Integrand size = 15, antiderivative size = 74 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=-\frac {2 a^3}{b^4 \sqrt {a+\frac {b}{x}}}-\frac {6 a^2 \sqrt {a+\frac {b}{x}}}{b^4}+\frac {2 a \left (a+\frac {b}{x}\right )^{3/2}}{b^4}-\frac {2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b^4} \] Output:
-2*a^3/b^4/(a+b/x)^(1/2)-6*a^2*(a+b/x)^(1/2)/b^4+2*a*(a+b/x)^(3/2)/b^4-2/5 *(a+b/x)^(5/2)/b^4
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=-\frac {2 \sqrt {\frac {b+a x}{x}} \left (b^3-2 a b^2 x+8 a^2 b x^2+16 a^3 x^3\right )}{5 b^4 x^2 (b+a x)} \] Input:
Integrate[1/((a + b/x)^(3/2)*x^5),x]
Output:
(-2*Sqrt[(b + a*x)/x]*(b^3 - 2*a*b^2*x + 8*a^2*b*x^2 + 16*a^3*x^3))/(5*b^4 *x^2*(b + a*x))
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {798, 53, 2009}
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.
\(\displaystyle \int \frac {1}{x^5 \left (a+\frac {b}{x}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\int \left (-\frac {a^3}{b^3 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {3 a^2}{b^3 \sqrt {a+\frac {b}{x}}}-\frac {3 \sqrt {a+\frac {b}{x}} a}{b^3}+\frac {\left (a+\frac {b}{x}\right )^{3/2}}{b^3}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^3}{b^4 \sqrt {a+\frac {b}{x}}}-\frac {6 a^2 \sqrt {a+\frac {b}{x}}}{b^4}+\frac {2 a \left (a+\frac {b}{x}\right )^{3/2}}{b^4}-\frac {2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b^4}\) |
Input:
Int[1/((a + b/x)^(3/2)*x^5),x]
Output:
(-2*a^3)/(b^4*Sqrt[a + b/x]) - (6*a^2*Sqrt[a + b/x])/b^4 + (2*a*(a + b/x)^ (3/2))/b^4 - (2*(a + b/x)^(5/2))/(5*b^4)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.69
method | result | size |
orering | \(-\frac {2 \left (16 a^{3} x^{3}+8 a^{2} b \,x^{2}-2 a \,b^{2} x +b^{3}\right ) \left (a x +b \right )}{5 b^{4} x^{4} \left (a +\frac {b}{x}\right )^{\frac {3}{2}}}\) | \(51\) |
gosper | \(-\frac {2 \left (a x +b \right ) \left (16 a^{3} x^{3}+8 a^{2} b \,x^{2}-2 a \,b^{2} x +b^{3}\right )}{5 x^{4} b^{4} \left (\frac {a x +b}{x}\right )^{\frac {3}{2}}}\) | \(53\) |
trager | \(-\frac {2 \left (16 a^{3} x^{3}+8 a^{2} b \,x^{2}-2 a \,b^{2} x +b^{3}\right ) \sqrt {-\frac {-a x -b}{x}}}{5 x^{2} b^{4} \left (a x +b \right )}\) | \(59\) |
risch | \(-\frac {2 \left (a x +b \right ) \left (11 a^{2} x^{2}-3 a b x +b^{2}\right )}{5 b^{4} x^{3} \sqrt {\frac {a x +b}{x}}}-\frac {2 a^{3}}{b^{4} \sqrt {\frac {a x +b}{x}}}\) | \(62\) |
default | \(\frac {\sqrt {\frac {a x +b}{x}}\, \left (10 \sqrt {x \left (a x +b \right )}\, a^{\frac {11}{2}} x^{6}+10 \sqrt {a \,x^{2}+b x}\, a^{\frac {11}{2}} x^{6}+5 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{5} b \,x^{6}-5 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{5} b \,x^{6}+10 a^{\frac {9}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} x^{4}+20 a^{\frac {9}{2}} \sqrt {x \left (a x +b \right )}\, b \,x^{5}-30 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {9}{2}} x^{4}+20 a^{\frac {9}{2}} \sqrt {a \,x^{2}+b x}\, b \,x^{5}+10 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{4} b^{2} x^{5}-10 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{4} b^{2} x^{5}+10 a^{\frac {7}{2}} \sqrt {x \left (a x +b \right )}\, b^{2} x^{4}-52 a^{\frac {7}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b \,x^{3}+10 a^{\frac {7}{2}} \sqrt {a \,x^{2}+b x}\, b^{2} x^{4}+5 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{3} x^{4}-5 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{3} x^{4}-16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2} x^{2}+4 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{3} x -2 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{4}\right )}{5 x^{3} \sqrt {x \left (a x +b \right )}\, b^{5} \left (a x +b \right )^{2} \sqrt {a}}\) | \(497\) |
Input:
int(1/(a+b/x)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-2/5*(16*a^3*x^3+8*a^2*b*x^2-2*a*b^2*x+b^3)/b^4/x^4*(a*x+b)/(a+b/x)^(3/2)
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=-\frac {2 \, {\left (16 \, a^{3} x^{3} + 8 \, a^{2} b x^{2} - 2 \, a b^{2} x + b^{3}\right )} \sqrt {\frac {a x + b}{x}}}{5 \, {\left (a b^{4} x^{3} + b^{5} x^{2}\right )}} \] Input:
integrate(1/(a+b/x)^(3/2)/x^5,x, algorithm="fricas")
Output:
-2/5*(16*a^3*x^3 + 8*a^2*b*x^2 - 2*a*b^2*x + b^3)*sqrt((a*x + b)/x)/(a*b^4 *x^3 + b^5*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 2032 vs. \(2 (63) = 126\).
Time = 2.04 (sec) , antiderivative size = 2032, normalized size of antiderivative = 27.46 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b/x)**(3/2)/x**5,x)
Output:
-32*a**(21/2)*b**(23/2)*x**8*sqrt(a*x/b + 1)/(5*a**(17/2)*b**15*x**(17/2) + 30*a**(15/2)*b**16*x**(15/2) + 75*a**(13/2)*b**17*x**(13/2) + 100*a**(11 /2)*b**18*x**(11/2) + 75*a**(9/2)*b**19*x**(9/2) + 30*a**(7/2)*b**20*x**(7 /2) + 5*a**(5/2)*b**21*x**(5/2)) - 176*a**(19/2)*b**(25/2)*x**7*sqrt(a*x/b + 1)/(5*a**(17/2)*b**15*x**(17/2) + 30*a**(15/2)*b**16*x**(15/2) + 75*a** (13/2)*b**17*x**(13/2) + 100*a**(11/2)*b**18*x**(11/2) + 75*a**(9/2)*b**19 *x**(9/2) + 30*a**(7/2)*b**20*x**(7/2) + 5*a**(5/2)*b**21*x**(5/2)) - 396* a**(17/2)*b**(27/2)*x**6*sqrt(a*x/b + 1)/(5*a**(17/2)*b**15*x**(17/2) + 30 *a**(15/2)*b**16*x**(15/2) + 75*a**(13/2)*b**17*x**(13/2) + 100*a**(11/2)* b**18*x**(11/2) + 75*a**(9/2)*b**19*x**(9/2) + 30*a**(7/2)*b**20*x**(7/2) + 5*a**(5/2)*b**21*x**(5/2)) - 462*a**(15/2)*b**(29/2)*x**5*sqrt(a*x/b + 1 )/(5*a**(17/2)*b**15*x**(17/2) + 30*a**(15/2)*b**16*x**(15/2) + 75*a**(13/ 2)*b**17*x**(13/2) + 100*a**(11/2)*b**18*x**(11/2) + 75*a**(9/2)*b**19*x** (9/2) + 30*a**(7/2)*b**20*x**(7/2) + 5*a**(5/2)*b**21*x**(5/2)) - 290*a**( 13/2)*b**(31/2)*x**4*sqrt(a*x/b + 1)/(5*a**(17/2)*b**15*x**(17/2) + 30*a** (15/2)*b**16*x**(15/2) + 75*a**(13/2)*b**17*x**(13/2) + 100*a**(11/2)*b**1 8*x**(11/2) + 75*a**(9/2)*b**19*x**(9/2) + 30*a**(7/2)*b**20*x**(7/2) + 5* a**(5/2)*b**21*x**(5/2)) - 92*a**(11/2)*b**(33/2)*x**3*sqrt(a*x/b + 1)/(5* a**(17/2)*b**15*x**(17/2) + 30*a**(15/2)*b**16*x**(15/2) + 75*a**(13/2)*b* *17*x**(13/2) + 100*a**(11/2)*b**18*x**(11/2) + 75*a**(9/2)*b**19*x**(9...
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=-\frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}}}{5 \, b^{4}} + \frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a}{b^{4}} - \frac {6 \, \sqrt {a + \frac {b}{x}} a^{2}}{b^{4}} - \frac {2 \, a^{3}}{\sqrt {a + \frac {b}{x}} b^{4}} \] Input:
integrate(1/(a+b/x)^(3/2)/x^5,x, algorithm="maxima")
Output:
-2/5*(a + b/x)^(5/2)/b^4 + 2*(a + b/x)^(3/2)*a/b^4 - 6*sqrt(a + b/x)*a^2/b ^4 - 2*a^3/(sqrt(a + b/x)*b^4)
\[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} x^{5}} \,d x } \] Input:
integrate(1/(a+b/x)^(3/2)/x^5,x, algorithm="giac")
Output:
integrate(1/((a + b/x)^(3/2)*x^5), x)
Time = 0.62 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=-\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (16\,a^3\,x^3+8\,a^2\,b\,x^2-2\,a\,b^2\,x+b^3\right )}{5\,b^4\,x^2\,\left (b+a\,x\right )} \] Input:
int(1/(x^5*(a + b/x)^(3/2)),x)
Output:
-(2*(a + b/x)^(1/2)*(b^3 + 16*a^3*x^3 + 8*a^2*b*x^2 - 2*a*b^2*x))/(5*b^4*x ^2*(b + a*x))
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^5} \, dx=\frac {\frac {32 \sqrt {a}\, \sqrt {a x +b}\, a^{2} x^{3}}{5}-\frac {32 \sqrt {x}\, a^{3} x^{3}}{5}-\frac {16 \sqrt {x}\, a^{2} b \,x^{2}}{5}+\frac {4 \sqrt {x}\, a \,b^{2} x}{5}-\frac {2 \sqrt {x}\, b^{3}}{5}}{\sqrt {a x +b}\, b^{4} x^{3}} \] Input:
int(1/(a+b/x)^(3/2)/x^5,x)
Output:
(2*(16*sqrt(a)*sqrt(a*x + b)*a**2*x**3 - 16*sqrt(x)*a**3*x**3 - 8*sqrt(x)* a**2*b*x**2 + 2*sqrt(x)*a*b**2*x - sqrt(x)*b**3))/(5*sqrt(a*x + b)*b**4*x* *3)