\(\int (a+\frac {b}{x})^{5/2} x \, dx\) [1714]

Optimal result

Integrand size = 13, antiderivative size = 84 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=-\frac {15}{4} b^2 \sqrt {a+\frac {b}{x}}+\frac {5}{4} b \left (a+\frac {b}{x}\right )^{3/2} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{5/2} x^2+\frac {15}{4} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \]

[Out]

5/4*b*(a+b/x)^(3/2)*x+1/2*(a+b/x)^(5/2)*x^2+15/4*b^2*arctanh((a+b/x)^(1/2)/a^(1/2))*a^(1/2)-15/4*b^2*(a+b/x)^(
1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {272, 43, 52, 65, 214} \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {15}{4} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {15}{4} b^2 \sqrt {a+\frac {b}{x}}+\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{5/2}+\frac {5}{4} b x \left (a+\frac {b}{x}\right )^{3/2} \]

[In]

Int[(a + b/x)^(5/2)*x,x]

[Out]

(-15*b^2*Sqrt[a + b/x])/4 + (5*b*(a + b/x)^(3/2)*x)/4 + ((a + b/x)^(5/2)*x^2)/2 + (15*Sqrt[a]*b^2*ArcTanh[Sqrt
[a + b/x]/Sqrt[a]])/4

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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^3} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{2} \left (a+\frac {b}{x}\right )^{5/2} x^2-\frac {1}{4} (5 b) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {5}{4} b \left (a+\frac {b}{x}\right )^{3/2} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{5/2} x^2-\frac {1}{8} \left (15 b^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {15}{4} b^2 \sqrt {a+\frac {b}{x}}+\frac {5}{4} b \left (a+\frac {b}{x}\right )^{3/2} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{5/2} x^2-\frac {1}{8} \left (15 a b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {15}{4} b^2 \sqrt {a+\frac {b}{x}}+\frac {5}{4} b \left (a+\frac {b}{x}\right )^{3/2} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{5/2} x^2-\frac {1}{4} (15 a b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right ) \\ & = -\frac {15}{4} b^2 \sqrt {a+\frac {b}{x}}+\frac {5}{4} b \left (a+\frac {b}{x}\right )^{3/2} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{5/2} x^2+\frac {15}{4} \sqrt {a} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {1}{4} \left (\sqrt {a+\frac {b}{x}} \left (-8 b^2+9 a b x+2 a^2 x^2\right )+15 \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\right ) \]

[In]

Integrate[(a + b/x)^(5/2)*x,x]

[Out]

(Sqrt[a + b/x]*(-8*b^2 + 9*a*b*x + 2*a^2*x^2) + 15*Sqrt[a]*b^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {\left (2 a^{2} x^{2}+9 a b x -8 b^{2}\right ) \sqrt {\frac {a x +b}{x}}}{4}+\frac {15 \sqrt {a}\, b^{2} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{8 \left (a x +b \right )}\)	\(93\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-4 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} x^{3}-34 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b \,x^{2}-15 a^{2} \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{2} x^{2}+16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b \right )}{8 x \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}}}\)	\(125\)

[In]

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

[Out]

1/4*(2*a^2*x^2+9*a*b*x-8*b^2)*((a*x+b)/x)^(1/2)+15/8*a^(1/2)*b^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 = 0.39 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.55 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\left [\frac {15}{8} \, \sqrt {a} b^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + \frac {1}{4} \, {\left (2 \, a^{2} x^{2} + 9 \, a b x - 8 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}, -\frac {15}{4} \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + \frac {1}{4} \, {\left (2 \, a^{2} x^{2} + 9 \, a b x - 8 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}\right ] \]

[In]

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

[Out]

[15/8*sqrt(a)*b^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 1/4*(2*a^2*x^2 + 9*a*b*x - 8*b^2)*sqrt((a*x
+ b)/x), -15/4*sqrt(-a)*b^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + 1/4*(2*a^2*x^2 + 9*a*b*x - 8*b^2)*sqrt((a*
x + b)/x)]

Sympy [A] (verification not implemented)

Time = 2.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {15 \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4} + \frac {a^{3} x^{\frac {5}{2}}}{2 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {11 a^{2} \sqrt {b} x^{\frac {3}{2}}}{4 \sqrt {\frac {a x}{b} + 1}} + \frac {a b^{\frac {3}{2}} \sqrt {x}}{4 \sqrt {\frac {a x}{b} + 1}} - \frac {2 b^{\frac {5}{2}}}{\sqrt {x} \sqrt {\frac {a x}{b} + 1}} \]

[In]

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

[Out]

15*sqrt(a)*b**2*asinh(sqrt(a)*sqrt(x)/sqrt(b))/4 + a**3*x**(5/2)/(2*sqrt(b)*sqrt(a*x/b + 1)) + 11*a**2*sqrt(b)
*x**(3/2)/(4*sqrt(a*x/b + 1)) + a*b**(3/2)*sqrt(x)/(4*sqrt(a*x/b + 1)) - 2*b**(5/2)/(sqrt(x)*sqrt(a*x/b + 1))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.37 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=-\frac {15}{8} \, \sqrt {a} b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - 2 \, \sqrt {a + \frac {b}{x}} b^{2} + \frac {9 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{2} - 7 \, \sqrt {a + \frac {b}{x}} a^{2} b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} - 2 \, {\left (a + \frac {b}{x}\right )} a + a^{2}\right )}} \]

[In]

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

[Out]

-15/8*sqrt(a)*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) - 2*sqrt(a + b/x)*b^2 + 1/4*(9*(a +
b/x)^(3/2)*a*b^2 - 7*sqrt(a + b/x)*a^2*b^2)/((a + b/x)^2 - 2*(a + b/x)*a + a^2)

Giac [F(-2)]

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

[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 = 5.92 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {9\,a\,x^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{4}-2\,b^2\,\sqrt {a+\frac {b}{x}}-\frac {7\,a^2\,x^2\,\sqrt {a+\frac {b}{x}}}{4}-\frac {\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{4} \]

[In]

int(x*(a + b/x)^(5/2),x)

[Out]

(9*a*x^2*(a + b/x)^(3/2))/4 - (a^(1/2)*b^2*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*15i)/4 - 2*b^2*(a + b/x)^(1/2) -
(7*a^2*x^2*(a + b/x)^(1/2))/4