Integrand size = 15, antiderivative size = 57 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{b^3}+\frac {2 a \left (a+\frac {b}{x^2}\right )^{3/2}}{3 b^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{5 b^3} \] Output:
-a^2*(a+b/x^2)^(1/2)/b^3+2/3*a*(a+b/x^2)^(3/2)/b^3-1/5*(a+b/x^2)^(5/2)/b^3
Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=-\frac {\sqrt {a+\frac {b}{x^2}} \left (3 b^2-4 a b x^2+8 a^2 x^4\right )}{15 b^3 x^4} \] Input:
Integrate[1/(Sqrt[a + b/x^2]*x^7),x]
Output:
-1/15*(Sqrt[a + b/x^2]*(3*b^2 - 4*a*b*x^2 + 8*a^2*x^4))/(b^3*x^4)
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {1}{x^7 \sqrt {a+\frac {b}{x^2}}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^4}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {a^2}{b^2 \sqrt {a+\frac {b}{x^2}}}-\frac {2 \sqrt {a+\frac {b}{x^2}} a}{b^2}+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{b^2}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 a^2 \sqrt {a+\frac {b}{x^2}}}{b^3}-\frac {2 \left (a+\frac {b}{x^2}\right )^{5/2}}{5 b^3}+\frac {4 a \left (a+\frac {b}{x^2}\right )^{3/2}}{3 b^3}\right )\) |
Input:
Int[1/(Sqrt[a + b/x^2]*x^7),x]
Output:
((-2*a^2*Sqrt[a + b/x^2])/b^3 + (4*a*(a + b/x^2)^(3/2))/(3*b^3) - (2*(a + b/x^2)^(5/2))/(5*b^3))/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 = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81
method | result | size |
orering | \(-\frac {\left (8 a^{2} x^{4}-4 a b \,x^{2}+3 b^{2}\right ) \left (a \,x^{2}+b \right )}{15 b^{3} x^{6} \sqrt {a +\frac {b}{x^{2}}}}\) | \(46\) |
trager | \(-\frac {\left (8 a^{2} x^{4}-4 a b \,x^{2}+3 b^{2}\right ) \sqrt {-\frac {-a \,x^{2}-b}{x^{2}}}}{15 x^{4} b^{3}}\) | \(47\) |
gosper | \(-\frac {\left (a \,x^{2}+b \right ) \left (8 a^{2} x^{4}-4 a b \,x^{2}+3 b^{2}\right )}{15 x^{6} b^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\) | \(50\) |
default | \(-\frac {\left (a \,x^{2}+b \right ) \left (8 a^{2} x^{4}-4 a b \,x^{2}+3 b^{2}\right )}{15 x^{6} b^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\) | \(50\) |
risch | \(-\frac {\left (a \,x^{2}+b \right ) \left (8 a^{2} x^{4}-4 a b \,x^{2}+3 b^{2}\right )}{15 x^{6} b^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\) | \(50\) |
Input:
int(1/(a+b/x^2)^(1/2)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/15*(8*a^2*x^4-4*a*b*x^2+3*b^2)/b^3*(a*x^2+b)/x^6/(a+b/x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=-\frac {{\left (8 \, a^{2} x^{4} - 4 \, a b x^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{15 \, b^{3} x^{4}} \] Input:
integrate(1/(a+b/x^2)^(1/2)/x^7,x, algorithm="fricas")
Output:
-1/15*(8*a^2*x^4 - 4*a*b*x^2 + 3*b^2)*sqrt((a*x^2 + b)/x^2)/(b^3*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 750 vs. \(2 (49) = 98\).
Time = 1.58 (sec) , antiderivative size = 750, normalized size of antiderivative = 13.16 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=- \frac {8 a^{\frac {15}{2}} b^{\frac {9}{2}} x^{10} \sqrt {\frac {a x^{2}}{b} + 1}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} - \frac {20 a^{\frac {13}{2}} b^{\frac {11}{2}} x^{8} \sqrt {\frac {a x^{2}}{b} + 1}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} - \frac {15 a^{\frac {11}{2}} b^{\frac {13}{2}} x^{6} \sqrt {\frac {a x^{2}}{b} + 1}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} - \frac {5 a^{\frac {9}{2}} b^{\frac {15}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} - \frac {5 a^{\frac {7}{2}} b^{\frac {17}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} - \frac {3 a^{\frac {5}{2}} b^{\frac {19}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} + \frac {8 a^{8} b^{4} x^{11}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} + \frac {24 a^{7} b^{5} x^{9}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} + \frac {24 a^{6} b^{6} x^{7}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} + \frac {8 a^{5} b^{7} x^{5}}{15 a^{\frac {11}{2}} b^{7} x^{11} + 45 a^{\frac {9}{2}} b^{8} x^{9} + 45 a^{\frac {7}{2}} b^{9} x^{7} + 15 a^{\frac {5}{2}} b^{10} x^{5}} \] Input:
integrate(1/(a+b/x**2)**(1/2)/x**7,x)
Output:
-8*a**(15/2)*b**(9/2)*x**10*sqrt(a*x**2/b + 1)/(15*a**(11/2)*b**7*x**11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) - 20*a**(13/2)*b**(11/2)*x**8*sqrt(a*x**2/b + 1)/(15*a**(11/2)*b**7*x**11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) - 15*a**(11/2)*b**(13/2)*x**6*sqrt(a*x**2/b + 1)/(15*a**(11/2)*b**7*x**11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) - 5*a**(9/2)*b**(15/2)*x**4*sqrt(a*x**2/b + 1)/(15*a**(11/2)*b**7*x**11 + 45 *a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) - 5* a**(7/2)*b**(17/2)*x**2*sqrt(a*x**2/b + 1)/(15*a**(11/2)*b**7*x**11 + 45*a **(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) - 3*a* *(5/2)*b**(19/2)*sqrt(a*x**2/b + 1)/(15*a**(11/2)*b**7*x**11 + 45*a**(9/2) *b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) + 8*a**8*b**4 *x**11/(15*a**(11/2)*b**7*x**11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9 *x**7 + 15*a**(5/2)*b**10*x**5) + 24*a**7*b**5*x**9/(15*a**(11/2)*b**7*x** 11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x** 5) + 24*a**6*b**6*x**7/(15*a**(11/2)*b**7*x**11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a**(5/2)*b**10*x**5) + 8*a**5*b**7*x**5/(15*a** (11/2)*b**7*x**11 + 45*a**(9/2)*b**8*x**9 + 45*a**(7/2)*b**9*x**7 + 15*a** (5/2)*b**10*x**5)
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=-\frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}}}{5 \, b^{3}} + \frac {2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a}{3 \, b^{3}} - \frac {\sqrt {a + \frac {b}{x^{2}}} a^{2}}{b^{3}} \] Input:
integrate(1/(a+b/x^2)^(1/2)/x^7,x, algorithm="maxima")
Output:
-1/5*(a + b/x^2)^(5/2)/b^3 + 2/3*(a + b/x^2)^(3/2)*a/b^3 - sqrt(a + b/x^2) *a^2/b^3
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=\frac {16 \, {\left (10 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{4} - 5 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} b + b^{2}\right )} a^{\frac {5}{2}}}{15 \, {\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b\right )}^{5} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a+b/x^2)^(1/2)/x^7,x, algorithm="giac")
Output:
16/15*(10*(sqrt(a)*x - sqrt(a*x^2 + b))^4 - 5*(sqrt(a)*x - sqrt(a*x^2 + b) )^2*b + b^2)*a^(5/2)/(((sqrt(a)*x - sqrt(a*x^2 + b))^2 - b)^5*sgn(x))
Time = 0.50 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=-\frac {3\,b^2\,\sqrt {a+\frac {b}{x^2}}+8\,a^2\,x^4\,\sqrt {a+\frac {b}{x^2}}-4\,a\,b\,x^2\,\sqrt {a+\frac {b}{x^2}}}{15\,b^3\,x^4} \] Input:
int(1/(x^7*(a + b/x^2)^(1/2)),x)
Output:
-(3*b^2*(a + b/x^2)^(1/2) + 8*a^2*x^4*(a + b/x^2)^(1/2) - 4*a*b*x^2*(a + b /x^2)^(1/2))/(15*b^3*x^4)
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^7} \, dx=\frac {-8 \sqrt {a \,x^{2}+b}\, a^{2} x^{4}+4 \sqrt {a \,x^{2}+b}\, a b \,x^{2}-3 \sqrt {a \,x^{2}+b}\, b^{2}+8 \sqrt {a}\, a^{2} x^{5}}{15 b^{3} x^{5}} \] Input:
int(1/(a+b/x^2)^(1/2)/x^7,x)
Output:
( - 8*sqrt(a*x**2 + b)*a**2*x**4 + 4*sqrt(a*x**2 + b)*a*b*x**2 - 3*sqrt(a* x**2 + b)*b**2 + 8*sqrt(a)*a**2*x**5)/(15*b**3*x**5)