Integrand size = 27, antiderivative size = 225 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=-\frac {(a B-(A c-a C) x) (d+e x)^2}{6 a c \left (a+c x^2\right )^3}-\frac {2 a e (4 A c d+2 a C d+a B e)+\left (3 a (A c+a C) e^2-c d (5 A c d+a C d+2 a B e)\right ) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac {\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) x}{16 a^3 c^2 \left (a+c x^2\right )}+\frac {\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}} \]
-1/6*(a*B-(A*c-C*a)*x)*(e*x+d)^2/a/c/(c*x^2+a)^3+1/24*(-2*a*e*(4*A*c*d+B*a *e+2*C*a*d)-(3*a*(A*c+C*a)*e^2-c*d*(5*A*c*d+2*B*a*e+C*a*d))*x)/a^2/c^2/(c* x^2+a)^2+1/16*(a*(A*c+C*a)*e^2+c*d*(5*A*c*d+2*B*a*e+C*a*d))*x/a^3/c^2/(c*x ^2+a)+1/16*(a*(A*c+C*a)*e^2+c*d*(5*A*c*d+2*B*a*e+C*a*d))*arctan(x*c^(1/2)/ a^(1/2))/a^(7/2)/c^(5/2)
Time = 0.10 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=\frac {\left (A c \left (5 c d^2+a e^2\right )+a \left (a C e^2+c d (C d+2 B e)\right )\right ) x}{16 a^3 c^2 \left (a+c x^2\right )}+\frac {5 A c^2 d^2 x+a c \left (C d^2+e (2 B d+A e)\right ) x-a^2 e (12 C d+6 B e+7 C e x)}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac {A c^2 d^2 x+a^2 e (2 C d+B e+C e x)-a c \left (C d^2 x+A e (2 d+e x)+B d (d+2 e x)\right )}{6 a c^2 \left (a+c x^2\right )^3}+\frac {\left (A c \left (5 c d^2+a e^2\right )+a \left (a C e^2+c d (C d+2 B e)\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}} \]
((A*c*(5*c*d^2 + a*e^2) + a*(a*C*e^2 + c*d*(C*d + 2*B*e)))*x)/(16*a^3*c^2* (a + c*x^2)) + (5*A*c^2*d^2*x + a*c*(C*d^2 + e*(2*B*d + A*e))*x - a^2*e*(1 2*C*d + 6*B*e + 7*C*e*x))/(24*a^2*c^2*(a + c*x^2)^2) + (A*c^2*d^2*x + a^2* e*(2*C*d + B*e + C*e*x) - a*c*(C*d^2*x + A*e*(2*d + e*x) + B*d*(d + 2*e*x) ))/(6*a*c^2*(a + c*x^2)^3) + ((A*c*(5*c*d^2 + a*e^2) + a*(a*C*e^2 + c*d*(C *d + 2*B*e)))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(5/2))
Time = 0.43 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2176, 25, 675, 215, 218}
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.
| \(\displaystyle \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle -\frac {\int -\frac {(d+e x) (5 A c d+a C d+2 a B e+3 (A c+a C) e x)}{\left (c x^2+a\right )^3}dx}{6 a c}-\frac {(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(d+e x) (5 A c d+a C d+2 a B e+3 (A c+a C) e x)}{\left (c x^2+a\right )^3}dx}{6 a c}-\frac {(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 675 |
| \(\displaystyle \frac {\frac {3 \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right ) \int \frac {1}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {x \left (3 a e^2 (a C+A c)-c d (2 a B e+a C d+5 A c d)\right )}{4 a c \left (a+c x^2\right )^2}-\frac {e (a B e+2 a C d+4 A c d)}{2 c \left (a+c x^2\right )^2}}{6 a c}-\frac {(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {3 \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right ) \left (\frac {\int \frac {1}{c x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+c x^2\right )}\right )}{4 a c}-\frac {x \left (3 a e^2 (a C+A c)-c d (2 a B e+a C d+5 A c d)\right )}{4 a c \left (a+c x^2\right )^2}-\frac {e (a B e+2 a C d+4 A c d)}{2 c \left (a+c x^2\right )^2}}{6 a c}-\frac {(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}+\frac {x}{2 a \left (a+c x^2\right )}\right ) \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right )}{4 a c}-\frac {x \left (3 a e^2 (a C+A c)-c d (2 a B e+a C d+5 A c d)\right )}{4 a c \left (a+c x^2\right )^2}-\frac {e (a B e+2 a C d+4 A c d)}{2 c \left (a+c x^2\right )^2}}{6 a c}-\frac {(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}\) |
-1/6*((a*B - (A*c - a*C)*x)*(d + e*x)^2)/(a*c*(a + c*x^2)^3) + (-1/2*(e*(4 *A*c*d + 2*a*C*d + a*B*e))/(c*(a + c*x^2)^2) - ((3*a*(A*c + a*C)*e^2 - c*d *(5*A*c*d + a*C*d + 2*a*B*e))*x)/(4*a*c*(a + c*x^2)^2) + (3*(a*(A*c + a*C) *e^2 + c*d*(5*A*c*d + a*C*d + 2*a*B*e))*(x/(2*a*(a + c*x^2)) + ArcTan[(Sqr t[c]*x)/Sqrt[a]]/(2*a^(3/2)*Sqrt[c])))/(4*a*c))/(6*a*c)
3.1.66.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)) Int[(a + c*x^2)^(p + 1), x], x]) / ; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && !(IntegerQ[p] && NiceSqrtQ [(-a)*c])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 , x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && !(IGtQ[m, 0] && R ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 0.60 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.19
| method | result | size |
| default | \(\frac {\frac {\left (A a c \,e^{2}+5 A \,c^{2} d^{2}+2 B a c d e +a^{2} C \,e^{2}+C a c \,d^{2}\right ) x^{5}}{16 a^{3}}+\frac {\left (A a c \,e^{2}+5 A \,c^{2} d^{2}+2 B a c d e -a^{2} C \,e^{2}+C a c \,d^{2}\right ) x^{3}}{6 c \,a^{2}}-\frac {e \left (B e +2 C d \right ) x^{2}}{4 c}-\frac {\left (A a c \,e^{2}-11 A \,c^{2} d^{2}+2 B a c d e +a^{2} C \,e^{2}+C a c \,d^{2}\right ) x}{16 a \,c^{2}}-\frac {4 A c d e +B a \,e^{2}+2 B c \,d^{2}+2 a d e C}{12 c^{2}}}{\left (c \,x^{2}+a \right )^{3}}+\frac {\left (A a c \,e^{2}+5 A \,c^{2} d^{2}+2 B a c d e +a^{2} C \,e^{2}+C a c \,d^{2}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 a^{3} c^{2} \sqrt {a c}}\) | \(268\) |
| risch | \(\frac {\frac {\left (A a c \,e^{2}+5 A \,c^{2} d^{2}+2 B a c d e +a^{2} C \,e^{2}+C a c \,d^{2}\right ) x^{5}}{16 a^{3}}+\frac {\left (A a c \,e^{2}+5 A \,c^{2} d^{2}+2 B a c d e -a^{2} C \,e^{2}+C a c \,d^{2}\right ) x^{3}}{6 c \,a^{2}}-\frac {e \left (B e +2 C d \right ) x^{2}}{4 c}-\frac {\left (A a c \,e^{2}-11 A \,c^{2} d^{2}+2 B a c d e +a^{2} C \,e^{2}+C a c \,d^{2}\right ) x}{16 a \,c^{2}}-\frac {4 A c d e +B a \,e^{2}+2 B c \,d^{2}+2 a d e C}{12 c^{2}}}{\left (c \,x^{2}+a \right )^{3}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) A \,e^{2}}{32 \sqrt {-a c}\, c \,a^{2}}-\frac {5 \ln \left (c x +\sqrt {-a c}\right ) A \,d^{2}}{32 \sqrt {-a c}\, a^{3}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) B d e}{16 \sqrt {-a c}\, c \,a^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) C \,e^{2}}{32 \sqrt {-a c}\, c^{2} a}-\frac {\ln \left (c x +\sqrt {-a c}\right ) C \,d^{2}}{32 \sqrt {-a c}\, c \,a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) A \,e^{2}}{32 \sqrt {-a c}\, c \,a^{2}}+\frac {5 \ln \left (-c x +\sqrt {-a c}\right ) A \,d^{2}}{32 \sqrt {-a c}\, a^{3}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) B d e}{16 \sqrt {-a c}\, c \,a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) C \,e^{2}}{32 \sqrt {-a c}\, c^{2} a}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) C \,d^{2}}{32 \sqrt {-a c}\, c \,a^{2}}\) | \(494\) |
(1/16*(A*a*c*e^2+5*A*c^2*d^2+2*B*a*c*d*e+C*a^2*e^2+C*a*c*d^2)/a^3*x^5+1/6* (A*a*c*e^2+5*A*c^2*d^2+2*B*a*c*d*e-C*a^2*e^2+C*a*c*d^2)/c/a^2*x^3-1/4*e*(B *e+2*C*d)*x^2/c-1/16*(A*a*c*e^2-11*A*c^2*d^2+2*B*a*c*d*e+C*a^2*e^2+C*a*c*d ^2)/a/c^2*x-1/12*(4*A*c*d*e+B*a*e^2+2*B*c*d^2+2*C*a*d*e)/c^2)/(c*x^2+a)^3+ 1/16*(A*a*c*e^2+5*A*c^2*d^2+2*B*a*c*d*e+C*a^2*e^2+C*a*c*d^2)/a^3/c^2/(a*c) ^(1/2)*arctan(c*x/(a*c)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (208) = 416\).
Time = 0.54 (sec) , antiderivative size = 1062, normalized size of antiderivative = 4.72 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=\left [-\frac {16 \, B a^{4} c^{2} d^{2} + 8 \, B a^{5} c e^{2} - 6 \, {\left (2 \, B a^{2} c^{4} d e + {\left (C a^{2} c^{4} + 5 \, A a c^{5}\right )} d^{2} + {\left (C a^{3} c^{3} + A a^{2} c^{4}\right )} e^{2}\right )} x^{5} - 16 \, {\left (2 \, B a^{3} c^{3} d e + {\left (C a^{3} c^{3} + 5 \, A a^{2} c^{4}\right )} d^{2} - {\left (C a^{4} c^{2} - A a^{3} c^{3}\right )} e^{2}\right )} x^{3} + 16 \, {\left (C a^{5} c + 2 \, A a^{4} c^{2}\right )} d e + 24 \, {\left (2 \, C a^{4} c^{2} d e + B a^{4} c^{2} e^{2}\right )} x^{2} + 3 \, {\left (2 \, B a^{4} c d e + {\left (2 \, B a c^{4} d e + {\left (C a c^{4} + 5 \, A c^{5}\right )} d^{2} + {\left (C a^{2} c^{3} + A a c^{4}\right )} e^{2}\right )} x^{6} + 3 \, {\left (2 \, B a^{2} c^{3} d e + {\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} d^{2} + {\left (C a^{3} c^{2} + A a^{2} c^{3}\right )} e^{2}\right )} x^{4} + {\left (C a^{4} c + 5 \, A a^{3} c^{2}\right )} d^{2} + {\left (C a^{5} + A a^{4} c\right )} e^{2} + 3 \, {\left (2 \, B a^{3} c^{2} d e + {\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} d^{2} + {\left (C a^{4} c + A a^{3} c^{2}\right )} e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 6 \, {\left (2 \, B a^{4} c^{2} d e + {\left (C a^{4} c^{2} - 11 \, A a^{3} c^{3}\right )} d^{2} + {\left (C a^{5} c + A a^{4} c^{2}\right )} e^{2}\right )} x}{96 \, {\left (a^{4} c^{6} x^{6} + 3 \, a^{5} c^{5} x^{4} + 3 \, a^{6} c^{4} x^{2} + a^{7} c^{3}\right )}}, -\frac {8 \, B a^{4} c^{2} d^{2} + 4 \, B a^{5} c e^{2} - 3 \, {\left (2 \, B a^{2} c^{4} d e + {\left (C a^{2} c^{4} + 5 \, A a c^{5}\right )} d^{2} + {\left (C a^{3} c^{3} + A a^{2} c^{4}\right )} e^{2}\right )} x^{5} - 8 \, {\left (2 \, B a^{3} c^{3} d e + {\left (C a^{3} c^{3} + 5 \, A a^{2} c^{4}\right )} d^{2} - {\left (C a^{4} c^{2} - A a^{3} c^{3}\right )} e^{2}\right )} x^{3} + 8 \, {\left (C a^{5} c + 2 \, A a^{4} c^{2}\right )} d e + 12 \, {\left (2 \, C a^{4} c^{2} d e + B a^{4} c^{2} e^{2}\right )} x^{2} - 3 \, {\left (2 \, B a^{4} c d e + {\left (2 \, B a c^{4} d e + {\left (C a c^{4} + 5 \, A c^{5}\right )} d^{2} + {\left (C a^{2} c^{3} + A a c^{4}\right )} e^{2}\right )} x^{6} + 3 \, {\left (2 \, B a^{2} c^{3} d e + {\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} d^{2} + {\left (C a^{3} c^{2} + A a^{2} c^{3}\right )} e^{2}\right )} x^{4} + {\left (C a^{4} c + 5 \, A a^{3} c^{2}\right )} d^{2} + {\left (C a^{5} + A a^{4} c\right )} e^{2} + 3 \, {\left (2 \, B a^{3} c^{2} d e + {\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} d^{2} + {\left (C a^{4} c + A a^{3} c^{2}\right )} e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 3 \, {\left (2 \, B a^{4} c^{2} d e + {\left (C a^{4} c^{2} - 11 \, A a^{3} c^{3}\right )} d^{2} + {\left (C a^{5} c + A a^{4} c^{2}\right )} e^{2}\right )} x}{48 \, {\left (a^{4} c^{6} x^{6} + 3 \, a^{5} c^{5} x^{4} + 3 \, a^{6} c^{4} x^{2} + a^{7} c^{3}\right )}}\right ] \]
[-1/96*(16*B*a^4*c^2*d^2 + 8*B*a^5*c*e^2 - 6*(2*B*a^2*c^4*d*e + (C*a^2*c^4 + 5*A*a*c^5)*d^2 + (C*a^3*c^3 + A*a^2*c^4)*e^2)*x^5 - 16*(2*B*a^3*c^3*d*e + (C*a^3*c^3 + 5*A*a^2*c^4)*d^2 - (C*a^4*c^2 - A*a^3*c^3)*e^2)*x^3 + 16*( C*a^5*c + 2*A*a^4*c^2)*d*e + 24*(2*C*a^4*c^2*d*e + B*a^4*c^2*e^2)*x^2 + 3* (2*B*a^4*c*d*e + (2*B*a*c^4*d*e + (C*a*c^4 + 5*A*c^5)*d^2 + (C*a^2*c^3 + A *a*c^4)*e^2)*x^6 + 3*(2*B*a^2*c^3*d*e + (C*a^2*c^3 + 5*A*a*c^4)*d^2 + (C*a ^3*c^2 + A*a^2*c^3)*e^2)*x^4 + (C*a^4*c + 5*A*a^3*c^2)*d^2 + (C*a^5 + A*a^ 4*c)*e^2 + 3*(2*B*a^3*c^2*d*e + (C*a^3*c^2 + 5*A*a^2*c^3)*d^2 + (C*a^4*c + A*a^3*c^2)*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 6*(2*B*a^4*c^2*d*e + (C*a^4*c^2 - 11*A*a^3*c^3)*d^2 + (C*a^5*c + A* a^4*c^2)*e^2)*x)/(a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 + a^7*c^3), -1/48*(8*B*a^4*c^2*d^2 + 4*B*a^5*c*e^2 - 3*(2*B*a^2*c^4*d*e + (C*a^2*c^4 + 5*A*a*c^5)*d^2 + (C*a^3*c^3 + A*a^2*c^4)*e^2)*x^5 - 8*(2*B*a^3*c^3*d*e + (C*a^3*c^3 + 5*A*a^2*c^4)*d^2 - (C*a^4*c^2 - A*a^3*c^3)*e^2)*x^3 + 8*(C*a^ 5*c + 2*A*a^4*c^2)*d*e + 12*(2*C*a^4*c^2*d*e + B*a^4*c^2*e^2)*x^2 - 3*(2*B *a^4*c*d*e + (2*B*a*c^4*d*e + (C*a*c^4 + 5*A*c^5)*d^2 + (C*a^2*c^3 + A*a*c ^4)*e^2)*x^6 + 3*(2*B*a^2*c^3*d*e + (C*a^2*c^3 + 5*A*a*c^4)*d^2 + (C*a^3*c ^2 + A*a^2*c^3)*e^2)*x^4 + (C*a^4*c + 5*A*a^3*c^2)*d^2 + (C*a^5 + A*a^4*c) *e^2 + 3*(2*B*a^3*c^2*d*e + (C*a^3*c^2 + 5*A*a^2*c^3)*d^2 + (C*a^4*c + A*a ^3*c^2)*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 3*(2*B*a^4*c^2*d*e ...
Timed out. \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=-\frac {8 \, B a^{3} c d^{2} + 4 \, B a^{4} e^{2} - 3 \, {\left (2 \, B a c^{3} d e + {\left (C a c^{3} + 5 \, A c^{4}\right )} d^{2} + {\left (C a^{2} c^{2} + A a c^{3}\right )} e^{2}\right )} x^{5} - 8 \, {\left (2 \, B a^{2} c^{2} d e + {\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} d^{2} - {\left (C a^{3} c - A a^{2} c^{2}\right )} e^{2}\right )} x^{3} + 8 \, {\left (C a^{4} + 2 \, A a^{3} c\right )} d e + 12 \, {\left (2 \, C a^{3} c d e + B a^{3} c e^{2}\right )} x^{2} + 3 \, {\left (2 \, B a^{3} c d e + {\left (C a^{3} c - 11 \, A a^{2} c^{2}\right )} d^{2} + {\left (C a^{4} + A a^{3} c\right )} e^{2}\right )} x}{48 \, {\left (a^{3} c^{5} x^{6} + 3 \, a^{4} c^{4} x^{4} + 3 \, a^{5} c^{3} x^{2} + a^{6} c^{2}\right )}} + \frac {{\left (2 \, B a c d e + {\left (C a c + 5 \, A c^{2}\right )} d^{2} + {\left (C a^{2} + A a c\right )} e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} \]
-1/48*(8*B*a^3*c*d^2 + 4*B*a^4*e^2 - 3*(2*B*a*c^3*d*e + (C*a*c^3 + 5*A*c^4 )*d^2 + (C*a^2*c^2 + A*a*c^3)*e^2)*x^5 - 8*(2*B*a^2*c^2*d*e + (C*a^2*c^2 + 5*A*a*c^3)*d^2 - (C*a^3*c - A*a^2*c^2)*e^2)*x^3 + 8*(C*a^4 + 2*A*a^3*c)*d *e + 12*(2*C*a^3*c*d*e + B*a^3*c*e^2)*x^2 + 3*(2*B*a^3*c*d*e + (C*a^3*c - 11*A*a^2*c^2)*d^2 + (C*a^4 + A*a^3*c)*e^2)*x)/(a^3*c^5*x^6 + 3*a^4*c^4*x^4 + 3*a^5*c^3*x^2 + a^6*c^2) + 1/16*(2*B*a*c*d*e + (C*a*c + 5*A*c^2)*d^2 + (C*a^2 + A*a*c)*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c^2)
Time = 0.27 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.47 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=\frac {{\left (C a c d^{2} + 5 \, A c^{2} d^{2} + 2 \, B a c d e + C a^{2} e^{2} + A a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} + \frac {3 \, C a c^{3} d^{2} x^{5} + 15 \, A c^{4} d^{2} x^{5} + 6 \, B a c^{3} d e x^{5} + 3 \, C a^{2} c^{2} e^{2} x^{5} + 3 \, A a c^{3} e^{2} x^{5} + 8 \, C a^{2} c^{2} d^{2} x^{3} + 40 \, A a c^{3} d^{2} x^{3} + 16 \, B a^{2} c^{2} d e x^{3} - 8 \, C a^{3} c e^{2} x^{3} + 8 \, A a^{2} c^{2} e^{2} x^{3} - 24 \, C a^{3} c d e x^{2} - 12 \, B a^{3} c e^{2} x^{2} - 3 \, C a^{3} c d^{2} x + 33 \, A a^{2} c^{2} d^{2} x - 6 \, B a^{3} c d e x - 3 \, C a^{4} e^{2} x - 3 \, A a^{3} c e^{2} x - 8 \, B a^{3} c d^{2} - 8 \, C a^{4} d e - 16 \, A a^{3} c d e - 4 \, B a^{4} e^{2}}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \]
1/16*(C*a*c*d^2 + 5*A*c^2*d^2 + 2*B*a*c*d*e + C*a^2*e^2 + A*a*c*e^2)*arcta n(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c^2) + 1/48*(3*C*a*c^3*d^2*x^5 + 15*A*c^4* d^2*x^5 + 6*B*a*c^3*d*e*x^5 + 3*C*a^2*c^2*e^2*x^5 + 3*A*a*c^3*e^2*x^5 + 8* C*a^2*c^2*d^2*x^3 + 40*A*a*c^3*d^2*x^3 + 16*B*a^2*c^2*d*e*x^3 - 8*C*a^3*c* e^2*x^3 + 8*A*a^2*c^2*e^2*x^3 - 24*C*a^3*c*d*e*x^2 - 12*B*a^3*c*e^2*x^2 - 3*C*a^3*c*d^2*x + 33*A*a^2*c^2*d^2*x - 6*B*a^3*c*d*e*x - 3*C*a^4*e^2*x - 3 *A*a^3*c*e^2*x - 8*B*a^3*c*d^2 - 8*C*a^4*d*e - 16*A*a^3*c*d*e - 4*B*a^4*e^ 2)/((c*x^2 + a)^3*a^3*c^2)
Time = 0.23 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2+5\,A\,c^2\,d^2\right )}{16\,a^{7/2}\,c^{5/2}}-\frac {\frac {B\,a\,e^2+2\,B\,c\,d^2+4\,A\,c\,d\,e+2\,C\,a\,d\,e}{12\,c^2}-\frac {x^5\,\left (C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2+5\,A\,c^2\,d^2\right )}{16\,a^3}+\frac {x^2\,\left (B\,e^2+2\,C\,d\,e\right )}{4\,c}+\frac {x\,\left (C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2-11\,A\,c^2\,d^2\right )}{16\,a\,c^2}-\frac {x^3\,\left (-C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2+5\,A\,c^2\,d^2\right )}{6\,a^2\,c}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6} \]
(atan((c^(1/2)*x)/a^(1/2))*(5*A*c^2*d^2 + C*a^2*e^2 + A*a*c*e^2 + C*a*c*d^ 2 + 2*B*a*c*d*e))/(16*a^(7/2)*c^(5/2)) - ((B*a*e^2 + 2*B*c*d^2 + 4*A*c*d*e + 2*C*a*d*e)/(12*c^2) - (x^5*(5*A*c^2*d^2 + C*a^2*e^2 + A*a*c*e^2 + C*a*c *d^2 + 2*B*a*c*d*e))/(16*a^3) + (x^2*(B*e^2 + 2*C*d*e))/(4*c) + (x*(C*a^2* e^2 - 11*A*c^2*d^2 + A*a*c*e^2 + C*a*c*d^2 + 2*B*a*c*d*e))/(16*a*c^2) - (x ^3*(5*A*c^2*d^2 - C*a^2*e^2 + A*a*c*e^2 + C*a*c*d^2 + 2*B*a*c*d*e))/(6*a^2 *c))/(a^3 + c^3*x^6 + 3*a^2*c*x^2 + 3*a*c^2*x^4)