Integrand size = 15, antiderivative size = 55 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\frac {2 a^2}{3 b^3 \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 a}{b^3 \sqrt {a+\frac {b}{x}}}-\frac {2 \sqrt {a+\frac {b}{x}}}{b^3} \] Output:
2/3*a^2/b^3/(a+b/x)^(3/2)-4*a/b^3/(a+b/x)^(1/2)-2*(a+b/x)^(1/2)/b^3
Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2 \sqrt {\frac {b+a x}{x}} \left (3 b^2+12 a b x+8 a^2 x^2\right )}{3 b^3 (b+a x)^2} \] Input:
Integrate[1/((a + b/x)^(5/2)*x^4),x]
Output:
(-2*Sqrt[(b + a*x)/x]*(3*b^2 + 12*a*b*x + 8*a^2*x^2))/(3*b^3*(b + a*x)^2)
Time = 0.32 (sec) , antiderivative size = 55, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a+\frac {b}{x}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\int \left (\frac {a^2}{b^2 \left (a+\frac {b}{x}\right )^{5/2}}-\frac {2 a}{b^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {1}{b^2 \sqrt {a+\frac {b}{x}}}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^2}{3 b^3 \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 a}{b^3 \sqrt {a+\frac {b}{x}}}-\frac {2 \sqrt {a+\frac {b}{x}}}{b^3}\) |
Input:
Int[1/((a + b/x)^(5/2)*x^4),x]
Output:
(2*a^2)/(3*b^3*(a + b/x)^(3/2)) - (4*a)/(b^3*Sqrt[a + b/x]) - (2*Sqrt[a + b/x])/b^3
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 = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76
method | result | size |
orering | \(-\frac {2 \left (8 a^{2} x^{2}+12 a b x +3 b^{2}\right ) \left (a x +b \right )}{3 b^{3} x^{3} \left (a +\frac {b}{x}\right )^{\frac {5}{2}}}\) | \(42\) |
gosper | \(-\frac {2 \left (a x +b \right ) \left (8 a^{2} x^{2}+12 a b x +3 b^{2}\right )}{3 x^{3} b^{3} \left (\frac {a x +b}{x}\right )^{\frac {5}{2}}}\) | \(44\) |
trager | \(-\frac {2 \left (8 a^{2} x^{2}+12 a b x +3 b^{2}\right ) \sqrt {-\frac {-a x -b}{x}}}{3 b^{3} \left (a x +b \right )^{2}}\) | \(47\) |
risch | \(-\frac {2 \left (a x +b \right )}{b^{3} x \sqrt {\frac {a x +b}{x}}}-\frac {2 a \left (5 a x +6 b \right )}{3 \left (a x +b \right ) b^{3} \sqrt {\frac {a x +b}{x}}}\) | \(58\) |
default | \(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (6 \sqrt {a \,x^{2}+b x}\, a^{\frac {9}{2}} x^{5}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{4} b \,x^{5}+6 a^{\frac {9}{2}} \sqrt {x \left (a x +b \right )}\, x^{5}-3 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{4} b \,x^{5}+12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{3}+18 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} b \,x^{4}+9 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{2} x^{4}-24 a^{\frac {7}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} x^{3}+18 a^{\frac {7}{2}} \sqrt {x \left (a x +b \right )}\, b \,x^{4}-9 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{2} x^{4}+36 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b \,x^{2}+18 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b^{2} x^{3}+9 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{3} x^{3}-28 a^{\frac {5}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} b \,x^{2}+18 a^{\frac {5}{2}} \sqrt {x \left (a x +b \right )}\, b^{2} x^{3}-9 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{3} x^{3}+36 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{2} x +6 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{3} x^{2}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4} x^{2}+6 a^{\frac {3}{2}} \sqrt {x \left (a x +b \right )}\, b^{3} x^{2}-3 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4} x^{2}+12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{3}\right )}{6 x \sqrt {x \left (a x +b \right )}\, b^{4} \sqrt {a}\, \left (a x +b \right )^{3}}\) | \(607\) |
Input:
int(1/(a+b/x)^(5/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-2/3*(8*a^2*x^2+12*a*b*x+3*b^2)/b^3/x^3*(a*x+b)/(a+b/x)^(5/2)
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2 \, {\left (8 \, a^{2} x^{2} + 12 \, a b x + 3 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{2} b^{3} x^{2} + 2 \, a b^{4} x + b^{5}\right )}} \] Input:
integrate(1/(a+b/x)^(5/2)/x^4,x, algorithm="fricas")
Output:
-2/3*(8*a^2*x^2 + 12*a*b*x + 3*b^2)*sqrt((a*x + b)/x)/(a^2*b^3*x^2 + 2*a*b ^4*x + b^5)
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (46) = 92\).
Time = 0.74 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.47 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\begin {cases} - \frac {16 a^{2} x^{2}}{3 a b^{3} x^{2} \sqrt {a + \frac {b}{x}} + 3 b^{4} x \sqrt {a + \frac {b}{x}}} - \frac {24 a b x}{3 a b^{3} x^{2} \sqrt {a + \frac {b}{x}} + 3 b^{4} x \sqrt {a + \frac {b}{x}}} - \frac {6 b^{2}}{3 a b^{3} x^{2} \sqrt {a + \frac {b}{x}} + 3 b^{4} x \sqrt {a + \frac {b}{x}}} & \text {for}\: b \neq 0 \\- \frac {1}{3 a^{\frac {5}{2}} x^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a+b/x)**(5/2)/x**4,x)
Output:
Piecewise((-16*a**2*x**2/(3*a*b**3*x**2*sqrt(a + b/x) + 3*b**4*x*sqrt(a + b/x)) - 24*a*b*x/(3*a*b**3*x**2*sqrt(a + b/x) + 3*b**4*x*sqrt(a + b/x)) - 6*b**2/(3*a*b**3*x**2*sqrt(a + b/x) + 3*b**4*x*sqrt(a + b/x)), Ne(b, 0)), (-1/(3*a**(5/2)*x**3), True))
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2 \, \sqrt {a + \frac {b}{x}}}{b^{3}} - \frac {4 \, a}{\sqrt {a + \frac {b}{x}} b^{3}} + \frac {2 \, a^{2}}{3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{3}} \] Input:
integrate(1/(a+b/x)^(5/2)/x^4,x, algorithm="maxima")
Output:
-2*sqrt(a + b/x)/b^3 - 4*a/(sqrt(a + b/x)*b^3) + 2/3*a^2/((a + b/x)^(3/2)* b^3)
\[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:
integrate(1/(a+b/x)^(5/2)/x^4,x, algorithm="giac")
Output:
integrate(1/((a + b/x)^(5/2)*x^4), x)
Time = 0.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (8\,a^2\,x^2+12\,a\,b\,x+3\,b^2\right )}{3\,b^3\,{\left (b+a\,x\right )}^2} \] Input:
int(1/(x^4*(a + b/x)^(5/2)),x)
Output:
-(2*(a + b/x)^(1/2)*(3*b^2 + 8*a^2*x^2 + 12*a*b*x))/(3*b^3*(b + a*x)^2)
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\frac {\frac {16 \sqrt {a}\, \sqrt {a x +b}\, a \,x^{2}}{3}+\frac {16 \sqrt {a}\, \sqrt {a x +b}\, b x}{3}-\frac {16 \sqrt {x}\, a^{2} x^{2}}{3}-8 \sqrt {x}\, a b x -2 \sqrt {x}\, b^{2}}{\sqrt {a x +b}\, b^{3} x \left (a x +b \right )} \] Input:
int(1/(a+b/x)^(5/2)/x^4,x)
Output:
(2*(8*sqrt(a)*sqrt(a*x + b)*a*x**2 + 8*sqrt(a)*sqrt(a*x + b)*b*x - 8*sqrt( x)*a**2*x**2 - 12*sqrt(x)*a*b*x - 3*sqrt(x)*b**2))/(3*sqrt(a*x + b)*b**3*x *(a*x + b))