Integrand size = 30, antiderivative size = 207 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {\left (b^3 c-5 a b^2 d+13 a^2 b e-25 a^3 f\right ) x}{4 a b^5}+\frac {(b e-3 a f) x^3}{3 b^4}+\frac {f x^5}{5 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^5}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x}{8 b^5 \left (a+b x^2\right )}+\frac {\left (3 b^3 c-15 a b^2 d+35 a^2 b e-63 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{11/2}} \]
-1/4*(-25*a^3*f+13*a^2*b*e-5*a*b^2*d+b^3*c)*x/a/b^5+1/3*(-3*a*f+b*e)*x^3/b ^4+1/5*f*x^5/b^3+1/4*(c-a*(a^2*f-a*b*e+b^2*d)/b^3)*x^5/a/(b*x^2+a)^2-1/8*( -13*a^3*f+9*a^2*b*e-5*a*b^2*d+b^3*c)*x/b^5/(b*x^2+a)+1/8*(-63*a^3*f+35*a^2 *b*e-15*a*b^2*d+3*b^3*c)*arctan(x*b^(1/2)/a^(1/2))/b^(11/2)/a^(1/2)
Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (945 a^4 f-525 a^3 b \left (e-3 f x^2\right )+a^2 b^2 \left (225 d-875 e x^2+504 f x^4\right )-a b^3 \left (45 c-375 d x^2+280 e x^4+72 f x^6\right )+b^4 x^2 \left (-75 c+8 \left (15 d x^2+5 e x^4+3 f x^6\right )\right )\right )}{120 b^5 \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c-15 a b^2 d+35 a^2 b e-63 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{11/2}} \]
(x*(945*a^4*f - 525*a^3*b*(e - 3*f*x^2) + a^2*b^2*(225*d - 875*e*x^2 + 504 *f*x^4) - a*b^3*(45*c - 375*d*x^2 + 280*e*x^4 + 72*f*x^6) + b^4*x^2*(-75*c + 8*(15*d*x^2 + 5*e*x^4 + 3*f*x^6))))/(120*b^5*(a + b*x^2)^2) + ((3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqr t[a]*b^(11/2))
Time = 0.64 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2335, 9, 1580, 2341, 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.
| \(\displaystyle \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {x^5 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x^3 \left (-4 a f x^5-4 a \left (e-\frac {a f}{b}\right ) x^3+\left (-\frac {5 f a^3}{b^2}+\frac {5 e a^2}{b}-5 d a+b c\right ) x\right )}{\left (b x^2+a\right )^2}dx}{4 a b}\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \frac {x^5 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x^4 \left (-4 a f x^4-4 a \left (e-\frac {a f}{b}\right ) x^2+b c-5 a d+\frac {5 a^2 e}{b}-\frac {5 a^3 f}{b^2}\right )}{\left (b x^2+a\right )^2}dx}{4 a b}\) |
| \(\Big \downarrow \) 1580 |
| \(\displaystyle \frac {x^5 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\frac {a x \left (-13 a^3 f+9 a^2 b e-5 a b^2 d+b^3 c\right )}{2 b^4 \left (a+b x^2\right )}-\frac {\int \frac {8 a b^3 f x^6+8 a b^2 (b e-2 a f) x^4-2 b \left (-13 f a^3+9 b e a^2-5 b^2 d a+b^3 c\right ) x^2+a \left (-13 f a^3+9 b e a^2-5 b^2 d a+b^3 c\right )}{b x^2+a}dx}{2 b^4}}{4 a b}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {x^5 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\frac {a x \left (-13 a^3 f+9 a^2 b e-5 a b^2 d+b^3 c\right )}{2 b^4 \left (a+b x^2\right )}-\frac {\int \left (8 a b^2 f x^4+8 a b (b e-3 a f) x^2-2 \left (-25 f a^3+13 b e a^2-5 b^2 d a+b^3 c\right )+\frac {-63 f a^4+35 b e a^3-15 b^2 d a^2+3 b^3 c a}{b x^2+a}\right )dx}{2 b^4}}{4 a b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^5 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\frac {a x \left (-13 a^3 f+9 a^2 b e-5 a b^2 d+b^3 c\right )}{2 b^4 \left (a+b x^2\right )}-\frac {\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-63 a^3 f+35 a^2 b e-15 a b^2 d+3 b^3 c\right )}{\sqrt {b}}-2 x \left (-25 a^3 f+13 a^2 b e-5 a b^2 d+b^3 c\right )+\frac {8}{5} a b^2 f x^5+\frac {8}{3} a b x^3 (b e-3 a f)}{2 b^4}}{4 a b}\) |
((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^5)/(4*a*(a + b*x^2)^2) - ((a*(b^3 *c - 5*a*b^2*d + 9*a^2*b*e - 13*a^3*f)*x)/(2*b^4*(a + b*x^2)) - (-2*(b^3*c - 5*a*b^2*d + 13*a^2*b*e - 25*a^3*f)*x + (8*a*b*(b*e - 3*a*f)*x^3)/3 + (8 *a*b^2*f*x^5)/5 + (Sqrt[a]*(3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)* ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[b])/(2*b^4))/(4*a*b)
3.2.35.3.1 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[(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]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 3.50 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86
| method | result | size |
| default | \(\frac {\frac {1}{5} f \,x^{5} b^{2}-a b f \,x^{3}+\frac {1}{3} b^{2} e \,x^{3}+6 a^{2} f x -3 a b e x +b^{2} d x}{b^{5}}-\frac {\frac {\left (-\frac {17}{8} a^{3} b f +\frac {13}{8} a^{2} e \,b^{2}-\frac {9}{8} a \,b^{3} d +\frac {5}{8} b^{4} c \right ) x^{3}-\frac {a \left (15 f \,a^{3}-11 a^{2} b e +7 a \,b^{2} d -3 b^{3} c \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (63 f \,a^{3}-35 a^{2} b e +15 a \,b^{2} d -3 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{5}}\) | \(177\) |
| risch | \(\frac {f \,x^{5}}{5 b^{3}}-\frac {a f \,x^{3}}{b^{4}}+\frac {e \,x^{3}}{3 b^{3}}+\frac {6 a^{2} f x}{b^{5}}-\frac {3 a e x}{b^{4}}+\frac {d x}{b^{3}}+\frac {\left (\frac {17}{8} a^{3} b f -\frac {13}{8} a^{2} e \,b^{2}+\frac {9}{8} a \,b^{3} d -\frac {5}{8} b^{4} c \right ) x^{3}+\frac {a \left (15 f \,a^{3}-11 a^{2} b e +7 a \,b^{2} d -3 b^{3} c \right ) x}{8}}{b^{5} \left (b \,x^{2}+a \right )^{2}}-\frac {63 \ln \left (b x -\sqrt {-a b}\right ) f \,a^{3}}{16 b^{5} \sqrt {-a b}}+\frac {35 \ln \left (b x -\sqrt {-a b}\right ) a^{2} e}{16 b^{4} \sqrt {-a b}}-\frac {15 \ln \left (b x -\sqrt {-a b}\right ) a d}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (b x -\sqrt {-a b}\right ) c}{16 b^{2} \sqrt {-a b}}+\frac {63 \ln \left (-b x -\sqrt {-a b}\right ) f \,a^{3}}{16 b^{5} \sqrt {-a b}}-\frac {35 \ln \left (-b x -\sqrt {-a b}\right ) a^{2} e}{16 b^{4} \sqrt {-a b}}+\frac {15 \ln \left (-b x -\sqrt {-a b}\right ) a d}{16 b^{3} \sqrt {-a b}}-\frac {3 \ln \left (-b x -\sqrt {-a b}\right ) c}{16 b^{2} \sqrt {-a b}}\) | \(351\) |
1/b^5*(1/5*f*x^5*b^2-a*b*f*x^3+1/3*b^2*e*x^3+6*a^2*f*x-3*a*b*e*x+b^2*d*x)- 1/b^5*(((-17/8*a^3*b*f+13/8*a^2*e*b^2-9/8*a*b^3*d+5/8*b^4*c)*x^3-1/8*a*(15 *a^3*f-11*a^2*b*e+7*a*b^2*d-3*b^3*c)*x)/(b*x^2+a)^2+1/8*(63*a^3*f-35*a^2*b *e+15*a*b^2*d-3*b^3*c)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 614, normalized size of antiderivative = 2.97 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {48 \, a b^{5} f x^{9} + 16 \, {\left (5 \, a b^{5} e - 9 \, a^{2} b^{4} f\right )} x^{7} + 16 \, {\left (15 \, a b^{5} d - 35 \, a^{2} b^{4} e + 63 \, a^{3} b^{3} f\right )} x^{5} - 50 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 35 \, a^{3} b^{3} e - 63 \, a^{4} b^{2} f\right )} x^{3} + 15 \, {\left (3 \, a^{2} b^{3} c - 15 \, a^{3} b^{2} d + 35 \, a^{4} b e - 63 \, a^{5} f + {\left (3 \, b^{5} c - 15 \, a b^{4} d + 35 \, a^{2} b^{3} e - 63 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c - 15 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 63 \, a^{4} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 30 \, {\left (3 \, a^{2} b^{4} c - 15 \, a^{3} b^{3} d + 35 \, a^{4} b^{2} e - 63 \, a^{5} b f\right )} x}{240 \, {\left (a b^{8} x^{4} + 2 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}}, \frac {24 \, a b^{5} f x^{9} + 8 \, {\left (5 \, a b^{5} e - 9 \, a^{2} b^{4} f\right )} x^{7} + 8 \, {\left (15 \, a b^{5} d - 35 \, a^{2} b^{4} e + 63 \, a^{3} b^{3} f\right )} x^{5} - 25 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 35 \, a^{3} b^{3} e - 63 \, a^{4} b^{2} f\right )} x^{3} + 15 \, {\left (3 \, a^{2} b^{3} c - 15 \, a^{3} b^{2} d + 35 \, a^{4} b e - 63 \, a^{5} f + {\left (3 \, b^{5} c - 15 \, a b^{4} d + 35 \, a^{2} b^{3} e - 63 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c - 15 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 63 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 15 \, {\left (3 \, a^{2} b^{4} c - 15 \, a^{3} b^{3} d + 35 \, a^{4} b^{2} e - 63 \, a^{5} b f\right )} x}{120 \, {\left (a b^{8} x^{4} + 2 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}}\right ] \]
[1/240*(48*a*b^5*f*x^9 + 16*(5*a*b^5*e - 9*a^2*b^4*f)*x^7 + 16*(15*a*b^5*d - 35*a^2*b^4*e + 63*a^3*b^3*f)*x^5 - 50*(3*a*b^5*c - 15*a^2*b^4*d + 35*a^ 3*b^3*e - 63*a^4*b^2*f)*x^3 + 15*(3*a^2*b^3*c - 15*a^3*b^2*d + 35*a^4*b*e - 63*a^5*f + (3*b^5*c - 15*a*b^4*d + 35*a^2*b^3*e - 63*a^3*b^2*f)*x^4 + 2* (3*a*b^4*c - 15*a^2*b^3*d + 35*a^3*b^2*e - 63*a^4*b*f)*x^2)*sqrt(-a*b)*log ((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 30*(3*a^2*b^4*c - 15*a^3*b^3* d + 35*a^4*b^2*e - 63*a^5*b*f)*x)/(a*b^8*x^4 + 2*a^2*b^7*x^2 + a^3*b^6), 1 /120*(24*a*b^5*f*x^9 + 8*(5*a*b^5*e - 9*a^2*b^4*f)*x^7 + 8*(15*a*b^5*d - 3 5*a^2*b^4*e + 63*a^3*b^3*f)*x^5 - 25*(3*a*b^5*c - 15*a^2*b^4*d + 35*a^3*b^ 3*e - 63*a^4*b^2*f)*x^3 + 15*(3*a^2*b^3*c - 15*a^3*b^2*d + 35*a^4*b*e - 63 *a^5*f + (3*b^5*c - 15*a*b^4*d + 35*a^2*b^3*e - 63*a^3*b^2*f)*x^4 + 2*(3*a *b^4*c - 15*a^2*b^3*d + 35*a^3*b^2*e - 63*a^4*b*f)*x^2)*sqrt(a*b)*arctan(s qrt(a*b)*x/a) - 15*(3*a^2*b^4*c - 15*a^3*b^3*d + 35*a^4*b^2*e - 63*a^5*b*f )*x)/(a*b^8*x^4 + 2*a^2*b^7*x^2 + a^3*b^6)]
Time = 15.94 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.35 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x^{3} \left (- \frac {a f}{b^{4}} + \frac {e}{3 b^{3}}\right ) + x \left (\frac {6 a^{2} f}{b^{5}} - \frac {3 a e}{b^{4}} + \frac {d}{b^{3}}\right ) + \frac {\sqrt {- \frac {1}{a b^{11}}} \cdot \left (63 a^{3} f - 35 a^{2} b e + 15 a b^{2} d - 3 b^{3} c\right ) \log {\left (- a b^{5} \sqrt {- \frac {1}{a b^{11}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{a b^{11}}} \cdot \left (63 a^{3} f - 35 a^{2} b e + 15 a b^{2} d - 3 b^{3} c\right ) \log {\left (a b^{5} \sqrt {- \frac {1}{a b^{11}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (17 a^{3} b f - 13 a^{2} b^{2} e + 9 a b^{3} d - 5 b^{4} c\right ) + x \left (15 a^{4} f - 11 a^{3} b e + 7 a^{2} b^{2} d - 3 a b^{3} c\right )}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} + \frac {f x^{5}}{5 b^{3}} \]
x**3*(-a*f/b**4 + e/(3*b**3)) + x*(6*a**2*f/b**5 - 3*a*e/b**4 + d/b**3) + sqrt(-1/(a*b**11))*(63*a**3*f - 35*a**2*b*e + 15*a*b**2*d - 3*b**3*c)*log( -a*b**5*sqrt(-1/(a*b**11)) + x)/16 - sqrt(-1/(a*b**11))*(63*a**3*f - 35*a* *2*b*e + 15*a*b**2*d - 3*b**3*c)*log(a*b**5*sqrt(-1/(a*b**11)) + x)/16 + ( x**3*(17*a**3*b*f - 13*a**2*b**2*e + 9*a*b**3*d - 5*b**4*c) + x*(15*a**4*f - 11*a**3*b*e + 7*a**2*b**2*d - 3*a*b**3*c))/(8*a**2*b**5 + 16*a*b**6*x** 2 + 8*b**7*x**4) + f*x**5/(5*b**3)
Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.93 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (5 \, b^{4} c - 9 \, a b^{3} d + 13 \, a^{2} b^{2} e - 17 \, a^{3} b f\right )} x^{3} + {\left (3 \, a b^{3} c - 7 \, a^{2} b^{2} d + 11 \, a^{3} b e - 15 \, a^{4} f\right )} x}{8 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} + \frac {{\left (3 \, b^{3} c - 15 \, a b^{2} d + 35 \, a^{2} b e - 63 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {3 \, b^{2} f x^{5} + 5 \, {\left (b^{2} e - 3 \, a b f\right )} x^{3} + 15 \, {\left (b^{2} d - 3 \, a b e + 6 \, a^{2} f\right )} x}{15 \, b^{5}} \]
-1/8*((5*b^4*c - 9*a*b^3*d + 13*a^2*b^2*e - 17*a^3*b*f)*x^3 + (3*a*b^3*c - 7*a^2*b^2*d + 11*a^3*b*e - 15*a^4*f)*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5) + 1/8*(3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*arctan(b*x/sqrt(a*b) )/(sqrt(a*b)*b^5) + 1/15*(3*b^2*f*x^5 + 5*(b^2*e - 3*a*b*f)*x^3 + 15*(b^2* d - 3*a*b*e + 6*a^2*f)*x)/b^5
Time = 0.29 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, b^{3} c - 15 \, a b^{2} d + 35 \, a^{2} b e - 63 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} - \frac {5 \, b^{4} c x^{3} - 9 \, a b^{3} d x^{3} + 13 \, a^{2} b^{2} e x^{3} - 17 \, a^{3} b f x^{3} + 3 \, a b^{3} c x - 7 \, a^{2} b^{2} d x + 11 \, a^{3} b e x - 15 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{5}} + \frac {3 \, b^{12} f x^{5} + 5 \, b^{12} e x^{3} - 15 \, a b^{11} f x^{3} + 15 \, b^{12} d x - 45 \, a b^{11} e x + 90 \, a^{2} b^{10} f x}{15 \, b^{15}} \]
1/8*(3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*arctan(b*x/sqrt(a*b))/( sqrt(a*b)*b^5) - 1/8*(5*b^4*c*x^3 - 9*a*b^3*d*x^3 + 13*a^2*b^2*e*x^3 - 17* a^3*b*f*x^3 + 3*a*b^3*c*x - 7*a^2*b^2*d*x + 11*a^3*b*e*x - 15*a^4*f*x)/((b *x^2 + a)^2*b^5) + 1/15*(3*b^12*f*x^5 + 5*b^12*e*x^3 - 15*a*b^11*f*x^3 + 1 5*b^12*d*x - 45*a*b^11*e*x + 90*a^2*b^10*f*x)/b^15
Time = 5.62 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x^3\,\left (\frac {e}{3\,b^3}-\frac {a\,f}{b^4}\right )-x\,\left (\frac {3\,a^2\,f}{b^5}-\frac {d}{b^3}+\frac {3\,a\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )}{b}\right )-\frac {x^3\,\left (-\frac {17\,f\,a^3\,b}{8}+\frac {13\,e\,a^2\,b^2}{8}-\frac {9\,d\,a\,b^3}{8}+\frac {5\,c\,b^4}{8}\right )-x\,\left (\frac {15\,f\,a^4}{8}-\frac {11\,e\,a^3\,b}{8}+\frac {7\,d\,a^2\,b^2}{8}-\frac {3\,c\,a\,b^3}{8}\right )}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}+\frac {f\,x^5}{5\,b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-63\,f\,a^3+35\,e\,a^2\,b-15\,d\,a\,b^2+3\,c\,b^3\right )}{8\,\sqrt {a}\,b^{11/2}} \]
x^3*(e/(3*b^3) - (a*f)/b^4) - x*((3*a^2*f)/b^5 - d/b^3 + (3*a*(e/b^3 - (3* a*f)/b^4))/b) - (x^3*((5*b^4*c)/8 + (13*a^2*b^2*e)/8 - (9*a*b^3*d)/8 - (17 *a^3*b*f)/8) - x*((15*a^4*f)/8 + (7*a^2*b^2*d)/8 - (3*a*b^3*c)/8 - (11*a^3 *b*e)/8))/(a^2*b^5 + b^7*x^4 + 2*a*b^6*x^2) + (f*x^5)/(5*b^3) + (atan((b^( 1/2)*x)/a^(1/2))*(3*b^3*c - 63*a^3*f - 15*a*b^2*d + 35*a^2*b*e))/(8*a^(1/2 )*b^(11/2))