3.1802 \(\int \frac {1}{(a+\frac {b}{x})^{5/2} x^{9/2}} \, dx\)

Optimal. Leaf size=99 \[ \frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{b^{7/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{b^3 \sqrt {x}}+\frac {10}{3 b^2 x^{3/2} \sqrt {a+\frac {b}{x}}}+\frac {2}{3 b x^{5/2} \left (a+\frac {b}{x}\right )^{3/2}} \]

[Out]

2/3/b/(a+b/x)^(3/2)/x^(5/2)+5*a*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2))/b^(7/2)+10/3/b^2/x^(3/2)/(a+b/x)^(1/2)-
5*(a+b/x)^(1/2)/b^3/x^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {337, 288, 321, 217, 206} \[ \frac {10}{3 b^2 x^{3/2} \sqrt {a+\frac {b}{x}}}-\frac {5 \sqrt {a+\frac {b}{x}}}{b^3 \sqrt {x}}+\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{b^{7/2}}+\frac {2}{3 b x^{5/2} \left (a+\frac {b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(5/2)*x^(9/2)),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{9/2}} \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {x^6}{\left (a+b x^2\right )^{5/2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}-\frac {10 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{3 b}\\ &=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}+\frac {10}{3 b^2 \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {10 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{b^2}\\ &=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}+\frac {10}{3 b^2 \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{b^3 \sqrt {x}}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{b^3}\\ &=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}+\frac {10}{3 b^2 \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{b^3 \sqrt {x}}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{b^3}\\ &=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}+\frac {10}{3 b^2 \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{b^3 \sqrt {x}}+\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.57 \[ -\frac {2 \sqrt {\frac {b}{a x}+1} \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};-\frac {b}{a x}\right )}{7 a^2 x^{7/2} \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[5/2, 7/2, 9/2, -(b/(a*x))])/(7*a^2*Sqrt[a + b/x]*x^(7/2))

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fricas [A]  time = 0.89, size = 249, normalized size = 2.52 \[ \left [\frac {15 \, {\left (a^{3} x^{3} + 2 \, a^{2} b x^{2} + a b^{2} x\right )} \sqrt {b} \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (15 \, a^{2} b x^{2} + 20 \, a b^{2} x + 3 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{6 \, {\left (a^{2} b^{4} x^{3} + 2 \, a b^{5} x^{2} + b^{6} x\right )}}, -\frac {15 \, {\left (a^{3} x^{3} + 2 \, a^{2} b x^{2} + a b^{2} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (15 \, a^{2} b x^{2} + 20 \, a b^{2} x + 3 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{2} b^{4} x^{3} + 2 \, a b^{5} x^{2} + b^{6} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(9/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 65, normalized size = 0.66 \[ -\frac {5 \, a \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} - \frac {2 \, {\left (6 \, {\left (a x + b\right )} a + a b\right )}}{3 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{3}} - \frac {\sqrt {a x + b}}{b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(9/2),x, algorithm="giac")

[Out]

-5*a*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^3) - 2/3*(6*(a*x + b)*a + a*b)/((a*x + b)^(3/2)*b^3) - sqrt(a*
x + b)/(b^3*x)

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maple [A]  time = 0.02, size = 102, normalized size = 1.03 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-15 \sqrt {a x +b}\, a^{2} x^{2} \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )+15 a^{2} \sqrt {b}\, x^{2}-15 \sqrt {a x +b}\, a b x \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )+20 a \,b^{\frac {3}{2}} x +3 b^{\frac {5}{2}}\right )}{3 \left (a x +b \right )^{2} b^{\frac {7}{2}} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(5/2)/x^(9/2),x)

[Out]

-1/3*((a*x+b)/x)^(1/2)*(-15*arctanh((a*x+b)^(1/2)/b^(1/2))*(a*x+b)^(1/2)*x^2*a^2+3*b^(5/2)+20*a*b^(3/2)*x+15*a
^2*b^(1/2)*x^2-15*arctanh((a*x+b)^(1/2)/b^(1/2))*x*a*b*(a*x+b)^(1/2))/x^(1/2)/(a*x+b)^2/b^(7/2)

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maxima [A]  time = 2.37, size = 119, normalized size = 1.20 \[ -\frac {15 \, {\left (a + \frac {b}{x}\right )}^{2} a x^{2} - 10 \, {\left (a + \frac {b}{x}\right )} a b x - 2 \, a b^{2}}{3 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} - {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{4} x^{\frac {3}{2}}\right )}} - \frac {5 \, a \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{2 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(9/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{9/2}\,{\left (a+\frac {b}{x}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(9/2)*(a + b/x)^(5/2)),x)

[Out]

int(1/(x^(9/2)*(a + b/x)^(5/2)), x)

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sympy [B]  time = 133.44, size = 818, normalized size = 8.26 \[ - \frac {15 a^{4} b^{13} x^{4} \log {\left (\frac {a x}{b} \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} + \frac {30 a^{4} b^{13} x^{4} \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {30 a^{3} b^{14} x^{3} \sqrt {\frac {a x}{b} + 1}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {45 a^{3} b^{14} x^{3} \log {\left (\frac {a x}{b} \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} + \frac {90 a^{3} b^{14} x^{3} \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {70 a^{2} b^{15} x^{2} \sqrt {\frac {a x}{b} + 1}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {45 a^{2} b^{15} x^{2} \log {\left (\frac {a x}{b} \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} + \frac {90 a^{2} b^{15} x^{2} \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {46 a b^{16} x \sqrt {\frac {a x}{b} + 1}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {15 a b^{16} x \log {\left (\frac {a x}{b} \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} + \frac {30 a b^{16} x \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} - \frac {6 b^{17} \sqrt {\frac {a x}{b} + 1}}{6 a^{3} b^{\frac {33}{2}} x^{4} + 18 a^{2} b^{\frac {35}{2}} x^{3} + 18 a b^{\frac {37}{2}} x^{2} + 6 b^{\frac {39}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(5/2)/x**(9/2),x)

[Out]

-15*a**4*b**13*x**4*log(a*x/b)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2)*x**2 + 6*b**(3
9/2)*x) + 30*a**4*b**13*x**4*log(sqrt(a*x/b + 1) + 1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b
**(37/2)*x**2 + 6*b**(39/2)*x) - 30*a**3*b**14*x**3*sqrt(a*x/b + 1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)
*x**3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x) - 45*a**3*b**14*x**3*log(a*x/b)/(6*a**3*b**(33/2)*x**4 + 18*a**2*
b**(35/2)*x**3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x) + 90*a**3*b**14*x**3*log(sqrt(a*x/b + 1) + 1)/(6*a**3*b*
*(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x) - 70*a**2*b**15*x**2*sqrt(a*x/b +
1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x) - 45*a**2*b**15*x**
2*log(a*x/b)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x) + 90*a**2*
b**15*x**2*log(sqrt(a*x/b + 1) + 1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2)*x**2 + 6*
b**(39/2)*x) - 46*a*b**16*x*sqrt(a*x/b + 1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2)*x
**2 + 6*b**(39/2)*x) - 15*a*b**16*x*log(a*x/b)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**3 + 18*a*b**(37/2
)*x**2 + 6*b**(39/2)*x) + 30*a*b**16*x*log(sqrt(a*x/b + 1) + 1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*x**
3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x) - 6*b**17*sqrt(a*x/b + 1)/(6*a**3*b**(33/2)*x**4 + 18*a**2*b**(35/2)*
x**3 + 18*a*b**(37/2)*x**2 + 6*b**(39/2)*x)

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