\(\int \frac {(a+\frac {b}{x})^{5/2}}{x} \, dx\) [1716]

Optimal result

Integrand size = 15, antiderivative size = 73 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=-2 a^2 \sqrt {a+\frac {b}{x}}-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2}+2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 52, 65, 214} \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-2 a^2 \sqrt {a+\frac {b}{x}}-\frac {2}{3} a \left (a+\frac {b}{x}\right )^{3/2}-\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} \]

[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 52

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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*} \text {integral}& = -\text {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 \text {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 \text {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 \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=-\frac {2 \sqrt {a+\frac {b}{x}} \left (3 b^2+11 a b x+23 a^2 x^2\right )}{15 x^2}+2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \]

[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]]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.26

method	result	size
risch	\(-\frac {2 \left (23 a^{2} x^{2}+11 a b x +3 b^{2}\right ) \sqrt {\frac {a x +b}{x}}}{15 x^{2}}+\frac {a^{\frac {5}{2}} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{a x +b}\)	\(92\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-30 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} x^{4}-15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b \,x^{4}+30 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} x^{2}+16 a^{\frac {3}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b x +6 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{2}\right )}{15 x^{3} b \sqrt {x \left (a x +b \right )}\, \sqrt {a}}\)	\(145\)

[In]

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

[Out]

-2/15*(23*a^2*x^2+11*a*b*x+3*b^2)/x^2*((a*x+b)/x)^(1/2)+a^(5/2)*ln((1/2*b+a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))*((a*
x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)/(a*x+b)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.95 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=\left [\frac {15 \, a^{\frac {5}{2}} x^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (23 \, a^{2} x^{2} + 11 \, a b x + 3 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}}{15 \, x^{2}}, -\frac {2 \, {\left (15 \, \sqrt {-a} a^{2} x^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (23 \, a^{2} x^{2} + 11 \, a b x + 3 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}\right )}}{15 \, x^{2}}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 2.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=- \frac {46 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}}{15} - a^{\frac {5}{2}} \log {\left (\frac {b}{a x} \right )} + 2 a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )} - \frac {22 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x}}}{15 x} - \frac {2 \sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x}}}{5 x^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=-a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - \frac {2}{5} \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} - \frac {2}{3} \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a - 2 \, \sqrt {a + \frac {b}{x}} a^{2} \]

[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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=\text {Exception raised: TypeError} \]

[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,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 6.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x} \, dx=-\frac {2\,a\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}-\frac {2\,{\left (a+\frac {b}{x}\right )}^{5/2}}{5}-2\,a^2\,\sqrt {a+\frac {b}{x}}-a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

[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)