Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )} \]
-1/17*a^5*((b*x^2+a)^2)^(1/2)/x^17/(b*x^2+a)-1/3*a^4*b*((b*x^2+a)^2)^(1/2) /x^15/(b*x^2+a)-10/13*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^13/(b*x^2+a)-10/11*a^2 *b^3*((b*x^2+a)^2)^(1/2)/x^11/(b*x^2+a)-5/9*a*b^4*((b*x^2+a)^2)^(1/2)/x^9/ (b*x^2+a)-1/7*b^5*((b*x^2+a)^2)^(1/2)/x^7/(b*x^2+a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (9009 a^5+51051 a^4 b x^2+117810 a^3 b^2 x^4+139230 a^2 b^3 x^6+85085 a b^4 x^8+21879 b^5 x^{10}\right )}{153153 x^{17} \left (a+b x^2\right )} \]
-1/153153*(Sqrt[(a + b*x^2)^2]*(9009*a^5 + 51051*a^4*b*x^2 + 117810*a^3*b^ 2*x^4 + 139230*a^2*b^3*x^6 + 85085*a*b^4*x^8 + 21879*b^5*x^10))/(x^17*(a + b*x^2))
Time = 0.24 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{18}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{18}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{18}}+\frac {5 b a^4}{x^{16}}+\frac {10 b^2 a^3}{x^{14}}+\frac {10 b^3 a^2}{x^{12}}+\frac {5 b^4 a}{x^{10}}+\frac {b^5}{x^8}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{17 x^{17}}-\frac {a^4 b}{3 x^{15}}-\frac {10 a^3 b^2}{13 x^{13}}-\frac {10 a^2 b^3}{11 x^{11}}-\frac {5 a b^4}{9 x^9}-\frac {b^5}{7 x^7}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}\) |
((-1/17*a^5/x^17 - (a^4*b)/(3*x^15) - (10*a^3*b^2)/(13*x^13) - (10*a^2*b^3 )/(11*x^11) - (5*a*b^4)/(9*x^9) - b^5/(7*x^7))*Sqrt[a^2 + 2*a*b*x^2 + b^2* x^4])/(a + b*x^2)
3.7.20.3.1 Defintions of rubi rules used
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 = 13.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{17} a^{5}-\frac {1}{3} x^{2} a^{4} b -\frac {10}{13} a^{3} x^{4} b^{2}-\frac {10}{11} a^{2} x^{6} b^{3}-\frac {5}{9} a \,x^{8} b^{4}-\frac {1}{7} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{17}}\) | \(79\) |
gosper | \(-\frac {\left (21879 x^{10} b^{5}+85085 a \,x^{8} b^{4}+139230 a^{2} x^{6} b^{3}+117810 a^{3} x^{4} b^{2}+51051 x^{2} a^{4} b +9009 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{153153 x^{17} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (21879 x^{10} b^{5}+85085 a \,x^{8} b^{4}+139230 a^{2} x^{6} b^{3}+117810 a^{3} x^{4} b^{2}+51051 x^{2} a^{4} b +9009 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{153153 x^{17} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
((b*x^2+a)^2)^(1/2)/(b*x^2+a)*(-1/17*a^5-1/3*x^2*a^4*b-10/13*a^3*x^4*b^2-1 0/11*a^2*x^6*b^3-5/9*a*x^8*b^4-1/7*x^10*b^5)/x^17
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=-\frac {21879 \, b^{5} x^{10} + 85085 \, a b^{4} x^{8} + 139230 \, a^{2} b^{3} x^{6} + 117810 \, a^{3} b^{2} x^{4} + 51051 \, a^{4} b x^{2} + 9009 \, a^{5}}{153153 \, x^{17}} \]
-1/153153*(21879*b^5*x^10 + 85085*a*b^4*x^8 + 139230*a^2*b^3*x^6 + 117810* a^3*b^2*x^4 + 51051*a^4*b*x^2 + 9009*a^5)/x^17
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{18}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=-\frac {b^{5}}{7 \, x^{7}} - \frac {5 \, a b^{4}}{9 \, x^{9}} - \frac {10 \, a^{2} b^{3}}{11 \, x^{11}} - \frac {10 \, a^{3} b^{2}}{13 \, x^{13}} - \frac {a^{4} b}{3 \, x^{15}} - \frac {a^{5}}{17 \, x^{17}} \]
-1/7*b^5/x^7 - 5/9*a*b^4/x^9 - 10/11*a^2*b^3/x^11 - 10/13*a^3*b^2/x^13 - 1 /3*a^4*b/x^15 - 1/17*a^5/x^17
Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=-\frac {21879 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 85085 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 139230 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 117810 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 51051 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 9009 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{153153 \, x^{17}} \]
-1/153153*(21879*b^5*x^10*sgn(b*x^2 + a) + 85085*a*b^4*x^8*sgn(b*x^2 + a) + 139230*a^2*b^3*x^6*sgn(b*x^2 + a) + 117810*a^3*b^2*x^4*sgn(b*x^2 + a) + 51051*a^4*b*x^2*sgn(b*x^2 + a) + 9009*a^5*sgn(b*x^2 + a))/x^17
Time = 13.35 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{18}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{17\,x^{17}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^{15}\,\left (b\,x^2+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )} \]
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(17*x^17*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(7*x^7*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2* x^4 + 2*a*b*x^2)^(1/2))/(9*x^9*(a + b*x^2)) - (a^4*b*(a^2 + b^2*x^4 + 2*a* b*x^2)^(1/2))/(3*x^15*(a + b*x^2)) - (10*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^ 2)^(1/2))/(11*x^11*(a + b*x^2)) - (10*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^ (1/2))/(13*x^13*(a + b*x^2))