Integrand size = 15, antiderivative size = 59 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=-\frac {2 a^2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b^3}+\frac {4 a \left (a+\frac {b}{x}\right )^{5/2}}{5 b^3}-\frac {2 \left (a+\frac {b}{x}\right )^{7/2}}{7 b^3} \]
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=-\frac {2 \sqrt {\frac {b+a x}{x}} \left (15 b^3+3 a b^2 x-4 a^2 b x^2+8 a^3 x^3\right )}{105 b^3 x^3} \]
Time = 0.19 (sec) , antiderivative size = 59, 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.
\(\displaystyle \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \frac {\sqrt {a+\frac {b}{x}}}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\int \left (\frac {\left (a+\frac {b}{x}\right )^{5/2}}{b^2}-\frac {2 a \left (a+\frac {b}{x}\right )^{3/2}}{b^2}+\frac {a^2 \sqrt {a+\frac {b}{x}}}{b^2}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b^3}-\frac {2 \left (a+\frac {b}{x}\right )^{7/2}}{7 b^3}+\frac {4 a \left (a+\frac {b}{x}\right )^{5/2}}{5 b^3}\) |
(-2*a^2*(a + b/x)^(3/2))/(3*b^3) + (4*a*(a + b/x)^(5/2))/(5*b^3) - (2*(a + b/x)^(7/2))/(7*b^3)
3.17.98.3.1 Defintions of rubi rules used
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 = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75
method | result | size |
gosper | \(-\frac {2 \left (a x +b \right ) \left (8 a^{2} x^{2}-12 a b x +15 b^{2}\right ) \sqrt {\frac {a x +b}{x}}}{105 b^{3} x^{3}}\) | \(44\) |
risch | \(-\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (8 a^{3} x^{3}-4 a^{2} b \,x^{2}+3 a \,b^{2} x +15 b^{3}\right )}{105 x^{3} b^{3}}\) | \(50\) |
trager | \(-\frac {2 \left (8 a^{3} x^{3}-4 a^{2} b \,x^{2}+3 a \,b^{2} x +15 b^{3}\right ) \sqrt {-\frac {-a x -b}{x}}}{105 x^{3} b^{3}}\) | \(54\) |
default | \(-\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \left (8 a^{2} x^{2}-12 a b x +15 b^{2}\right )}{105 x^{4} \sqrt {x \left (a x +b \right )}\, b^{3}}\) | \(59\) |
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=-\frac {2 \, {\left (8 \, a^{3} x^{3} - 4 \, a^{2} b x^{2} + 3 \, a b^{2} x + 15 \, b^{3}\right )} \sqrt {\frac {a x + b}{x}}}{105 \, b^{3} x^{3}} \]
Leaf count of result is larger than twice the leaf count of optimal. 899 vs. \(2 (49) = 98\).
Time = 1.06 (sec) , antiderivative size = 899, normalized size of antiderivative = 15.24 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=- \frac {16 a^{\frac {19}{2}} b^{\frac {9}{2}} x^{6} \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} - \frac {40 a^{\frac {17}{2}} b^{\frac {11}{2}} x^{5} \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} - \frac {30 a^{\frac {15}{2}} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} - \frac {40 a^{\frac {13}{2}} b^{\frac {15}{2}} x^{3} \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} - \frac {100 a^{\frac {11}{2}} b^{\frac {17}{2}} x^{2} \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} - \frac {96 a^{\frac {9}{2}} b^{\frac {19}{2}} x \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} - \frac {30 a^{\frac {7}{2}} b^{\frac {21}{2}} \sqrt {\frac {a x}{b} + 1}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} + \frac {16 a^{10} b^{4} x^{\frac {13}{2}}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} + \frac {48 a^{9} b^{5} x^{\frac {11}{2}}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} + \frac {48 a^{8} b^{6} x^{\frac {9}{2}}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} + \frac {16 a^{7} b^{7} x^{\frac {7}{2}}}{105 a^{\frac {13}{2}} b^{7} x^{\frac {13}{2}} + 315 a^{\frac {11}{2}} b^{8} x^{\frac {11}{2}} + 315 a^{\frac {9}{2}} b^{9} x^{\frac {9}{2}} + 105 a^{\frac {7}{2}} b^{10} x^{\frac {7}{2}}} \]
-16*a**(19/2)*b**(9/2)*x**6*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2) *b**10*x**(7/2)) - 40*a**(17/2)*b**(11/2)*x**5*sqrt(a*x/b + 1)/(105*a**(13 /2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**( 9/2) + 105*a**(7/2)*b**10*x**(7/2)) - 30*a**(15/2)*b**(13/2)*x**4*sqrt(a*x /b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315 *a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) - 40*a**(13/2)*b**( 15/2)*x**3*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b **8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) - 100*a**(11/2)*b**(17/2)*x**2*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/ 2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7 /2)*b**10*x**(7/2)) - 96*a**(9/2)*b**(19/2)*x*sqrt(a*x/b + 1)/(105*a**(13/ 2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9 /2) + 105*a**(7/2)*b**10*x**(7/2)) - 30*a**(7/2)*b**(21/2)*sqrt(a*x/b + 1) /(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/ 2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) + 16*a**10*b**4*x**(13/2)/ (105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2 )*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) + 48*a**9*b**5*x**(11/2)/(1 05*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)* b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) + 48*a**8*b**6*x**(9/2)/(1...
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=-\frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}}}{7 \, b^{3}} + \frac {4 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a}{5 \, b^{3}} - \frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2}}{3 \, b^{3}} \]
Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (47) = 94\).
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.47 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=\frac {2 \, {\left (140 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{4} a^{2} \mathrm {sgn}\left (x\right ) + 315 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{3} a^{\frac {3}{2}} b \mathrm {sgn}\left (x\right ) + 273 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{2} a b^{2} \mathrm {sgn}\left (x\right ) + 105 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} b^{3} \mathrm {sgn}\left (x\right ) + 15 \, b^{4} \mathrm {sgn}\left (x\right )\right )}}{105 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{7}} \]
2/105*(140*(sqrt(a)*x - sqrt(a*x^2 + b*x))^4*a^2*sgn(x) + 315*(sqrt(a)*x - sqrt(a*x^2 + b*x))^3*a^(3/2)*b*sgn(x) + 273*(sqrt(a)*x - sqrt(a*x^2 + b*x ))^2*a*b^2*sgn(x) + 105*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*b^3*sgn(x) + 15*b^4*sgn(x))/(sqrt(a)*x - sqrt(a*x^2 + b*x))^7
Time = 6.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^4} \, dx=\frac {8\,a^2\,\sqrt {a+\frac {b}{x}}}{105\,b^2\,x}-\frac {16\,a^3\,\sqrt {a+\frac {b}{x}}}{105\,b^3}-\frac {2\,a\,\sqrt {a+\frac {b}{x}}}{35\,b\,x^2}-\frac {2\,\sqrt {a+\frac {b}{x}}}{7\,x^3} \]