Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 x^{15} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )} \] Output:
-1/15*a^5*((b*x^2+a)^2)^(1/2)/x^15/(b*x^2+a)-5/13*a^4*b*((b*x^2+a)^2)^(1/2 )/x^13/(b*x^2+a)-10/11*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^11/(b*x^2+a)-10/9*a^2 *b^3*((b*x^2+a)^2)^(1/2)/x^9/(b*x^2+a)-5/7*a*b^4*((b*x^2+a)^2)^(1/2)/x^7/( b*x^2+a)-1/5*b^5*((b*x^2+a)^2)^(1/2)/x^5/(b*x^2+a)
Time = 1.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (3003 a^5+17325 a^4 b x^2+40950 a^3 b^2 x^4+50050 a^2 b^3 x^6+32175 a b^4 x^8+9009 b^5 x^{10}\right )}{45045 x^{15} \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^16,x]
Output:
-1/45045*(Sqrt[(a + b*x^2)^2]*(3003*a^5 + 17325*a^4*b*x^2 + 40950*a^3*b^2* x^4 + 50050*a^2*b^3*x^6 + 32175*a*b^4*x^8 + 9009*b^5*x^10))/(x^15*(a + b*x ^2))
Time = 0.40 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1384, 27, 244, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{16}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{16}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{16}}+\frac {5 b a^4}{x^{14}}+\frac {10 b^2 a^3}{x^{12}}+\frac {10 b^3 a^2}{x^{10}}+\frac {5 b^4 a}{x^8}+\frac {b^5}{x^6}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{15 x^{15}}-\frac {5 a^4 b}{13 x^{13}}-\frac {10 a^3 b^2}{11 x^{11}}-\frac {10 a^2 b^3}{9 x^9}-\frac {5 a b^4}{7 x^7}-\frac {b^5}{5 x^5}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^16,x]
Output:
((-1/15*a^5/x^15 - (5*a^4*b)/(13*x^13) - (10*a^3*b^2)/(11*x^11) - (10*a^2* b^3)/(9*x^9) - (5*a*b^4)/(7*x^7) - b^5/(5*x^5))*Sqrt[a^2 + 2*a*b*x^2 + b^2 *x^4])/(a + b*x^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 10.40 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{15} a^{5}-\frac {5}{13} x^{2} a^{4} b -\frac {10}{11} a^{3} x^{4} b^{2}-\frac {10}{9} a^{2} x^{6} b^{3}-\frac {5}{7} a \,x^{8} b^{4}-\frac {1}{5} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{15}}\) | \(79\) |
gosper | \(-\frac {\left (9009 x^{10} b^{5}+32175 a \,x^{8} b^{4}+50050 a^{2} x^{6} b^{3}+40950 a^{3} x^{4} b^{2}+17325 x^{2} a^{4} b +3003 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{45045 x^{15} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (9009 x^{10} b^{5}+32175 a \,x^{8} b^{4}+50050 a^{2} x^{6} b^{3}+40950 a^{3} x^{4} b^{2}+17325 x^{2} a^{4} b +3003 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{45045 x^{15} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (9009 x^{10} b^{5}+32175 a \,x^{8} b^{4}+50050 a^{2} x^{6} b^{3}+40950 a^{3} x^{4} b^{2}+17325 x^{2} a^{4} b +3003 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{45045 x^{15} \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^16,x,method=_RETURNVERBOSE)
Output:
((b*x^2+a)^2)^(1/2)/(b*x^2+a)*(-1/15*a^5-5/13*x^2*a^4*b-10/11*a^3*x^4*b^2- 10/9*a^2*x^6*b^3-5/7*a*x^8*b^4-1/5*x^10*b^5)/x^15
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=-\frac {9009 \, b^{5} x^{10} + 32175 \, a b^{4} x^{8} + 50050 \, a^{2} b^{3} x^{6} + 40950 \, a^{3} b^{2} x^{4} + 17325 \, a^{4} b x^{2} + 3003 \, a^{5}}{45045 \, x^{15}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^16,x, algorithm="fricas")
Output:
-1/45045*(9009*b^5*x^10 + 32175*a*b^4*x^8 + 50050*a^2*b^3*x^6 + 40950*a^3* b^2*x^4 + 17325*a^4*b*x^2 + 3003*a^5)/x^15
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{16}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**16,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**16, x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=-\frac {b^{5}}{5 \, x^{5}} - \frac {5 \, a b^{4}}{7 \, x^{7}} - \frac {10 \, a^{2} b^{3}}{9 \, x^{9}} - \frac {10 \, a^{3} b^{2}}{11 \, x^{11}} - \frac {5 \, a^{4} b}{13 \, x^{13}} - \frac {a^{5}}{15 \, x^{15}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^16,x, algorithm="maxima")
Output:
-1/5*b^5/x^5 - 5/7*a*b^4/x^7 - 10/9*a^2*b^3/x^9 - 10/11*a^3*b^2/x^11 - 5/1 3*a^4*b/x^13 - 1/15*a^5/x^15
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=-\frac {9009 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 32175 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 50050 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 40950 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 17325 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3003 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{45045 \, x^{15}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^16,x, algorithm="giac")
Output:
-1/45045*(9009*b^5*x^10*sgn(b*x^2 + a) + 32175*a*b^4*x^8*sgn(b*x^2 + a) + 50050*a^2*b^3*x^6*sgn(b*x^2 + a) + 40950*a^3*b^2*x^4*sgn(b*x^2 + a) + 1732 5*a^4*b*x^2*sgn(b*x^2 + a) + 3003*a^5*sgn(b*x^2 + a))/x^15
Time = 20.04 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{15\,x^{15}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{5\,x^5\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^16,x)
Output:
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(15*x^15*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(5*x^5*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2* x^4 + 2*a*b*x^2)^(1/2))/(7*x^7*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2* a*b*x^2)^(1/2))/(13*x^13*(a + b*x^2)) - (10*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b *x^2)^(1/2))/(9*x^9*(a + b*x^2)) - (10*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2) ^(1/2))/(11*x^11*(a + b*x^2))
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{16}} \, dx=\frac {-9009 b^{5} x^{10}-32175 a \,b^{4} x^{8}-50050 a^{2} b^{3} x^{6}-40950 a^{3} b^{2} x^{4}-17325 a^{4} b \,x^{2}-3003 a^{5}}{45045 x^{15}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^16,x)
Output:
( - 3003*a**5 - 17325*a**4*b*x**2 - 40950*a**3*b**2*x**4 - 50050*a**2*b**3 *x**6 - 32175*a*b**4*x**8 - 9009*b**5*x**10)/(45045*x**15)