\(\int \frac {1}{(a+\frac {b}{x})^{5/2} x^{11/2}} \, dx\) [255]

Optimal result

Integrand size = 17, antiderivative size = 129 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{7/2}}+\frac {14}{3 b^2 \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {35 \sqrt {a+\frac {b}{x}}}{6 b^3 x^{3/2}}+\frac {35 a \sqrt {a+\frac {b}{x}}}{4 b^4 \sqrt {x}}-\frac {35 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{9/2}} \] Output:

2/3/b/(a+b/x)^(3/2)/x^(7/2)+14/3/b^2/(a+b/x)^(1/2)/x^(5/2)-35/6*(a+b/x)^(1 
/2)/b^3/x^(3/2)+35/4*a*(a+b/x)^(1/2)/b^4/x^(1/2)-35/4*a^2*arctanh(b^(1/2)/ 
(a+b/x)^(1/2)/x^(1/2))/b^(9/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=-\frac {2 \sqrt {1+\frac {b}{a x}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {9}{2},\frac {11}{2},-\frac {b}{a x}\right )}{9 a^2 \sqrt {a+\frac {b}{x}} x^{9/2}} \] Input:

Integrate[1/((a + b/x)^(5/2)*x^(11/2)),x]
 

Output:

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[5/2, 9/2, 11/2, -(b/(a*x))])/(9*a^ 
2*Sqrt[a + b/x]*x^(9/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {860, 252, 252, 262, 262, 224, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{x^{11/2} \left (a+\frac {b}{x}\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 860

\(\displaystyle -2 \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4}d\frac {1}{\sqrt {x}}\)

\(\Big \downarrow \) 252

\(\displaystyle -2 \left (\frac {7 \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^3}d\frac {1}{\sqrt {x}}}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\)

\(\Big \downarrow \) 252

\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \int \frac {1}{\sqrt {a+\frac {b}{x}} x^2}d\frac {1}{\sqrt {x}}}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \int \frac {1}{\sqrt {a+\frac {b}{x}} x}d\frac {1}{\sqrt {x}}}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x}}}d\frac {1}{\sqrt {x}}}{2 b}\right )}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{1-\frac {b}{x}}d\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}}{2 b}\right )}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{2 b^{3/2}}\right )}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\)

Input:

Int[1/((a + b/x)^(5/2)*x^(11/2)),x]
 

Output:

-2*(-1/3*1/(b*(a + b/x)^(3/2)*x^(7/2)) + (7*(-(1/(b*Sqrt[a + b/x]*x^(5/2)) 
) + (5*(Sqrt[a + b/x]/(4*b*x^(3/2)) - (3*a*(Sqrt[a + b/x]/(2*b*Sqrt[x]) - 
(a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(2*b^(3/2))))/(4*b)))/b))/(3* 
b))
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 860

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[-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]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\left (a x +b \right ) \left (11 a x -2 b \right )}{4 b^{4} x^{\frac {5}{2}} \sqrt {\frac {a x +b}{x}}}+\frac {a^{2} \left (-\frac {70 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )}{\sqrt {b}}+\frac {48}{\sqrt {a x +b}}+\frac {16 b}{3 \left (a x +b \right )^{\frac {3}{2}}}\right ) \sqrt {a x +b}}{8 b^{4} \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}\)	\(100\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (105 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {a x +b}\, a^{3} x^{3}+105 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{2} b \,x^{2} \sqrt {a x +b}-105 a^{3} x^{3} \sqrt {b}-140 a^{2} b^{\frac {3}{2}} x^{2}-21 b^{\frac {5}{2}} a x +6 b^{\frac {7}{2}}\right )}{12 x^{\frac {3}{2}} \left (a x +b \right )^{2} b^{\frac {9}{2}}}\)	\(117\)

Input:

int(1/(a+b/x)^(5/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

1/4*(a*x+b)*(11*a*x-2*b)/b^4/x^(5/2)/((a*x+b)/x)^(1/2)+1/8/b^4*a^2*(-70/b^ 
(1/2)*arctanh((a*x+b)^(1/2)/b^(1/2))+48/(a*x+b)^(1/2)+16/3*b/(a*x+b)^(3/2) 
)/((a*x+b)/x)^(1/2)/x^(1/2)*(a*x+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.22 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\left [\frac {105 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {b} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + 2 \, {\left (105 \, a^{3} b x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a b^{3} x - 6 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{24 \, {\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}}, \frac {105 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) + {\left (105 \, a^{3} b x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a b^{3} x - 6 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{12 \, {\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}}\right ] \] Input:

integrate(1/(a+b/x)^(5/2)/x^(11/2),x, algorithm="fricas")
 

Output:

[1/24*(105*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x^2)*sqrt(b)*log((a*x - 2*sqrt 
(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2*(105*a^3*b*x^3 + 140*a^2*b^2*x 
^2 + 21*a*b^3*x - 6*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a^2*b^5*x^4 + 2*a*b^6 
*x^3 + b^7*x^2), 1/12*(105*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x^2)*sqrt(-b)* 
arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/(a*x + b)) + (105*a^3*b*x^3 + 14 
0*a^2*b^2*x^2 + 21*a*b^3*x - 6*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a^2*b^5*x^ 
4 + 2*a*b^6*x^3 + b^7*x^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b/x)**(5/2)/x**(11/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {105 \, {\left (a + \frac {b}{x}\right )}^{3} a^{2} x^{3} - 175 \, {\left (a + \frac {b}{x}\right )}^{2} a^{2} b x^{2} + 56 \, {\left (a + \frac {b}{x}\right )} a^{2} b^{2} x + 8 \, a^{2} b^{3}}{12 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} x^{\frac {7}{2}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{5} x^{\frac {5}{2}} + {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{8 \, b^{\frac {9}{2}}} \] Input:

integrate(1/(a+b/x)^(5/2)/x^(11/2),x, algorithm="maxima")
 

Output:

1/12*(105*(a + b/x)^3*a^2*x^3 - 175*(a + b/x)^2*a^2*b*x^2 + 56*(a + b/x)*a 
^2*b^2*x + 8*a^2*b^3)/((a + b/x)^(7/2)*b^4*x^(7/2) - 2*(a + b/x)^(5/2)*b^5 
*x^(5/2) + (a + b/x)^(3/2)*b^6*x^(3/2)) + 35/8*a^2*log((sqrt(a + b/x)*sqrt 
(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b)))/b^(9/2)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {35 \, a^{2} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (9 \, {\left (a x + b\right )} a^{2} + a^{2} b\right )}}{3 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{4} \mathrm {sgn}\left (x\right )} + \frac {11 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{2} - 13 \, \sqrt {a x + b} a^{2} b}{4 \, a^{2} b^{4} x^{2} \mathrm {sgn}\left (x\right )} \] Input:

integrate(1/(a+b/x)^(5/2)/x^(11/2),x, algorithm="giac")
 

Output:

35/4*a^2*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^4*sgn(x)) + 2/3*(9*(a* 
x + b)*a^2 + a^2*b)/((a*x + b)^(3/2)*b^4*sgn(x)) + 1/4*(11*(a*x + b)^(3/2) 
*a^2 - 13*sqrt(a*x + b)*a^2*b)/(a^2*b^4*x^2*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\int \frac {1}{x^{11/2}\,{\left (a+\frac {b}{x}\right )}^{5/2}} \,d x \] Input:

int(1/(x^(11/2)*(a + b/x)^(5/2)),x)
 

Output:

int(1/(x^(11/2)*(a + b/x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {105 \sqrt {b}\, \sqrt {a x +b}\, \mathrm {log}\left (\sqrt {a x +b}-\sqrt {b}\right ) a^{3} x^{3}+105 \sqrt {b}\, \sqrt {a x +b}\, \mathrm {log}\left (\sqrt {a x +b}-\sqrt {b}\right ) a^{2} b \,x^{2}-105 \sqrt {b}\, \sqrt {a x +b}\, \mathrm {log}\left (\sqrt {a x +b}+\sqrt {b}\right ) a^{3} x^{3}-105 \sqrt {b}\, \sqrt {a x +b}\, \mathrm {log}\left (\sqrt {a x +b}+\sqrt {b}\right ) a^{2} b \,x^{2}+210 a^{3} b \,x^{3}+280 a^{2} b^{2} x^{2}+42 a \,b^{3} x -12 b^{4}}{24 \sqrt {a x +b}\, b^{5} x^{2} \left (a x +b \right )} \] Input:

int(1/(a+b/x)^(5/2)/x^(11/2),x)
 

Output:

(105*sqrt(b)*sqrt(a*x + b)*log(sqrt(a*x + b) - sqrt(b))*a**3*x**3 + 105*sq 
rt(b)*sqrt(a*x + b)*log(sqrt(a*x + b) - sqrt(b))*a**2*b*x**2 - 105*sqrt(b) 
*sqrt(a*x + b)*log(sqrt(a*x + b) + sqrt(b))*a**3*x**3 - 105*sqrt(b)*sqrt(a 
*x + b)*log(sqrt(a*x + b) + sqrt(b))*a**2*b*x**2 + 210*a**3*b*x**3 + 280*a 
**2*b**2*x**2 + 42*a*b**3*x - 12*b**4)/(24*sqrt(a*x + b)*b**5*x**2*(a*x + 
b))