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

Optimal. Leaf size=79 \[ -\frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 b}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}} \]

[Out]

5/3*b/a^2/(a+b/x)^(3/2)+x/a/(a+b/x)^(3/2)-5*b*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)+5*b/a^3/(a+b/x)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 82, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {242, 51, 63, 208} \[ \frac {5 x \sqrt {a+\frac {b}{x}}}{a^3}-\frac {10 x}{3 a^2 \sqrt {a+\frac {b}{x}}}-\frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {2 x}{3 a \left (a+\frac {b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(-5/2),x]

[Out]

(-2*x)/(3*a*(a + b/x)^(3/2)) - (10*x)/(3*a^2*Sqrt[a + b/x]) + (5*Sqrt[a + b/x]*x)/a^3 - (5*b*ArcTanh[Sqrt[a +
b/x]/Sqrt[a]])/a^(7/2)

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 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 \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2 x}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{3 a}\\ &=-\frac {2 x}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {10 x}{3 a^2 \sqrt {a+\frac {b}{x}}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{a^2}\\ &=-\frac {2 x}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {10 x}{3 a^2 \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x}{a^3}+\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{2 a^3}\\ &=-\frac {2 x}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {10 x}{3 a^2 \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x}{a^3}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{a^3}\\ &=-\frac {2 x}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {10 x}{3 a^2 \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x}{a^3}-\frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 38, normalized size = 0.48 \[ \frac {2 b \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {a+\frac {b}{x}}{a}\right )}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(-5/2),x]

[Out]

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

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fricas [A]  time = 0.59, size = 225, normalized size = 2.85 \[ \left [\frac {15 \, {\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (3 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{6 \, {\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}, \frac {15 \, {\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (3 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a^2*b*x^2 + 2*a*b^2*x + b^3)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(3*a^3*x^3 +
20*a^2*b*x^2 + 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2), 1/3*(15*(a^2*b*x^2 + 2*a*b^2*x
+ b^3)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a^3*x^3 + 20*a^2*b*x^2 + 15*a*b^2*x)*sqrt((a*x + b)
/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2)]

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giac [A]  time = 0.24, size = 98, normalized size = 1.24 \[ \frac {1}{3} \, b {\left (\frac {2 \, {\left (a + \frac {6 \, {\left (a x + b\right )}}{x}\right )} x}{{\left (a x + b\right )} a^{3} \sqrt {\frac {a x + b}{x}}} + \frac {15 \, \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {3 \, \sqrt {\frac {a x + b}{x}}}{{\left (a - \frac {a x + b}{x}\right )} a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*(2*(a + 6*(a*x + b)/x)*x/((a*x + b)*a^3*sqrt((a*x + b)/x)) + 15*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt
(-a)*a^3) - 3*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a^3))

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maple [B]  time = 0.07, size = 271, normalized size = 3.43 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (-15 a^{3} b \,x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-45 a^{2} b^{2} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+30 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}} x^{3}-45 a \,b^{3} x \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+90 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} b \,x^{2}-15 b^{4} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+90 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{2} x -24 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {5}{2}} x +30 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b^{3}-20 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b \right ) x}{6 \sqrt {\left (a x +b \right ) x}\, \left (a x +b \right )^{3} a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((a*x+b)/x)^(1/2)*x*(30*a^(7/2)*((a*x+b)*x)^(1/2)*x^3-24*a^(5/2)*((a*x+b)*x)^(3/2)*x+90*a^(5/2)*((a*x+b)*x
)^(1/2)*x^2*b-15*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*a^3*b-20*b*a^(3/2)*((a*x+b)*x)^(3/2
)+90*a^(3/2)*((a*x+b)*x)^(1/2)*x*b^2-45*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^2*a^2*b^2-45*l
n(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x*a*b^3+30*a^(1/2)*((a*x+b)*x)^(1/2)*b^3-15*ln(1/2*(2*a*x
+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*b^4)/a^(7/2)/((a*x+b)*x)^(1/2)/(a*x+b)^3

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maxima [A]  time = 1.29, size = 101, normalized size = 1.28 \[ \frac {15 \, {\left (a + \frac {b}{x}\right )}^{2} b - 10 \, {\left (a + \frac {b}{x}\right )} a b - 2 \, a^{2} b}{3 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} - {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4}\right )}} + \frac {5 \, b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{2 \, a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(15*(a + b/x)^2*b - 10*(a + b/x)*a*b - 2*a^2*b)/((a + b/x)^(5/2)*a^3 - (a + b/x)^(3/2)*a^4) + 5/2*b*log((s
qrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2)

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mupad [B]  time = 1.72, size = 34, normalized size = 0.43 \[ \frac {2\,x\,{\left (\frac {a\,x}{b}+1\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {a\,x}{b}\right )}{7\,{\left (a+\frac {b}{x}\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/x)^(5/2),x)

[Out]

(2*x*((a*x)/b + 1)^(5/2)*hypergeom([5/2, 7/2], 9/2, -(a*x)/b))/(7*(a + b/x)^(5/2))

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sympy [B]  time = 7.92, size = 774, normalized size = 9.80 \[ \frac {6 a^{17} x^{4} \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {46 a^{16} b x^{3} \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {15 a^{16} b x^{3} \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {30 a^{16} b x^{3} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {70 a^{15} b^{2} x^{2} \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {45 a^{15} b^{2} x^{2} \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {90 a^{15} b^{2} x^{2} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {30 a^{14} b^{3} x \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {45 a^{14} b^{3} x \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {90 a^{14} b^{3} x \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {15 a^{13} b^{4} \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {30 a^{13} b^{4} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*a**17*x**4*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**
3) + 46*a**16*b*x**3*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(3
3/2)*b**3) + 15*a**16*b*x**3*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a*
*(33/2)*b**3) - 30*a**16*b*x**3*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35
/2)*b**2*x + 6*a**(33/2)*b**3) + 70*a**15*b**2*x**2*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2
+ 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**15*b**2*x**2*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b
*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**15*b**2*x**2*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x
**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 30*a**14*b**3*x*sqrt(1 + b/(a*x))/(6*a**
(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**14*b**3*x*log(b/(a*x))/(6*
a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**14*b**3*x*log(sqrt(1 +
b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 15*a**13*b**
4*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 30*a**13*b*
*4*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3
)

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