3.1745 \(\int \frac {1}{(a+\frac {b}{x})^{5/2} x} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2}{a^2 \sqrt {a+\frac {b}{x}}}-\frac {2}{3 a \left (a+\frac {b}{x}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac {2}{a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2}{3 a \left (a+\frac {b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(5/2)*x),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{\left (a+\frac {b}{x}\right )^{5/2} x} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {2}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2}{a^2 \sqrt {a+\frac {b}{x}}}-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{a^2}\\ &=-\frac {2}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2}{a^2 \sqrt {a+\frac {b}{x}}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{a^2 b}\\ &=-\frac {2}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2}{a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 36, normalized size = 0.60 \[ -\frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b}{a x}+1\right )}{3 a \left (a+\frac {b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(5/2)*x),x]

[Out]

(-2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + b/(a*x)])/(3*a*(a + b/x)^(3/2))

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fricas [B]  time = 1.04, size = 197, normalized size = 3.28 \[ \left [\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (4 \, a^{2} x^{2} + 3 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}, -\frac {2 \, {\left (3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (4 \, a^{2} x^{2} + 3 \, a b x\right )} \sqrt {\frac {a x + b}{x}}\right )}}{3 \, {\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 65, normalized size = 1.08 \[ -\frac {2 \, {\left (a + \frac {3 \, {\left (a x + b\right )}}{x}\right )} x}{3 \, {\left (a x + b\right )} a^{2} \sqrt {\frac {a x + b}{x}}} - \frac {2 \, \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 274, normalized size = 4.57 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (3 a^{3} b \,x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+9 a^{2} b^{2} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-6 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}} x^{3}+9 a \,b^{3} x \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-18 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} b \,x^{2}+3 b^{4} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-18 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{2} x +6 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {5}{2}} x -6 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b^{3}+4 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b \right ) x}{3 \sqrt {\left (a x +b \right ) x}\, \left (a x +b \right )^{3} a^{\frac {5}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(5/2)/x,x)

[Out]

1/3*((a*x+b)/x)^(1/2)*x/a^(5/2)*(3*a^3*b*x^3*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))-6*((a*x+b)*
x)^(1/2)*a^(7/2)*x^3+9*a^2*b^2*x^2*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))+6*((a*x+b)*x)^(3/2)*a
^(5/2)*x-18*((a*x+b)*x)^(1/2)*a^(5/2)*b*x^2+9*a*b^3*x*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))+4*
((a*x+b)*x)^(3/2)*a^(3/2)*b-18*((a*x+b)*x)^(1/2)*a^(3/2)*b^2*x+3*b^4*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/
2))/a^(1/2))-6*((a*x+b)*x)^(1/2)*a^(1/2)*b^3)/((a*x+b)*x)^(1/2)/(a*x+b)^3/b

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maxima [A]  time = 2.21, size = 62, normalized size = 1.03 \[ -\frac {\log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {2 \, {\left (4 \, a + \frac {3 \, b}{x}\right )}}{3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.58, size = 49, normalized size = 0.82 \[ \frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {2\,\left (a+\frac {b}{x}\right )}{a^2}+\frac {2}{3\,a}}{{\left (a+\frac {b}{x}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x)^(5/2)),x)

[Out]

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

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sympy [B]  time = 3.56, size = 700, normalized size = 11.67 \[ - \frac {8 a^{7} x^{3} \sqrt {1 + \frac {b}{a x}}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} - \frac {3 a^{7} x^{3} \log {\left (\frac {b}{a x} \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} + \frac {6 a^{7} x^{3} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} - \frac {14 a^{6} b x^{2} \sqrt {1 + \frac {b}{a x}}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} - \frac {9 a^{6} b x^{2} \log {\left (\frac {b}{a x} \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} + \frac {18 a^{6} b x^{2} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} - \frac {6 a^{5} b^{2} x \sqrt {1 + \frac {b}{a x}}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} - \frac {9 a^{5} b^{2} x \log {\left (\frac {b}{a x} \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} + \frac {18 a^{5} b^{2} x \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} - \frac {3 a^{4} b^{3} \log {\left (\frac {b}{a x} \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} + \frac {6 a^{4} b^{3} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{3 a^{\frac {19}{2}} x^{3} + 9 a^{\frac {17}{2}} b x^{2} + 9 a^{\frac {15}{2}} b^{2} x + 3 a^{\frac {13}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(5/2)/x,x)

[Out]

-8*a**7*x**3*sqrt(1 + b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3)
- 3*a**7*x**3*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) +
6*a**7*x**3*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/
2)*b**3) - 14*a**6*b*x**2*sqrt(1 + b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a*
*(13/2)*b**3) - 9*a**6*b*x**2*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**
(13/2)*b**3) + 18*a**6*b*x**2*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*
b**2*x + 3*a**(13/2)*b**3) - 6*a**5*b**2*x*sqrt(1 + b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15
/2)*b**2*x + 3*a**(13/2)*b**3) - 9*a**5*b**2*x*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/
2)*b**2*x + 3*a**(13/2)*b**3) + 18*a**5*b**2*x*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x*
*2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) - 3*a**4*b**3*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2
+ 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) + 6*a**4*b**3*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**(17
/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3)

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