Integrand size = 30, antiderivative size = 408 \[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=-\frac {5 b^2-14 a c}{6 a^2 \left (b^2-4 a c\right ) e (d+e x)^3}+\frac {b \left (5 b^2-19 a c\right )}{2 a^3 \left (b^2-4 a c\right ) e (d+e x)}+\frac {b^2-2 a c+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) e (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac {\sqrt {c} \left (5 b^4-29 a b^2 c+28 a^2 c^2+b \left (5 b^2-19 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^3 \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}} e}-\frac {\sqrt {c} \left (5 b^4-29 a b^2 c+28 a^2 c^2-b \left (5 b^2-19 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a^3 \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}} e} \]
1/6*(14*a*c-5*b^2)/a^2/(-4*a*c+b^2)/e/(e*x+d)^3+1/2*b*(-19*a*c+5*b^2)/a^3/ (-4*a*c+b^2)/e/(e*x+d)+1/2*(b^2-2*a*c+b*c*(e*x+d)^2)/a/(-4*a*c+b^2)/e/(e*x +d)^3/(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)*(5*b^4-29*a*b^2*c+28*a^2*c^2+b*(-19*a*c+5 *b^2)*(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+b^2)^(3/2)/e*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)*(5*b^4-29*a*b^2*c+28*a^2*c^2-b*(-19*a*c+5*b^2)*(-4*a*c+b^2)^( 1/2))/a^3/(-4*a*c+b^2)^(3/2)/e*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 1.78 (sec) , antiderivative size = 384, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\frac {-\frac {4 a}{(d+e x)^3}+\frac {24 b}{d+e x}+\frac {6 (d+e x) \left (b^4-4 a b^2 c+2 a^2 c^2+b^3 c (d+e x)^2-3 a b c^2 (d+e x)^2\right )}{\left (b^2-4 a c\right ) \left (a+(d+e x)^2 \left (b+c (d+e x)^2\right )\right )}+\frac {3 \sqrt {2} \sqrt {c} \left (5 b^4-29 a b^2 c+28 a^2 c^2+5 b^3 \sqrt {b^2-4 a c}-19 a b c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {3 \sqrt {2} \sqrt {c} \left (-5 b^4+29 a b^2 c-28 a^2 c^2+5 b^3 \sqrt {b^2-4 a c}-19 a b c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{12 a^3 e} \]
((-4*a)/(d + e*x)^3 + (24*b)/(d + e*x) + (6*(d + e*x)*(b^4 - 4*a*b^2*c + 2 *a^2*c^2 + b^3*c*(d + e*x)^2 - 3*a*b*c^2*(d + e*x)^2))/((b^2 - 4*a*c)*(a + (d + e*x)^2*(b + c*(d + e*x)^2))) + (3*Sqrt[2]*Sqrt[c]*(5*b^4 - 29*a*b^2* c + 28*a^2*c^2 + 5*b^3*Sqrt[b^2 - 4*a*c] - 19*a*b*c*Sqrt[b^2 - 4*a*c])*Arc Tan[(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]]) + (3*Sqrt[2]*Sqrt[c]*(-5*b^4 + 29*a* b^2*c - 28*a^2*c^2 + 5*b^3*Sqrt[b^2 - 4*a*c] - 19*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]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/(12*a^3*e)
Time = 1.14 (sec) , antiderivative size = 394, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1462, 1441, 25, 1604, 27, 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.
\(\displaystyle \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1462 |
\(\displaystyle \frac {\int \frac {1}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )^2}d(d+e x)}{e}\) |
\(\Big \downarrow \) 1441 |
\(\displaystyle \frac {\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {\int -\frac {5 b^2+5 c (d+e x)^2 b-14 a c}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{2 a \left (b^2-4 a c\right )}}{e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {5 b^2+5 c (d+e x)^2 b-14 a c}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e}\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {3 \left (c \left (5 b^2-14 a c\right ) (d+e x)^2+b \left (5 b^2-19 a c\right )\right )}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{3 a}-\frac {5 b^2-14 a c}{3 a (d+e x)^3}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {c \left (5 b^2-14 a c\right ) (d+e x)^2+b \left (5 b^2-19 a c\right )}{(d+e x)^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)}{a}-\frac {5 b^2-14 a c}{3 a (d+e x)^3}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e}\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {5 b^4-24 a c b^2+c \left (5 b^2-19 a c\right ) (d+e x)^2 b+14 a^2 c^2}{c (d+e x)^4+b (d+e x)^2+a}d(d+e x)}{a}-\frac {b \left (5 b^2-19 a c\right )}{a (d+e x)}}{a}-\frac {5 b^2-14 a c}{3 a (d+e x)^3}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {c \left (28 a^2 c^2-29 a b^2 c+b \left (5 b^2-19 a c\right ) \sqrt {b^2-4 a c}+5 b^4\right ) \int \frac {1}{c (d+e x)^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}d(d+e x)}{2 \sqrt {b^2-4 a c}}-\frac {c \left (28 a^2 c^2-29 a b^2 c-b \left (5 b^2-19 a c\right ) \sqrt {b^2-4 a c}+5 b^4\right ) \int \frac {1}{c (d+e x)^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}d(d+e x)}{2 \sqrt {b^2-4 a c}}}{a}-\frac {b \left (5 b^2-19 a c\right )}{a (d+e x)}}{a}-\frac {5 b^2-14 a c}{3 a (d+e x)^3}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {-\frac {-\frac {\frac {\sqrt {c} \left (28 a^2 c^2-29 a b^2 c+b \left (5 b^2-19 a c\right ) \sqrt {b^2-4 a c}+5 b^4\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (28 a^2 c^2-29 a b^2 c-b \left (5 b^2-19 a c\right ) \sqrt {b^2-4 a c}+5 b^4\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}}}{a}-\frac {b \left (5 b^2-19 a c\right )}{a (d+e x)}}{a}-\frac {5 b^2-14 a c}{3 a (d+e x)^3}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{e}\) |
((b^2 - 2*a*c + b*c*(d + e*x)^2)/(2*a*(b^2 - 4*a*c)*(d + e*x)^3*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (-1/3*(5*b^2 - 14*a*c)/(a*(d + e*x)^3) - (-(( b*(5*b^2 - 19*a*c))/(a*(d + e*x))) - ((Sqrt[c]*(5*b^4 - 29*a*b^2*c + 28*a^ 2*c^2 + b*(5*b^2 - 19*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 - S qrt[b^2 - 4*a*c]]) - (Sqrt[c]*(5*b^4 - 29*a*b^2*c + 28*a^2*c^2 - b*(5*b^2 - 19*a*c)*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b + S qrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]] ))/a)/a)/(2*a*(b^2 - 4*a*c)))/e
3.7.29.3.1 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[((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 = 0.78 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.20
method | result | size |
default | \(-\frac {\frac {-\frac {b c \,e^{2} \left (3 a c -b^{2}\right ) x^{3}}{2 \left (4 a c -b^{2}\right )}-\frac {3 d b c e \left (3 a c -b^{2}\right ) x^{2}}{2 \left (4 a c -b^{2}\right )}+\frac {\left (-9 b \,c^{2} d^{2} a +3 b^{3} c \,d^{2}+2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right ) x}{8 a c -2 b^{2}}+\frac {d \left (-3 b \,c^{2} d^{2} a +b^{3} c \,d^{2}+2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )}{2 e \left (4 a c -b^{2}\right )}}{c \,x^{4} e^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+d^{4} c +2 b d e x +b \,d^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 d^{3} e c +2 b d e \right ) \textit {\_Z} +d^{4} c +b \,d^{2}+a \right )}{\sum }\frac {\left (b c \,e^{2} \left (-19 a c +5 b^{2}\right ) \textit {\_R}^{2}+2 b c d e \left (-19 a c +5 b^{2}\right ) \textit {\_R} -19 b \,c^{2} d^{2} a +5 b^{3} c \,d^{2}+14 a^{2} c^{2}-24 a \,b^{2} c +5 b^{4}\right ) \ln \left (x -\textit {\_R} \right )}{2 e^{3} c \,\textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 d^{3} c +b e \textit {\_R} +b d}}{4 \left (4 a c -b^{2}\right ) e}}{a^{3}}-\frac {1}{3 a^{2} e \left (e x +d \right )^{3}}+\frac {2 b}{a^{3} e \left (e x +d \right )}\) | \(489\) |
risch | \(\text {Expression too large to display}\) | \(1394\) |
-1/a^3*((-1/2*b*c*e^2*(3*a*c-b^2)/(4*a*c-b^2)*x^3-3/2*d*b*c*e*(3*a*c-b^2)/ (4*a*c-b^2)*x^2+1/2*(-9*a*b*c^2*d^2+3*b^3*c*d^2+2*a^2*c^2-4*a*b^2*c+b^4)/( 4*a*c-b^2)*x+1/2*d/e*(-3*a*b*c^2*d^2+b^3*c*d^2+2*a^2*c^2-4*a*b^2*c+b^4)/(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((b*c*e^2*(-19*a*c+5*b^2)*_R ^2+2*b*c*d*e*(-19*a*c+5*b^2)*_R-19*b*c^2*d^2*a+5*b^3*c*d^2+14*a^2*c^2-24*a *b^2*c+5*b^4)/(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)))-1/3/a^2/e/(e*x+d)^3+2/a^3*b/e/(e*x+d )
Leaf count of result is larger than twice the leaf count of optimal. 5734 vs. \(2 (358) = 716\).
Time = 0.64 (sec) , antiderivative size = 5734, normalized size of antiderivative = 14.05 \[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\int { \frac {1}{{\left ({\left (e x + d\right )}^{4} c + {\left (e x + d\right )}^{2} b + a\right )}^{2} {\left (e x + d\right )}^{4}} \,d x } \]
1/6*(3*(5*b^3*c - 19*a*b*c^2)*e^6*x^6 + 18*(5*b^3*c - 19*a*b*c^2)*d*e^5*x^ 5 + (15*b^4 - 62*a*b^2*c + 14*a^2*c^2 + 45*(5*b^3*c - 19*a*b*c^2)*d^2)*e^4 *x^4 + 3*(5*b^3*c - 19*a*b*c^2)*d^6 + 4*(15*(5*b^3*c - 19*a*b*c^2)*d^3 + ( 15*b^4 - 62*a*b^2*c + 14*a^2*c^2)*d)*e^3*x^3 + (15*b^4 - 62*a*b^2*c + 14*a ^2*c^2)*d^4 + (45*(5*b^3*c - 19*a*b*c^2)*d^4 + 10*a*b^3 - 40*a^2*b*c + 6*( 15*b^4 - 62*a*b^2*c + 14*a^2*c^2)*d^2)*e^2*x^2 - 2*a^2*b^2 + 8*a^3*c + 10* (a*b^3 - 4*a^2*b*c)*d^2 + 2*(9*(5*b^3*c - 19*a*b*c^2)*d^5 + 2*(15*b^4 - 62 *a*b^2*c + 14*a^2*c^2)*d^3 + 10*(a*b^3 - 4*a^2*b*c)*d)*e*x)/((a^3*b^2*c - 4*a^4*c^2)*e^8*x^7 + 7*(a^3*b^2*c - 4*a^4*c^2)*d*e^7*x^6 + (a^3*b^3 - 4*a^ 4*b*c + 21*(a^3*b^2*c - 4*a^4*c^2)*d^2)*e^6*x^5 + 5*(7*(a^3*b^2*c - 4*a^4* c^2)*d^3 + (a^3*b^3 - 4*a^4*b*c)*d)*e^5*x^4 + (a^4*b^2 - 4*a^5*c + 35*(a^3 *b^2*c - 4*a^4*c^2)*d^4 + 10*(a^3*b^3 - 4*a^4*b*c)*d^2)*e^4*x^3 + (21*(a^3 *b^2*c - 4*a^4*c^2)*d^5 + 10*(a^3*b^3 - 4*a^4*b*c)*d^3 + 3*(a^4*b^2 - 4*a^ 5*c)*d)*e^3*x^2 + (7*(a^3*b^2*c - 4*a^4*c^2)*d^6 + 5*(a^3*b^3 - 4*a^4*b*c) *d^4 + 3*(a^4*b^2 - 4*a^5*c)*d^2)*e^2*x + ((a^3*b^2*c - 4*a^4*c^2)*d^7 + ( a^3*b^3 - 4*a^4*b*c)*d^5 + (a^4*b^2 - 4*a^5*c)*d^3)*e) + 1/2*integrate(((5 *b^3*c - 19*a*b*c^2)*e^2*x^2 + 5*b^4 - 24*a*b^2*c + 14*a^2*c^2 + 2*(5*b^3* c - 19*a*b*c^2)*d*e*x + (5*b^3*c - 19*a*b*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...
Leaf count of result is larger than twice the leaf count of optimal. 2122 vs. \(2 (358) = 716\).
Time = 0.36 (sec) , antiderivative size = 2122, normalized size of antiderivative = 5.20 \[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \]
-1/4*((5*b^3*c*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4 )) + d/e)^2 - 19*a*b*c^2*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e ^2)/(c*e^4)) + d/e)^2 - 10*b^3*c*d*e*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e) + 38*a*b*c^2*d*e*(sqrt(1/2)*sqrt(-(b*e^2 + sqr t(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e) + 5*b^3*c*d^2 - 19*a*b*c^2*d^2 + 5*b^4 - 24*a*b^2*c + 14*a^2*c^2)*log(x + sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4* a*c)*e^2)/(c*e^4)) + d/e)/(2*c*e^4*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4* a*c)*e^2)/(c*e^4)) + d/e)^3 - 6*c*d*e^3*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e)^2 - 2*c*d^3*e - b*d*e + (6*c*d^2*e^2 + b*e^ 2)*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e)) - (5* b^3*c*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) - d/e) ^2 - 19*a*b*c^2*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^ 4)) - d/e)^2 + 10*b^3*c*d*e*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^ 2)/(c*e^4)) - d/e) - 38*a*b*c^2*d*e*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4 *a*c)*e^2)/(c*e^4)) - d/e) + 5*b^3*c*d^2 - 19*a*b*c^2*d^2 + 5*b^4 - 24*a*b ^2*c + 14*a^2*c^2)*log(x - sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2) /(c*e^4)) + d/e)/(2*c*e^4*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2) /(c*e^4)) - d/e)^3 + 6*c*d*e^3*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c) *e^2)/(c*e^4)) - d/e)^2 + 2*c*d^3*e + b*d*e + (6*c*d^2*e^2 + b*e^2)*(sqrt( 1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) - d/e)) + (5*b^3*c*...
Time = 13.05 (sec) , antiderivative size = 12239, normalized size of antiderivative = 30.00 \[ \int \frac {1}{(d+e x)^4 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \]
atan(((-(25*b^15 - 25*b^6*(-(4*a*c - b^2)^9)^(1/2) - 80640*a^7*b*c^7 + 636 6*a^2*b^11*c^2 - 35767*a^3*b^9*c^3 + 116928*a^4*b^7*c^4 - 219744*a^5*b^5*c ^5 + 215040*a^6*b^3*c^6 + 49*a^3*c^3*(-(4*a*c - b^2)^9)^(1/2) - 615*a*b^13 *c - 246*a^2*b^2*c^2*(-(4*a*c - b^2)^9)^(1/2) + 165*a*b^4*c*(-(4*a*c - b^2 )^9)^(1/2))/(32*(a^7*b^12*e^2 + 4096*a^13*c^6*e^2 - 24*a^8*b^10*c*e^2 + 24 0*a^9*b^8*c^2*e^2 - 1280*a^10*b^6*c^3*e^2 + 3840*a^11*b^4*c^4*e^2 - 6144*a ^12*b^2*c^5*e^2)))^(1/2)*((-(25*b^15 - 25*b^6*(-(4*a*c - b^2)^9)^(1/2) - 8 0640*a^7*b*c^7 + 6366*a^2*b^11*c^2 - 35767*a^3*b^9*c^3 + 116928*a^4*b^7*c^ 4 - 219744*a^5*b^5*c^5 + 215040*a^6*b^3*c^6 + 49*a^3*c^3*(-(4*a*c - b^2)^9 )^(1/2) - 615*a*b^13*c - 246*a^2*b^2*c^2*(-(4*a*c - b^2)^9)^(1/2) + 165*a* b^4*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(a^7*b^12*e^2 + 4096*a^13*c^6*e^2 - 24 *a^8*b^10*c*e^2 + 240*a^9*b^8*c^2*e^2 - 1280*a^10*b^6*c^3*e^2 + 3840*a^11* b^4*c^4*e^2 - 6144*a^12*b^2*c^5*e^2)))^(1/2)*((-(25*b^15 - 25*b^6*(-(4*a*c - b^2)^9)^(1/2) - 80640*a^7*b*c^7 + 6366*a^2*b^11*c^2 - 35767*a^3*b^9*c^3 + 116928*a^4*b^7*c^4 - 219744*a^5*b^5*c^5 + 215040*a^6*b^3*c^6 + 49*a^3*c ^3*(-(4*a*c - b^2)^9)^(1/2) - 615*a*b^13*c - 246*a^2*b^2*c^2*(-(4*a*c - b^ 2)^9)^(1/2) + 165*a*b^4*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(a^7*b^12*e^2 + 40 96*a^13*c^6*e^2 - 24*a^8*b^10*c*e^2 + 240*a^9*b^8*c^2*e^2 - 1280*a^10*b^6* c^3*e^2 + 3840*a^11*b^4*c^4*e^2 - 6144*a^12*b^2*c^5*e^2)))^(1/2)*(x*(10485 76*a^21*b*c^8*e^14 + 256*a^15*b^13*c^2*e^14 - 6144*a^16*b^11*c^3*e^14 +...