Integrand size = 30, antiderivative size = 167 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {(b e-3 a f) x}{b^4}+\frac {f x^3}{3 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{9/2}} \]
(-3*a*f+b*e)*x/b^4+1/3*f*x^3/b^3+1/4*(c-a*(a^2*f-a*b*e+b^2*d)/b^3)*x^3/a/( b*x^2+a)^2-1/8*(11*a^3*f-7*a^2*b*e+3*a*b^2*d+b^3*c)*x/a/b^4/(b*x^2+a)+1/8* (35*a^3*f-15*a^2*b*e+3*a*b^2*d+b^3*c)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/b^ (9/2)
Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (-105 a^4 f+3 b^4 c x^2+5 a^3 b \left (9 e-35 f x^2\right )+a^2 b^2 \left (-9 d+75 e x^2-56 f x^4\right )+a b^3 \left (-3 c-15 d x^2+24 e x^4+8 f x^6\right )\right )}{24 a b^4 \left (a+b x^2\right )^2}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{9/2}} \]
(x*(-105*a^4*f + 3*b^4*c*x^2 + 5*a^3*b*(9*e - 35*f*x^2) + a^2*b^2*(-9*d + 75*e*x^2 - 56*f*x^4) + a*b^3*(-3*c - 15*d*x^2 + 24*e*x^4 + 8*f*x^6)))/(24* a*b^4*(a + b*x^2)^2) + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3*f)*ArcTan [(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(9/2))
Time = 0.52 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2335, 9, 25, 1580, 25, 1467, 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.
| \(\displaystyle \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\int -\frac {x \left (4 a f x^5+4 a \left (e-\frac {a f}{b}\right ) x^3+\left (\frac {3 f a^3}{b^2}-\frac {3 e a^2}{b}+3 d a+b c\right ) x\right )}{\left (b x^2+a\right )^2}dx}{4 a b}\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\int -\frac {x^2 \left (4 a f x^4+4 a \left (e-\frac {a f}{b}\right ) x^2+b c+3 a d-\frac {3 a^2 e}{b}+\frac {3 a^3 f}{b^2}\right )}{\left (b x^2+a\right )^2}dx}{4 a b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^2 \left (4 a f x^4+4 a \left (e-\frac {a f}{b}\right ) x^2+b c+3 a d-\frac {3 a^2 e}{b}+\frac {3 a^3 f}{b^2}\right )}{\left (b x^2+a\right )^2}dx}{4 a b}+\frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 1580 |
| \(\displaystyle \frac {-\frac {\int -\frac {8 a b^2 f x^4+8 a b (b e-2 a f) x^2+b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f}{b x^2+a}dx}{2 b^3}-\frac {x \left (11 a^3 f-7 a^2 b e+3 a b^2 d+b^3 c\right )}{2 b^3 \left (a+b x^2\right )}}{4 a b}+\frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {8 a b^2 f x^4+8 a b (b e-2 a f) x^2+b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f}{b x^2+a}dx}{2 b^3}-\frac {x \left (11 a^3 f-7 a^2 b e+3 a b^2 d+b^3 c\right )}{2 b^3 \left (a+b x^2\right )}}{4 a b}+\frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 1467 |
| \(\displaystyle \frac {\frac {\int \left (8 a b f x^2+8 a (b e-3 a f)+\frac {35 f a^3-15 b e a^2+3 b^2 d a+b^3 c}{b x^2+a}\right )dx}{2 b^3}-\frac {x \left (11 a^3 f-7 a^2 b e+3 a b^2 d+b^3 c\right )}{2 b^3 \left (a+b x^2\right )}}{4 a b}+\frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}+\frac {\frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (35 a^3 f-15 a^2 b e+3 a b^2 d+b^3 c\right )}{\sqrt {a} \sqrt {b}}+8 a x (b e-3 a f)+\frac {8}{3} a b f x^3}{2 b^3}-\frac {x \left (11 a^3 f-7 a^2 b e+3 a b^2 d+b^3 c\right )}{2 b^3 \left (a+b x^2\right )}}{4 a b}\) |
((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^3)/(4*a*(a + b*x^2)^2) + (-1/2*(( b^3*c + 3*a*b^2*d - 7*a^2*b*e + 11*a^3*f)*x)/(b^3*(a + b*x^2)) + (8*a*(b*e - 3*a*f)*x + (8*a*b*f*x^3)/3 + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3* f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]))/(2*b^3))/(4*a*b)
3.2.36.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) ^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* (q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b *d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e }, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
Time = 3.49 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90
| method | result | size |
| default | \(-\frac {-\frac {1}{3} f \,x^{3} b +3 a f x -b e x}{b^{4}}+\frac {\frac {-\frac {b \left (13 f \,a^{3}-9 a^{2} b e +5 a \,b^{2} d -b^{3} c \right ) x^{3}}{8 a}+\left (-\frac {11}{8} f \,a^{3}+\frac {7}{8} a^{2} b e -\frac {3}{8} a \,b^{2} d -\frac {1}{8} b^{3} c \right ) x}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (35 f \,a^{3}-15 a^{2} b e +3 a \,b^{2} d +b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}}{b^{4}}\) | \(151\) |
| risch | \(\frac {f \,x^{3}}{3 b^{3}}-\frac {3 a f x}{b^{4}}+\frac {e x}{b^{3}}+\frac {-\frac {b \left (13 f \,a^{3}-9 a^{2} b e +5 a \,b^{2} d -b^{3} c \right ) x^{3}}{8 a}+\left (-\frac {11}{8} f \,a^{3}+\frac {7}{8} a^{2} b e -\frac {3}{8} a \,b^{2} d -\frac {1}{8} b^{3} c \right ) x}{b^{4} \left (b \,x^{2}+a \right )^{2}}-\frac {35 a^{2} \ln \left (b x +\sqrt {-a b}\right ) f}{16 b^{4} \sqrt {-a b}}+\frac {15 a \ln \left (b x +\sqrt {-a b}\right ) e}{16 b^{3} \sqrt {-a b}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) d}{16 b^{2} \sqrt {-a b}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) c}{16 b \sqrt {-a b}\, a}+\frac {35 a^{2} \ln \left (-b x +\sqrt {-a b}\right ) f}{16 b^{4} \sqrt {-a b}}-\frac {15 a \ln \left (-b x +\sqrt {-a b}\right ) e}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) d}{16 b^{2} \sqrt {-a b}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) c}{16 b \sqrt {-a b}\, a}\) | \(306\) |
-1/b^4*(-1/3*f*x^3*b+3*a*f*x-b*e*x)+1/b^4*((-1/8*b*(13*a^3*f-9*a^2*b*e+5*a *b^2*d-b^3*c)/a*x^3+(-11/8*f*a^3+7/8*a^2*b*e-3/8*a*b^2*d-1/8*b^3*c)*x)/(b* x^2+a)^2+1/8*(35*a^3*f-15*a^2*b*e+3*a*b^2*d+b^3*c)/a/(a*b)^(1/2)*arctan(b* x/(a*b)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 555, normalized size of antiderivative = 3.32 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, a^{2} b^{4} f x^{7} + 16 \, {\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + 2 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} - 3 \, {\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f + {\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 6 \, {\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{48 \, {\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, \frac {8 \, a^{2} b^{4} f x^{7} + 8 \, {\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} + 3 \, {\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f + {\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 3 \, {\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{24 \, {\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \]
[1/48*(16*a^2*b^4*f*x^7 + 16*(3*a^2*b^4*e - 7*a^3*b^3*f)*x^5 + 2*(3*a*b^5* c - 15*a^2*b^4*d + 75*a^3*b^3*e - 175*a^4*b^2*f)*x^3 - 3*(a^2*b^3*c + 3*a^ 3*b^2*d - 15*a^4*b*e + 35*a^5*f + (b^5*c + 3*a*b^4*d - 15*a^2*b^3*e + 35*a ^3*b^2*f)*x^4 + 2*(a*b^4*c + 3*a^2*b^3*d - 15*a^3*b^2*e + 35*a^4*b*f)*x^2) *sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 6*(a^2*b^4*c + 3*a^3*b^3*d - 15*a^4*b^2*e + 35*a^5*b*f)*x)/(a^2*b^7*x^4 + 2*a^3*b^6*x^2 + a^4*b^5), 1/24*(8*a^2*b^4*f*x^7 + 8*(3*a^2*b^4*e - 7*a^3*b^3*f)*x^5 + (3 *a*b^5*c - 15*a^2*b^4*d + 75*a^3*b^3*e - 175*a^4*b^2*f)*x^3 + 3*(a^2*b^3*c + 3*a^3*b^2*d - 15*a^4*b*e + 35*a^5*f + (b^5*c + 3*a*b^4*d - 15*a^2*b^3*e + 35*a^3*b^2*f)*x^4 + 2*(a*b^4*c + 3*a^2*b^3*d - 15*a^3*b^2*e + 35*a^4*b* f)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - 3*(a^2*b^4*c + 3*a^3*b^3*d - 15* a^4*b^2*e + 35*a^5*b*f)*x)/(a^2*b^7*x^4 + 2*a^3*b^6*x^2 + a^4*b^5)]
Time = 5.71 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.56 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x \left (- \frac {3 a f}{b^{4}} + \frac {e}{b^{3}}\right ) - \frac {\sqrt {- \frac {1}{a^{3} b^{9}}} \cdot \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log {\left (- a^{2} b^{4} \sqrt {- \frac {1}{a^{3} b^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{9}}} \cdot \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log {\left (a^{2} b^{4} \sqrt {- \frac {1}{a^{3} b^{9}}} + x \right )}}{16} + \frac {x^{3} \left (- 13 a^{3} b f + 9 a^{2} b^{2} e - 5 a b^{3} d + b^{4} c\right ) + x \left (- 11 a^{4} f + 7 a^{3} b e - 3 a^{2} b^{2} d - a b^{3} c\right )}{8 a^{3} b^{4} + 16 a^{2} b^{5} x^{2} + 8 a b^{6} x^{4}} + \frac {f x^{3}}{3 b^{3}} \]
x*(-3*a*f/b**4 + e/b**3) - sqrt(-1/(a**3*b**9))*(35*a**3*f - 15*a**2*b*e + 3*a*b**2*d + b**3*c)*log(-a**2*b**4*sqrt(-1/(a**3*b**9)) + x)/16 + sqrt(- 1/(a**3*b**9))*(35*a**3*f - 15*a**2*b*e + 3*a*b**2*d + b**3*c)*log(a**2*b* *4*sqrt(-1/(a**3*b**9)) + x)/16 + (x**3*(-13*a**3*b*f + 9*a**2*b**2*e - 5* a*b**3*d + b**4*c) + x*(-11*a**4*f + 7*a**3*b*e - 3*a**2*b**2*d - a*b**3*c ))/(8*a**3*b**4 + 16*a**2*b**5*x**2 + 8*a*b**6*x**4) + f*x**3/(3*b**3)
Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (b^{4} c - 5 \, a b^{3} d + 9 \, a^{2} b^{2} e - 13 \, a^{3} b f\right )} x^{3} - {\left (a b^{3} c + 3 \, a^{2} b^{2} d - 7 \, a^{3} b e + 11 \, a^{4} f\right )} x}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} + \frac {b f x^{3} + 3 \, {\left (b e - 3 \, a f\right )} x}{3 \, b^{4}} + \frac {{\left (b^{3} c + 3 \, a b^{2} d - 15 \, a^{2} b e + 35 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{4}} \]
1/8*((b^4*c - 5*a*b^3*d + 9*a^2*b^2*e - 13*a^3*b*f)*x^3 - (a*b^3*c + 3*a^2 *b^2*d - 7*a^3*b*e + 11*a^4*f)*x)/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4) + 1/3*(b*f*x^3 + 3*(b*e - 3*a*f)*x)/b^4 + 1/8*(b^3*c + 3*a*b^2*d - 15*a^2*b* e + 35*a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^4)
Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (b^{3} c + 3 \, a b^{2} d - 15 \, a^{2} b e + 35 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{4}} + \frac {b^{4} c x^{3} - 5 \, a b^{3} d x^{3} + 9 \, a^{2} b^{2} e x^{3} - 13 \, a^{3} b f x^{3} - a b^{3} c x - 3 \, a^{2} b^{2} d x + 7 \, a^{3} b e x - 11 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{4}} + \frac {b^{6} f x^{3} + 3 \, b^{6} e x - 9 \, a b^{5} f x}{3 \, b^{9}} \]
1/8*(b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3*f)*arctan(b*x/sqrt(a*b))/(sqr t(a*b)*a*b^4) + 1/8*(b^4*c*x^3 - 5*a*b^3*d*x^3 + 9*a^2*b^2*e*x^3 - 13*a^3* b*f*x^3 - a*b^3*c*x - 3*a^2*b^2*d*x + 7*a^3*b*e*x - 11*a^4*f*x)/((b*x^2 + a)^2*a*b^4) + 1/3*(b^6*f*x^3 + 3*b^6*e*x - 9*a*b^5*f*x)/b^9
Time = 5.51 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )-\frac {x\,\left (\frac {11\,f\,a^3}{8}-\frac {7\,e\,a^2\,b}{8}+\frac {3\,d\,a\,b^2}{8}+\frac {c\,b^3}{8}\right )-\frac {x^3\,\left (-13\,f\,a^3\,b+9\,e\,a^2\,b^2-5\,d\,a\,b^3+c\,b^4\right )}{8\,a}}{a^2\,b^4+2\,a\,b^5\,x^2+b^6\,x^4}+\frac {f\,x^3}{3\,b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (35\,f\,a^3-15\,e\,a^2\,b+3\,d\,a\,b^2+c\,b^3\right )}{8\,a^{3/2}\,b^{9/2}} \]
x*(e/b^3 - (3*a*f)/b^4) - (x*((b^3*c)/8 + (11*a^3*f)/8 + (3*a*b^2*d)/8 - ( 7*a^2*b*e)/8) - (x^3*(b^4*c + 9*a^2*b^2*e - 5*a*b^3*d - 13*a^3*b*f))/(8*a) )/(a^2*b^4 + b^6*x^4 + 2*a*b^5*x^2) + (f*x^3)/(3*b^3) + (atan((b^(1/2)*x)/ a^(1/2))*(b^3*c + 35*a^3*f + 3*a*b^2*d - 15*a^2*b*e))/(8*a^(3/2)*b^(9/2))