Integrand size = 24, antiderivative size = 250 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^2 x}{e^4}+\frac {\left (c d^2-b d e+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\left (19 c d^2-7 b d e-5 a e^2\right ) \left (c d^2-b d e+a e^2\right ) x}{24 d^2 e^4 \left (d+e x^2\right )^2}+\frac {\left (29 c^2 d^4-2 c d^2 e (11 b d-a e)+e^2 \left (b^2 d^2+2 a b d e+5 a^2 e^2\right )\right ) x}{16 d^3 e^4 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 c d^2 e (5 b d+a e)-e^2 \left (b^2 d^2+2 a b d e+5 a^2 e^2\right )\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \]
c^2*x/e^4+1/6*(a*e^2-b*d*e+c*d^2)^2*x/d/e^4/(e*x^2+d)^3-1/24*(-5*a*e^2-7*b *d*e+19*c*d^2)*(a*e^2-b*d*e+c*d^2)*x/d^2/e^4/(e*x^2+d)^2+1/16*(29*c^2*d^4- 2*c*d^2*e*(-a*e+11*b*d)+e^2*(5*a^2*e^2+2*a*b*d*e+b^2*d^2))*x/d^3/e^4/(e*x^ 2+d)-1/16*(35*c^2*d^4-2*c*d^2*e*(a*e+5*b*d)-e^2*(5*a^2*e^2+2*a*b*d*e+b^2*d ^2))*arctan(x*e^(1/2)/d^(1/2))/d^(7/2)/e^(9/2)
Time = 0.10 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^2 x}{e^4}+\frac {\left (c d^2+e (-b d+a e)\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\left (19 c^2 d^4+2 c d^2 e (-13 b d+7 a e)+e^2 \left (7 b^2 d^2-2 a b d e-5 a^2 e^2\right )\right ) x}{24 d^2 e^4 \left (d+e x^2\right )^2}+\frac {\left (29 c^2 d^4+2 c d^2 e (-11 b d+a e)+e^2 \left (b^2 d^2+2 a b d e+5 a^2 e^2\right )\right ) x}{16 d^3 e^4 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 c d^2 e (5 b d+a e)-e^2 \left (b^2 d^2+2 a b d e+5 a^2 e^2\right )\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \]
(c^2*x)/e^4 + ((c*d^2 + e*(-(b*d) + a*e))^2*x)/(6*d*e^4*(d + e*x^2)^3) - ( (19*c^2*d^4 + 2*c*d^2*e*(-13*b*d + 7*a*e) + e^2*(7*b^2*d^2 - 2*a*b*d*e - 5 *a^2*e^2))*x)/(24*d^2*e^4*(d + e*x^2)^2) + ((29*c^2*d^4 + 2*c*d^2*e*(-11*b *d + a*e) + e^2*(b^2*d^2 + 2*a*b*d*e + 5*a^2*e^2))*x)/(16*d^3*e^4*(d + e*x ^2)) - ((35*c^2*d^4 - 2*c*d^2*e*(5*b*d + a*e) - e^2*(b^2*d^2 + 2*a*b*d*e + 5*a^2*e^2))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*e^(9/2))
Time = 0.77 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1471, 2345, 27, 1471, 27, 299, 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.
| \(\displaystyle \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\int \frac {-\frac {6 c^2 d x^6}{e}+\frac {6 c d (c d-2 b e) x^4}{e^2}-\frac {6 d \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right ) x^2}{e^3}+\frac {c^2 d^4-2 c e (b d-a e) d^2+e^2 \left (b^2 d^2-2 a b e d-5 a^2 e^2\right )}{e^4}}{\left (e x^2+d\right )^3}dx}{6 d}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (-5 a e^2-7 b d e+19 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\int \frac {3 \left (\frac {8 c^2 d^2 x^4}{e^2}-\frac {16 c d^2 (c d-b e) x^2}{e^3}+\frac {5 c^2 d^4-2 c e (3 b d-a e) d^2+e^2 \left (b^2 d^2+2 a b e d+5 a^2 e^2\right )}{e^4}\right )}{\left (e x^2+d\right )^2}dx}{4 d}}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (-5 a e^2-7 b d e+19 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \int \frac {\frac {8 c^2 d^2 x^4}{e^2}-\frac {16 c d^2 (c d-b e) x^2}{e^3}+\frac {5 c^2 d^4-2 c e (3 b d-a e) d^2+e^2 \left (b^2 d^2+2 a b e d+5 a^2 e^2\right )}{e^4}}{\left (e x^2+d\right )^2}dx}{4 d}}{6 d}\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (-5 a e^2-7 b d e+19 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (e^2 \left (5 a^2 e^2+2 a b d e+b^2 d^2\right )-2 c d^2 e (11 b d-a e)+29 c^2 d^4\right )}{2 d e^4 \left (d+e x^2\right )}-\frac {\int \frac {19 c^2 d^4-16 c^2 e x^2 d^3-2 c e (5 b d+a e) d^2-e^2 \left (b^2 d^2+2 a b e d+5 a^2 e^2\right )}{e^4 \left (e x^2+d\right )}dx}{2 d}\right )}{4 d}}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (-5 a e^2-7 b d e+19 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (e^2 \left (5 a^2 e^2+2 a b d e+b^2 d^2\right )-2 c d^2 e (11 b d-a e)+29 c^2 d^4\right )}{2 d e^4 \left (d+e x^2\right )}-\frac {\int \frac {19 c^2 d^4-16 c^2 e x^2 d^3-2 c e (5 b d+a e) d^2-e^2 \left (b^2 d^2+2 a b e d+5 a^2 e^2\right )}{e x^2+d}dx}{2 d e^4}\right )}{4 d}}{6 d}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (-5 a e^2-7 b d e+19 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (e^2 \left (5 a^2 e^2+2 a b d e+b^2 d^2\right )-2 c d^2 e (11 b d-a e)+29 c^2 d^4\right )}{2 d e^4 \left (d+e x^2\right )}-\frac {\left (-e^2 \left (5 a^2 e^2+2 a b d e+b^2 d^2\right )-2 c d^2 e (a e+5 b d)+35 c^2 d^4\right ) \int \frac {1}{e x^2+d}dx-16 c^2 d^3 x}{2 d e^4}\right )}{4 d}}{6 d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (-5 a e^2-7 b d e+19 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (e^2 \left (5 a^2 e^2+2 a b d e+b^2 d^2\right )-2 c d^2 e (11 b d-a e)+29 c^2 d^4\right )}{2 d e^4 \left (d+e x^2\right )}-\frac {\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-e^2 \left (5 a^2 e^2+2 a b d e+b^2 d^2\right )-2 c d^2 e (a e+5 b d)+35 c^2 d^4\right )}{\sqrt {d} \sqrt {e}}-16 c^2 d^3 x}{2 d e^4}\right )}{4 d}}{6 d}\) |
((c*d^2 - b*d*e + a*e^2)^2*x)/(6*d*e^4*(d + e*x^2)^3) - (((19*c*d^2 - 7*b* d*e - 5*a*e^2)*(c*d^2 - b*d*e + a*e^2)*x)/(4*d*e^4*(d + e*x^2)^2) - (3*((( 29*c^2*d^4 - 2*c*d^2*e*(11*b*d - a*e) + e^2*(b^2*d^2 + 2*a*b*d*e + 5*a^2*e ^2))*x)/(2*d*e^4*(d + e*x^2)) - (-16*c^2*d^3*x + ((35*c^2*d^4 - 2*c*d^2*e* (5*b*d + a*e) - e^2*(b^2*d^2 + 2*a*b*d*e + 5*a^2*e^2))*ArcTan[(Sqrt[e]*x)/ Sqrt[d]])/(Sqrt[d]*Sqrt[e]))/(2*d*e^4)))/(4*d))/(6*d)
3.3.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], 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] && LtQ[q, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[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[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.26 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.14
| method | result | size |
| default | \(\frac {c^{2} x}{e^{4}}+\frac {\frac {\frac {e^{2} \left (5 a^{2} e^{4}+2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-22 b c \,d^{3} e +29 c^{2} d^{4}\right ) x^{5}}{16 d^{3}}+\frac {e \left (5 a^{2} e^{4}+2 a b d \,e^{3}-2 a c \,d^{2} e^{2}-b^{2} d^{2} e^{2}-10 b c \,d^{3} e +17 c^{2} d^{4}\right ) x^{3}}{6 d^{2}}+\frac {\left (11 a^{2} e^{4}-2 a b d \,e^{3}-2 a c \,d^{2} e^{2}-b^{2} d^{2} e^{2}-10 b c \,d^{3} e +19 c^{2} d^{4}\right ) x}{16 d}}{\left (e \,x^{2}+d \right )^{3}}+\frac {\left (5 a^{2} e^{4}+2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+10 b c \,d^{3} e -35 c^{2} d^{4}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{16 d^{3} \sqrt {e d}}}{e^{4}}\) | \(285\) |
| risch | \(\frac {c^{2} x}{e^{4}}+\frac {\frac {e^{2} \left (5 a^{2} e^{4}+2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-22 b c \,d^{3} e +29 c^{2} d^{4}\right ) x^{5}}{16 d^{3}}+\frac {e \left (5 a^{2} e^{4}+2 a b d \,e^{3}-2 a c \,d^{2} e^{2}-b^{2} d^{2} e^{2}-10 b c \,d^{3} e +17 c^{2} d^{4}\right ) x^{3}}{6 d^{2}}+\frac {\left (11 a^{2} e^{4}-2 a b d \,e^{3}-2 a c \,d^{2} e^{2}-b^{2} d^{2} e^{2}-10 b c \,d^{3} e +19 c^{2} d^{4}\right ) x}{16 d}}{e^{4} \left (e \,x^{2}+d \right )^{3}}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) a^{2}}{32 \sqrt {-e d}\, d^{3}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) a b}{16 e \sqrt {-e d}\, d^{2}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) a c}{16 e^{2} \sqrt {-e d}\, d}-\frac {\ln \left (e x +\sqrt {-e d}\right ) b^{2}}{32 e^{2} \sqrt {-e d}\, d}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) b c}{16 e^{3} \sqrt {-e d}}+\frac {35 d \ln \left (e x +\sqrt {-e d}\right ) c^{2}}{32 e^{4} \sqrt {-e d}}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) a^{2}}{32 \sqrt {-e d}\, d^{3}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) a b}{16 e \sqrt {-e d}\, d^{2}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) a c}{16 e^{2} \sqrt {-e d}\, d}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) b^{2}}{32 e^{2} \sqrt {-e d}\, d}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) b c}{16 e^{3} \sqrt {-e d}}-\frac {35 d \ln \left (-e x +\sqrt {-e d}\right ) c^{2}}{32 e^{4} \sqrt {-e d}}\) | \(531\) |
c^2*x/e^4+1/e^4*((1/16*e^2*(5*a^2*e^4+2*a*b*d*e^3+2*a*c*d^2*e^2+b^2*d^2*e^ 2-22*b*c*d^3*e+29*c^2*d^4)/d^3*x^5+1/6*e*(5*a^2*e^4+2*a*b*d*e^3-2*a*c*d^2* e^2-b^2*d^2*e^2-10*b*c*d^3*e+17*c^2*d^4)/d^2*x^3+1/16*(11*a^2*e^4-2*a*b*d* e^3-2*a*c*d^2*e^2-b^2*d^2*e^2-10*b*c*d^3*e+19*c^2*d^4)/d*x)/(e*x^2+d)^3+1/ 16*(5*a^2*e^4+2*a*b*d*e^3+2*a*c*d^2*e^2+b^2*d^2*e^2+10*b*c*d^3*e-35*c^2*d^ 4)/d^3/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (234) = 468\).
Time = 0.27 (sec) , antiderivative size = 1016, normalized size of antiderivative = 4.06 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\left [\frac {96 \, c^{2} d^{4} e^{4} x^{7} + 6 \, {\left (77 \, c^{2} d^{5} e^{3} - 22 \, b c d^{4} e^{4} + 2 \, a b d^{2} e^{6} + 5 \, a^{2} d e^{7} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{5}\right )} x^{5} + 16 \, {\left (35 \, c^{2} d^{6} e^{2} - 10 \, b c d^{5} e^{3} + 2 \, a b d^{3} e^{5} + 5 \, a^{2} d^{2} e^{6} - {\left (b^{2} + 2 \, a c\right )} d^{4} e^{4}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{7} - 10 \, b c d^{6} e - 2 \, a b d^{4} e^{3} - 5 \, a^{2} d^{3} e^{4} - {\left (b^{2} + 2 \, a c\right )} d^{5} e^{2} + {\left (35 \, c^{2} d^{4} e^{3} - 10 \, b c d^{3} e^{4} - 2 \, a b d e^{6} - 5 \, a^{2} e^{7} - {\left (b^{2} + 2 \, a c\right )} d^{2} e^{5}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 10 \, b c d^{4} e^{3} - 2 \, a b d^{2} e^{5} - 5 \, a^{2} d e^{6} - {\left (b^{2} + 2 \, a c\right )} d^{3} e^{4}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 10 \, b c d^{5} e^{2} - 2 \, a b d^{3} e^{4} - 5 \, a^{2} d^{2} e^{5} - {\left (b^{2} + 2 \, a c\right )} d^{4} e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 6 \, {\left (35 \, c^{2} d^{7} e - 10 \, b c d^{6} e^{2} - 2 \, a b d^{4} e^{4} + 11 \, a^{2} d^{3} e^{5} - {\left (b^{2} + 2 \, a c\right )} d^{5} e^{3}\right )} x}{96 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}, \frac {48 \, c^{2} d^{4} e^{4} x^{7} + 3 \, {\left (77 \, c^{2} d^{5} e^{3} - 22 \, b c d^{4} e^{4} + 2 \, a b d^{2} e^{6} + 5 \, a^{2} d e^{7} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{5}\right )} x^{5} + 8 \, {\left (35 \, c^{2} d^{6} e^{2} - 10 \, b c d^{5} e^{3} + 2 \, a b d^{3} e^{5} + 5 \, a^{2} d^{2} e^{6} - {\left (b^{2} + 2 \, a c\right )} d^{4} e^{4}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{7} - 10 \, b c d^{6} e - 2 \, a b d^{4} e^{3} - 5 \, a^{2} d^{3} e^{4} - {\left (b^{2} + 2 \, a c\right )} d^{5} e^{2} + {\left (35 \, c^{2} d^{4} e^{3} - 10 \, b c d^{3} e^{4} - 2 \, a b d e^{6} - 5 \, a^{2} e^{7} - {\left (b^{2} + 2 \, a c\right )} d^{2} e^{5}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 10 \, b c d^{4} e^{3} - 2 \, a b d^{2} e^{5} - 5 \, a^{2} d e^{6} - {\left (b^{2} + 2 \, a c\right )} d^{3} e^{4}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 10 \, b c d^{5} e^{2} - 2 \, a b d^{3} e^{4} - 5 \, a^{2} d^{2} e^{5} - {\left (b^{2} + 2 \, a c\right )} d^{4} e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + 3 \, {\left (35 \, c^{2} d^{7} e - 10 \, b c d^{6} e^{2} - 2 \, a b d^{4} e^{4} + 11 \, a^{2} d^{3} e^{5} - {\left (b^{2} + 2 \, a c\right )} d^{5} e^{3}\right )} x}{48 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}\right ] \]
[1/96*(96*c^2*d^4*e^4*x^7 + 6*(77*c^2*d^5*e^3 - 22*b*c*d^4*e^4 + 2*a*b*d^2 *e^6 + 5*a^2*d*e^7 + (b^2 + 2*a*c)*d^3*e^5)*x^5 + 16*(35*c^2*d^6*e^2 - 10* b*c*d^5*e^3 + 2*a*b*d^3*e^5 + 5*a^2*d^2*e^6 - (b^2 + 2*a*c)*d^4*e^4)*x^3 + 3*(35*c^2*d^7 - 10*b*c*d^6*e - 2*a*b*d^4*e^3 - 5*a^2*d^3*e^4 - (b^2 + 2*a *c)*d^5*e^2 + (35*c^2*d^4*e^3 - 10*b*c*d^3*e^4 - 2*a*b*d*e^6 - 5*a^2*e^7 - (b^2 + 2*a*c)*d^2*e^5)*x^6 + 3*(35*c^2*d^5*e^2 - 10*b*c*d^4*e^3 - 2*a*b*d ^2*e^5 - 5*a^2*d*e^6 - (b^2 + 2*a*c)*d^3*e^4)*x^4 + 3*(35*c^2*d^6*e - 10*b *c*d^5*e^2 - 2*a*b*d^3*e^4 - 5*a^2*d^2*e^5 - (b^2 + 2*a*c)*d^4*e^3)*x^2)*s qrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) + 6*(35*c^2*d^7*e - 10*b*c*d^6*e^2 - 2*a*b*d^4*e^4 + 11*a^2*d^3*e^5 - (b^2 + 2*a*c)*d^5*e^3) *x)/(d^4*e^8*x^6 + 3*d^5*e^7*x^4 + 3*d^6*e^6*x^2 + d^7*e^5), 1/48*(48*c^2* d^4*e^4*x^7 + 3*(77*c^2*d^5*e^3 - 22*b*c*d^4*e^4 + 2*a*b*d^2*e^6 + 5*a^2*d *e^7 + (b^2 + 2*a*c)*d^3*e^5)*x^5 + 8*(35*c^2*d^6*e^2 - 10*b*c*d^5*e^3 + 2 *a*b*d^3*e^5 + 5*a^2*d^2*e^6 - (b^2 + 2*a*c)*d^4*e^4)*x^3 - 3*(35*c^2*d^7 - 10*b*c*d^6*e - 2*a*b*d^4*e^3 - 5*a^2*d^3*e^4 - (b^2 + 2*a*c)*d^5*e^2 + ( 35*c^2*d^4*e^3 - 10*b*c*d^3*e^4 - 2*a*b*d*e^6 - 5*a^2*e^7 - (b^2 + 2*a*c)* d^2*e^5)*x^6 + 3*(35*c^2*d^5*e^2 - 10*b*c*d^4*e^3 - 2*a*b*d^2*e^5 - 5*a^2* d*e^6 - (b^2 + 2*a*c)*d^3*e^4)*x^4 + 3*(35*c^2*d^6*e - 10*b*c*d^5*e^2 - 2* a*b*d^3*e^4 - 5*a^2*d^2*e^5 - (b^2 + 2*a*c)*d^4*e^3)*x^2)*sqrt(d*e)*arctan (sqrt(d*e)*x/d) + 3*(35*c^2*d^7*e - 10*b*c*d^6*e^2 - 2*a*b*d^4*e^4 + 11...
Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.29 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^{2} x}{e^{4}} - \frac {{\left (35 \, c^{2} d^{4} - 10 \, b c d^{3} e - b^{2} d^{2} e^{2} - 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} - 5 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{4}} + \frac {87 \, c^{2} d^{4} e^{2} x^{5} - 66 \, b c d^{3} e^{3} x^{5} + 3 \, b^{2} d^{2} e^{4} x^{5} + 6 \, a c d^{2} e^{4} x^{5} + 6 \, a b d e^{5} x^{5} + 15 \, a^{2} e^{6} x^{5} + 136 \, c^{2} d^{5} e x^{3} - 80 \, b c d^{4} e^{2} x^{3} - 8 \, b^{2} d^{3} e^{3} x^{3} - 16 \, a c d^{3} e^{3} x^{3} + 16 \, a b d^{2} e^{4} x^{3} + 40 \, a^{2} d e^{5} x^{3} + 57 \, c^{2} d^{6} x - 30 \, b c d^{5} e x - 3 \, b^{2} d^{4} e^{2} x - 6 \, a c d^{4} e^{2} x - 6 \, a b d^{3} e^{3} x + 33 \, a^{2} d^{2} e^{4} x}{48 \, {\left (e x^{2} + d\right )}^{3} d^{3} e^{4}} \]
c^2*x/e^4 - 1/16*(35*c^2*d^4 - 10*b*c*d^3*e - b^2*d^2*e^2 - 2*a*c*d^2*e^2 - 2*a*b*d*e^3 - 5*a^2*e^4)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3*e^4) + 1/4 8*(87*c^2*d^4*e^2*x^5 - 66*b*c*d^3*e^3*x^5 + 3*b^2*d^2*e^4*x^5 + 6*a*c*d^2 *e^4*x^5 + 6*a*b*d*e^5*x^5 + 15*a^2*e^6*x^5 + 136*c^2*d^5*e*x^3 - 80*b*c*d ^4*e^2*x^3 - 8*b^2*d^3*e^3*x^3 - 16*a*c*d^3*e^3*x^3 + 16*a*b*d^2*e^4*x^3 + 40*a^2*d*e^5*x^3 + 57*c^2*d^6*x - 30*b*c*d^5*e*x - 3*b^2*d^4*e^2*x - 6*a* c*d^4*e^2*x - 6*a*b*d^3*e^3*x + 33*a^2*d^2*e^4*x)/((e*x^2 + d)^3*d^3*e^4)
Time = 7.72 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {\frac {x^5\,\left (5\,a^2\,e^6+2\,a\,b\,d\,e^5+2\,a\,c\,d^2\,e^4+b^2\,d^2\,e^4-22\,b\,c\,d^3\,e^3+29\,c^2\,d^4\,e^2\right )}{16\,d^3}-\frac {x\,\left (-11\,a^2\,e^4+2\,a\,b\,d\,e^3+2\,a\,c\,d^2\,e^2+b^2\,d^2\,e^2+10\,b\,c\,d^3\,e-19\,c^2\,d^4\right )}{16\,d}+\frac {x^3\,\left (5\,a^2\,e^5+2\,a\,b\,d\,e^4-2\,a\,c\,d^2\,e^3-b^2\,d^2\,e^3-10\,b\,c\,d^3\,e^2+17\,c^2\,d^4\,e\right )}{6\,d^2}}{d^3\,e^4+3\,d^2\,e^5\,x^2+3\,d\,e^6\,x^4+e^7\,x^6}+\frac {c^2\,x}{e^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (5\,a^2\,e^4+2\,a\,b\,d\,e^3+2\,a\,c\,d^2\,e^2+b^2\,d^2\,e^2+10\,b\,c\,d^3\,e-35\,c^2\,d^4\right )}{16\,d^{7/2}\,e^{9/2}} \]
((x^5*(5*a^2*e^6 + b^2*d^2*e^4 + 29*c^2*d^4*e^2 + 2*a*b*d*e^5 + 2*a*c*d^2* e^4 - 22*b*c*d^3*e^3))/(16*d^3) - (x*(b^2*d^2*e^2 - 19*c^2*d^4 - 11*a^2*e^ 4 + 2*a*b*d*e^3 + 10*b*c*d^3*e + 2*a*c*d^2*e^2))/(16*d) + (x^3*(5*a^2*e^5 + 17*c^2*d^4*e - b^2*d^2*e^3 + 2*a*b*d*e^4 - 2*a*c*d^2*e^3 - 10*b*c*d^3*e^ 2))/(6*d^2))/(d^3*e^4 + e^7*x^6 + 3*d*e^6*x^4 + 3*d^2*e^5*x^2) + (c^2*x)/e ^4 + (atan((e^(1/2)*x)/d^(1/2))*(5*a^2*e^4 - 35*c^2*d^4 + b^2*d^2*e^2 + 2* a*b*d*e^3 + 10*b*c*d^3*e + 2*a*c*d^2*e^2))/(16*d^(7/2)*e^(9/2))