Integrand size = 33, antiderivative size = 360 \[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=-\frac {3 b^2-10 a c}{2 a^2 \left (b^2-4 a c\right ) e f^2 (d+e x)}+\frac {b^2-2 a c+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) e f^2 (d+e x) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {\sqrt {c} \left (3 b^3-16 a b c+\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}} e f^2}+\frac {\sqrt {c} \left (3 b^3-16 a b c-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}} e f^2} \]
1/2*(10*a*c-3*b^2)/a^2/(-4*a*c+b^2)/e/f^2/(e*x+d)+1/2*(b^2-2*a*c+b*c*(e*x+ d)^2)/a/(-4*a*c+b^2)/e/f^2/(e*x+d)/(a+b*(e*x+d)^2+c*(e*x+d)^4)-1/4*arctan( (e*x+d)*2^(1/2)*c^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(3*b^3-16*a* b*c+(-10*a*c+3*b^2)*(-4*a*c+b^2)^(1/2))/a^2/(-4*a*c+b^2)^(3/2)/e/f^2*2^(1/ 2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)+1/4*arctan((e*x+d)*2^(1/2)*c^(1/2)/(b+(-4* a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(3*b^3-16*a*b*c-(-10*a*c+3*b^2)*(-4*a*c+b^2 )^(1/2))/a^2/(-4*a*c+b^2)^(3/2)/e/f^2*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.97 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\frac {-\frac {4}{d+e x}+\frac {2 (d+e x) \left (b^3-3 a b c+b^2 c (d+e x)^2-2 a c^2 (d+e x)^2\right )}{\left (-b^2+4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac {\sqrt {2} \sqrt {c} \left (-3 b^3+16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \sqrt {c} \left (3 b^3-16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{4 a^2 e f^2} \]
(-4/(d + e*x) + (2*(d + e*x)*(b^3 - 3*a*b*c + b^2*c*(d + e*x)^2 - 2*a*c^2* (d + e*x)^2))/((-b^2 + 4*a*c)*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (Sqrt [2]*Sqrt[c]*(-3*b^3 + 16*a*b*c - 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]] )/((b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*Sqrt[c]*(3* b^3 - 16*a*b*c - 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c])*ArcTa n[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c) ^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/(4*a^2*e*f^2)
Time = 0.62 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1462, 1441, 25, 1604, 1480, 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.
| \(\displaystyle \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1462 |
| \(\displaystyle \frac {\int \frac {1}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )^2}d(d+e x)}{e f^2}\) |
| \(\Big \downarrow \) 1441 |
| \(\displaystyle \frac {\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {\int -\frac {3 b^2+3 c (d+e x)^2 b-10 a c}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{2 a \left (b^2-4 a c\right )}}{e f^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 b^2+3 c (d+e x)^2 b-10 a c}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x) \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e f^2}\) |
| \(\Big \downarrow \) 1604 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {c \left (3 b^2-10 a c\right ) (d+e x)^2+b \left (3 b^2-13 a c\right )}{c (d+e x)^4+b (d+e x)^2+a}d(d+e x)}{a}-\frac {3 b^2-10 a c}{a (d+e x)}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x) \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e f^2}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {-\frac {\frac {1}{2} c \left (-\frac {16 a b c}{\sqrt {b^2-4 a c}}+\frac {3 b^3}{\sqrt {b^2-4 a c}}-10 a c+3 b^2\right ) \int \frac {1}{c (d+e x)^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}d(d+e x)-\frac {c \left (-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}-16 a b c+3 b^3\right ) \int \frac {1}{c (d+e x)^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}d(d+e x)}{2 \sqrt {b^2-4 a c}}}{a}-\frac {3 b^2-10 a c}{a (d+e x)}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x) \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e f^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {-\frac {\frac {\sqrt {c} \left (-\frac {16 a b c}{\sqrt {b^2-4 a c}}+\frac {3 b^3}{\sqrt {b^2-4 a c}}-10 a c+3 b^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (-\left (3 b^2-10 a c\right ) \sqrt {b^2-4 a c}-16 a b c+3 b^3\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}}}{a}-\frac {3 b^2-10 a c}{a (d+e x)}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x) \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e f^2}\) |
((b^2 - 2*a*c + b*c*(d + e*x)^2)/(2*a*(b^2 - 4*a*c)*(d + e*x)*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (-((3*b^2 - 10*a*c)/(a*(d + e*x))) - ((Sqrt[c]* (3*b^2 - 10*a*c + (3*b^3)/Sqrt[b^2 - 4*a*c] - (16*a*b*c)/Sqrt[b^2 - 4*a*c] )*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2 ]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (Sqrt[c]*(3*b^3 - 16*a*b*c - (3*b^2 - 10* a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b + Sqrt[b ^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/a) /(2*a*(b^2 - 4*a*c)))/(e*f^2)
3.7.51.3.1 Defintions of rubi rules used
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_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-(d*x)^(m + 1))*(b^2 - 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*d*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m + 4*p + 5) + b*c*(m + 4*p + 7)*x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x ] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Si mp[u^m/(Coefficient[v, x, 1]*v^m) Subst[Int[x^m*(a + b*x^2 + c*x^(2*2))^p , x], x, v], x] /; FreeQ[{a, b, c, m, p}, x] && LinearPairQ[u, v, x]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) /(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1)) Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x ], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.75 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.24
| method | result | size |
| default | \(\frac {-\frac {\frac {\frac {c \,e^{2} \left (2 a c -b^{2}\right ) x^{3}}{8 a c -2 b^{2}}+\frac {3 d c e \left (2 a c -b^{2}\right ) x^{2}}{2 \left (4 a c -b^{2}\right )}+\frac {\left (6 a \,c^{2} d^{2}-3 b^{2} c \,d^{2}+3 a b c -b^{3}\right ) x}{8 a c -2 b^{2}}+\frac {d \left (2 a \,c^{2} d^{2}-b^{2} c \,d^{2}+3 a b c -b^{3}\right )}{2 e \left (4 a c -b^{2}\right )}}{c \,x^{4} e^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+d^{4} c +2 b d e x +b \,d^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 d^{3} e c +2 b d e \right ) \textit {\_Z} +d^{4} c +b \,d^{2}+a \right )}{\sum }\frac {\left (c \,e^{2} \left (10 a c -3 b^{2}\right ) \textit {\_R}^{2}+2 d c e \left (10 a c -3 b^{2}\right ) \textit {\_R} +10 a \,c^{2} d^{2}-3 b^{2} c \,d^{2}+13 a b c -3 b^{3}\right ) \ln \left (x -\textit {\_R} \right )}{2 e^{3} c \,\textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 d^{3} c +b e \textit {\_R} +b d}}{4 \left (4 a c -b^{2}\right ) e}}{a^{2}}-\frac {1}{a^{2} e \left (e x +d \right )}}{f^{2}}\) | \(445\) |
| risch | \(\text {Expression too large to display}\) | \(1268\) |
1/f^2*(-1/a^2*((1/2*c*e^2*(2*a*c-b^2)/(4*a*c-b^2)*x^3+3/2*d*c*e*(2*a*c-b^2 )/(4*a*c-b^2)*x^2+1/2*(6*a*c^2*d^2-3*b^2*c*d^2+3*a*b*c-b^3)/(4*a*c-b^2)*x+ 1/2*d/e*(2*a*c^2*d^2-b^2*c*d^2+3*a*b*c-b^3)/(4*a*c-b^2))/(c*e^4*x^4+4*c*d* e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2+c*d^4+2*b*d*e*x+b*d^2+a)+1/4 /(4*a*c-b^2)/e*sum((c*e^2*(10*a*c-3*b^2)*_R^2+2*d*c*e*(10*a*c-3*b^2)*_R+10 *a*c^2*d^2-3*b^2*c*d^2+13*a*b*c-3*b^3)/(2*_R^3*c*e^3+6*_R^2*c*d*e^2+6*_R*c *d^2*e+2*c*d^3+_R*b*e+b*d)*ln(x-_R),_R=RootOf(c*e^4*_Z^4+4*c*d*e^3*_Z^3+(6 *c*d^2*e^2+b*e^2)*_Z^2+(4*c*d^3*e+2*b*d*e)*_Z+d^4*c+b*d^2+a)))-1/a^2/e/(e* x+d))
Leaf count of result is larger than twice the leaf count of optimal. 4520 vs. \(2 (312) = 624\).
Time = 0.39 (sec) , antiderivative size = 4520, normalized size of antiderivative = 12.56 \[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \]
-1/4*(2*(3*b^2*c - 10*a*c^2)*e^4*x^4 + 8*(3*b^2*c - 10*a*c^2)*d*e^3*x^3 + 2*(3*b^2*c - 10*a*c^2)*d^4 + 2*(3*b^3 - 11*a*b*c + 6*(3*b^2*c - 10*a*c^2)* d^2)*e^2*x^2 + 4*a*b^2 - 16*a^2*c + 2*(3*b^3 - 11*a*b*c)*d^2 + 4*(2*(3*b^2 *c - 10*a*c^2)*d^3 + (3*b^3 - 11*a*b*c)*d)*e*x + sqrt(1/2)*((a^2*b^2*c - 4 *a^3*c^2)*e^6*f^2*x^5 + 5*(a^2*b^2*c - 4*a^3*c^2)*d*e^5*f^2*x^4 + (a^2*b^3 - 4*a^3*b*c + 10*(a^2*b^2*c - 4*a^3*c^2)*d^2)*e^4*f^2*x^3 + (10*(a^2*b^2* c - 4*a^3*c^2)*d^3 + 3*(a^2*b^3 - 4*a^3*b*c)*d)*e^3*f^2*x^2 + (a^3*b^2 - 4 *a^4*c + 5*(a^2*b^2*c - 4*a^3*c^2)*d^4 + 3*(a^2*b^3 - 4*a^3*b*c)*d^2)*e^2* f^2*x + ((a^2*b^2*c - 4*a^3*c^2)*d^5 + (a^2*b^3 - 4*a^3*b*c)*d^3 + (a^3*b^ 2 - 4*a^4*c)*d)*e*f^2)*sqrt(-((a^5*b^6 - 12*a^6*b^4*c + 48*a^7*b^2*c^2 - 6 4*a^8*c^3)*e^2*f^4*sqrt((81*b^8 - 918*a*b^6*c + 3051*a^2*b^4*c^2 - 2550*a^ 3*b^2*c^3 + 625*a^4*c^4)/((a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64 *a^13*c^3)*e^4*f^8)) + 9*b^7 - 105*a*b^5*c + 385*a^2*b^3*c^2 - 420*a^3*b*c ^3)/((a^5*b^6 - 12*a^6*b^4*c + 48*a^7*b^2*c^2 - 64*a^8*c^3)*e^2*f^4))*log( -(189*b^6*c^3 - 1971*a*b^4*c^4 + 5625*a^2*b^2*c^5 - 2500*a^3*c^6)*e*x - (1 89*b^6*c^3 - 1971*a*b^4*c^4 + 5625*a^2*b^2*c^5 - 2500*a^3*c^6)*d + 1/2*sqr t(1/2)*((3*a^5*b^10 - 55*a^6*b^8*c + 392*a^7*b^6*c^2 - 1344*a^8*b^4*c^3 + 2176*a^9*b^2*c^4 - 1280*a^10*c^5)*e^3*f^6*sqrt((81*b^8 - 918*a*b^6*c + 305 1*a^2*b^4*c^2 - 2550*a^3*b^2*c^3 + 625*a^4*c^4)/((a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)*e^4*f^8)) - (27*b^11 - 486*a*b^9*c + ...
Timed out. \[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\int { \frac {1}{{\left ({\left (e x + d\right )}^{4} c + {\left (e x + d\right )}^{2} b + a\right )}^{2} {\left (e f x + d f\right )}^{2}} \,d x } \]
-1/2*((3*b^2*c - 10*a*c^2)*e^4*x^4 + 4*(3*b^2*c - 10*a*c^2)*d*e^3*x^3 + (3 *b^2*c - 10*a*c^2)*d^4 + (3*b^3 - 11*a*b*c + 6*(3*b^2*c - 10*a*c^2)*d^2)*e ^2*x^2 + 2*a*b^2 - 8*a^2*c + (3*b^3 - 11*a*b*c)*d^2 + 2*(2*(3*b^2*c - 10*a *c^2)*d^3 + (3*b^3 - 11*a*b*c)*d)*e*x)/((a^2*b^2*c - 4*a^3*c^2)*e^6*f^2*x^ 5 + 5*(a^2*b^2*c - 4*a^3*c^2)*d*e^5*f^2*x^4 + (a^2*b^3 - 4*a^3*b*c + 10*(a ^2*b^2*c - 4*a^3*c^2)*d^2)*e^4*f^2*x^3 + (10*(a^2*b^2*c - 4*a^3*c^2)*d^3 + 3*(a^2*b^3 - 4*a^3*b*c)*d)*e^3*f^2*x^2 + (a^3*b^2 - 4*a^4*c + 5*(a^2*b^2* c - 4*a^3*c^2)*d^4 + 3*(a^2*b^3 - 4*a^3*b*c)*d^2)*e^2*f^2*x + ((a^2*b^2*c - 4*a^3*c^2)*d^5 + (a^2*b^3 - 4*a^3*b*c)*d^3 + (a^3*b^2 - 4*a^4*c)*d)*e*f^ 2) - 1/2*integrate(((3*b^2*c - 10*a*c^2)*e^2*x^2 + 2*(3*b^2*c - 10*a*c^2)* d*e*x + 3*b^3 - 13*a*b*c + (3*b^2*c - 10*a*c^2)*d^2)/((b^2*c - 4*a*c^2)*e^ 4*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^3*x^3 + (b^2*c - 4*a*c^2)*d^4 + (b^3 - 4*a *b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^2*x^2 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c )*d^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*d)*e*x), x)/(a^2*f^2)
Leaf count of result is larger than twice the leaf count of optimal. 1031 vs. \(2 (312) = 624\).
Time = 0.34 (sec) , antiderivative size = 1031, normalized size of antiderivative = 2.86 \[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=-\frac {\frac {b^{2} c}{{\left (e f x + d f\right )} e f} - \frac {2 \, a c^{2}}{{\left (e f x + d f\right )} e f} + \frac {b^{3} f}{{\left (e f x + d f\right )}^{3} e} - \frac {3 \, a b c f}{{\left (e f x + d f\right )}^{3} e}}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} {\left (c + \frac {b f^{2}}{{\left (e f x + d f\right )}^{2}} + \frac {a f^{4}}{{\left (e f x + d f\right )}^{4}}\right )}} - \frac {1}{{\left (e f x + d f\right )} a^{2} e f} + \frac {{\left ({\left (3 \, a^{4} b^{7} - 31 \, a^{5} b^{5} c + 96 \, a^{6} b^{3} c^{2} - 80 \, a^{7} b c^{3}\right )} \sqrt {2 \, a b + 2 \, \sqrt {b^{2} - 4 \, a c} a} e^{4} f^{8} + 2 \, {\left (3 \, a^{3} b^{2} c - 10 \, a^{4} c^{2}\right )} \sqrt {2 \, a b + 2 \, \sqrt {b^{2} - 4 \, a c} a} \sqrt {b^{2} - 4 \, a c} e^{2} f^{4} {\left | a^{2} b^{2} e^{2} f^{4} - 4 \, a^{3} c e^{2} f^{4} \right |} - {\left (a^{2} b^{2} e^{2} f^{4} - 4 \, a^{3} c e^{2} f^{4}\right )}^{2} {\left (3 \, b^{3} - 13 \, a b c\right )} \sqrt {2 \, a b + 2 \, \sqrt {b^{2} - 4 \, a c} a}\right )} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}}}{{\left (e f x + d f\right )} e f \sqrt {\frac {a^{2} b^{3} e^{2} f^{4} - 4 \, a^{3} b c e^{2} f^{4} + \sqrt {{\left (a^{2} b^{3} e^{2} f^{4} - 4 \, a^{3} b c e^{2} f^{4}\right )}^{2} - 4 \, {\left (a^{3} b^{2} e^{4} f^{8} - 4 \, a^{4} c e^{4} f^{8}\right )} {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )}}}{a^{3} b^{2} e^{4} f^{8} - 4 \, a^{4} c e^{4} f^{8}}}}\right )}{16 \, {\left (a^{5} b^{2} c - 4 \, a^{6} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} e^{3} f^{6} {\left | a^{2} b^{2} e^{2} f^{4} - 4 \, a^{3} c e^{2} f^{4} \right |} {\left | a \right |}} - \frac {{\left ({\left (3 \, a^{4} b^{7} - 31 \, a^{5} b^{5} c + 96 \, a^{6} b^{3} c^{2} - 80 \, a^{7} b c^{3}\right )} \sqrt {2 \, a b - 2 \, \sqrt {b^{2} - 4 \, a c} a} e^{4} f^{8} - 2 \, {\left (3 \, a^{3} b^{2} c - 10 \, a^{4} c^{2}\right )} \sqrt {2 \, a b - 2 \, \sqrt {b^{2} - 4 \, a c} a} \sqrt {b^{2} - 4 \, a c} e^{2} f^{4} {\left | a^{2} b^{2} e^{2} f^{4} - 4 \, a^{3} c e^{2} f^{4} \right |} - {\left (a^{2} b^{2} e^{2} f^{4} - 4 \, a^{3} c e^{2} f^{4}\right )}^{2} {\left (3 \, b^{3} - 13 \, a b c\right )} \sqrt {2 \, a b - 2 \, \sqrt {b^{2} - 4 \, a c} a}\right )} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}}}{{\left (e f x + d f\right )} e f \sqrt {\frac {a^{2} b^{3} e^{2} f^{4} - 4 \, a^{3} b c e^{2} f^{4} - \sqrt {{\left (a^{2} b^{3} e^{2} f^{4} - 4 \, a^{3} b c e^{2} f^{4}\right )}^{2} - 4 \, {\left (a^{3} b^{2} e^{4} f^{8} - 4 \, a^{4} c e^{4} f^{8}\right )} {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )}}}{a^{3} b^{2} e^{4} f^{8} - 4 \, a^{4} c e^{4} f^{8}}}}\right )}{16 \, {\left (a^{5} b^{2} c - 4 \, a^{6} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} e^{3} f^{6} {\left | a^{2} b^{2} e^{2} f^{4} - 4 \, a^{3} c e^{2} f^{4} \right |} {\left | a \right |}} \]
-1/2*(b^2*c/((e*f*x + d*f)*e*f) - 2*a*c^2/((e*f*x + d*f)*e*f) + b^3*f/((e* f*x + d*f)^3*e) - 3*a*b*c*f/((e*f*x + d*f)^3*e))/((a^2*b^2 - 4*a^3*c)*(c + b*f^2/(e*f*x + d*f)^2 + a*f^4/(e*f*x + d*f)^4)) - 1/((e*f*x + d*f)*a^2*e* f) + 1/16*((3*a^4*b^7 - 31*a^5*b^5*c + 96*a^6*b^3*c^2 - 80*a^7*b*c^3)*sqrt (2*a*b + 2*sqrt(b^2 - 4*a*c)*a)*e^4*f^8 + 2*(3*a^3*b^2*c - 10*a^4*c^2)*sqr t(2*a*b + 2*sqrt(b^2 - 4*a*c)*a)*sqrt(b^2 - 4*a*c)*e^2*f^4*abs(a^2*b^2*e^2 *f^4 - 4*a^3*c*e^2*f^4) - (a^2*b^2*e^2*f^4 - 4*a^3*c*e^2*f^4)^2*(3*b^3 - 1 3*a*b*c)*sqrt(2*a*b + 2*sqrt(b^2 - 4*a*c)*a))*arctan(2*sqrt(1/2)/((e*f*x + d*f)*e*f*sqrt((a^2*b^3*e^2*f^4 - 4*a^3*b*c*e^2*f^4 + sqrt((a^2*b^3*e^2*f^ 4 - 4*a^3*b*c*e^2*f^4)^2 - 4*(a^3*b^2*e^4*f^8 - 4*a^4*c*e^4*f^8)*(a^2*b^2* c - 4*a^3*c^2)))/(a^3*b^2*e^4*f^8 - 4*a^4*c*e^4*f^8))))/((a^5*b^2*c - 4*a^ 6*c^2)*sqrt(b^2 - 4*a*c)*e^3*f^6*abs(a^2*b^2*e^2*f^4 - 4*a^3*c*e^2*f^4)*ab s(a)) - 1/16*((3*a^4*b^7 - 31*a^5*b^5*c + 96*a^6*b^3*c^2 - 80*a^7*b*c^3)*s qrt(2*a*b - 2*sqrt(b^2 - 4*a*c)*a)*e^4*f^8 - 2*(3*a^3*b^2*c - 10*a^4*c^2)* sqrt(2*a*b - 2*sqrt(b^2 - 4*a*c)*a)*sqrt(b^2 - 4*a*c)*e^2*f^4*abs(a^2*b^2* e^2*f^4 - 4*a^3*c*e^2*f^4) - (a^2*b^2*e^2*f^4 - 4*a^3*c*e^2*f^4)^2*(3*b^3 - 13*a*b*c)*sqrt(2*a*b - 2*sqrt(b^2 - 4*a*c)*a))*arctan(2*sqrt(1/2)/((e*f* x + d*f)*e*f*sqrt((a^2*b^3*e^2*f^4 - 4*a^3*b*c*e^2*f^4 - sqrt((a^2*b^3*e^2 *f^4 - 4*a^3*b*c*e^2*f^4)^2 - 4*(a^3*b^2*e^4*f^8 - 4*a^4*c*e^4*f^8)*(a^2*b ^2*c - 4*a^3*c^2)))/(a^3*b^2*e^4*f^8 - 4*a^4*c*e^4*f^8))))/((a^5*b^2*c ...
Time = 11.87 (sec) , antiderivative size = 12008, normalized size of antiderivative = 33.36 \[ \int \frac {1}{(d f+e f x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \]
- atan(((-(9*b^13 - 9*b^4*(-(4*a*c - b^2)^9)^(1/2) + 26880*a^6*b*c^6 + 207 7*a^2*b^9*c^2 - 10656*a^3*b^7*c^3 + 30240*a^4*b^5*c^4 - 44800*a^5*b^3*c^5 - 25*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) - 213*a*b^11*c + 51*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(a^5*b^12*e^2*f^4 + 4096*a^11*c^6*e^2*f^4 + 240*a^7* b^8*c^2*e^2*f^4 - 1280*a^8*b^6*c^3*e^2*f^4 + 3840*a^9*b^4*c^4*e^2*f^4 - 61 44*a^10*b^2*c^5*e^2*f^4 - 24*a^6*b^10*c*e^2*f^4)))^(1/2)*((-(9*b^13 - 9*b^ 4*(-(4*a*c - b^2)^9)^(1/2) + 26880*a^6*b*c^6 + 2077*a^2*b^9*c^2 - 10656*a^ 3*b^7*c^3 + 30240*a^4*b^5*c^4 - 44800*a^5*b^3*c^5 - 25*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) - 213*a*b^11*c + 51*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(a ^5*b^12*e^2*f^4 + 4096*a^11*c^6*e^2*f^4 + 240*a^7*b^8*c^2*e^2*f^4 - 1280*a ^8*b^6*c^3*e^2*f^4 + 3840*a^9*b^4*c^4*e^2*f^4 - 6144*a^10*b^2*c^5*e^2*f^4 - 24*a^6*b^10*c*e^2*f^4)))^(1/2)*((-(9*b^13 - 9*b^4*(-(4*a*c - b^2)^9)^(1/ 2) + 26880*a^6*b*c^6 + 2077*a^2*b^9*c^2 - 10656*a^3*b^7*c^3 + 30240*a^4*b^ 5*c^4 - 44800*a^5*b^3*c^5 - 25*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) - 213*a*b^ 11*c + 51*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(a^5*b^12*e^2*f^4 + 4096*a ^11*c^6*e^2*f^4 + 240*a^7*b^8*c^2*e^2*f^4 - 1280*a^8*b^6*c^3*e^2*f^4 + 384 0*a^9*b^4*c^4*e^2*f^4 - 6144*a^10*b^2*c^5*e^2*f^4 - 24*a^6*b^10*c*e^2*f^4) ))^(1/2)*(x*(256*a^10*b^13*c^2*e^14*f^10 - 6144*a^11*b^11*c^3*e^14*f^10 + 61440*a^12*b^9*c^4*e^14*f^10 - 327680*a^13*b^7*c^5*e^14*f^10 + 983040*a^14 *b^5*c^6*e^14*f^10 - 1572864*a^15*b^3*c^7*e^14*f^10 + 1048576*a^16*b*c^...