Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 x^{23} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{21 x^{21} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )} \]
-1/23*a^5*((b*x^2+a)^2)^(1/2)/x^23/(b*x^2+a)-5/21*a^4*b*((b*x^2+a)^2)^(1/2 )/x^21/(b*x^2+a)-10/19*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^19/(b*x^2+a)-10/17*a^ 2*b^3*((b*x^2+a)^2)^(1/2)/x^17/(b*x^2+a)-1/3*a*b^4*((b*x^2+a)^2)^(1/2)/x^1 5/(b*x^2+a)-1/13*b^5*((b*x^2+a)^2)^(1/2)/x^13/(b*x^2+a)
Time = 1.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (88179 a^5+482885 a^4 b x^2+1067430 a^3 b^2 x^4+1193010 a^2 b^3 x^6+676039 a b^4 x^8+156009 b^5 x^{10}\right )}{2028117 x^{23} \left (a+b x^2\right )} \]
-1/2028117*(Sqrt[(a + b*x^2)^2]*(88179*a^5 + 482885*a^4*b*x^2 + 1067430*a^ 3*b^2*x^4 + 1193010*a^2*b^3*x^6 + 676039*a*b^4*x^8 + 156009*b^5*x^10))/(x^ 23*(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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{24}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{24}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{24}}+\frac {5 b a^4}{x^{22}}+\frac {10 b^2 a^3}{x^{20}}+\frac {10 b^3 a^2}{x^{18}}+\frac {5 b^4 a}{x^{16}}+\frac {b^5}{x^{14}}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{23 x^{23}}-\frac {5 a^4 b}{21 x^{21}}-\frac {10 a^3 b^2}{19 x^{19}}-\frac {10 a^2 b^3}{17 x^{17}}-\frac {a b^4}{3 x^{15}}-\frac {b^5}{13 x^{13}}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}\) |
((-1/23*a^5/x^23 - (5*a^4*b)/(21*x^21) - (10*a^3*b^2)/(19*x^19) - (10*a^2* b^3)/(17*x^17) - (a*b^4)/(3*x^15) - b^5/(13*x^13))*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b*x^2)
3.7.23.3.1 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 = 29.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{23} a^{5}-\frac {5}{21} x^{2} a^{4} b -\frac {10}{19} a^{3} x^{4} b^{2}-\frac {10}{17} a^{2} x^{6} b^{3}-\frac {1}{3} a \,x^{8} b^{4}-\frac {1}{13} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{23}}\) | \(79\) |
gosper | \(-\frac {\left (156009 x^{10} b^{5}+676039 a \,x^{8} b^{4}+1193010 a^{2} x^{6} b^{3}+1067430 a^{3} x^{4} b^{2}+482885 x^{2} a^{4} b +88179 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{2028117 x^{23} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (156009 x^{10} b^{5}+676039 a \,x^{8} b^{4}+1193010 a^{2} x^{6} b^{3}+1067430 a^{3} x^{4} b^{2}+482885 x^{2} a^{4} b +88179 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{2028117 x^{23} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
((b*x^2+a)^2)^(1/2)/(b*x^2+a)*(-1/23*a^5-5/21*x^2*a^4*b-10/19*a^3*x^4*b^2- 10/17*a^2*x^6*b^3-1/3*a*x^8*b^4-1/13*x^10*b^5)/x^23
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=-\frac {156009 \, b^{5} x^{10} + 676039 \, a b^{4} x^{8} + 1193010 \, a^{2} b^{3} x^{6} + 1067430 \, a^{3} b^{2} x^{4} + 482885 \, a^{4} b x^{2} + 88179 \, a^{5}}{2028117 \, x^{23}} \]
-1/2028117*(156009*b^5*x^10 + 676039*a*b^4*x^8 + 1193010*a^2*b^3*x^6 + 106 7430*a^3*b^2*x^4 + 482885*a^4*b*x^2 + 88179*a^5)/x^23
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{24}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=-\frac {b^{5}}{13 \, x^{13}} - \frac {a b^{4}}{3 \, x^{15}} - \frac {10 \, a^{2} b^{3}}{17 \, x^{17}} - \frac {10 \, a^{3} b^{2}}{19 \, x^{19}} - \frac {5 \, a^{4} b}{21 \, x^{21}} - \frac {a^{5}}{23 \, x^{23}} \]
-1/13*b^5/x^13 - 1/3*a*b^4/x^15 - 10/17*a^2*b^3/x^17 - 10/19*a^3*b^2/x^19 - 5/21*a^4*b/x^21 - 1/23*a^5/x^23
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=-\frac {156009 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 676039 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 1193010 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 1067430 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 482885 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 88179 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{2028117 \, x^{23}} \]
-1/2028117*(156009*b^5*x^10*sgn(b*x^2 + a) + 676039*a*b^4*x^8*sgn(b*x^2 + a) + 1193010*a^2*b^3*x^6*sgn(b*x^2 + a) + 1067430*a^3*b^2*x^4*sgn(b*x^2 + a) + 482885*a^4*b*x^2*sgn(b*x^2 + a) + 88179*a^5*sgn(b*x^2 + a))/x^23
Time = 13.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{24}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{23\,x^{23}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )}-\frac {a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^{15}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{21\,x^{21}\,\left (b\,x^2+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{17\,x^{17}\,\left (b\,x^2+a\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{19\,x^{19}\,\left (b\,x^2+a\right )} \]
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(23*x^23*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(13*x^13*(a + b*x^2)) - (a*b^4*(a^2 + b^2* x^4 + 2*a*b*x^2)^(1/2))/(3*x^15*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2 *a*b*x^2)^(1/2))/(21*x^21*(a + b*x^2)) - (10*a^2*b^3*(a^2 + b^2*x^4 + 2*a* b*x^2)^(1/2))/(17*x^17*(a + b*x^2)) - (10*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x ^2)^(1/2))/(19*x^19*(a + b*x^2))