Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{21 x^{21} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )} \] Output:
-1/21*a^5*((b*x^2+a)^2)^(1/2)/x^21/(b*x^2+a)-5/19*a^4*b*((b*x^2+a)^2)^(1/2 )/x^19/(b*x^2+a)-10/17*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^17/(b*x^2+a)-2/3*a^2* b^3*((b*x^2+a)^2)^(1/2)/x^15/(b*x^2+a)-5/13*a*b^4*((b*x^2+a)^2)^(1/2)/x^13 /(b*x^2+a)-1/11*b^5*((b*x^2+a)^2)^(1/2)/x^11/(b*x^2+a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (46189 a^5+255255 a^4 b x^2+570570 a^3 b^2 x^4+646646 a^2 b^3 x^6+373065 a b^4 x^8+88179 b^5 x^{10}\right )}{969969 x^{21} \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^22,x]
Output:
-1/969969*(Sqrt[(a + b*x^2)^2]*(46189*a^5 + 255255*a^4*b*x^2 + 570570*a^3* b^2*x^4 + 646646*a^2*b^3*x^6 + 373065*a*b^4*x^8 + 88179*b^5*x^10))/(x^21*( 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{22}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{22}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{22}}+\frac {5 b a^4}{x^{20}}+\frac {10 b^2 a^3}{x^{18}}+\frac {10 b^3 a^2}{x^{16}}+\frac {5 b^4 a}{x^{14}}+\frac {b^5}{x^{12}}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{21 x^{21}}-\frac {5 a^4 b}{19 x^{19}}-\frac {10 a^3 b^2}{17 x^{17}}-\frac {2 a^2 b^3}{3 x^{15}}-\frac {5 a b^4}{13 x^{13}}-\frac {b^5}{11 x^{11}}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^22,x]
Output:
((-1/21*a^5/x^21 - (5*a^4*b)/(19*x^19) - (10*a^3*b^2)/(17*x^17) - (2*a^2*b ^3)/(3*x^15) - (5*a*b^4)/(13*x^13) - b^5/(11*x^11))*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 = 27.44 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{21} a^{5}-\frac {5}{19} x^{2} a^{4} b -\frac {10}{17} a^{3} x^{4} b^{2}-\frac {2}{3} a^{2} x^{6} b^{3}-\frac {5}{13} a \,x^{8} b^{4}-\frac {1}{11} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{21}}\) | \(79\) |
gosper | \(-\frac {\left (88179 x^{10} b^{5}+373065 a \,x^{8} b^{4}+646646 a^{2} x^{6} b^{3}+570570 a^{3} x^{4} b^{2}+255255 x^{2} a^{4} b +46189 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{969969 x^{21} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (88179 x^{10} b^{5}+373065 a \,x^{8} b^{4}+646646 a^{2} x^{6} b^{3}+570570 a^{3} x^{4} b^{2}+255255 x^{2} a^{4} b +46189 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{969969 x^{21} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (88179 x^{10} b^{5}+373065 a \,x^{8} b^{4}+646646 a^{2} x^{6} b^{3}+570570 a^{3} x^{4} b^{2}+255255 x^{2} a^{4} b +46189 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{969969 x^{21} \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^22,x,method=_RETURNVERBOSE)
Output:
((b*x^2+a)^2)^(1/2)/(b*x^2+a)*(-1/21*a^5-5/19*x^2*a^4*b-10/17*a^3*x^4*b^2- 2/3*a^2*x^6*b^3-5/13*a*x^8*b^4-1/11*x^10*b^5)/x^21
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=-\frac {88179 \, b^{5} x^{10} + 373065 \, a b^{4} x^{8} + 646646 \, a^{2} b^{3} x^{6} + 570570 \, a^{3} b^{2} x^{4} + 255255 \, a^{4} b x^{2} + 46189 \, a^{5}}{969969 \, x^{21}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^22,x, algorithm="fricas")
Output:
-1/969969*(88179*b^5*x^10 + 373065*a*b^4*x^8 + 646646*a^2*b^3*x^6 + 570570 *a^3*b^2*x^4 + 255255*a^4*b*x^2 + 46189*a^5)/x^21
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{22}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**22,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**22, x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=-\frac {b^{5}}{11 \, x^{11}} - \frac {5 \, a b^{4}}{13 \, x^{13}} - \frac {2 \, a^{2} b^{3}}{3 \, x^{15}} - \frac {10 \, a^{3} b^{2}}{17 \, x^{17}} - \frac {5 \, a^{4} b}{19 \, x^{19}} - \frac {a^{5}}{21 \, x^{21}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^22,x, algorithm="maxima")
Output:
-1/11*b^5/x^11 - 5/13*a*b^4/x^13 - 2/3*a^2*b^3/x^15 - 10/17*a^3*b^2/x^17 - 5/19*a^4*b/x^19 - 1/21*a^5/x^21
Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=-\frac {88179 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 373065 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 646646 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 570570 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 255255 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 46189 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{969969 \, x^{21}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^22,x, algorithm="giac")
Output:
-1/969969*(88179*b^5*x^10*sgn(b*x^2 + a) + 373065*a*b^4*x^8*sgn(b*x^2 + a) + 646646*a^2*b^3*x^6*sgn(b*x^2 + a) + 570570*a^3*b^2*x^4*sgn(b*x^2 + a) + 255255*a^4*b*x^2*sgn(b*x^2 + a) + 46189*a^5*sgn(b*x^2 + a))/x^21
Time = 17.35 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{21\,x^{21}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{19\,x^{19}\,\left (b\,x^2+a\right )}-\frac {2\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^{15}\,\left (b\,x^2+a\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{17\,x^{17}\,\left (b\,x^2+a\right )} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^22,x)
Output:
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(21*x^21*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(11*x^11*(a + b*x^2)) - (5*a*b^4*(a^2 + b^ 2*x^4 + 2*a*b*x^2)^(1/2))/(13*x^13*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(19*x^19*(a + b*x^2)) - (2*a^2*b^3*(a^2 + b^2*x^4 + 2* a*b*x^2)^(1/2))/(3*x^15*(a + b*x^2)) - (10*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b* x^2)^(1/2))/(17*x^17*(a + b*x^2))
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{22}} \, dx=\frac {-88179 b^{5} x^{10}-373065 a \,b^{4} x^{8}-646646 a^{2} b^{3} x^{6}-570570 a^{3} b^{2} x^{4}-255255 a^{4} b \,x^{2}-46189 a^{5}}{969969 x^{21}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^22,x)
Output:
( - 46189*a**5 - 255255*a**4*b*x**2 - 570570*a**3*b**2*x**4 - 646646*a**2* b**3*x**6 - 373065*a*b**4*x**8 - 88179*b**5*x**10)/(969969*x**21)