Integrand size = 30, antiderivative size = 234 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=-\frac {c}{7 a^3 x^7}+\frac {3 b c-a d}{5 a^4 x^5}-\frac {6 b^2 c-3 a b d+a^2 e}{3 a^5 x^3}+\frac {10 b^3 c-6 a b^2 d+3 a^2 b e-a^3 f}{a^6 x}+\frac {b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^5 \left (a+b x^2\right )^2}+\frac {b \left (19 b^3 c-15 a b^2 d+11 a^2 b e-7 a^3 f\right ) x}{8 a^6 \left (a+b x^2\right )}+\frac {\sqrt {b} \left (99 b^3 c-63 a b^2 d+35 a^2 b e-15 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{13/2}} \]
-1/7*c/a^3/x^7+1/5*(-a*d+3*b*c)/a^4/x^5+1/3*(-a^2*e+3*a*b*d-6*b^2*c)/a^5/x ^3+(-a^3*f+3*a^2*b*e-6*a*b^2*d+10*b^3*c)/a^6/x+1/4*b*(-a^3*f+a^2*b*e-a*b^2 *d+b^3*c)*x/a^5/(b*x^2+a)^2+1/8*b*(-7*a^3*f+11*a^2*b*e-15*a*b^2*d+19*b^3*c )*x/a^6/(b*x^2+a)+1/8*(-15*a^3*f+35*a^2*b*e-63*a*b^2*d+99*b^3*c)*arctan(x* b^(1/2)/a^(1/2))*b^(1/2)/a^(13/2)
Time = 0.09 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=-\frac {c}{7 a^3 x^7}+\frac {3 b c-a d}{5 a^4 x^5}-\frac {6 b^2 c-3 a b d+a^2 e}{3 a^5 x^3}+\frac {10 b^3 c-6 a b^2 d+3 a^2 b e-a^3 f}{a^6 x}+\frac {b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^5 \left (a+b x^2\right )^2}+\frac {b \left (19 b^3 c-15 a b^2 d+11 a^2 b e-7 a^3 f\right ) x}{8 a^6 \left (a+b x^2\right )}+\frac {\sqrt {b} \left (99 b^3 c-63 a b^2 d+35 a^2 b e-15 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{13/2}} \]
-1/7*c/(a^3*x^7) + (3*b*c - a*d)/(5*a^4*x^5) - (6*b^2*c - 3*a*b*d + a^2*e) /(3*a^5*x^3) + (10*b^3*c - 6*a*b^2*d + 3*a^2*b*e - a^3*f)/(a^6*x) + (b*(b^ 3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/(4*a^5*(a + b*x^2)^2) + (b*(19*b^3*c - 15*a*b^2*d + 11*a^2*b*e - 7*a^3*f)*x)/(8*a^6*(a + b*x^2)) + (Sqrt[b]*(99* b^3*c - 63*a*b^2*d + 35*a^2*b*e - 15*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/( 8*a^(13/2))
Time = 1.06 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2336, 25, 2336, 25, 2333, 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.
| \(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {b x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^5 \left (a+b x^2\right )^2}-\frac {\int -\frac {\frac {3 b \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x^8}{a^4}-\frac {4 \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x^6}{a^3}+\frac {4 \left (e a^2-b d a+b^2 c\right ) x^4}{a^2}-4 \left (\frac {b c}{a}-d\right ) x^2+4 c}{x^8 \left (b x^2+a\right )^2}dx}{4 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\frac {3 b \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x^8}{a^4}-\frac {4 \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x^6}{a^3}+\frac {4 \left (e a^2-b d a+b^2 c\right ) x^4}{a^2}-4 \left (\frac {b c}{a}-d\right ) x^2+4 c}{x^8 \left (b x^2+a\right )^2}dx}{4 a}+\frac {b x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^5 \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {b x \left (-7 a^3 f+11 a^2 b e-15 a b^2 d+19 b^3 c\right )}{2 a^5 \left (a+b x^2\right )}-\frac {\int -\frac {\frac {b \left (-7 f a^3+11 b e a^2-15 b^2 d a+19 b^3 c\right ) x^8}{a^4}-\frac {8 \left (-f a^3+2 b e a^2-3 b^2 d a+4 b^3 c\right ) x^6}{a^3}+\frac {8 \left (e a^2-2 b d a+3 b^2 c\right ) x^4}{a^2}-8 \left (\frac {2 b c}{a}-d\right ) x^2+8 c}{x^8 \left (b x^2+a\right )}dx}{2 a}}{4 a}+\frac {b x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^5 \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {b \left (-7 f a^3+11 b e a^2-15 b^2 d a+19 b^3 c\right ) x^8}{a^4}-\frac {8 \left (-f a^3+2 b e a^2-3 b^2 d a+4 b^3 c\right ) x^6}{a^3}+\frac {8 \left (e a^2-2 b d a+3 b^2 c\right ) x^4}{a^2}-8 \left (\frac {2 b c}{a}-d\right ) x^2+8 c}{x^8 \left (b x^2+a\right )}dx}{2 a}+\frac {b x \left (-7 a^3 f+11 a^2 b e-15 a b^2 d+19 b^3 c\right )}{2 a^5 \left (a+b x^2\right )}}{4 a}+\frac {b x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^5 \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\frac {\int \left (\frac {8 c}{a x^8}-\frac {b \left (15 f a^3-35 b e a^2+63 b^2 d a-99 b^3 c\right )}{a^4 \left (b x^2+a\right )}+\frac {8 \left (f a^3-3 b e a^2+6 b^2 d a-10 b^3 c\right )}{a^4 x^2}+\frac {8 \left (e a^2-3 b d a+6 b^2 c\right )}{a^3 x^4}+\frac {8 (a d-3 b c)}{a^2 x^6}\right )dx}{2 a}+\frac {b x \left (-7 a^3 f+11 a^2 b e-15 a b^2 d+19 b^3 c\right )}{2 a^5 \left (a+b x^2\right )}}{4 a}+\frac {b x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^5 \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^5 \left (a+b x^2\right )^2}+\frac {\frac {b x \left (-7 a^3 f+11 a^2 b e-15 a b^2 d+19 b^3 c\right )}{2 a^5 \left (a+b x^2\right )}+\frac {\frac {8 (3 b c-a d)}{5 a^2 x^5}-\frac {8 \left (a^2 e-3 a b d+6 b^2 c\right )}{3 a^3 x^3}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-15 a^3 f+35 a^2 b e-63 a b^2 d+99 b^3 c\right )}{a^{9/2}}+\frac {8 \left (a^3 (-f)+3 a^2 b e-6 a b^2 d+10 b^3 c\right )}{a^4 x}-\frac {8 c}{7 a x^7}}{2 a}}{4 a}\) |
(b*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/(4*a^5*(a + b*x^2)^2) + ((b*(19* b^3*c - 15*a*b^2*d + 11*a^2*b*e - 7*a^3*f)*x)/(2*a^5*(a + b*x^2)) + ((-8*c )/(7*a*x^7) + (8*(3*b*c - a*d))/(5*a^2*x^5) - (8*(6*b^2*c - 3*a*b*d + a^2* e))/(3*a^3*x^3) + (8*(10*b^3*c - 6*a*b^2*d + 3*a^2*b*e - a^3*f))/(a^4*x) + (Sqrt[b]*(99*b^3*c - 63*a*b^2*d + 35*a^2*b*e - 15*a^3*f)*ArcTan[(Sqrt[b]* x)/Sqrt[a]])/a^(9/2))/(2*a))/(4*a)
3.2.41.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Time = 3.64 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.91
| method | result | size |
| default | \(-\frac {c}{7 a^{3} x^{7}}-\frac {a d -3 b c}{5 a^{4} x^{5}}-\frac {a^{2} e -3 a b d +6 b^{2} c}{3 a^{5} x^{3}}-\frac {f \,a^{3}-3 a^{2} b e +6 a \,b^{2} d -10 b^{3} c}{a^{6} x}-\frac {b \left (\frac {\left (\frac {7}{8} a^{3} b f -\frac {11}{8} a^{2} e \,b^{2}+\frac {15}{8} a \,b^{3} d -\frac {19}{8} b^{4} c \right ) x^{3}+\frac {a \left (9 f \,a^{3}-13 a^{2} b e +17 a \,b^{2} d -21 b^{3} c \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (15 f \,a^{3}-35 a^{2} b e +63 a \,b^{2} d -99 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{6}}\) | \(212\) |
| risch | \(\frac {-\frac {b^{2} \left (15 f \,a^{3}-35 a^{2} b e +63 a \,b^{2} d -99 b^{3} c \right ) x^{10}}{8 a^{6}}-\frac {5 b \left (15 f \,a^{3}-35 a^{2} b e +63 a \,b^{2} d -99 b^{3} c \right ) x^{8}}{24 a^{5}}-\frac {\left (15 f \,a^{3}-35 a^{2} b e +63 a \,b^{2} d -99 b^{3} c \right ) x^{6}}{15 a^{4}}-\frac {\left (35 a^{2} e -63 a b d +99 b^{2} c \right ) x^{4}}{105 a^{3}}-\frac {\left (7 a d -11 b c \right ) x^{2}}{35 a^{2}}-\frac {c}{7 a}}{x^{7} \left (b \,x^{2}+a \right )^{2}}+\frac {15 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) f}{16 a^{4}}-\frac {35 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) b e}{16 a^{5}}+\frac {63 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) b^{2} d}{16 a^{6}}-\frac {99 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) b^{3} c}{16 a^{7}}-\frac {15 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) f}{16 a^{4}}+\frac {35 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) b e}{16 a^{5}}-\frac {63 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) b^{2} d}{16 a^{6}}+\frac {99 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) b^{3} c}{16 a^{7}}\) | \(388\) |
-1/7*c/a^3/x^7-1/5*(a*d-3*b*c)/a^4/x^5-1/3*(a^2*e-3*a*b*d+6*b^2*c)/a^5/x^3 -(a^3*f-3*a^2*b*e+6*a*b^2*d-10*b^3*c)/a^6/x-b/a^6*(((7/8*a^3*b*f-11/8*a^2* e*b^2+15/8*a*b^3*d-19/8*b^4*c)*x^3+1/8*a*(9*a^3*f-13*a^2*b*e+17*a*b^2*d-21 *b^3*c)*x)/(b*x^2+a)^2+1/8*(15*a^3*f-35*a^2*b*e+63*a*b^2*d-99*b^3*c)/(a*b) ^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 678, normalized size of antiderivative = 2.90 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=\left [\frac {210 \, {\left (99 \, b^{5} c - 63 \, a b^{4} d + 35 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{10} + 350 \, {\left (99 \, a b^{4} c - 63 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{8} + 112 \, {\left (99 \, a^{2} b^{3} c - 63 \, a^{3} b^{2} d + 35 \, a^{4} b e - 15 \, a^{5} f\right )} x^{6} - 240 \, a^{5} c - 16 \, {\left (99 \, a^{3} b^{2} c - 63 \, a^{4} b d + 35 \, a^{5} e\right )} x^{4} + 48 \, {\left (11 \, a^{4} b c - 7 \, a^{5} d\right )} x^{2} - 105 \, {\left ({\left (99 \, b^{5} c - 63 \, a b^{4} d + 35 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{11} + 2 \, {\left (99 \, a b^{4} c - 63 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{9} + {\left (99 \, a^{2} b^{3} c - 63 \, a^{3} b^{2} d + 35 \, a^{4} b e - 15 \, a^{5} f\right )} x^{7}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{1680 \, {\left (a^{6} b^{2} x^{11} + 2 \, a^{7} b x^{9} + a^{8} x^{7}\right )}}, \frac {105 \, {\left (99 \, b^{5} c - 63 \, a b^{4} d + 35 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{10} + 175 \, {\left (99 \, a b^{4} c - 63 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{8} + 56 \, {\left (99 \, a^{2} b^{3} c - 63 \, a^{3} b^{2} d + 35 \, a^{4} b e - 15 \, a^{5} f\right )} x^{6} - 120 \, a^{5} c - 8 \, {\left (99 \, a^{3} b^{2} c - 63 \, a^{4} b d + 35 \, a^{5} e\right )} x^{4} + 24 \, {\left (11 \, a^{4} b c - 7 \, a^{5} d\right )} x^{2} + 105 \, {\left ({\left (99 \, b^{5} c - 63 \, a b^{4} d + 35 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{11} + 2 \, {\left (99 \, a b^{4} c - 63 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{9} + {\left (99 \, a^{2} b^{3} c - 63 \, a^{3} b^{2} d + 35 \, a^{4} b e - 15 \, a^{5} f\right )} x^{7}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{840 \, {\left (a^{6} b^{2} x^{11} + 2 \, a^{7} b x^{9} + a^{8} x^{7}\right )}}\right ] \]
[1/1680*(210*(99*b^5*c - 63*a*b^4*d + 35*a^2*b^3*e - 15*a^3*b^2*f)*x^10 + 350*(99*a*b^4*c - 63*a^2*b^3*d + 35*a^3*b^2*e - 15*a^4*b*f)*x^8 + 112*(99* a^2*b^3*c - 63*a^3*b^2*d + 35*a^4*b*e - 15*a^5*f)*x^6 - 240*a^5*c - 16*(99 *a^3*b^2*c - 63*a^4*b*d + 35*a^5*e)*x^4 + 48*(11*a^4*b*c - 7*a^5*d)*x^2 - 105*((99*b^5*c - 63*a*b^4*d + 35*a^2*b^3*e - 15*a^3*b^2*f)*x^11 + 2*(99*a* b^4*c - 63*a^2*b^3*d + 35*a^3*b^2*e - 15*a^4*b*f)*x^9 + (99*a^2*b^3*c - 63 *a^3*b^2*d + 35*a^4*b*e - 15*a^5*f)*x^7)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqr t(-b/a) - a)/(b*x^2 + a)))/(a^6*b^2*x^11 + 2*a^7*b*x^9 + a^8*x^7), 1/840*( 105*(99*b^5*c - 63*a*b^4*d + 35*a^2*b^3*e - 15*a^3*b^2*f)*x^10 + 175*(99*a *b^4*c - 63*a^2*b^3*d + 35*a^3*b^2*e - 15*a^4*b*f)*x^8 + 56*(99*a^2*b^3*c - 63*a^3*b^2*d + 35*a^4*b*e - 15*a^5*f)*x^6 - 120*a^5*c - 8*(99*a^3*b^2*c - 63*a^4*b*d + 35*a^5*e)*x^4 + 24*(11*a^4*b*c - 7*a^5*d)*x^2 + 105*((99*b^ 5*c - 63*a*b^4*d + 35*a^2*b^3*e - 15*a^3*b^2*f)*x^11 + 2*(99*a*b^4*c - 63* a^2*b^3*d + 35*a^3*b^2*e - 15*a^4*b*f)*x^9 + (99*a^2*b^3*c - 63*a^3*b^2*d + 35*a^4*b*e - 15*a^5*f)*x^7)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^6*b^2*x^11 + 2*a^7*b*x^9 + a^8*x^7)]
Timed out. \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=\frac {105 \, {\left (99 \, b^{5} c - 63 \, a b^{4} d + 35 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{10} + 175 \, {\left (99 \, a b^{4} c - 63 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{8} + 56 \, {\left (99 \, a^{2} b^{3} c - 63 \, a^{3} b^{2} d + 35 \, a^{4} b e - 15 \, a^{5} f\right )} x^{6} - 120 \, a^{5} c - 8 \, {\left (99 \, a^{3} b^{2} c - 63 \, a^{4} b d + 35 \, a^{5} e\right )} x^{4} + 24 \, {\left (11 \, a^{4} b c - 7 \, a^{5} d\right )} x^{2}}{840 \, {\left (a^{6} b^{2} x^{11} + 2 \, a^{7} b x^{9} + a^{8} x^{7}\right )}} + \frac {{\left (99 \, b^{4} c - 63 \, a b^{3} d + 35 \, a^{2} b^{2} e - 15 \, a^{3} b f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{6}} \]
1/840*(105*(99*b^5*c - 63*a*b^4*d + 35*a^2*b^3*e - 15*a^3*b^2*f)*x^10 + 17 5*(99*a*b^4*c - 63*a^2*b^3*d + 35*a^3*b^2*e - 15*a^4*b*f)*x^8 + 56*(99*a^2 *b^3*c - 63*a^3*b^2*d + 35*a^4*b*e - 15*a^5*f)*x^6 - 120*a^5*c - 8*(99*a^3 *b^2*c - 63*a^4*b*d + 35*a^5*e)*x^4 + 24*(11*a^4*b*c - 7*a^5*d)*x^2)/(a^6* b^2*x^11 + 2*a^7*b*x^9 + a^8*x^7) + 1/8*(99*b^4*c - 63*a*b^3*d + 35*a^2*b^ 2*e - 15*a^3*b*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^6)
Time = 0.30 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.05 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=\frac {{\left (99 \, b^{4} c - 63 \, a b^{3} d + 35 \, a^{2} b^{2} e - 15 \, a^{3} b f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{6}} + \frac {19 \, b^{5} c x^{3} - 15 \, a b^{4} d x^{3} + 11 \, a^{2} b^{3} e x^{3} - 7 \, a^{3} b^{2} f x^{3} + 21 \, a b^{4} c x - 17 \, a^{2} b^{3} d x + 13 \, a^{3} b^{2} e x - 9 \, a^{4} b f x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{6}} + \frac {1050 \, b^{3} c x^{6} - 630 \, a b^{2} d x^{6} + 315 \, a^{2} b e x^{6} - 105 \, a^{3} f x^{6} - 210 \, a b^{2} c x^{4} + 105 \, a^{2} b d x^{4} - 35 \, a^{3} e x^{4} + 63 \, a^{2} b c x^{2} - 21 \, a^{3} d x^{2} - 15 \, a^{3} c}{105 \, a^{6} x^{7}} \]
1/8*(99*b^4*c - 63*a*b^3*d + 35*a^2*b^2*e - 15*a^3*b*f)*arctan(b*x/sqrt(a* b))/(sqrt(a*b)*a^6) + 1/8*(19*b^5*c*x^3 - 15*a*b^4*d*x^3 + 11*a^2*b^3*e*x^ 3 - 7*a^3*b^2*f*x^3 + 21*a*b^4*c*x - 17*a^2*b^3*d*x + 13*a^3*b^2*e*x - 9*a ^4*b*f*x)/((b*x^2 + a)^2*a^6) + 1/105*(1050*b^3*c*x^6 - 630*a*b^2*d*x^6 + 315*a^2*b*e*x^6 - 105*a^3*f*x^6 - 210*a*b^2*c*x^4 + 105*a^2*b*d*x^4 - 35*a ^3*e*x^4 + 63*a^2*b*c*x^2 - 21*a^3*d*x^2 - 15*a^3*c)/(a^6*x^7)
Time = 5.61 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^8 \left (a+b x^2\right )^3} \, dx=\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-15\,f\,a^3+35\,e\,a^2\,b-63\,d\,a\,b^2+99\,c\,b^3\right )}{8\,a^{13/2}}-\frac {\frac {c}{7\,a}-\frac {x^6\,\left (-15\,f\,a^3+35\,e\,a^2\,b-63\,d\,a\,b^2+99\,c\,b^3\right )}{15\,a^4}+\frac {x^2\,\left (7\,a\,d-11\,b\,c\right )}{35\,a^2}+\frac {x^4\,\left (35\,e\,a^2-63\,d\,a\,b+99\,c\,b^2\right )}{105\,a^3}-\frac {5\,b\,x^8\,\left (-15\,f\,a^3+35\,e\,a^2\,b-63\,d\,a\,b^2+99\,c\,b^3\right )}{24\,a^5}-\frac {b^2\,x^{10}\,\left (-15\,f\,a^3+35\,e\,a^2\,b-63\,d\,a\,b^2+99\,c\,b^3\right )}{8\,a^6}}{a^2\,x^7+2\,a\,b\,x^9+b^2\,x^{11}} \]
(b^(1/2)*atan((b^(1/2)*x)/a^(1/2))*(99*b^3*c - 15*a^3*f - 63*a*b^2*d + 35* a^2*b*e))/(8*a^(13/2)) - (c/(7*a) - (x^6*(99*b^3*c - 15*a^3*f - 63*a*b^2*d + 35*a^2*b*e))/(15*a^4) + (x^2*(7*a*d - 11*b*c))/(35*a^2) + (x^4*(99*b^2* c + 35*a^2*e - 63*a*b*d))/(105*a^3) - (5*b*x^8*(99*b^3*c - 15*a^3*f - 63*a *b^2*d + 35*a^2*b*e))/(24*a^5) - (b^2*x^10*(99*b^3*c - 15*a^3*f - 63*a*b^2 *d + 35*a^2*b*e))/(8*a^6))/(a^2*x^7 + b^2*x^11 + 2*a*b*x^9)