Integrand size = 22, antiderivative size = 745 \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {1}{a^2 d x}-\frac {c x \left (a e+c d x^2\right )}{4 a^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac {e^{9/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2} \left (c d^2+a e^2\right )^2}+\frac {c^{3/4} \left (\sqrt {c} d+3 \sqrt {a} e\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )}+\frac {c^{3/4} \left (a^{3/2} e^3+\sqrt {c} d \left (c d^2+2 a e^2\right )\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )^2}-\frac {c^{3/4} \left (\sqrt {c} d+3 \sqrt {a} e\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )}-\frac {c^{3/4} \left (a^{3/2} e^3+\sqrt {c} d \left (c d^2+2 a e^2\right )\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )^2}-\frac {c^{3/4} \left (\sqrt {c} d-3 \sqrt {a} e\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )}+\frac {c^{3/4} \left (a^{3/2} e^3-\sqrt {c} d \left (c d^2+2 a e^2\right )\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )^2}+\frac {c^{3/4} \left (\sqrt {c} d-3 \sqrt {a} e\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )}-\frac {c^{3/4} \left (a^{3/2} e^3-\sqrt {c} d \left (c d^2+2 a e^2\right )\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )^2} \]
-1/a^2/d/x-1/4*c*x*(c*d*x^2+a*e)/a^2/(a*e^2+c*d^2)/(c*x^4+a)-e^(9/2)*arcta n(x*e^(1/2)/d^(1/2))/d^(3/2)/(a*e^2+c*d^2)^2-1/32*c^(3/4)*ln(-a^(1/4)*c^(1 /4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(-3*e*a^(1/2)+d*c^(1/2))/a^(9/4)/(a*e^2 +c*d^2)*2^(1/2)+1/32*c^(3/4)*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1 /2))*(-3*e*a^(1/2)+d*c^(1/2))/a^(9/4)/(a*e^2+c*d^2)*2^(1/2)-1/16*c^(3/4)*a rctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))*(3*e*a^(1/2)+d*c^(1/2))/a^(9/4)/(a*e^2 +c*d^2)*2^(1/2)-1/16*c^(3/4)*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))*(3*e*a^(1 /2)+d*c^(1/2))/a^(9/4)/(a*e^2+c*d^2)*2^(1/2)+1/8*c^(3/4)*ln(-a^(1/4)*c^(1/ 4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(a^(3/2)*e^3-d*(2*a*e^2+c*d^2)*c^(1/2))/ a^(9/4)/(a*e^2+c*d^2)^2*2^(1/2)-1/8*c^(3/4)*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a ^(1/2)+x^2*c^(1/2))*(a^(3/2)*e^3-d*(2*a*e^2+c*d^2)*c^(1/2))/a^(9/4)/(a*e^2 +c*d^2)^2*2^(1/2)-1/4*c^(3/4)*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))*(a^(3/2 )*e^3+d*(2*a*e^2+c*d^2)*c^(1/2))/a^(9/4)/(a*e^2+c*d^2)^2*2^(1/2)-1/4*c^(3/ 4)*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))*(a^(3/2)*e^3+d*(2*a*e^2+c*d^2)*c^(1 /2))/a^(9/4)/(a*e^2+c*d^2)^2*2^(1/2)
Time = 0.24 (sec) , antiderivative size = 499, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {1}{32} \left (-\frac {32}{a^2 d x}-\frac {8 c x \left (a e+c d x^2\right )}{a^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac {32 e^{9/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2} \left (c d^2+a e^2\right )^2}+\frac {2 \sqrt {2} c^{3/4} \left (5 c^{3/2} d^3+3 \sqrt {a} c d^2 e+9 a \sqrt {c} d e^2+7 a^{3/2} e^3\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{9/4} \left (c d^2+a e^2\right )^2}-\frac {2 \sqrt {2} c^{3/4} \left (5 c^{3/2} d^3+3 \sqrt {a} c d^2 e+9 a \sqrt {c} d e^2+7 a^{3/2} e^3\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{9/4} \left (c d^2+a e^2\right )^2}+\frac {\sqrt {2} c^{3/4} \left (-5 c^{3/2} d^3+3 \sqrt {a} c d^2 e-9 a \sqrt {c} d e^2+7 a^{3/2} e^3\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{9/4} \left (c d^2+a e^2\right )^2}+\frac {\sqrt {2} c^{3/4} \left (5 c^{3/2} d^3-3 \sqrt {a} c d^2 e+9 a \sqrt {c} d e^2-7 a^{3/2} e^3\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{9/4} \left (c d^2+a e^2\right )^2}\right ) \]
(-32/(a^2*d*x) - (8*c*x*(a*e + c*d*x^2))/(a^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (32*e^(9/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(d^(3/2)*(c*d^2 + a*e^2)^2) + (2*Sqrt[2]*c^(3/4)*(5*c^(3/2)*d^3 + 3*Sqrt[a]*c*d^2*e + 9*a*Sqrt[c]*d*e^2 + 7*a^(3/2)*e^3)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(a^(9/4)*(c*d^2 + a*e^2)^2) - (2*Sqrt[2]*c^(3/4)*(5*c^(3/2)*d^3 + 3*Sqrt[a]*c*d^2*e + 9*a* Sqrt[c]*d*e^2 + 7*a^(3/2)*e^3)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(a ^(9/4)*(c*d^2 + a*e^2)^2) + (Sqrt[2]*c^(3/4)*(-5*c^(3/2)*d^3 + 3*Sqrt[a]*c *d^2*e - 9*a*Sqrt[c]*d*e^2 + 7*a^(3/2)*e^3)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)* c^(1/4)*x + Sqrt[c]*x^2])/(a^(9/4)*(c*d^2 + a*e^2)^2) + (Sqrt[2]*c^(3/4)*( 5*c^(3/2)*d^3 - 3*Sqrt[a]*c*d^2*e + 9*a*Sqrt[c]*d*e^2 - 7*a^(3/2)*e^3)*Log [Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(a^(9/4)*(c*d^2 + a*e ^2)^2))/32
Time = 0.95 (sec) , antiderivative size = 745, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1675, 2009}
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.
| \(\displaystyle \int \frac {1}{x^2 \left (a+c x^4\right )^2 \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1675 |
| \(\displaystyle \int \left (\frac {c \left (-a^2 e^3-c d x^2 \left (2 a e^2+c d^2\right )\right )}{a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )^2}+\frac {1}{a^2 d x^2}-\frac {c \left (a e+c d x^2\right )}{a \left (a+c x^4\right )^2 \left (a e^2+c d^2\right )}-\frac {e^5}{d \left (d+e x^2\right ) \left (a e^2+c d^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (3 \sqrt {a} e+\sqrt {c} d\right )}{8 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )}-\frac {c^{3/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (3 \sqrt {a} e+\sqrt {c} d\right )}{8 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )}+\frac {c^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (a^{3/2} e^3+\sqrt {c} d \left (2 a e^2+c d^2\right )\right )}{2 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )^2}-\frac {c^{3/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (a^{3/2} e^3+\sqrt {c} d \left (2 a e^2+c d^2\right )\right )}{2 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )^2}-\frac {c^{3/4} \left (\sqrt {c} d-3 \sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )}+\frac {c^{3/4} \left (\sqrt {c} d-3 \sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )}+\frac {c^{3/4} \left (a^{3/2} e^3-\sqrt {c} d \left (2 a e^2+c d^2\right )\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )^2}-\frac {c^{3/4} \left (a^{3/2} e^3-\sqrt {c} d \left (2 a e^2+c d^2\right )\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{9/4} \left (a e^2+c d^2\right )^2}-\frac {c x \left (a e+c d x^2\right )}{4 a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}-\frac {1}{a^2 d x}-\frac {e^{9/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2} \left (a e^2+c d^2\right )^2}\) |
-(1/(a^2*d*x)) - (c*x*(a*e + c*d*x^2))/(4*a^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (e^(9/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(d^(3/2)*(c*d^2 + a*e^2)^2) + (c^ (3/4)*(Sqrt[c]*d + 3*Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/( 8*Sqrt[2]*a^(9/4)*(c*d^2 + a*e^2)) + (c^(3/4)*(a^(3/2)*e^3 + Sqrt[c]*d*(c* d^2 + 2*a*e^2))*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(9/4 )*(c*d^2 + a*e^2)^2) - (c^(3/4)*(Sqrt[c]*d + 3*Sqrt[a]*e)*ArcTan[1 + (Sqrt [2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(9/4)*(c*d^2 + a*e^2)) - (c^(3/4)*(a ^(3/2)*e^3 + Sqrt[c]*d*(c*d^2 + 2*a*e^2))*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a ^(1/4)])/(2*Sqrt[2]*a^(9/4)*(c*d^2 + a*e^2)^2) - (c^(3/4)*(Sqrt[c]*d - 3*S qrt[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt [2]*a^(9/4)*(c*d^2 + a*e^2)) + (c^(3/4)*(a^(3/2)*e^3 - Sqrt[c]*d*(c*d^2 + 2*a*e^2))*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[ 2]*a^(9/4)*(c*d^2 + a*e^2)^2) + (c^(3/4)*(Sqrt[c]*d - 3*Sqrt[a]*e)*Log[Sqr t[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(9/4)*(c*d^ 2 + a*e^2)) - (c^(3/4)*(a^(3/2)*e^3 - Sqrt[c]*d*(c*d^2 + 2*a*e^2))*Log[Sqr t[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(9/4)*(c*d^2 + a*e^2)^2)
3.3.57.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q, 0] | | IntegersQ[m, q])
Time = 0.50 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.48
| method | result | size |
| default | \(-\frac {1}{a^{2} d x}-\frac {c \left (\frac {\left (\frac {1}{4} a c d \,e^{2}+\frac {1}{4} c^{2} d^{3}\right ) x^{3}+\left (\frac {1}{4} e^{3} a^{2}+\frac {1}{4} a c \,d^{2} e \right ) x}{c \,x^{4}+a}+\frac {\left (7 e^{3} a^{2}+3 a c \,d^{2} e \right ) \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a}+\frac {\left (9 a c d \,e^{2}+5 c^{2} d^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} a^{2}}-\frac {e^{5} \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{d \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {e d}}\) | \(355\) |
| risch | \(\text {Expression too large to display}\) | \(1781\) |
-1/a^2/d/x-c/(a*e^2+c*d^2)^2/a^2*(((1/4*a*c*d*e^2+1/4*c^2*d^3)*x^3+(1/4*e^ 3*a^2+1/4*a*c*d^2*e)*x)/(c*x^4+a)+1/32*(7*a^2*e^3+3*a*c*d^2*e)*(a/c)^(1/4) /a*2^(1/2)*(ln((x^2+(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2))/(x^2-(a/c)^(1/4)*x* 2^(1/2)+(a/c)^(1/2)))+2*arctan(2^(1/2)/(a/c)^(1/4)*x+1)+2*arctan(2^(1/2)/( a/c)^(1/4)*x-1))+1/32*(9*a*c*d*e^2+5*c^2*d^3)/c/(a/c)^(1/4)*2^(1/2)*(ln((x ^2-(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2))/(x^2+(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/ 2)))+2*arctan(2^(1/2)/(a/c)^(1/4)*x+1)+2*arctan(2^(1/2)/(a/c)^(1/4)*x-1))) -1/d*e^5/(a*e^2+c*d^2)^2/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 5082 vs. \(2 (573) = 1146\).
Time = 29.01 (sec) , antiderivative size = 10188, normalized size of antiderivative = 13.68 \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \]
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 = 659, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {e^{5} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{{\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4}\right )} \sqrt {d e}} - \frac {{\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d^{2} e + 7 \, \left (a c^{3}\right )^{\frac {1}{4}} a^{2} c e^{3} + 5 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{3} + 9 \, \left (a c^{3}\right )^{\frac {3}{4}} a d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{8 \, {\left (\sqrt {2} a^{3} c^{3} d^{4} + 2 \, \sqrt {2} a^{4} c^{2} d^{2} e^{2} + \sqrt {2} a^{5} c e^{4}\right )}} - \frac {{\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d^{2} e + 7 \, \left (a c^{3}\right )^{\frac {1}{4}} a^{2} c e^{3} + 5 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{3} + 9 \, \left (a c^{3}\right )^{\frac {3}{4}} a d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{8 \, {\left (\sqrt {2} a^{3} c^{3} d^{4} + 2 \, \sqrt {2} a^{4} c^{2} d^{2} e^{2} + \sqrt {2} a^{5} c e^{4}\right )}} - \frac {{\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d^{2} e + 7 \, \left (a c^{3}\right )^{\frac {1}{4}} a^{2} c e^{3} - 5 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{3} - 9 \, \left (a c^{3}\right )^{\frac {3}{4}} a d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{16 \, {\left (\sqrt {2} a^{3} c^{3} d^{4} + 2 \, \sqrt {2} a^{4} c^{2} d^{2} e^{2} + \sqrt {2} a^{5} c e^{4}\right )}} + \frac {{\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d^{2} e + 7 \, \left (a c^{3}\right )^{\frac {1}{4}} a^{2} c e^{3} - 5 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{3} - 9 \, \left (a c^{3}\right )^{\frac {3}{4}} a d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{16 \, {\left (\sqrt {2} a^{3} c^{3} d^{4} + 2 \, \sqrt {2} a^{4} c^{2} d^{2} e^{2} + \sqrt {2} a^{5} c e^{4}\right )}} - \frac {5 \, c^{2} d^{2} x^{4} + 4 \, a c e^{2} x^{4} + a c d e x^{2} + 4 \, a c d^{2} + 4 \, a^{2} e^{2}}{4 \, {\left (a^{2} c d^{3} + a^{3} d e^{2}\right )} {\left (c x^{5} + a x\right )}} \]
-e^5*arctan(e*x/sqrt(d*e))/((c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4)*sqrt(d*e )) - 1/8*(3*(a*c^3)^(1/4)*a*c^2*d^2*e + 7*(a*c^3)^(1/4)*a^2*c*e^3 + 5*(a*c ^3)^(3/4)*c*d^3 + 9*(a*c^3)^(3/4)*a*d*e^2)*arctan(1/2*sqrt(2)*(2*x + sqrt( 2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt(2)*a^3*c^3*d^4 + 2*sqrt(2)*a^4*c^2*d^2* e^2 + sqrt(2)*a^5*c*e^4) - 1/8*(3*(a*c^3)^(1/4)*a*c^2*d^2*e + 7*(a*c^3)^(1 /4)*a^2*c*e^3 + 5*(a*c^3)^(3/4)*c*d^3 + 9*(a*c^3)^(3/4)*a*d*e^2)*arctan(1/ 2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt(2)*a^3*c^3*d^4 + 2*sqrt(2)*a^4*c^2*d^2*e^2 + sqrt(2)*a^5*c*e^4) - 1/16*(3*(a*c^3)^(1/4)*a*c ^2*d^2*e + 7*(a*c^3)^(1/4)*a^2*c*e^3 - 5*(a*c^3)^(3/4)*c*d^3 - 9*(a*c^3)^( 3/4)*a*d*e^2)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^3*c^ 3*d^4 + 2*sqrt(2)*a^4*c^2*d^2*e^2 + sqrt(2)*a^5*c*e^4) + 1/16*(3*(a*c^3)^( 1/4)*a*c^2*d^2*e + 7*(a*c^3)^(1/4)*a^2*c*e^3 - 5*(a*c^3)^(3/4)*c*d^3 - 9*( a*c^3)^(3/4)*a*d*e^2)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2 )*a^3*c^3*d^4 + 2*sqrt(2)*a^4*c^2*d^2*e^2 + sqrt(2)*a^5*c*e^4) - 1/4*(5*c^ 2*d^2*x^4 + 4*a*c*e^2*x^4 + a*c*d*e*x^2 + 4*a*c*d^2 + 4*a^2*e^2)/((a^2*c*d ^3 + a^3*d*e^2)*(c*x^5 + a*x))
Time = 10.73 (sec) , antiderivative size = 24015, normalized size of antiderivative = 32.23 \[ \int \frac {1}{x^2 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \]
- (1/(a*d) + (c*e*x^2)/(4*a*(a*e^2 + c*d^2)) + (c*x^4*(4*a*e^2 + 5*c*d^2)) /(4*a^2*d*(a*e^2 + c*d^2)))/(a*x + c*x^5) - atan(((11875*a^5*c^10*d^15*e - a^9*c^3*(72128*a^3*d*e^15 + 265655*c^3*d^7*e^9 - 76440*a*c^2*d^5*e^11 - 1 78585*a^2*c*d^3*e^13) + 68800*a^6*c^9*d^13*e^3 + 89403*a^7*c^8*d^11*e^5 - 126488*a^8*c^7*d^9*e^7)*(a^25*d^2*e^19*x*(-(49*a^3*e^6*(-a^9*c^3)^(1/2) - 25*c^3*d^6*(-a^9*c^3)^(1/2) + 30*a^5*c^4*d^5*e + 126*a^7*c^2*d*e^5 + 124*a ^6*c^3*d^3*e^3 - 81*a*c^2*d^4*e^2*(-a^9*c^3)^(1/2) - 39*a^2*c*d^2*e^4*(-a^ 9*c^3)^(1/2))/(a^13*e^8 + a^9*c^4*d^8 + 4*a^12*c*d^2*e^6 + 4*a^10*c^3*d^6* e^2 + 6*a^11*c^2*d^4*e^4))^(5/2)*2i - a^15*c^2*e^17*x*(-(49*a^3*e^6*(-a^9* c^3)^(1/2) - 25*c^3*d^6*(-a^9*c^3)^(1/2) + 30*a^5*c^4*d^5*e + 126*a^7*c^2* d*e^5 + 124*a^6*c^3*d^3*e^3 - 81*a*c^2*d^4*e^2*(-a^9*c^3)^(1/2) - 39*a^2*c *d^2*e^4*(-a^9*c^3)^(1/2))/(a^13*e^8 + a^9*c^4*d^8 + 4*a^12*c*d^2*e^6 + 4* a^10*c^3*d^6*e^2 + 6*a^11*c^2*d^4*e^4))^(1/2)*3136i - a^11*c^10*d^19*x*(-( 49*a^3*e^6*(-a^9*c^3)^(1/2) - 25*c^3*d^6*(-a^9*c^3)^(1/2) + 30*a^5*c^4*d^5 *e + 126*a^7*c^2*d*e^5 + 124*a^6*c^3*d^3*e^3 - 81*a*c^2*d^4*e^2*(-a^9*c^3) ^(1/2) - 39*a^2*c*d^2*e^4*(-a^9*c^3)^(1/2))/(a^13*e^8 + a^9*c^4*d^8 + 4*a^ 12*c*d^2*e^6 + 4*a^10*c^3*d^6*e^2 + 6*a^11*c^2*d^4*e^4))^(3/2)*25i - a^16* c^9*d^20*e*x*(-(49*a^3*e^6*(-a^9*c^3)^(1/2) - 25*c^3*d^6*(-a^9*c^3)^(1/2) + 30*a^5*c^4*d^5*e + 126*a^7*c^2*d*e^5 + 124*a^6*c^3*d^3*e^3 - 81*a*c^2*d^ 4*e^2*(-a^9*c^3)^(1/2) - 39*a^2*c*d^2*e^4*(-a^9*c^3)^(1/2))/(a^13*e^8 +...