]%%%}+%%%{2,[1,3,1]%%%}+%%%{-2,[1,1,1]%%%}+%%%{1,[0,2,2]%%%}] at parameters
values [86,-97,-82]Warning, choosing root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[0,2,2]%%%}] a
t parameters values [82.1195442914,26,-89]Warning, choosing root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%}
,0,%%%{1,[0,2,2]%%%}] at parameters values [85.3561567818,-64,-30]Warning, choosing root of [1,0,%%%{-2,[1,0,1
]%%%}+%%%{-4,[0,1,0]%%%},0,%%%{1,[2,0,2]%%%}] at parameters values [42,43.9628838282,-9]Sign error (%%%{-b,0%%
%}+%%%{2*sqrt(a)*sqrt(b),1/2%%%}+%%%{-2*a,1%%%}+%%%{a*sqrt(a)*sqrt(b)/b,3/2%%%}+%%%{-a^2*sqrt(a)*sqrt(b)/(4*b^
2),5/2%%%}+%%%{undef,7/2%%%})Limit: Max order reached or unable to make series expansion Error: Bad Argument V
alue

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maple [B]  time = 0.01, size = 100, normalized size = 1.85 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-3 a b \,x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-6 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} x^{2}+4 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\right )}{2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, x} \]

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*a*b*x^2*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/
2))/a^(1/2))+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 [A]  time = 2.36, size = 63, normalized size = 1.17 \[ \sqrt {a + \frac {b}{x}} a x - \frac {3}{2} \, \sqrt {a} b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - 2 \, \sqrt {a + \frac {b}{x}} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.25, size = 34, normalized size = 0.63 \[ -\frac {2\,x\,{\left (a+\frac {b}{x}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {a\,x}{b}\right )}{{\left (\frac {a\,x}{b}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(3/2),x)

[Out]

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

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sympy [B]  time = 3.75, size = 92, normalized size = 1.70 \[ 3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )} + \frac {a^{2} x^{\frac {3}{2}}}{\sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {a \sqrt {b} \sqrt {x}}{\sqrt {\frac {a x}{b} + 1}} - \frac {2 b^{\frac {3}{2}}}{\sqrt {x} \sqrt {\frac {a x}{b} + 1}} \]

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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