Integrand size = 15, antiderivative size = 38 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {2 a \left (a+\frac {b}{x}\right )^{7/2}}{7 b^2}-\frac {2 \left (a+\frac {b}{x}\right )^{9/2}}{9 b^2} \] Output:
2/7*a*(a+b/x)^(7/2)/b^2-2/9*(a+b/x)^(9/2)/b^2
Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {2 (b+a x)^3 \sqrt {\frac {b+a x}{x}} (-7 b+2 a x)}{63 b^2 x^4} \] Input:
Integrate[(a + b/x)^(5/2)/x^3,x]
Output:
(2*(b + a*x)^3*Sqrt[(b + a*x)/x]*(-7*b + 2*a*x))/(63*b^2*x^4)
Time = 0.28 (sec) , antiderivative size = 38, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x}d\frac {1}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\int \left (\frac {\left (a+\frac {b}{x}\right )^{7/2}}{b}-\frac {a \left (a+\frac {b}{x}\right )^{5/2}}{b}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a \left (a+\frac {b}{x}\right )^{7/2}}{7 b^2}-\frac {2 \left (a+\frac {b}{x}\right )^{9/2}}{9 b^2}\) |
Input:
Int[(a + b/x)^(5/2)/x^3,x]
Output:
(2*a*(a + b/x)^(7/2))/(7*b^2) - (2*(a + b/x)^(9/2))/(9*b^2)
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 = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82
method | result | size |
orering | \(\frac {2 \left (2 a x -7 b \right ) \left (a x +b \right ) \left (a +\frac {b}{x}\right )^{\frac {5}{2}}}{63 b^{2} x^{2}}\) | \(31\) |
gosper | \(\frac {2 \left (a x +b \right ) \left (2 a x -7 b \right ) \left (\frac {a x +b}{x}\right )^{\frac {5}{2}}}{63 b^{2} x^{2}}\) | \(33\) |
risch | \(\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (2 a^{4} x^{4}-a^{3} b \,x^{3}-15 a^{2} b^{2} x^{2}-19 a \,b^{3} x -7 b^{4}\right )}{63 x^{4} b^{2}}\) | \(61\) |
trager | \(\frac {2 \left (2 a^{4} x^{4}-a^{3} b \,x^{3}-15 a^{2} b^{2} x^{2}-19 a \,b^{3} x -7 b^{4}\right ) \sqrt {-\frac {-a x -b}{x}}}{63 x^{4} b^{2}}\) | \(65\) |
default | \(\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \left (2 a^{3} x^{3}-3 a^{2} b \,x^{2}-12 a \,b^{2} x -7 b^{3}\right )}{63 x^{5} b^{2} \sqrt {x \left (a x +b \right )}}\) | \(70\) |
Input:
int((a+b/x)^(5/2)/x^3,x,method=_RETURNVERBOSE)
Output:
2/63*(2*a*x-7*b)/b^2/x^2*(a*x+b)*(a+b/x)^(5/2)
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.58 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {2 \, {\left (2 \, a^{4} x^{4} - a^{3} b x^{3} - 15 \, a^{2} b^{2} x^{2} - 19 \, a b^{3} x - 7 \, b^{4}\right )} \sqrt {\frac {a x + b}{x}}}{63 \, b^{2} x^{4}} \] Input:
integrate((a+b/x)^(5/2)/x^3,x, algorithm="fricas")
Output:
2/63*(2*a^4*x^4 - a^3*b*x^3 - 15*a^2*b^2*x^2 - 19*a*b^3*x - 7*b^4)*sqrt((a *x + b)/x)/(b^2*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (31) = 62\).
Time = 1.21 (sec) , antiderivative size = 416, normalized size of antiderivative = 10.95 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {4 a^{\frac {19}{2}} b^{\frac {3}{2}} x^{5} \sqrt {\frac {a x}{b} + 1}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} + \frac {2 a^{\frac {17}{2}} b^{\frac {5}{2}} x^{4} \sqrt {\frac {a x}{b} + 1}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} - \frac {32 a^{\frac {15}{2}} b^{\frac {7}{2}} x^{3} \sqrt {\frac {a x}{b} + 1}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} - \frac {68 a^{\frac {13}{2}} b^{\frac {9}{2}} x^{2} \sqrt {\frac {a x}{b} + 1}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} - \frac {52 a^{\frac {11}{2}} b^{\frac {11}{2}} x \sqrt {\frac {a x}{b} + 1}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} - \frac {14 a^{\frac {9}{2}} b^{\frac {13}{2}} \sqrt {\frac {a x}{b} + 1}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} - \frac {4 a^{10} b x^{\frac {11}{2}}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} - \frac {4 a^{9} b^{2} x^{\frac {9}{2}}}{63 a^{\frac {11}{2}} b^{3} x^{\frac {11}{2}} + 63 a^{\frac {9}{2}} b^{4} x^{\frac {9}{2}}} \] Input:
integrate((a+b/x)**(5/2)/x**3,x)
Output:
4*a**(19/2)*b**(3/2)*x**5*sqrt(a*x/b + 1)/(63*a**(11/2)*b**3*x**(11/2) + 6 3*a**(9/2)*b**4*x**(9/2)) + 2*a**(17/2)*b**(5/2)*x**4*sqrt(a*x/b + 1)/(63* a**(11/2)*b**3*x**(11/2) + 63*a**(9/2)*b**4*x**(9/2)) - 32*a**(15/2)*b**(7 /2)*x**3*sqrt(a*x/b + 1)/(63*a**(11/2)*b**3*x**(11/2) + 63*a**(9/2)*b**4*x **(9/2)) - 68*a**(13/2)*b**(9/2)*x**2*sqrt(a*x/b + 1)/(63*a**(11/2)*b**3*x **(11/2) + 63*a**(9/2)*b**4*x**(9/2)) - 52*a**(11/2)*b**(11/2)*x*sqrt(a*x/ b + 1)/(63*a**(11/2)*b**3*x**(11/2) + 63*a**(9/2)*b**4*x**(9/2)) - 14*a**( 9/2)*b**(13/2)*sqrt(a*x/b + 1)/(63*a**(11/2)*b**3*x**(11/2) + 63*a**(9/2)* b**4*x**(9/2)) - 4*a**10*b*x**(11/2)/(63*a**(11/2)*b**3*x**(11/2) + 63*a** (9/2)*b**4*x**(9/2)) - 4*a**9*b**2*x**(9/2)/(63*a**(11/2)*b**3*x**(11/2) + 63*a**(9/2)*b**4*x**(9/2))
Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=-\frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{2}}}{9 \, b^{2}} + \frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} a}{7 \, b^{2}} \] Input:
integrate((a+b/x)^(5/2)/x^3,x, algorithm="maxima")
Output:
-2/9*(a + b/x)^(9/2)/b^2 + 2/7*(a + b/x)^(7/2)*a/b^2
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (30) = 60\).
Time = 0.15 (sec) , antiderivative size = 239, normalized size of antiderivative = 6.29 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {2 \, {\left (63 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{7} a^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 273 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{6} a^{3} b \mathrm {sgn}\left (x\right ) + 567 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{5} a^{\frac {5}{2}} b^{2} \mathrm {sgn}\left (x\right ) + 693 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{4} a^{2} b^{3} \mathrm {sgn}\left (x\right ) + 525 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{3} a^{\frac {3}{2}} b^{4} \mathrm {sgn}\left (x\right ) + 243 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{2} a b^{5} \mathrm {sgn}\left (x\right ) + 63 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} b^{6} \mathrm {sgn}\left (x\right ) + 7 \, b^{7} \mathrm {sgn}\left (x\right )\right )}}{63 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{9}} \] Input:
integrate((a+b/x)^(5/2)/x^3,x, algorithm="giac")
Output:
2/63*(63*(sqrt(a)*x - sqrt(a*x^2 + b*x))^7*a^(7/2)*sgn(x) + 273*(sqrt(a)*x - sqrt(a*x^2 + b*x))^6*a^3*b*sgn(x) + 567*(sqrt(a)*x - sqrt(a*x^2 + b*x)) ^5*a^(5/2)*b^2*sgn(x) + 693*(sqrt(a)*x - sqrt(a*x^2 + b*x))^4*a^2*b^3*sgn( x) + 525*(sqrt(a)*x - sqrt(a*x^2 + b*x))^3*a^(3/2)*b^4*sgn(x) + 243*(sqrt( a)*x - sqrt(a*x^2 + b*x))^2*a*b^5*sgn(x) + 63*(sqrt(a)*x - sqrt(a*x^2 + b* x))*sqrt(a)*b^6*sgn(x) + 7*b^7*sgn(x))/(sqrt(a)*x - sqrt(a*x^2 + b*x))^9
Time = 1.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.32 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {4\,a^4\,\sqrt {a+\frac {b}{x}}}{63\,b^2}-\frac {10\,a^2\,\sqrt {a+\frac {b}{x}}}{21\,x^2}-\frac {2\,b^2\,\sqrt {a+\frac {b}{x}}}{9\,x^4}-\frac {2\,a^3\,\sqrt {a+\frac {b}{x}}}{63\,b\,x}-\frac {38\,a\,b\,\sqrt {a+\frac {b}{x}}}{63\,x^3} \] Input:
int((a + b/x)^(5/2)/x^3,x)
Output:
(4*a^4*(a + b/x)^(1/2))/(63*b^2) - (10*a^2*(a + b/x)^(1/2))/(21*x^2) - (2* b^2*(a + b/x)^(1/2))/(9*x^4) - (2*a^3*(a + b/x)^(1/2))/(63*b*x) - (38*a*b* (a + b/x)^(1/2))/(63*x^3)
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.61 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{x^3} \, dx=\frac {\frac {4 \sqrt {x}\, \sqrt {a x +b}\, a^{4} x^{4}}{63}-\frac {2 \sqrt {x}\, \sqrt {a x +b}\, a^{3} b \,x^{3}}{63}-\frac {10 \sqrt {x}\, \sqrt {a x +b}\, a^{2} b^{2} x^{2}}{21}-\frac {38 \sqrt {x}\, \sqrt {a x +b}\, a \,b^{3} x}{63}-\frac {2 \sqrt {x}\, \sqrt {a x +b}\, b^{4}}{9}-\frac {4 \sqrt {a}\, a^{4} x^{5}}{63}}{b^{2} x^{5}} \] Input:
int((a+b/x)^(5/2)/x^3,x)
Output:
(2*(2*sqrt(x)*sqrt(a*x + b)*a**4*x**4 - sqrt(x)*sqrt(a*x + b)*a**3*b*x**3 - 15*sqrt(x)*sqrt(a*x + b)*a**2*b**2*x**2 - 19*sqrt(x)*sqrt(a*x + b)*a*b** 3*x - 7*sqrt(x)*sqrt(a*x + b)*b**4 - 2*sqrt(a)*a**4*x**5))/(63*b**2*x**5)