\(\int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx\) [235]

Optimal result

Integrand size = 22, antiderivative size = 270 \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\frac {d \left (3 e^2+\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b+\sqrt {b^2-4 a c}}}-\frac {e \left (6 c d^2-b e^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {e^3 \log \left (a+b x^2+c x^4\right )}{4 c} \] Output:

1/2*d*(3*e^2+(-3*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)/c^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1 
/2)+1/2*d*(3*e^2-(-3*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)/c^(1/2)/(b+(-4*a*c+b^2)^(1/2) 
)^(1/2)-1/2*e*(-b*e^2+6*c*d^2)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c/( 
-4*a*c+b^2)^(1/2)+1/4*e^3*ln(c*x^4+b*x^2+a)/c
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\frac {\frac {2 \sqrt {2} \sqrt {c} d \left (2 c d^2+3 \left (-b+\sqrt {b^2-4 a c}\right ) e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}+\frac {2 \sqrt {2} \sqrt {c} \left (-2 c d^3+3 \left (b+\sqrt {b^2-4 a c}\right ) d e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b+\sqrt {b^2-4 a c}}}+\left (6 c d^2 e+\left (-b+\sqrt {b^2-4 a c}\right ) e^3\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )+\left (-6 c d^2 e+\left (b+\sqrt {b^2-4 a c}\right ) e^3\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{4 c \sqrt {b^2-4 a c}} \] Input:

Integrate[(d + e*x)^3/(a + b*x^2 + c*x^4),x]
 

Output:

((2*Sqrt[2]*Sqrt[c]*d*(2*c*d^2 + 3*(-b + Sqrt[b^2 - 4*a*c])*e^2)*ArcTan[(S 
qrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/Sqrt[b - Sqrt[b^2 - 4*a*c] 
] + (2*Sqrt[2]*Sqrt[c]*(-2*c*d^3 + 3*(b + Sqrt[b^2 - 4*a*c])*d*e^2)*ArcTan 
[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/Sqrt[b + Sqrt[b^2 - 4*a 
*c]] + (6*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] + (-6*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])/(4*c*Sqrt[b^2 - 4*a*c])
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {2202, 1480, 218, 1576, 27, 1142, 1083, 219, 1103}

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.

\(\displaystyle \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 2202

\(\displaystyle \int \frac {d^3+3 e^2 x^2 d}{c x^4+b x^2+a}dx+\int \frac {x \left (x^2 e^3+3 d^2 e\right )}{c x^4+b x^2+a}dx\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {1}{2} d \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} d \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx+\int \frac {x \left (x^2 e^3+3 d^2 e\right )}{c x^4+b x^2+a}dx\)

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 1576

\(\displaystyle \frac {1}{2} \int \frac {e \left (3 d^2+e^2 x^2\right )}{c x^4+b x^2+a}dx^2+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} e \int \frac {3 d^2+e^2 x^2}{c x^4+b x^2+a}dx^2+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{2} e \left (\frac {\left (6 c d^2-b e^2\right ) \int \frac {1}{c x^4+b x^2+a}dx^2}{2 c}+\frac {e^2 \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}\right )+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} e \left (\frac {e^2 \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}-\frac {\left (6 c d^2-b e^2\right ) \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{c}\right )+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} e \left (\frac {e^2 \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}-\frac {\left (6 c d^2-b e^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}+3 e^2\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (3 e^2-\frac {2 c d^2-3 b e^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}+\frac {1}{2} e \left (\frac {e^2 \log \left (a+b x^2+c x^4\right )}{2 c}-\frac {\left (6 c d^2-b e^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )\)

Input:

Int[(d + e*x)^3/(a + b*x^2 + c*x^4),x]
 

Output:

(d*(3*e^2 + (2*c*d^2 - 3*b*e^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c] 
*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sqrt[b^2 - 4*a 
*c]]) + (d*(3*e^2 - (2*c*d^2 - 3*b*e^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2] 
*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b + Sqrt[b 
^2 - 4*a*c]]) + (e*(-(((6*c*d^2 - b*e^2)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 
4*a*c]])/(c*Sqrt[b^2 - 4*a*c])) + (e^2*Log[a + b*x^2 + c*x^4])/(2*c)))/2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1480

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]
 

rule 1576

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( 
p_.), x_Symbol] :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] 
, x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
 

rule 2202

Int[(Pn_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{n 
 = Expon[Pn, x], k}, Int[Sum[Coeff[Pn, x, 2*k]*x^(2*k), {k, 0, n/2}]*(a + b 
*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pn, x, 2*k + 1]*x^(2*k), {k, 0, (n - 
1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pn, x] 
 &&  !PolyQ[Pn, x^2]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.24

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{3} e^{3}+3 \textit {\_R}^{2} d \,e^{2}+3 \textit {\_R} \,d^{2} e +d^{3}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} c +b \textit {\_R}}\right )}{2}\)	\(65\)
default	\(4 c \left (-\frac {\sqrt {-4 a c +b^{2}}\, \left (-\frac {\left (-\sqrt {-4 a c +b^{2}}\, e^{3}+b \,e^{3}-6 d^{2} e c \right ) \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )}{4 c}+\frac {\left (-3 \sqrt {-4 a c +b^{2}}\, d \,e^{2}+3 b d \,e^{2}-2 c \,d^{3}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 \left (4 a c -b^{2}\right ) c}-\frac {\sqrt {-4 a c +b^{2}}\, \left (\frac {\left (\sqrt {-4 a c +b^{2}}\, e^{3}+b \,e^{3}-6 d^{2} e c \right ) \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )}{4 c}+\frac {\left (3 \sqrt {-4 a c +b^{2}}\, d \,e^{2}+3 b d \,e^{2}-2 c \,d^{3}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 \left (4 a c -b^{2}\right ) c}\right )\)	\(318\)

Input:

int((e*x+d)^3/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

1/2*sum((_R^3*e^3+3*_R^2*d*e^2+3*_R*d^2*e+d^3)/(2*_R^3*c+_R*b)*ln(x-_R),_R 
=RootOf(_Z^4*c+_Z^2*b+a))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)^3/(c*x^4+b*x^2+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x + d\right )}^{3}}{c x^{4} + b x^{2} + a} \,d x } \] Input:

integrate((e*x+d)^3/(c*x^4+b*x^2+a),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3317 vs. \(2 (228) = 456\).

Time = 0.99 (sec) , antiderivative size = 3317, normalized size of antiderivative = 12.29 \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\text {Too large to display} \] Input:

integrate((e*x+d)^3/(c*x^4+b*x^2+a),x, algorithm="giac")
 

Output:

1/4*e^3*log(abs(c*x^4 + b*x^2 + a))/c + 1/8*(3*(2*b^4*c^2 - 16*a*b^2*c^3 + 
 32*a^2*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^ 
4 + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c + 
2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c - 16*sqr 
t(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^2 - 8*sqrt(2) 
*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^2 - sqrt(2)*sqrt( 
b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^2 + 4*sqrt(2)*sqrt(b^2 
- 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^3 - 2*(b^2 - 4*a*c)*b^2*c^2 + 
 8*(b^2 - 4*a*c)*a*c^3)*c^2*d*e^2 + 2*(sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c 
)*c)*b^4*c^2 - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^3 - 2*sqr 
t(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^3 - 2*b^4*c^3 + 16*sqrt(2)*sqrt 
(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^4 + 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a* 
c)*c)*a*b*c^4 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^4 + 16*a*b^2 
*c^4 - 4*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^5 - 32*a^2*c^5 + 2*(b 
^2 - 4*a*c)*b^2*c^3 - 8*(b^2 - 4*a*c)*a*c^4)*d^3*abs(c) + 2*(2*b^3*c^5 - 8 
*a*b*c^6 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c 
^3 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^4 + 
 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^4 - sqr 
t(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b*c^5 - 2*(b^2 - 4* 
a*c)*b*c^5)*d^3 - 3*(2*b^4*c^4 - 8*a*b^2*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c...
 

Mupad [B] (verification not implemented)

Time = 23.06 (sec) , antiderivative size = 9076, normalized size of antiderivative = 33.61 \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

symsum(log(b^2*d^3*e^6 + 6*c^2*d^7*e^2 - 8*root(128*a^2*b^2*c^3*z^4 - 16*a 
*b^4*c^2*z^4 - 256*a^3*c^4*z^4 + 16*a*b^4*c*e^3*z^3 - 128*a^2*b^2*c^2*e^3* 
z^3 + 256*a^3*c^3*e^3*z^3 + 240*a^2*b*c^2*d^2*e^4*z^2 + 120*a*b^2*c^2*d^4* 
e^2*z^2 - 60*a*b^3*c*d^2*e^4*z^2 + 40*a^2*b^2*c*e^6*z^2 + 16*a*b*c^3*d^6*z 
^2 - 480*a^2*c^3*d^4*e^2*z^2 - 96*a^3*c^2*e^6*z^2 - 4*b^3*c^2*d^6*z^2 - 4* 
a*b^4*e^6*z^2 - 48*a^2*b*c*d^2*e^7*z + 48*a*b^2*c*d^4*e^5*z - 16*a*b*c^2*d 
^6*e^3*z - 192*a^2*c^2*d^4*e^5*z - 12*b^2*c^2*d^8*e*z + 4*b^3*c*d^6*e^3*z 
+ 12*a*b^3*d^2*e^7*z + 48*a*c^3*d^8*e*z + 16*a^3*c*e^9*z - 4*a^2*b^2*e^9*z 
 - 6*a*b*c*d^6*e^6 - 3*b^2*c*d^8*e^4 - 3*b*c^2*d^10*e^2 - 3*a^2*c*d^4*e^8 
- 3*a*c^2*d^8*e^4 - 3*a^2*b*d^2*e^10 - 3*a*b^2*d^4*e^8 - b^3*d^6*e^6 - c^3 
*d^12 - a^3*e^12, z, k)^3*b^3*c^2*x + 3*b^2*d^2*e^7*x + 10*c^2*d^6*e^3*x - 
 3*a*b*d*e^8 + 4*root(128*a^2*b^2*c^3*z^4 - 16*a*b^4*c^2*z^4 - 256*a^3*c^4 
*z^4 + 16*a*b^4*c*e^3*z^3 - 128*a^2*b^2*c^2*e^3*z^3 + 256*a^3*c^3*e^3*z^3 
+ 240*a^2*b*c^2*d^2*e^4*z^2 + 120*a*b^2*c^2*d^4*e^2*z^2 - 60*a*b^3*c*d^2*e 
^4*z^2 + 40*a^2*b^2*c*e^6*z^2 + 16*a*b*c^3*d^6*z^2 - 480*a^2*c^3*d^4*e^2*z 
^2 - 96*a^3*c^2*e^6*z^2 - 4*b^3*c^2*d^6*z^2 - 4*a*b^4*e^6*z^2 - 48*a^2*b*c 
*d^2*e^7*z + 48*a*b^2*c*d^4*e^5*z - 16*a*b*c^2*d^6*e^3*z - 192*a^2*c^2*d^4 
*e^5*z - 12*b^2*c^2*d^8*e*z + 4*b^3*c*d^6*e^3*z + 12*a*b^3*d^2*e^7*z + 48* 
a*c^3*d^8*e*z + 16*a^3*c*e^9*z - 4*a^2*b^2*e^9*z - 6*a*b*c*d^6*e^6 - 3*b^2 
*c*d^8*e^4 - 3*b*c^2*d^10*e^2 - 3*a^2*c*d^4*e^8 - 3*a*c^2*d^8*e^4 - 3*a...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1120, normalized size of antiderivative = 4.15 \[ \int \frac {(d+e x)^3}{a+b x^2+c x^4} \, dx =\text {Too large to display} \] Input:

int((e*x+d)^3/(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 - 
12*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*c*d**2*e 
- 12*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 
*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqr 
t(2*sqrt(c)*sqrt(a) + b))*b*c*d**3 + 6*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*s 
qrt(2*sqrt(c)*sqrt(a) + b)*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*e**3 - 12*s 
qrt(2*sqrt(c)*sqrt(a) + b)*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**2*e + 12 
*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*sqr 
t(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 - 6*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*ata 
n((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + ...