3.1716 \(\int \frac {(a+\frac {b}{x})^{5/2}}{x} \, dx\)

Optimal. Leaf size=73 \[ 2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-2 a^2 \sqrt {a+\frac {b}{x}}-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ -2 a^2 \sqrt {a+\frac {b}{x}}+2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/x,x]

[Out]

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

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 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 \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx &=-\operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2}-a \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2}-a^2 \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\frac {1}{x}\right )\\ &=-2 a^2 \sqrt {a+\frac {b}{x}}-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2}-a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=-2 a^2 \sqrt {a+\frac {b}{x}}-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2}-\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{b}\\ &=-2 a^2 \sqrt {a+\frac {b}{x}}-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2}+2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 63, normalized size = 0.86 \[ 2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {2 \sqrt {a+\frac {b}{x}} \left (23 a^2 x^2+11 a b x+3 b^2\right )}{15 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)/x,x]

[Out]

(-2*Sqrt[a + b/x]*(3*b^2 + 11*a*b*x + 23*a^2*x^2))/(15*x^2) + 2*a^(5/2)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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fricas [A]  time = 1.10, size = 142, normalized size = 1.95 \[ \left [\frac {15 \, a^{\frac {5}{2}} x^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (23 \, a^{2} x^{2} + 11 \, a b x + 3 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}}{15 \, x^{2}}, -\frac {2 \, {\left (15 \, \sqrt {-a} a^{2} x^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (23 \, a^{2} x^{2} + 11 \, a b x + 3 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}\right )}}{15 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x,x, algorithm="fricas")

[Out]

[1/15*(15*a^(5/2)*x^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(23*a^2*x^2 + 11*a*b*x + 3*b^2)*sqrt(
(a*x + b)/x))/x^2, -2/15*(15*sqrt(-a)*a^2*x^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (23*a^2*x^2 + 11*a*b*x +
3*b^2)*sqrt((a*x + b)/x))/x^2]

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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)^(5/2)/x,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 = 145, normalized size = 1.99 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-15 a^{3} b \,x^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-30 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} x^{4}+30 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} x^{2}+16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b x +6 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{2}\right )}{15 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)/x,x)

[Out]

-1/15*((a*x+b)/x)^(1/2)/x^3/b*(-30*(a*x^2+b*x)^(1/2)*a^(7/2)*x^4-15*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2
))/a^(1/2))*x^4*a^3*b+30*(a*x^2+b*x)^(3/2)*a^(5/2)*x^2+16*a^(3/2)*(a*x^2+b*x)^(3/2)*b*x+6*(a*x^2+b*x)^(3/2)*a^
(1/2)*b^2)/((a*x+b)*x)^(1/2)/a^(1/2)

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maxima [A]  time = 2.43, size = 75, normalized size = 1.03 \[ -a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - \frac {2}{5} \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} - \frac {2}{3} \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a - 2 \, \sqrt {a + \frac {b}{x}} a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x,x, algorithm="maxima")

[Out]

-a^(5/2)*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) - 2/5*(a + b/x)^(5/2) - 2/3*(a + b/x)^(3/2)*
a - 2*sqrt(a + b/x)*a^2

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mupad [B]  time = 1.75, size = 60, normalized size = 0.82 \[ -\frac {2\,a\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}-\frac {2\,{\left (a+\frac {b}{x}\right )}^{5/2}}{5}-2\,a^2\,\sqrt {a+\frac {b}{x}}-a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(5/2)/x,x)

[Out]

- a^(5/2)*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*2i - (2*a*(a + b/x)^(3/2))/3 - (2*(a + b/x)^(5/2))/5 - 2*a^2*(a +
b/x)^(1/2)

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sympy [A]  time = 4.42, size = 97, normalized size = 1.33 \[ - \frac {46 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}}{15} - a^{\frac {5}{2}} \log {\left (\frac {b}{a x} \right )} + 2 a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )} - \frac {22 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x}}}{15 x} - \frac {2 \sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x}}}{5 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)/x,x)

[Out]

-46*a**(5/2)*sqrt(1 + b/(a*x))/15 - a**(5/2)*log(b/(a*x)) + 2*a**(5/2)*log(sqrt(1 + b/(a*x)) + 1) - 22*a**(3/2
)*b*sqrt(1 + b/(a*x))/(15*x) - 2*sqrt(a)*b**2*sqrt(1 + b/(a*x))/(5*x**2)

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