\(\int \frac {1}{(a+\frac {b}{x})^{3/2} x^{9/2}} \, dx\) [1793]

Optimal result

Integrand size = 17, antiderivative size = 104 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=\frac {2}{b \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{2 b^2 x^{3/2}}+\frac {15 a \sqrt {a+\frac {b}{x}}}{4 b^3 \sqrt {x}}-\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {344, 294, 327, 223, 212} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=-\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{4 b^{7/2}}+\frac {15 a \sqrt {a+\frac {b}{x}}}{4 b^3 \sqrt {x}}-\frac {5 \sqrt {a+\frac {b}{x}}}{2 b^2 x^{3/2}}+\frac {2}{b x^{5/2} \sqrt {a+\frac {b}{x}}} \]

[In]

Int[1/((a + b/x)^(3/2)*x^(9/2)),x]

[Out]

2/(b*Sqrt[a + b/x]*x^(5/2)) - (5*Sqrt[a + b/x])/(2*b^2*x^(3/2)) + (15*a*Sqrt[a + b/x])/(4*b^3*Sqrt[x]) - (15*a
^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(4*b^(7/2))

Rule 212

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

Rule 223

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 294

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 327

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 344

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*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {x^6}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = \frac {2}{b \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {10 \text {Subst}\left (\int \frac {x^4}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{b} \\ & = \frac {2}{b \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{2 b^2 x^{3/2}}+\frac {(15 a) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{2 b^2} \\ & = \frac {2}{b \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{2 b^2 x^{3/2}}+\frac {15 a \sqrt {a+\frac {b}{x}}}{4 b^3 \sqrt {x}}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{4 b^3} \\ & = \frac {2}{b \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{2 b^2 x^{3/2}}+\frac {15 a \sqrt {a+\frac {b}{x}}}{4 b^3 \sqrt {x}}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^3} \\ & = \frac {2}{b \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {5 \sqrt {a+\frac {b}{x}}}{2 b^2 x^{3/2}}+\frac {15 a \sqrt {a+\frac {b}{x}}}{4 b^3 \sqrt {x}}-\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{7/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=-\frac {2 \sqrt {1+\frac {b}{a x}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},-\frac {b}{a x}\right )}{7 a \sqrt {a+\frac {b}{x}} x^{7/2}} \]

[In]

Integrate[1/((a + b/x)^(3/2)*x^(9/2)),x]

[Out]

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[3/2, 7/2, 9/2, -(b/(a*x))])/(7*a*Sqrt[a + b/x]*x^(7/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.75

method	result	size
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (15 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {a x +b}\, a^{2} x^{2}-5 b^{\frac {3}{2}} a x -15 a^{2} x^{2} \sqrt {b}+2 b^{\frac {5}{2}}\right )}{4 x^{\frac {3}{2}} \left (a x +b \right ) b^{\frac {7}{2}}}\)	\(78\)
risch	\(\frac {\left (a x +b \right ) \left (7 a x -2 b \right )}{4 b^{3} x^{\frac {5}{2}} \sqrt {\frac {a x +b}{x}}}+\frac {a^{2} \left (\frac {16}{\sqrt {a x +b}}-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {a x +b}}{8 b^{3} \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}\)	\(90\)

[In]

int(1/(a+b/x)^(3/2)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.09 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=\left [\frac {15 \, {\left (a^{3} x^{3} + a^{2} b x^{2}\right )} \sqrt {b} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + 2 \, {\left (15 \, a^{2} b x^{2} + 5 \, a b^{2} x - 2 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{8 \, {\left (a b^{4} x^{3} + b^{5} x^{2}\right )}}, \frac {15 \, {\left (a^{3} x^{3} + a^{2} b x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (15 \, a^{2} b x^{2} + 5 \, a b^{2} x - 2 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{4 \, {\left (a b^{4} x^{3} + b^{5} x^{2}\right )}}\right ] \]

[In]

integrate(1/(a+b/x)^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

[1/8*(15*(a^3*x^3 + a^2*b*x^2)*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 + 5*a*b^2*x - 2*b^3)*sqrt(x)*sqrt((a*x + b)/x))/(a*b^4*x^3 + b^5*x^2), 1/4*(15*(a^3*x^3 + a^2*b*x^2)*sqrt(
-b)*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (15*a^2*b*x^2 + 5*a*b^2*x - 2*b^3)*sqrt(x)*sqrt((a*x + b)/x
))/(a*b^4*x^3 + b^5*x^2)]

Sympy [A] (verification not implemented)

Time = 45.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=\frac {15 a^{\frac {3}{2}}}{4 b^{3} \sqrt {x} \sqrt {1 + \frac {b}{a x}}} + \frac {5 \sqrt {a}}{4 b^{2} x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} - \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{4 b^{\frac {7}{2}}} - \frac {1}{2 \sqrt {a} b x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} \]

[In]

integrate(1/(a+b/x)**(3/2)/x**(9/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=\frac {15 \, {\left (a + \frac {b}{x}\right )}^{2} a^{2} x^{2} - 25 \, {\left (a + \frac {b}{x}\right )} a^{2} b x + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{4} x^{\frac {3}{2}} + \sqrt {a + \frac {b}{x}} b^{5} \sqrt {x}\right )}} + \frac {15 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

integrate(1/(a+b/x)^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

1/4*(15*(a + b/x)^2*a^2*x^2 - 25*(a + b/x)*a^2*b*x + 8*a^2*b^2)/((a + b/x)^(5/2)*b^3*x^(5/2) - 2*(a + b/x)^(3/
2)*b^4*x^(3/2) + sqrt(a + b/x)*b^5*sqrt(x)) + 15/8*a^2*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sq
rt(x) + sqrt(b)))/b^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=\frac {15 \, a^{2} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{3}} + \frac {2 \, a^{2}}{\sqrt {a x + b} b^{3}} + \frac {7 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{2} - 9 \, \sqrt {a x + b} a^{2} b}{4 \, a^{2} b^{3} x^{2}} \]

[In]

integrate(1/(a+b/x)^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{9/2}} \, dx=\int \frac {1}{x^{9/2}\,{\left (a+\frac {b}{x}\right )}^{3/2}} \,d x \]

[In]

int(1/(x^(9/2)*(a + b/x)^(3/2)),x)

[Out]

int(1/(x^(9/2)*(a + b/x)^(3/2)), x)