Integrand size = 24, antiderivative size = 317 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {\left (c d^2-b d e+a e^2\right )^2 x}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\left (25 c d^2-9 b d e-7 a e^2\right ) \left (c d^2-b d e+a e^2\right ) x}{48 d^2 e^4 \left (d+e x^2\right )^3}+\frac {\left (163 c^2 d^4-2 c d^2 e (59 b d-3 a e)+e^2 \left (3 b^2 d^2+10 a b d e+35 a^2 e^2\right )\right ) x}{192 d^3 e^4 \left (d+e x^2\right )^2}-\frac {\left (93 c^2 d^4-2 c d^2 e (5 b d+3 a e)-e^2 \left (3 b^2 d^2+10 a b d e+35 a^2 e^2\right )\right ) x}{128 d^4 e^4 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4+2 c d^2 e (5 b d+3 a e)+e^2 \left (3 b^2 d^2+10 a b d e+35 a^2 e^2\right )\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{128 d^{9/2} e^{9/2}} \]
1/8*(a*e^2-b*d*e+c*d^2)^2*x/d/e^4/(e*x^2+d)^4-1/48*(-7*a*e^2-9*b*d*e+25*c* d^2)*(a*e^2-b*d*e+c*d^2)*x/d^2/e^4/(e*x^2+d)^3+1/192*(163*c^2*d^4-2*c*d^2* e*(-3*a*e+59*b*d)+e^2*(35*a^2*e^2+10*a*b*d*e+3*b^2*d^2))*x/d^3/e^4/(e*x^2+ d)^2-1/128*(93*c^2*d^4-2*c*d^2*e*(3*a*e+5*b*d)-e^2*(35*a^2*e^2+10*a*b*d*e+ 3*b^2*d^2))*x/d^4/e^4/(e*x^2+d)+1/128*(35*c^2*d^4+2*c*d^2*e*(3*a*e+5*b*d)+ e^2*(35*a^2*e^2+10*a*b*d*e+3*b^2*d^2))*arctan(x*e^(1/2)/d^(1/2))/d^(9/2)/e ^(9/2)
Time = 0.15 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {\frac {48 d^{7/2} \sqrt {e} \left (c d^2+e (-b d+a e)\right )^2 x}{\left (d+e x^2\right )^4}-\frac {8 d^{5/2} \sqrt {e} \left (25 c^2 d^4+2 c d^2 e (-17 b d+9 a e)+e^2 \left (9 b^2 d^2-2 a b d e-7 a^2 e^2\right )\right ) x}{\left (d+e x^2\right )^3}+\frac {2 d^{3/2} \sqrt {e} \left (163 c^2 d^4+2 c d^2 e (-59 b d+3 a e)+e^2 \left (3 b^2 d^2+10 a b d e+35 a^2 e^2\right )\right ) x}{\left (d+e x^2\right )^2}-\frac {3 \sqrt {d} \sqrt {e} \left (93 c^2 d^4-2 c d^2 e (5 b d+3 a e)-e^2 \left (3 b^2 d^2+10 a b d e+35 a^2 e^2\right )\right ) x}{d+e x^2}+3 \left (35 c^2 d^4+2 c d^2 e (5 b d+3 a e)+e^2 \left (3 b^2 d^2+10 a b d e+35 a^2 e^2\right )\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{384 d^{9/2} e^{9/2}} \]
((48*d^(7/2)*Sqrt[e]*(c*d^2 + e*(-(b*d) + a*e))^2*x)/(d + e*x^2)^4 - (8*d^ (5/2)*Sqrt[e]*(25*c^2*d^4 + 2*c*d^2*e*(-17*b*d + 9*a*e) + e^2*(9*b^2*d^2 - 2*a*b*d*e - 7*a^2*e^2))*x)/(d + e*x^2)^3 + (2*d^(3/2)*Sqrt[e]*(163*c^2*d^ 4 + 2*c*d^2*e*(-59*b*d + 3*a*e) + e^2*(3*b^2*d^2 + 10*a*b*d*e + 35*a^2*e^2 ))*x)/(d + e*x^2)^2 - (3*Sqrt[d]*Sqrt[e]*(93*c^2*d^4 - 2*c*d^2*e*(5*b*d + 3*a*e) - e^2*(3*b^2*d^2 + 10*a*b*d*e + 35*a^2*e^2))*x)/(d + e*x^2) + 3*(35 *c^2*d^4 + 2*c*d^2*e*(5*b*d + 3*a*e) + e^2*(3*b^2*d^2 + 10*a*b*d*e + 35*a^ 2*e^2))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(384*d^(9/2)*e^(9/2))
Time = 0.78 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1471, 2345, 1471, 27, 298, 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.
\(\displaystyle \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\int \frac {-\frac {8 c^2 d x^6}{e}+\frac {8 c d (c d-2 b e) x^4}{e^2}-\frac {8 d \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right ) x^2}{e^3}+\frac {c^2 d^4-2 c e (b d-a e) d^2+e^2 \left (b^2 d^2-2 a b e d-7 a^2 e^2\right )}{e^4}}{\left (e x^2+d\right )^4}dx}{8 d}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\frac {x \left (-7 a e^2-9 b d e+25 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\int \frac {\frac {48 c^2 d^2 x^4}{e^2}-\frac {96 c d^2 (c d-b e) x^2}{e^3}+\frac {19 c^2 d^4-2 c e (11 b d-3 a e) d^2+e^2 \left (3 b^2 d^2+10 a b e d+35 a^2 e^2\right )}{e^4}}{\left (e x^2+d\right )^3}dx}{6 d}}{8 d}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\frac {x \left (-7 a e^2-9 b d e+25 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )-2 c d^2 e (59 b d-3 a e)+163 c^2 d^4\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\int \frac {3 \left (29 c^2 d^4-64 c^2 e x^2 d^3-2 c e (5 b d+3 a e) d^2-e^2 \left (3 b^2 d^2+10 a b e d+35 a^2 e^2\right )\right )}{e^4 \left (e x^2+d\right )^2}dx}{4 d}}{6 d}}{8 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\frac {x \left (-7 a e^2-9 b d e+25 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )-2 c d^2 e (59 b d-3 a e)+163 c^2 d^4\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \int \frac {29 c^2 d^4-64 c^2 e x^2 d^3-2 c e (5 b d+3 a e) d^2-e^2 \left (3 b^2 d^2+10 a b e d+35 a^2 e^2\right )}{\left (e x^2+d\right )^2}dx}{4 d e^4}}{6 d}}{8 d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\frac {x \left (-7 a e^2-9 b d e+25 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )-2 c d^2 e (59 b d-3 a e)+163 c^2 d^4\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (-e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )-2 c d^2 e (3 a e+5 b d)+93 c^2 d^4\right )}{2 d \left (d+e x^2\right )}-\frac {\left (e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )+2 c d^2 e (3 a e+5 b d)+35 c^2 d^4\right ) \int \frac {1}{e x^2+d}dx}{2 d}\right )}{4 d e^4}}{6 d}}{8 d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\frac {x \left (-7 a e^2-9 b d e+25 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\frac {x \left (e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )-2 c d^2 e (59 b d-3 a e)+163 c^2 d^4\right )}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (-e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )-2 c d^2 e (3 a e+5 b d)+93 c^2 d^4\right )}{2 d \left (d+e x^2\right )}-\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e^2 \left (35 a^2 e^2+10 a b d e+3 b^2 d^2\right )+2 c d^2 e (3 a e+5 b d)+35 c^2 d^4\right )}{2 d^{3/2} \sqrt {e}}\right )}{4 d e^4}}{6 d}}{8 d}\) |
((c*d^2 - b*d*e + a*e^2)^2*x)/(8*d*e^4*(d + e*x^2)^4) - (((25*c*d^2 - 9*b* d*e - 7*a*e^2)*(c*d^2 - b*d*e + a*e^2)*x)/(6*d*e^4*(d + e*x^2)^3) - (((163 *c^2*d^4 - 2*c*d^2*e*(59*b*d - 3*a*e) + e^2*(3*b^2*d^2 + 10*a*b*d*e + 35*a ^2*e^2))*x)/(4*d*e^4*(d + e*x^2)^2) - (3*(((93*c^2*d^4 - 2*c*d^2*e*(5*b*d + 3*a*e) - e^2*(3*b^2*d^2 + 10*a*b*d*e + 35*a^2*e^2))*x)/(2*d*(d + e*x^2)) - ((35*c^2*d^4 + 2*c*d^2*e*(5*b*d + 3*a*e) + e^2*(3*b^2*d^2 + 10*a*b*d*e + 35*a^2*e^2))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*Sqrt[e])))/(4*d*e^4 ))/(6*d))/(8*d)
3.3.60.3.1 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[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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 = 0.30 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {\frac {\left (35 a^{2} e^{4}+10 a b d \,e^{3}+6 a c \,d^{2} e^{2}+3 b^{2} d^{2} e^{2}+10 b c \,d^{3} e -93 c^{2} d^{4}\right ) x^{7}}{128 d^{4} e}+\frac {\left (385 a^{2} e^{4}+110 a b d \,e^{3}+66 a c \,d^{2} e^{2}+33 b^{2} d^{2} e^{2}-146 b c \,d^{3} e -511 c^{2} d^{4}\right ) x^{5}}{384 d^{3} e^{2}}+\frac {\left (511 a^{2} e^{4}+146 a b d \,e^{3}-66 a c \,d^{2} e^{2}-33 b^{2} d^{2} e^{2}-110 b c \,d^{3} e -385 c^{2} d^{4}\right ) x^{3}}{384 d^{2} e^{3}}+\frac {\left (93 a^{2} e^{4}-10 a b d \,e^{3}-6 a c \,d^{2} e^{2}-3 b^{2} d^{2} e^{2}-10 b c \,d^{3} e -35 c^{2} d^{4}\right ) x}{128 e^{4} d}}{\left (e \,x^{2}+d \right )^{4}}+\frac {\left (35 a^{2} e^{4}+10 a b d \,e^{3}+6 a c \,d^{2} e^{2}+3 b^{2} d^{2} e^{2}+10 b c \,d^{3} e +35 c^{2} d^{4}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{128 d^{4} e^{4} \sqrt {e d}}\) | \(347\) |
risch | \(\frac {\frac {\left (35 a^{2} e^{4}+10 a b d \,e^{3}+6 a c \,d^{2} e^{2}+3 b^{2} d^{2} e^{2}+10 b c \,d^{3} e -93 c^{2} d^{4}\right ) x^{7}}{128 d^{4} e}+\frac {\left (385 a^{2} e^{4}+110 a b d \,e^{3}+66 a c \,d^{2} e^{2}+33 b^{2} d^{2} e^{2}-146 b c \,d^{3} e -511 c^{2} d^{4}\right ) x^{5}}{384 d^{3} e^{2}}+\frac {\left (511 a^{2} e^{4}+146 a b d \,e^{3}-66 a c \,d^{2} e^{2}-33 b^{2} d^{2} e^{2}-110 b c \,d^{3} e -385 c^{2} d^{4}\right ) x^{3}}{384 d^{2} e^{3}}+\frac {\left (93 a^{2} e^{4}-10 a b d \,e^{3}-6 a c \,d^{2} e^{2}-3 b^{2} d^{2} e^{2}-10 b c \,d^{3} e -35 c^{2} d^{4}\right ) x}{128 e^{4} d}}{\left (e \,x^{2}+d \right )^{4}}-\frac {35 \ln \left (e x +\sqrt {-e d}\right ) a^{2}}{256 \sqrt {-e d}\, d^{4}}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) a b}{128 \sqrt {-e d}\, e \,d^{3}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a c}{128 \sqrt {-e d}\, e^{2} d^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) b^{2}}{256 \sqrt {-e d}\, e^{2} d^{2}}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) b c}{128 \sqrt {-e d}\, e^{3} d}-\frac {35 \ln \left (e x +\sqrt {-e d}\right ) c^{2}}{256 \sqrt {-e d}\, e^{4}}+\frac {35 \ln \left (-e x +\sqrt {-e d}\right ) a^{2}}{256 \sqrt {-e d}\, d^{4}}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) a b}{128 \sqrt {-e d}\, e \,d^{3}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a c}{128 \sqrt {-e d}\, e^{2} d^{2}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) b^{2}}{256 \sqrt {-e d}\, e^{2} d^{2}}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) b c}{128 \sqrt {-e d}\, e^{3} d}+\frac {35 \ln \left (-e x +\sqrt {-e d}\right ) c^{2}}{256 \sqrt {-e d}\, e^{4}}\) | \(595\) |
(1/128*(35*a^2*e^4+10*a*b*d*e^3+6*a*c*d^2*e^2+3*b^2*d^2*e^2+10*b*c*d^3*e-9 3*c^2*d^4)/d^4/e*x^7+1/384*(385*a^2*e^4+110*a*b*d*e^3+66*a*c*d^2*e^2+33*b^ 2*d^2*e^2-146*b*c*d^3*e-511*c^2*d^4)/d^3/e^2*x^5+1/384*(511*a^2*e^4+146*a* b*d*e^3-66*a*c*d^2*e^2-33*b^2*d^2*e^2-110*b*c*d^3*e-385*c^2*d^4)/d^2/e^3*x ^3+1/128*(93*a^2*e^4-10*a*b*d*e^3-6*a*c*d^2*e^2-3*b^2*d^2*e^2-10*b*c*d^3*e -35*c^2*d^4)/e^4/d*x)/(e*x^2+d)^4+1/128*(35*a^2*e^4+10*a*b*d*e^3+6*a*c*d^2 *e^2+3*b^2*d^2*e^2+10*b*c*d^3*e+35*c^2*d^4)/d^4/e^4/(e*d)^(1/2)*arctan(e*x /(e*d)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 622 vs. \(2 (299) = 598\).
Time = 0.29 (sec) , antiderivative size = 1266, normalized size of antiderivative = 3.99 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\text {Too large to display} \]
[-1/768*(6*(93*c^2*d^5*e^4 - 10*b*c*d^4*e^5 - 10*a*b*d^2*e^7 - 35*a^2*d*e^ 8 - 3*(b^2 + 2*a*c)*d^3*e^6)*x^7 + 2*(511*c^2*d^6*e^3 + 146*b*c*d^5*e^4 - 110*a*b*d^3*e^6 - 385*a^2*d^2*e^7 - 33*(b^2 + 2*a*c)*d^4*e^5)*x^5 + 2*(385 *c^2*d^7*e^2 + 110*b*c*d^6*e^3 - 146*a*b*d^4*e^5 - 511*a^2*d^3*e^6 + 33*(b ^2 + 2*a*c)*d^5*e^4)*x^3 + 3*(35*c^2*d^8 + 10*b*c*d^7*e + 10*a*b*d^5*e^3 + 35*a^2*d^4*e^4 + 3*(b^2 + 2*a*c)*d^6*e^2 + (35*c^2*d^4*e^4 + 10*b*c*d^3*e ^5 + 10*a*b*d*e^7 + 35*a^2*e^8 + 3*(b^2 + 2*a*c)*d^2*e^6)*x^8 + 4*(35*c^2* d^5*e^3 + 10*b*c*d^4*e^4 + 10*a*b*d^2*e^6 + 35*a^2*d*e^7 + 3*(b^2 + 2*a*c) *d^3*e^5)*x^6 + 6*(35*c^2*d^6*e^2 + 10*b*c*d^5*e^3 + 10*a*b*d^3*e^5 + 35*a ^2*d^2*e^6 + 3*(b^2 + 2*a*c)*d^4*e^4)*x^4 + 4*(35*c^2*d^7*e + 10*b*c*d^6*e ^2 + 10*a*b*d^4*e^4 + 35*a^2*d^3*e^5 + 3*(b^2 + 2*a*c)*d^5*e^3)*x^2)*sqrt( -d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) + 6*(35*c^2*d^8*e + 10 *b*c*d^7*e^2 + 10*a*b*d^5*e^4 - 93*a^2*d^4*e^5 + 3*(b^2 + 2*a*c)*d^6*e^3)* x)/(d^5*e^9*x^8 + 4*d^6*e^8*x^6 + 6*d^7*e^7*x^4 + 4*d^8*e^6*x^2 + d^9*e^5) , -1/384*(3*(93*c^2*d^5*e^4 - 10*b*c*d^4*e^5 - 10*a*b*d^2*e^7 - 35*a^2*d*e ^8 - 3*(b^2 + 2*a*c)*d^3*e^6)*x^7 + (511*c^2*d^6*e^3 + 146*b*c*d^5*e^4 - 1 10*a*b*d^3*e^6 - 385*a^2*d^2*e^7 - 33*(b^2 + 2*a*c)*d^4*e^5)*x^5 + (385*c^ 2*d^7*e^2 + 110*b*c*d^6*e^3 - 146*a*b*d^4*e^5 - 511*a^2*d^3*e^6 + 33*(b^2 + 2*a*c)*d^5*e^4)*x^3 - 3*(35*c^2*d^8 + 10*b*c*d^7*e + 10*a*b*d^5*e^3 + 35 *a^2*d^4*e^4 + 3*(b^2 + 2*a*c)*d^6*e^2 + (35*c^2*d^4*e^4 + 10*b*c*d^3*e...
Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\text {Exception raised: ValueError} \]
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 = 0.28 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {{\left (35 \, c^{2} d^{4} + 10 \, b c d^{3} e + 3 \, b^{2} d^{2} e^{2} + 6 \, a c d^{2} e^{2} + 10 \, a b d e^{3} + 35 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{128 \, \sqrt {d e} d^{4} e^{4}} - \frac {279 \, c^{2} d^{4} e^{3} x^{7} - 30 \, b c d^{3} e^{4} x^{7} - 9 \, b^{2} d^{2} e^{5} x^{7} - 18 \, a c d^{2} e^{5} x^{7} - 30 \, a b d e^{6} x^{7} - 105 \, a^{2} e^{7} x^{7} + 511 \, c^{2} d^{5} e^{2} x^{5} + 146 \, b c d^{4} e^{3} x^{5} - 33 \, b^{2} d^{3} e^{4} x^{5} - 66 \, a c d^{3} e^{4} x^{5} - 110 \, a b d^{2} e^{5} x^{5} - 385 \, a^{2} d e^{6} x^{5} + 385 \, c^{2} d^{6} e x^{3} + 110 \, b c d^{5} e^{2} x^{3} + 33 \, b^{2} d^{4} e^{3} x^{3} + 66 \, a c d^{4} e^{3} x^{3} - 146 \, a b d^{3} e^{4} x^{3} - 511 \, a^{2} d^{2} e^{5} x^{3} + 105 \, c^{2} d^{7} x + 30 \, b c d^{6} e x + 9 \, b^{2} d^{5} e^{2} x + 18 \, a c d^{5} e^{2} x + 30 \, a b d^{4} e^{3} x - 279 \, a^{2} d^{3} e^{4} x}{384 \, {\left (e x^{2} + d\right )}^{4} d^{4} e^{4}} \]
1/128*(35*c^2*d^4 + 10*b*c*d^3*e + 3*b^2*d^2*e^2 + 6*a*c*d^2*e^2 + 10*a*b* d*e^3 + 35*a^2*e^4)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^4*e^4) - 1/384*(279 *c^2*d^4*e^3*x^7 - 30*b*c*d^3*e^4*x^7 - 9*b^2*d^2*e^5*x^7 - 18*a*c*d^2*e^5 *x^7 - 30*a*b*d*e^6*x^7 - 105*a^2*e^7*x^7 + 511*c^2*d^5*e^2*x^5 + 146*b*c* d^4*e^3*x^5 - 33*b^2*d^3*e^4*x^5 - 66*a*c*d^3*e^4*x^5 - 110*a*b*d^2*e^5*x^ 5 - 385*a^2*d*e^6*x^5 + 385*c^2*d^6*e*x^3 + 110*b*c*d^5*e^2*x^3 + 33*b^2*d ^4*e^3*x^3 + 66*a*c*d^4*e^3*x^3 - 146*a*b*d^3*e^4*x^3 - 511*a^2*d^2*e^5*x^ 3 + 105*c^2*d^7*x + 30*b*c*d^6*e*x + 9*b^2*d^5*e^2*x + 18*a*c*d^5*e^2*x + 30*a*b*d^4*e^3*x - 279*a^2*d^3*e^4*x)/((e*x^2 + d)^4*d^4*e^4)
Time = 7.68 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (35\,a^2\,e^4+10\,a\,b\,d\,e^3+6\,a\,c\,d^2\,e^2+3\,b^2\,d^2\,e^2+10\,b\,c\,d^3\,e+35\,c^2\,d^4\right )}{128\,d^{9/2}\,e^{9/2}}-\frac {\frac {x\,\left (-93\,a^2\,e^4+10\,a\,b\,d\,e^3+6\,a\,c\,d^2\,e^2+3\,b^2\,d^2\,e^2+10\,b\,c\,d^3\,e+35\,c^2\,d^4\right )}{128\,d\,e^4}-\frac {x^7\,\left (35\,a^2\,e^4+10\,a\,b\,d\,e^3+6\,a\,c\,d^2\,e^2+3\,b^2\,d^2\,e^2+10\,b\,c\,d^3\,e-93\,c^2\,d^4\right )}{128\,d^4\,e}+\frac {x^3\,\left (-511\,a^2\,e^4-146\,a\,b\,d\,e^3+66\,a\,c\,d^2\,e^2+33\,b^2\,d^2\,e^2+110\,b\,c\,d^3\,e+385\,c^2\,d^4\right )}{384\,d^2\,e^3}-\frac {x^5\,\left (385\,a^2\,e^4+110\,a\,b\,d\,e^3+66\,a\,c\,d^2\,e^2+33\,b^2\,d^2\,e^2-146\,b\,c\,d^3\,e-511\,c^2\,d^4\right )}{384\,d^3\,e^2}}{d^4+4\,d^3\,e\,x^2+6\,d^2\,e^2\,x^4+4\,d\,e^3\,x^6+e^4\,x^8} \]
(atan((e^(1/2)*x)/d^(1/2))*(35*a^2*e^4 + 35*c^2*d^4 + 3*b^2*d^2*e^2 + 10*a *b*d*e^3 + 10*b*c*d^3*e + 6*a*c*d^2*e^2))/(128*d^(9/2)*e^(9/2)) - ((x*(35* c^2*d^4 - 93*a^2*e^4 + 3*b^2*d^2*e^2 + 10*a*b*d*e^3 + 10*b*c*d^3*e + 6*a*c *d^2*e^2))/(128*d*e^4) - (x^7*(35*a^2*e^4 - 93*c^2*d^4 + 3*b^2*d^2*e^2 + 1 0*a*b*d*e^3 + 10*b*c*d^3*e + 6*a*c*d^2*e^2))/(128*d^4*e) + (x^3*(385*c^2*d ^4 - 511*a^2*e^4 + 33*b^2*d^2*e^2 - 146*a*b*d*e^3 + 110*b*c*d^3*e + 66*a*c *d^2*e^2))/(384*d^2*e^3) - (x^5*(385*a^2*e^4 - 511*c^2*d^4 + 33*b^2*d^2*e^ 2 + 110*a*b*d*e^3 - 146*b*c*d^3*e + 66*a*c*d^2*e^2))/(384*d^3*e^2))/(d^4 + e^4*x^8 + 4*d^3*e*x^2 + 4*d*e^3*x^6 + 6*d^2*e^2*x^4)