\(\int \frac {1}{(d+e x) (a+b x^2+c x^4)} \, dx\) [239]

Optimal result

Integrand size = 22, antiderivative size = 372 \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\frac {\sqrt {c} d \left (e^2+\frac {2 c d^2+b e^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}} \left (c d^4+b d^2 e^2+a e^4\right )}+\frac {\sqrt {c} d \left (e^2-\frac {2 c d^2+b e^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b+\sqrt {b^2-4 a c}} \left (c d^4+b d^2 e^2+a e^4\right )}+\frac {e \left (2 c d^2+b e^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 \sqrt {b^2-4 a c} \left (c d^4+b d^2 e^2+a e^4\right )}+\frac {e^3 \log (d+e x)}{c d^4+b d^2 e^2+a e^4}-\frac {e^3 \log \left (a+b x^2+c x^4\right )}{4 \left (c d^4+b d^2 e^2+a e^4\right )} \] Output:

1/2*c^(1/2)*d*(e^2+(b*e^2+2*c*d^2)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/2)*c^(1 
/2)*x/(b-(-4*a*c+b^2)^(1/2))^(1/2))*2^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)/( 
a*e^4+b*d^2*e^2+c*d^4)+1/2*c^(1/2)*d*(e^2-(b*e^2+2*c*d^2)/(-4*a*c+b^2)^(1/ 
2))*arctan(2^(1/2)*c^(1/2)*x/(b+(-4*a*c+b^2)^(1/2))^(1/2))*2^(1/2)/(b+(-4* 
a*c+b^2)^(1/2))^(1/2)/(a*e^4+b*d^2*e^2+c*d^4)+1/2*e*(b*e^2+2*c*d^2)*arctan 
h((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)/(a*e^4+b*d^2*e^2+c*d^ 
4)+e^3*ln(e*x+d)/(a*e^4+b*d^2*e^2+c*d^4)-e^3*ln(c*x^4+b*x^2+a)/(4*a*e^4+4* 
b*d^2*e^2+4*c*d^4)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\frac {\frac {2 \sqrt {2} \sqrt {c} d \left (2 c d^2+\left (b+\sqrt {b^2-4 a c}\right ) e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {2 \sqrt {2} \sqrt {c} d \left (2 c d^2+\left (b-\sqrt {b^2-4 a c}\right ) e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}}}+4 e^3 \log (d+e x)-\frac {\left (2 c d^2 e+\left (b+\sqrt {b^2-4 a c}\right ) e^3\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (2 c d^2 e+\left (b-\sqrt {b^2-4 a c}\right ) e^3\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}}{4 \left (c d^4+b d^2 e^2+a e^4\right )} \] Input:

Integrate[1/((d + e*x)*(a + b*x^2 + c*x^4)),x]
 

Output:

((2*Sqrt[2]*Sqrt[c]*d*(2*c*d^2 + (b + Sqrt[b^2 - 4*a*c])*e^2)*ArcTan[(Sqrt 
[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b - S 
qrt[b^2 - 4*a*c]]) - (2*Sqrt[2]*Sqrt[c]*d*(2*c*d^2 + (b - Sqrt[b^2 - 4*a*c 
])*e^2)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 
 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + 4*e^3*Log[d + e*x] - ((2*c*d^2*e 
+ (b + Sqrt[b^2 - 4*a*c])*e^3)*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2])/Sqrt 
[b^2 - 4*a*c] + ((2*c*d^2*e + (b - Sqrt[b^2 - 4*a*c])*e^3)*Log[b + Sqrt[b^ 
2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c])/(4*(c*d^4 + b*d^2*e^2 + a*e^4))
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 372, 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 = {7279, 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.

\(\displaystyle \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {(d-e x) \left (b e^2+c d^2+c e^2 x^2\right )}{\left (a+b x^2+c x^4\right ) \left (a e^4+b d^2 e^2+c d^4\right )}+\frac {e^4}{(d+e x) \left (a e^4+b d^2 e^2+c d^4\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {c} d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {b e^2+2 c d^2}{\sqrt {b^2-4 a c}}+e^2\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}} \left (a e^4+b d^2 e^2+c d^4\right )}+\frac {\sqrt {c} d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (e^2-\frac {b e^2+2 c d^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {\sqrt {b^2-4 a c}+b} \left (a e^4+b d^2 e^2+c d^4\right )}+\frac {e \left (b e^2+2 c d^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 \sqrt {b^2-4 a c} \left (a e^4+b d^2 e^2+c d^4\right )}-\frac {e^3 \log \left (a+b x^2+c x^4\right )}{4 \left (a e^4+b d^2 e^2+c d^4\right )}+\frac {e^3 \log (d+e x)}{a e^4+b d^2 e^2+c d^4}\)

Input:

Int[1/((d + e*x)*(a + b*x^2 + c*x^4)),x]
 

Output:

(Sqrt[c]*d*(e^2 + (2*c*d^2 + b*e^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqr 
t[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b - Sqrt[b^2 - 4*a*c]] 
*(c*d^4 + b*d^2*e^2 + a*e^4)) + (Sqrt[c]*d*(e^2 - (2*c*d^2 + b*e^2)/Sqrt[b 
^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqr 
t[2]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^4 + b*d^2*e^2 + a*e^4)) + (e*(2*c*d^ 
2 + b*e^2)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*Sqrt[b^2 - 4*a*c]* 
(c*d^4 + b*d^2*e^2 + a*e^4)) + (e^3*Log[d + e*x])/(c*d^4 + b*d^2*e^2 + a*e 
^4) - (e^3*Log[a + b*x^2 + c*x^4])/(4*(c*d^4 + b*d^2*e^2 + a*e^4))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {4 c \left (-\frac {\sqrt {-4 a c +b^{2}}\, \left (\sqrt {-4 a c +b^{2}}\, e^{2}+b \,e^{2}+2 c \,d^{2}\right ) \left (-\frac {e \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )}{4 c}-\frac {d \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{16 a c -4 b^{2}}-\frac {\sqrt {-4 a c +b^{2}}\, \left (\sqrt {-4 a c +b^{2}}\, e^{2}-b \,e^{2}-2 c \,d^{2}\right ) \left (-\frac {e \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )}{4 c}+\frac {d \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{16 a c -4 b^{2}}\right )}{e^{4} a +b \,d^{2} e^{2}+c \,d^{4}}+\frac {e^{3} \ln \left (e x +d \right )}{e^{4} a +b \,d^{2} e^{2}+c \,d^{4}}\)	\(307\)
risch	\(\frac {e^{3} \ln \left (e x +d \right )}{e^{4} a +b \,d^{2} e^{2}+c \,d^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{4} c^{2} e^{4}-8 a^{3} b^{2} c \,e^{4}+16 a^{3} b \,c^{2} d^{2} e^{2}+16 a^{3} c^{3} d^{4}+b^{4} a^{2} e^{4}-8 a^{2} b^{3} c \,d^{2} e^{2}-8 a^{2} b^{2} c^{2} d^{4}+a \,b^{5} d^{2} e^{2}+a \,b^{4} c \,d^{4}\right ) \textit {\_Z}^{4}+\left (32 a^{3} c^{2} e^{3}-16 a^{2} b^{2} c \,e^{3}+2 a \,b^{4} e^{3}\right ) \textit {\_Z}^{3}+\left (24 a^{2} c^{2} e^{2}-10 a \,b^{2} c \,e^{2}-4 a b \,c^{2} d^{2}+b^{4} e^{2}+b^{3} c \,d^{2}\right ) \textit {\_Z}^{2}+\left (8 a \,c^{2} e -2 b^{2} c e \right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (40 a^{3} c^{2} e^{6}-22 a^{2} b^{2} c \,e^{6}-20 a^{2} b \,c^{2} d^{2} e^{4}-24 a^{2} c^{3} d^{4} e^{2}+3 a \,b^{4} e^{6}+9 a \,b^{3} c \,d^{2} e^{4}+14 a \,b^{2} c^{2} d^{4} e^{2}+4 a b \,c^{3} d^{6}-b^{5} d^{2} e^{4}-2 b^{4} c \,d^{4} e^{2}-b^{3} c^{2} d^{6}\right ) \textit {\_R}^{3}+\left (60 a^{2} c^{2} e^{5}-27 a \,b^{2} c \,e^{5}-24 a b \,c^{2} d^{2} e^{3}-12 a \,c^{3} d^{4} e +3 b^{4} e^{5}+6 b^{3} c \,d^{2} e^{3}+3 b^{2} c^{2} d^{4} e \right ) \textit {\_R}^{2}+\left (30 a \,e^{4} c^{2}-8 b^{2} c \,e^{4}-2 b \,c^{2} d^{2} e^{2}-2 d^{4} c^{3}\right ) \textit {\_R} +5 e^{3} c^{2}\right ) x +\left (48 a^{3} c^{2} d \,e^{5}-28 a^{2} b^{2} c d \,e^{5}-16 a^{2} b \,c^{2} d^{3} e^{3}-16 a^{2} c^{3} d^{5} e +4 a \,b^{4} d \,e^{5}+4 a \,b^{3} c \,d^{3} e^{3}+4 a \,b^{2} c^{2} d^{5} e \right ) \textit {\_R}^{3}+\left (52 a^{2} c^{2} d \,e^{4}-17 a \,b^{2} c d \,e^{4}-8 a b \,c^{2} d^{3} e^{2}-4 a \,c^{3} d^{5}+b^{4} d \,e^{4}+2 b^{3} c \,d^{3} e^{2}+b^{2} c^{2} d^{5}\right ) \textit {\_R}^{2}+\left (16 a \,c^{2} d \,e^{3}-4 b^{2} c d \,e^{3}\right ) \textit {\_R} +c^{2} d \,e^{2}\right )\right )}{2}\)	\(740\)

Input:

int(1/(e*x+d)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

4/(a*e^4+b*d^2*e^2+c*d^4)*c*(-(-4*a*c+b^2)^(1/2)*((-4*a*c+b^2)^(1/2)*e^2+b 
*e^2+2*c*d^2)/(16*a*c-4*b^2)*(-1/4*e/c*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)-1 
/2*d*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x*2^(1/2)/((-b+(- 
4*a*c+b^2)^(1/2))*c)^(1/2)))-(-4*a*c+b^2)^(1/2)*((-4*a*c+b^2)^(1/2)*e^2-b* 
e^2-2*c*d^2)/(16*a*c-4*b^2)*(-1/4*e/c*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)+1/2 
*d*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c 
+b^2)^(1/2))*c)^(1/2))))+e^3*ln(e*x+d)/(a*e^4+b*d^2*e^2+c*d^4)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x+d)/(c*x^4+b*x^2+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x+d)/(c*x**4+b*x**2+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} {\left (e x + d\right )}} \,d x } \] Input:

integrate(1/(e*x+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")
 

Output:

e^3*log(e*x + d)/(c*d^4 + b*d^2*e^2 + a*e^4) - integrate((c*e^3*x^3 - c*d* 
e^2*x^2 - c*d^3 - b*d*e^2 + (c*d^2*e + b*e^3)*x)/(c*x^4 + b*x^2 + a), x)/( 
c*d^4 + b*d^2*e^2 + a*e^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19589 vs. \(2 (330) = 660\).

Time = 2.97 (sec) , antiderivative size = 19589, normalized size of antiderivative = 52.66 \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/(e*x+d)/(c*x^4+b*x^2+a),x, algorithm="giac")
 

Output:

e^4*log(abs(e*x + d))/(c*d^4*e + b*d^2*e^3 + a*e^5) - 1/4*e^3*log(abs(c*x^ 
4 + b*x^2 + a))/(c*d^4 + b*d^2*e^2 + a*e^4) + 1/8*(2*(2*b^3*c^9 - 8*a*b*c^ 
10 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^7 + 4 
*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^8 + 2*sqr 
t(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^8 - sqrt(2)*s 
qrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b*c^9 - 2*(b^2 - 4*a*c)*b 
*c^9)*d^19 + 9*(2*b^4*c^8 - 8*a*b^2*c^9 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b 
*c + sqrt(b^2 - 4*a*c)*c)*b^4*c^6 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + 
 sqrt(b^2 - 4*a*c)*c)*a*b^2*c^7 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + s 
qrt(b^2 - 4*a*c)*c)*b^3*c^7 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^ 
2 - 4*a*c)*c)*b^2*c^8 - 2*(b^2 - 4*a*c)*b^2*c^8)*d^17*e^2 + 8*(4*b^5*c^7 - 
 14*a*b^3*c^8 - 8*a^2*b*c^9 - 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt( 
b^2 - 4*a*c)*c)*b^5*c^5 + 7*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 
- 4*a*c)*c)*a*b^3*c^6 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 
4*a*c)*c)*b^4*c^6 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a* 
c)*c)*a^2*b*c^7 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c) 
*c)*a*b^2*c^7 - 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c 
)*b^3*c^7 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b* 
c^8 - 4*(b^2 - 4*a*c)*b^3*c^7 - 2*(b^2 - 4*a*c)*a*b*c^8)*d^15*e^4 + 14*(2* 
b^6*c^6 - 4*a*b^4*c^7 - 16*a^2*b^2*c^8 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt...
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 22.13 (sec) , antiderivative size = 1881, normalized size of antiderivative = 5.06 \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \] Input:

int(1/((d + e*x)*(a + b*x^2 + c*x^4)),x)
 

Output:

symsum(log(root(256*a^3*b*c^2*d^2*e^2*z^4 - 128*a^2*b^3*c*d^2*e^2*z^4 + 16 
*a*b^5*d^2*e^2*z^4 - 128*a^3*b^2*c*e^4*z^4 + 16*a*b^4*c*d^4*z^4 - 128*a^2* 
b^2*c^2*d^4*z^4 + 256*a^4*c^2*e^4*z^4 + 256*a^3*c^3*d^4*z^4 + 16*a^2*b^4*e 
^4*z^4 - 128*a^2*b^2*c*e^3*z^3 + 256*a^3*c^2*e^3*z^3 + 16*a*b^4*e^3*z^3 - 
40*a*b^2*c*e^2*z^2 - 16*a*b*c^2*d^2*z^2 + 96*a^2*c^2*e^2*z^2 + 4*b^3*c*d^2 
*z^2 + 4*b^4*e^2*z^2 - 4*b^2*c*e*z + 16*a*c^2*e*z + c^2, z, k)*(root(256*a 
^3*b*c^2*d^2*e^2*z^4 - 128*a^2*b^3*c*d^2*e^2*z^4 + 16*a*b^5*d^2*e^2*z^4 - 
128*a^3*b^2*c*e^4*z^4 + 16*a*b^4*c*d^4*z^4 - 128*a^2*b^2*c^2*d^4*z^4 + 256 
*a^4*c^2*e^4*z^4 + 256*a^3*c^3*d^4*z^4 + 16*a^2*b^4*e^4*z^4 - 128*a^2*b^2* 
c*e^3*z^3 + 256*a^3*c^2*e^3*z^3 + 16*a*b^4*e^3*z^3 - 40*a*b^2*c*e^2*z^2 - 
16*a*b*c^2*d^2*z^2 + 96*a^2*c^2*e^2*z^2 + 4*b^3*c*d^2*z^2 + 4*b^4*e^2*z^2 
- 4*b^2*c*e*z + 16*a*c^2*e*z + c^2, z, k)*(root(256*a^3*b*c^2*d^2*e^2*z^4 
- 128*a^2*b^3*c*d^2*e^2*z^4 + 16*a*b^5*d^2*e^2*z^4 - 128*a^3*b^2*c*e^4*z^4 
 + 16*a*b^4*c*d^4*z^4 - 128*a^2*b^2*c^2*d^4*z^4 + 256*a^4*c^2*e^4*z^4 + 25 
6*a^3*c^3*d^4*z^4 + 16*a^2*b^4*e^4*z^4 - 128*a^2*b^2*c*e^3*z^3 + 256*a^3*c 
^2*e^3*z^3 + 16*a*b^4*e^3*z^3 - 40*a*b^2*c*e^2*z^2 - 16*a*b*c^2*d^2*z^2 + 
96*a^2*c^2*e^2*z^2 + 4*b^3*c*d^2*z^2 + 4*b^4*e^2*z^2 - 4*b^2*c*e*z + 16*a* 
c^2*e*z + c^2, z, k)*(x*(240*a^2*c^4*e^6 + 12*b^4*c^2*e^6 - 108*a*b^2*c^3* 
e^6 - 48*a*c^5*d^4*e^2 + 12*b^2*c^4*d^4*e^2 + 24*b^3*c^3*d^2*e^4 - 96*a*b* 
c^4*d^2*e^4) - root(256*a^3*b*c^2*d^2*e^2*z^4 - 128*a^2*b^3*c*d^2*e^2*z...
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1383, normalized size of antiderivative = 3.72 \[ \int \frac {1}{(d+e x) \left (a+b x^2+c x^4\right )} \, dx =\text {Too large to display} \] Input:

int(1/(e*x+d)/(c*x^4+b*x^2+a),x)
 

Output:

(2*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sq 
rt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*e**3 + 
4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqr 
t(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d**2*e - 
 4*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 
 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d*e**2 + 2*sqrt(a)*sqrt(2*s 
qrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt( 
2*sqrt(c)*sqrt(a) + b))*b**2*d*e**2 + 2*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b 
)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) 
+ b))*b*c*d**3 - 2*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c 
)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*d*e**2 - 4* 
sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2* 
sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d**3 + 2*sqrt(2*sqrt(c)*sqrt(a 
) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*s 
qrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*e**3 + 4*sqrt(2*sqrt(c)*sqrt(a) 
 + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sq 
rt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*d**2*e + 4*sqrt(a)*sqrt(2*sqrt(c 
)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqr 
t(c)*sqrt(a) + b))*a*c*d*e**2 - 2*sqrt(a)*sqrt(2*sqrt(c)*sqrt(a) + b)*atan 
((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b...