\(\int \frac {x^{3/2}}{(a+\frac {b}{x})^3} \, dx\) [134]

Optimal result

Integrand size = 15, antiderivative size = 112 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=\frac {12 b^2 \sqrt {x}}{a^5}-\frac {2 b x^{3/2}}{a^4}+\frac {2 x^{5/2}}{5 a^3}-\frac {b^4 \sqrt {x}}{2 a^5 (b+a x)^2}+\frac {17 b^3 \sqrt {x}}{4 a^5 (b+a x)}-\frac {63 b^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{11/2}} \] Output:

12*b^2*x^(1/2)/a^5-2*b*x^(3/2)/a^4+2/5*x^(5/2)/a^3-1/2*b^4*x^(1/2)/a^5/(a* 
x+b)^2+17/4*b^3*x^(1/2)/a^5/(a*x+b)-63/4*b^(5/2)*arctan(a^(1/2)*x^(1/2)/b^ 
(1/2))/a^(11/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.82 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=\frac {\sqrt {x} \left (315 b^4+525 a b^3 x+168 a^2 b^2 x^2-24 a^3 b x^3+8 a^4 x^4\right )}{20 a^5 (b+a x)^2}-\frac {63 b^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{11/2}} \] Input:

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

Output:

(Sqrt[x]*(315*b^4 + 525*a*b^3*x + 168*a^2*b^2*x^2 - 24*a^3*b*x^3 + 8*a^4*x 
^4))/(20*a^5*(b + a*x)^2) - (63*b^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]) 
/(4*a^(11/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {795, 51, 51, 60, 60, 60, 73, 218}

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.

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

\(\Big \downarrow \) 795

\(\displaystyle \int \frac {x^{9/2}}{(a x+b)^3}dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {9 \int \frac {x^{7/2}}{(b+a x)^2}dx}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {9 \left (\frac {7 \int \frac {x^{5/2}}{b+a x}dx}{2 a}-\frac {x^{7/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {2 x^{5/2}}{5 a}-\frac {b \int \frac {x^{3/2}}{b+a x}dx}{a}\right )}{2 a}-\frac {x^{7/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \int \frac {\sqrt {x}}{b+a x}dx}{a}\right )}{a}\right )}{2 a}-\frac {x^{7/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {b \int \frac {1}{\sqrt {x} (b+a x)}dx}{a}\right )}{a}\right )}{a}\right )}{2 a}-\frac {x^{7/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {2 b \int \frac {1}{b+a x}d\sqrt {x}}{a}\right )}{a}\right )}{a}\right )}{2 a}-\frac {x^{7/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {9 \left (\frac {7 \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}}\right )}{a}\right )}{a}\right )}{2 a}-\frac {x^{7/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{9/2}}{2 a (a x+b)^2}\)

Input:

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

Output:

-1/2*x^(9/2)/(a*(b + a*x)^2) + (9*(-(x^(7/2)/(a*(b + a*x))) + (7*((2*x^(5/ 
2))/(5*a) - (b*((2*x^(3/2))/(3*a) - (b*((2*Sqrt[x])/a - (2*Sqrt[b]*ArcTan[ 
(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(3/2)))/a))/a))/(2*a)))/(4*a)
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 795

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* 
(b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {2 \left (a^{2} x^{2}-5 a b x +30 b^{2}\right ) \sqrt {x}}{5 a^{5}}-\frac {b^{3} \left (\frac {-\frac {17 a \,x^{\frac {3}{2}}}{4}-\frac {15 b \sqrt {x}}{4}}{\left (a x +b \right )^{2}}+\frac {63 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{a^{5}}\)	\(77\)
derivativedivides	\(\frac {\frac {2 a^{2} x^{\frac {5}{2}}}{5}-2 a b \,x^{\frac {3}{2}}+12 b^{2} \sqrt {x}}{a^{5}}-\frac {2 b^{3} \left (\frac {-\frac {17 a \,x^{\frac {3}{2}}}{8}-\frac {15 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {63 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}\)	\(79\)
default	\(\frac {\frac {2 a^{2} x^{\frac {5}{2}}}{5}-2 a b \,x^{\frac {3}{2}}+12 b^{2} \sqrt {x}}{a^{5}}-\frac {2 b^{3} \left (\frac {-\frac {17 a \,x^{\frac {3}{2}}}{8}-\frac {15 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {63 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}\)	\(79\)

Input:

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

Output:

2/5*(a^2*x^2-5*a*b*x+30*b^2)*x^(1/2)/a^5-b^3/a^5*(2*(-17/8*a*x^(3/2)-15/8* 
b*x^(1/2))/(a*x+b)^2+63/4/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.27 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=\left [\frac {315 \, {\left (a^{2} b^{2} x^{2} + 2 \, a b^{3} x + b^{4}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (8 \, a^{4} x^{4} - 24 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} + 525 \, a b^{3} x + 315 \, b^{4}\right )} \sqrt {x}}{40 \, {\left (a^{7} x^{2} + 2 \, a^{6} b x + a^{5} b^{2}\right )}}, -\frac {315 \, {\left (a^{2} b^{2} x^{2} + 2 \, a b^{3} x + b^{4}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - {\left (8 \, a^{4} x^{4} - 24 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} + 525 \, a b^{3} x + 315 \, b^{4}\right )} \sqrt {x}}{20 \, {\left (a^{7} x^{2} + 2 \, a^{6} b x + a^{5} b^{2}\right )}}\right ] \] Input:

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

Output:

[1/40*(315*(a^2*b^2*x^2 + 2*a*b^3*x + b^4)*sqrt(-b/a)*log((a*x - 2*a*sqrt( 
x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(8*a^4*x^4 - 24*a^3*b*x^3 + 168*a^2*b^2* 
x^2 + 525*a*b^3*x + 315*b^4)*sqrt(x))/(a^7*x^2 + 2*a^6*b*x + a^5*b^2), -1/ 
20*(315*(a^2*b^2*x^2 + 2*a*b^3*x + b^4)*sqrt(b/a)*arctan(a*sqrt(x)*sqrt(b/ 
a)/b) - (8*a^4*x^4 - 24*a^3*b*x^3 + 168*a^2*b^2*x^2 + 525*a*b^3*x + 315*b^ 
4)*sqrt(x))/(a^7*x^2 + 2*a^6*b*x + a^5*b^2)]
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 835 vs. \(2 (107) = 214\).

Time = 27.00 (sec) , antiderivative size = 835, normalized size of antiderivative = 7.46 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*x**(11/2), Eq(a, 0) & Eq(b, 0)), (2*x**(11/2)/(11*b**3), Eq 
(a, 0)), (2*x**(5/2)/(5*a**3), Eq(b, 0)), (16*a**5*x**(9/2)*sqrt(-b/a)/(40 
*a**8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)) 
- 48*a**4*b*x**(7/2)*sqrt(-b/a)/(40*a**8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqr 
t(-b/a) + 40*a**6*b**2*sqrt(-b/a)) + 336*a**3*b**2*x**(5/2)*sqrt(-b/a)/(40 
*a**8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)) 
+ 1050*a**2*b**3*x**(3/2)*sqrt(-b/a)/(40*a**8*x**2*sqrt(-b/a) + 80*a**7*b* 
x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)) - 315*a**2*b**3*x**2*log(sqrt(x) - 
 sqrt(-b/a))/(40*a**8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) + 40*a**6*b 
**2*sqrt(-b/a)) + 315*a**2*b**3*x**2*log(sqrt(x) + sqrt(-b/a))/(40*a**8*x* 
*2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)) + 630*a* 
b**4*sqrt(x)*sqrt(-b/a)/(40*a**8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) 
+ 40*a**6*b**2*sqrt(-b/a)) - 630*a*b**4*x*log(sqrt(x) - sqrt(-b/a))/(40*a* 
*8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)) + 6 
30*a*b**4*x*log(sqrt(x) + sqrt(-b/a))/(40*a**8*x**2*sqrt(-b/a) + 80*a**7*b 
*x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)) - 315*b**5*log(sqrt(x) - sqrt(-b/ 
a))/(40*a**8*x**2*sqrt(-b/a) + 80*a**7*b*x*sqrt(-b/a) + 40*a**6*b**2*sqrt( 
-b/a)) + 315*b**5*log(sqrt(x) + sqrt(-b/a))/(40*a**8*x**2*sqrt(-b/a) + 80* 
a**7*b*x*sqrt(-b/a) + 40*a**6*b**2*sqrt(-b/a)), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=\frac {8 \, a^{4} - \frac {24 \, a^{3} b}{x} + \frac {168 \, a^{2} b^{2}}{x^{2}} + \frac {525 \, a b^{3}}{x^{3}} + \frac {315 \, b^{4}}{x^{4}}}{20 \, {\left (\frac {a^{7}}{x^{\frac {5}{2}}} + \frac {2 \, a^{6} b}{x^{\frac {7}{2}}} + \frac {a^{5} b^{2}}{x^{\frac {9}{2}}}\right )}} + \frac {63 \, b^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a^{5}} \] Input:

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

Output:

1/20*(8*a^4 - 24*a^3*b/x + 168*a^2*b^2/x^2 + 525*a*b^3/x^3 + 315*b^4/x^4)/ 
(a^7/x^(5/2) + 2*a^6*b/x^(7/2) + a^5*b^2/x^(9/2)) + 63/4*b^3*arctan(b/(sqr 
t(a*b)*sqrt(x)))/(sqrt(a*b)*a^5)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=-\frac {63 \, b^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} + \frac {17 \, a b^{3} x^{\frac {3}{2}} + 15 \, b^{4} \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a^{5}} + \frac {2 \, {\left (a^{12} x^{\frac {5}{2}} - 5 \, a^{11} b x^{\frac {3}{2}} + 30 \, a^{10} b^{2} \sqrt {x}\right )}}{5 \, a^{15}} \] Input:

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

Output:

-63/4*b^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/4*(17*a*b^3*x^(3 
/2) + 15*b^4*sqrt(x))/((a*x + b)^2*a^5) + 2/5*(a^12*x^(5/2) - 5*a^11*b*x^( 
3/2) + 30*a^10*b^2*sqrt(x))/a^15
 

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=\frac {\frac {15\,b^4\,\sqrt {x}}{4}+\frac {17\,a\,b^3\,x^{3/2}}{4}}{a^7\,x^2+2\,a^6\,b\,x+a^5\,b^2}+\frac {2\,x^{5/2}}{5\,a^3}-\frac {2\,b\,x^{3/2}}{a^4}+\frac {12\,b^2\,\sqrt {x}}{a^5}-\frac {63\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{11/2}} \] Input:

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

Output:

((15*b^4*x^(1/2))/4 + (17*a*b^3*x^(3/2))/4)/(a^5*b^2 + a^7*x^2 + 2*a^6*b*x 
) + (2*x^(5/2))/(5*a^3) - (2*b*x^(3/2))/a^4 + (12*b^2*x^(1/2))/a^5 - (63*b 
^(5/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/(4*a^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.35 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^3} \, dx=\frac {-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}-630 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x -315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) b^{4}+8 \sqrt {x}\, a^{5} x^{4}-24 \sqrt {x}\, a^{4} b \,x^{3}+168 \sqrt {x}\, a^{3} b^{2} x^{2}+525 \sqrt {x}\, a^{2} b^{3} x +315 \sqrt {x}\, a \,b^{4}}{20 a^{6} \left (a^{2} x^{2}+2 a b x +b^{2}\right )} \] Input:

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

Output:

( - 315*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**2 
 - 630*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a*b**3*x - 315* 
sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*b**4 + 8*sqrt(x)*a**5* 
x**4 - 24*sqrt(x)*a**4*b*x**3 + 168*sqrt(x)*a**3*b**2*x**2 + 525*sqrt(x)*a 
**2*b**3*x + 315*sqrt(x)*a*b**4)/(20*a**6*(a**2*x**2 + 2*a*b*x + b**2))