\(\int \frac {1}{(a+\frac {b}{x})^{3/2} x^{11/2}} \, dx\) [246]

Optimal result

Integrand size = 17, antiderivative size = 130 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=\frac {2}{b \sqrt {a+\frac {b}{x}} x^{7/2}}-\frac {7 \sqrt {a+\frac {b}{x}}}{3 b^2 x^{5/2}}+\frac {35 a \sqrt {a+\frac {b}{x}}}{12 b^3 x^{3/2}}-\frac {35 a^2 \sqrt {a+\frac {b}{x}}}{8 b^4 \sqrt {x}}+\frac {35 a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{8 b^{9/2}} \] Output:

2/b/(a+b/x)^(1/2)/x^(7/2)-7/3*(a+b/x)^(1/2)/b^2/x^(5/2)+35/12*a*(a+b/x)^(1 
/2)/b^3/x^(3/2)-35/8*a^2*(a+b/x)^(1/2)/b^4/x^(1/2)+35/8*a^3*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 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=-\frac {2 \sqrt {1+\frac {b}{a x}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {9}{2},\frac {11}{2},-\frac {b}{a x}\right )}{9 a \sqrt {a+\frac {b}{x}} x^{9/2}} \] Input:

Integrate[1/((a + b/x)^(3/2)*x^(11/2)),x]
 

Output:

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[3/2, 9/2, 11/2, -(b/(a*x))])/(9*a* 
Sqrt[a + b/x]*x^(9/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {860, 252, 262, 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.

\(\displaystyle \int \frac {1}{x^{11/2} \left (a+\frac {b}{x}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 860

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

\(\displaystyle -2 \left (\frac {7 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \int \frac {1}{\sqrt {a+\frac {b}{x}} x^2}d\frac {1}{\sqrt {x}}}{6 b}\right )}{b}-\frac {1}{b x^{7/2} \sqrt {a+\frac {b}{x}}}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle -2 \left (\frac {7 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \int \frac {1}{\sqrt {a+\frac {b}{x}} x}d\frac {1}{\sqrt {x}}}{4 b}\right )}{6 b}\right )}{b}-\frac {1}{b x^{7/2} \sqrt {a+\frac {b}{x}}}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle -2 \left (\frac {7 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x}}}d\frac {1}{\sqrt {x}}}{2 b}\right )}{4 b}\right )}{6 b}\right )}{b}-\frac {1}{b x^{7/2} \sqrt {a+\frac {b}{x}}}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle -2 \left (\frac {7 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{1-\frac {b}{x}}d\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}}{2 b}\right )}{4 b}\right )}{6 b}\right )}{b}-\frac {1}{b x^{7/2} \sqrt {a+\frac {b}{x}}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle -2 \left (\frac {7 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{2 b^{3/2}}\right )}{4 b}\right )}{6 b}\right )}{b}-\frac {1}{b x^{7/2} \sqrt {a+\frac {b}{x}}}\right )\)

Input:

Int[1/((a + b/x)^(3/2)*x^(11/2)),x]
 

Output:

-2*(-(1/(b*Sqrt[a + b/x]*x^(7/2))) + (7*(Sqrt[a + b/x]/(6*b*x^(5/2)) - (5* 
a*(Sqrt[a + b/x]/(4*b*x^(3/2)) - (3*a*(Sqrt[a + b/x]/(2*b*Sqrt[x]) - (a*Ar 
cTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(2*b^(3/2))))/(4*b)))/(6*b)))/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 = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-105 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {a x +b}\, a^{3} x^{3}-14 b^{\frac {5}{2}} a x +35 a^{2} b^{\frac {3}{2}} x^{2}+105 a^{3} x^{3} \sqrt {b}+8 b^{\frac {7}{2}}\right )}{24 x^{\frac {5}{2}} \left (a x +b \right ) b^{\frac {9}{2}}}\)	\(89\)
risch	\(-\frac {\left (a x +b \right ) \left (57 a^{2} x^{2}-22 a b x +8 b^{2}\right )}{24 b^{4} x^{\frac {7}{2}} \sqrt {\frac {a x +b}{x}}}-\frac {a^{3} \left (\frac {32}{\sqrt {a x +b}}-\frac {70 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {a x +b}}{16 b^{4} \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}\)	\(101\)

Input:

int(1/(a+b/x)^(3/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*((a*x+b)/x)^(1/2)*(-105*arctanh((a*x+b)^(1/2)/b^(1/2))*(a*x+b)^(1/2) 
*a^3*x^3-14*b^(5/2)*a*x+35*a^2*b^(3/2)*x^2+105*a^3*x^3*b^(1/2)+8*b^(7/2))/ 
x^(5/2)/(a*x+b)/b^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.87 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=\left [\frac {105 \, {\left (a^{4} x^{4} + a^{3} b x^{3}\right )} \sqrt {b} \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (105 \, a^{3} b x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{48 \, {\left (a b^{5} x^{4} + b^{6} x^{3}\right )}}, -\frac {105 \, {\left (a^{4} x^{4} + a^{3} b x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) + {\left (105 \, a^{3} b x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{24 \, {\left (a b^{5} x^{4} + b^{6} x^{3}\right )}}\right ] \] Input:

integrate(1/(a+b/x)^(3/2)/x^(11/2),x, algorithm="fricas")
 

Output:

[1/48*(105*(a^4*x^4 + a^3*b*x^3)*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 + 35*a^2*b^2*x^2 - 14*a*b^3*x + 
 8*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a*b^5*x^4 + b^6*x^3), -1/24*(105*(a^4* 
x^4 + a^3*b*x^3)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/(a*x + 
 b)) + (105*a^3*b*x^3 + 35*a^2*b^2*x^2 - 14*a*b^3*x + 8*b^4)*sqrt(x)*sqrt( 
(a*x + b)/x))/(a*b^5*x^4 + b^6*x^3)]
 

Sympy [A] (verification not implemented)

Time = 179.62 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=- \frac {35 a^{\frac {5}{2}}}{8 b^{4} \sqrt {x} \sqrt {1 + \frac {b}{a x}}} - \frac {35 a^{\frac {3}{2}}}{24 b^{3} x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} + \frac {7 \sqrt {a}}{12 b^{2} x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} + \frac {35 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{8 b^{\frac {9}{2}}} - \frac {1}{3 \sqrt {a} b x^{\frac {7}{2}} \sqrt {1 + \frac {b}{a x}}} \] Input:

integrate(1/(a+b/x)**(3/2)/x**(11/2),x)
 

Output:

-35*a**(5/2)/(8*b**4*sqrt(x)*sqrt(1 + b/(a*x))) - 35*a**(3/2)/(24*b**3*x** 
(3/2)*sqrt(1 + b/(a*x))) + 7*sqrt(a)/(12*b**2*x**(5/2)*sqrt(1 + b/(a*x))) 
+ 35*a**3*asinh(sqrt(b)/(sqrt(a)*sqrt(x)))/(8*b**(9/2)) - 1/(3*sqrt(a)*b*x 
**(7/2)*sqrt(1 + b/(a*x)))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=-\frac {105 \, {\left (a + \frac {b}{x}\right )}^{3} a^{3} x^{3} - 280 \, {\left (a + \frac {b}{x}\right )}^{2} a^{3} b x^{2} + 231 \, {\left (a + \frac {b}{x}\right )} a^{3} b^{2} x - 48 \, a^{3} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} x^{\frac {7}{2}} - 3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{5} x^{\frac {5}{2}} + 3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{6} x^{\frac {3}{2}} - \sqrt {a + \frac {b}{x}} b^{7} \sqrt {x}\right )}} - \frac {35 \, a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{16 \, b^{\frac {9}{2}}} \] Input:

integrate(1/(a+b/x)^(3/2)/x^(11/2),x, algorithm="maxima")
 

Output:

-1/24*(105*(a + b/x)^3*a^3*x^3 - 280*(a + b/x)^2*a^3*b*x^2 + 231*(a + b/x) 
*a^3*b^2*x - 48*a^3*b^3)/((a + b/x)^(7/2)*b^4*x^(7/2) - 3*(a + b/x)^(5/2)* 
b^5*x^(5/2) + 3*(a + b/x)^(3/2)*b^6*x^(3/2) - sqrt(a + b/x)*b^7*sqrt(x)) - 
 35/16*a^3*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 = 107, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=-\frac {35 \, a^{3} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4} \mathrm {sgn}\left (x\right )} - \frac {2 \, a^{3}}{\sqrt {a x + b} b^{4} \mathrm {sgn}\left (x\right )} - \frac {57 \, {\left (a x + b\right )}^{\frac {5}{2}} a^{3} - 136 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{3} b + 87 \, \sqrt {a x + b} a^{3} b^{2}}{24 \, a^{3} b^{4} x^{3} \mathrm {sgn}\left (x\right )} \] Input:

integrate(1/(a+b/x)^(3/2)/x^(11/2),x, algorithm="giac")
 

Output:

-35/8*a^3*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^4*sgn(x)) - 2*a^3/(sq 
rt(a*x + b)*b^4*sgn(x)) - 1/24*(57*(a*x + b)^(5/2)*a^3 - 136*(a*x + b)^(3/ 
2)*a^3*b + 87*sqrt(a*x + b)*a^3*b^2)/(a^3*b^4*x^3*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=\int \frac {1}{x^{11/2}\,{\left (a+\frac {b}{x}\right )}^{3/2}} \,d x \] Input:

int(1/(x^(11/2)*(a + b/x)^(3/2)),x)
 

Output:

int(1/(x^(11/2)*(a + b/x)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{11/2}} \, dx=\frac {-105 \sqrt {b}\, \sqrt {a x +b}\, \mathrm {log}\left (\sqrt {a x +b}-\sqrt {b}\right ) a^{3} x^{3}+105 \sqrt {b}\, \sqrt {a x +b}\, \mathrm {log}\left (\sqrt {a x +b}+\sqrt {b}\right ) a^{3} x^{3}-210 a^{3} b \,x^{3}-70 a^{2} b^{2} x^{2}+28 a \,b^{3} x -16 b^{4}}{48 \sqrt {a x +b}\, b^{5} x^{3}} \] Input:

int(1/(a+b/x)^(3/2)/x^(11/2),x)
 

Output:

( - 105*sqrt(b)*sqrt(a*x + b)*log(sqrt(a*x + b) - sqrt(b))*a**3*x**3 + 105 
*sqrt(b)*sqrt(a*x + b)*log(sqrt(a*x + b) + sqrt(b))*a**3*x**3 - 210*a**3*b 
*x**3 - 70*a**2*b**2*x**2 + 28*a*b**3*x - 16*b**4)/(48*sqrt(a*x + b)*b**5* 
x**3)