\(\int \frac {(d f+e f x)^2}{(a+b (d+e x)^2+c (d+e x)^4)^2} \, dx\) [277]

Optimal result

Integrand size = 33, antiderivative size = 263 \[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=-\frac {f^2 (d+e x) \left (b+2 c (d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac {\sqrt {c} \left (2 b-\sqrt {b^2-4 a c}\right ) f^2 \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}} e}-\frac {\sqrt {c} \left (2 b+\sqrt {b^2-4 a c}\right ) f^2 \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}} e} \] Output:

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

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.95 \[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=-\frac {f^2 \left (\frac {b (d+e x)+2 c (d+e x)^3}{\left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac {\sqrt {2} \sqrt {c} \left (-2 b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \sqrt {c} \left (2 b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 e} \] Input:

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

Output:

-1/2*(f^2*((b*(d + e*x) + 2*c*(d + e*x)^3)/((b^2 - 4*a*c)*(a + b*(d + e*x) 
^2 + c*(d + e*x)^4)) + (Sqrt[2]*Sqrt[c]*(-2*b + 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]*(2*b + 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]])))/e
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1462, 1439, 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.

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

\(\Big \downarrow \) 1462

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

\(\Big \downarrow \) 1439

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(f^2*(-1/2*((d + e*x)*(b + 2*c*(d + e*x)^2))/((b^2 - 4*a*c)*(a + b*(d + e* 
x)^2 + c*(d + e*x)^4)) + (-((Sqrt[2]*Sqrt[c]*(1 - (2*b)/Sqrt[b^2 - 4*a*c]) 
*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/Sqrt[b - 
 Sqrt[b^2 - 4*a*c]]) - (Sqrt[2]*Sqrt[c]*(1 + (2*b)/Sqrt[b^2 - 4*a*c])*ArcT 
an[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/Sqrt[b + Sqrt 
[b^2 - 4*a*c]])/(2*(b^2 - 4*a*c))))/e
 

Defintions of rubi rules used

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 1439

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[d*(d*x)^(m - 1)*(b + 2*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*(p + 
1)*(b^2 - 4*a*c))), x] - Simp[d^2/(2*(p + 1)*(b^2 - 4*a*c))   Int[(d*x)^(m 
- 2)*(b*(m - 1) + 2*c*(m + 4*p + 5)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x 
] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 
1] && LeQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1462

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]
 

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]
 

Maple [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.23

method	result	size
default	\(f^{2} \left (\frac {\frac {c \,e^{2} x^{3}}{4 a c -b^{2}}+\frac {3 d c e \,x^{2}}{4 a c -b^{2}}+\frac {\left (6 c \,d^{2}+b \right ) x}{8 a c -2 b^{2}}+\frac {d \left (2 c \,d^{2}+b \right )}{2 e \left (4 a c -b^{2}\right )}}{c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 b d e \right ) \textit {\_Z} +c \,d^{4}+b \,d^{2}+a \right )}{\sum }\frac {\left (2 \textit {\_R}^{2} c \,e^{2}+4 \textit {\_R} c d e +2 c \,d^{2}-b \right ) \ln \left (x -\textit {\_R} \right )}{2 c \,e^{3} \textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 c \,d^{3}+e b \textit {\_R} +b d}}{4 \left (4 a c -b^{2}\right ) e}\right )\)	\(323\)
risch	\(\frac {\frac {c \,e^{2} f^{2} x^{3}}{4 a c -b^{2}}+\frac {3 d c e \,f^{2} x^{2}}{4 a c -b^{2}}+\frac {f^{2} \left (6 c \,d^{2}+b \right ) x}{8 a c -2 b^{2}}+\frac {d \,f^{2} \left (2 c \,d^{2}+b \right )}{2 e \left (4 a c -b^{2}\right )}}{c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a}+\frac {f^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 b d e \right ) \textit {\_Z} +c \,d^{4}+b \,d^{2}+a \right )}{\sum }\frac {\left (\frac {2 c \,e^{2} \textit {\_R}^{2}}{4 a c -b^{2}}+\frac {4 c d e \textit {\_R}}{4 a c -b^{2}}-\frac {-2 c \,d^{2}+b}{4 a c -b^{2}}\right ) \ln \left (x -\textit {\_R} \right )}{2 c \,e^{3} \textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 c \,d^{3}+e b \textit {\_R} +b d}\right )}{4 e}\)	\(359\)

Input:

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

Output:

f^2*((c*e^2/(4*a*c-b^2)*x^3+3*d*c*e/(4*a*c-b^2)*x^2+1/2*(6*c*d^2+b)/(4*a*c 
-b^2)*x+1/2*d/e*(2*c*d^2+b)/(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*s 
um((2*_R^2*c*e^2+4*_R*c*d*e+2*c*d^2-b)/(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+c*d^4+b*d^2+a)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2600 vs. \(2 (222) = 444\).

Time = 0.27 (sec) , antiderivative size = 2600, normalized size of antiderivative = 9.89 \[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(4*c*e^3*f^2*x^3 + 12*c*d*e^2*f^2*x^2 + 2*(6*c*d^2 + b)*e*f^2*x + 2*( 
2*c*d^3 + b*d)*f^2 + sqrt(1/2)*((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a 
*c^2)*d*e^4*x^3 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2 
*(b^2*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 
 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*d^2)*e)*sqrt(-((b^3 + 12*a*b*c)*f^4 + 
 (a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*sqrt(f^8/((a^2*b^6 - 
 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4))*e^2)/((a*b^6 - 12*a^2*b 
^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))*log((3*b^2*c + 4*a*c^2)*e*f^6*x 
+ (3*b^2*c + 4*a*c^2)*d*f^6 + 1/2*sqrt(1/2)*((b^5 - 8*a*b^3*c + 16*a^2*b*c 
^2)*e*f^4 - (a*b^8 - 8*a^2*b^6*c + 128*a^4*b^2*c^3 - 256*a^5*c^4)*sqrt(f^8 
/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4))*e^3)*sqrt(- 
((b^3 + 12*a*b*c)*f^4 + (a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^ 
3)*sqrt(f^8/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4))* 
e^2)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))) - sqrt(1 
/2)*((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^4*x^3 + (b^3 - 4* 
a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 + (b 
^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 - 4*a^2*c + (b^3 - 
 4*a*b*c)*d^2)*e)*sqrt(-((b^3 + 12*a*b*c)*f^4 + (a*b^6 - 12*a^2*b^4*c + 48 
*a^3*b^2*c^2 - 64*a^4*c^3)*sqrt(f^8/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2* 
c^2 - 64*a^5*c^3)*e^4))*e^2)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - ...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 646 vs. \(2 (240) = 480\).

Time = 11.83 (sec) , antiderivative size = 646, normalized size of antiderivative = 2.46 \[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\frac {b d f^{2} + 2 c d^{3} f^{2} + 6 c d e^{2} f^{2} x^{2} + 2 c e^{3} f^{2} x^{3} + x \left (b e f^{2} + 6 c d^{2} e f^{2}\right )}{8 a^{2} c e - 2 a b^{2} e + 8 a b c d^{2} e + 8 a c^{2} d^{4} e - 2 b^{3} d^{2} e - 2 b^{2} c d^{4} e + x^{4} \cdot \left (8 a c^{2} e^{5} - 2 b^{2} c e^{5}\right ) + x^{3} \cdot \left (32 a c^{2} d e^{4} - 8 b^{2} c d e^{4}\right ) + x^{2} \cdot \left (8 a b c e^{3} + 48 a c^{2} d^{2} e^{3} - 2 b^{3} e^{3} - 12 b^{2} c d^{2} e^{3}\right ) + x \left (16 a b c d e^{2} + 32 a c^{2} d^{3} e^{2} - 4 b^{3} d e^{2} - 8 b^{2} c d^{3} e^{2}\right )} + \operatorname {RootSum} {\left (t^{4} \cdot \left (1048576 a^{7} c^{6} e^{4} - 1572864 a^{6} b^{2} c^{5} e^{4} + 983040 a^{5} b^{4} c^{4} e^{4} - 327680 a^{4} b^{6} c^{3} e^{4} + 61440 a^{3} b^{8} c^{2} e^{4} - 6144 a^{2} b^{10} c e^{4} + 256 a b^{12} e^{4}\right ) + t^{2} \left (- 12288 a^{4} b c^{4} e^{2} f^{4} + 8192 a^{3} b^{3} c^{3} e^{2} f^{4} - 1536 a^{2} b^{5} c^{2} e^{2} f^{4} + 16 b^{9} e^{2} f^{4}\right ) + 16 a^{2} c^{3} f^{8} + 24 a b^{2} c^{2} f^{8} + 9 b^{4} c f^{8}, \left ( t \mapsto t \log {\left (x + \frac {16384 t^{3} a^{5} c^{4} e^{3} - 8192 t^{3} a^{4} b^{2} c^{3} e^{3} + 512 t^{3} a^{2} b^{6} c e^{3} - 64 t^{3} a b^{8} e^{3} - 128 t a^{2} b c^{2} e f^{4} - 16 t a b^{3} c e f^{4} - 4 t b^{5} e f^{4} + 4 a c^{2} d f^{6} + 3 b^{2} c d f^{6}}{4 a c^{2} e f^{6} + 3 b^{2} c e f^{6}} \right )} \right )\right )} \] Input:

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

Output:

(b*d*f**2 + 2*c*d**3*f**2 + 6*c*d*e**2*f**2*x**2 + 2*c*e**3*f**2*x**3 + x* 
(b*e*f**2 + 6*c*d**2*e*f**2))/(8*a**2*c*e - 2*a*b**2*e + 8*a*b*c*d**2*e + 
8*a*c**2*d**4*e - 2*b**3*d**2*e - 2*b**2*c*d**4*e + x**4*(8*a*c**2*e**5 - 
2*b**2*c*e**5) + x**3*(32*a*c**2*d*e**4 - 8*b**2*c*d*e**4) + x**2*(8*a*b*c 
*e**3 + 48*a*c**2*d**2*e**3 - 2*b**3*e**3 - 12*b**2*c*d**2*e**3) + x*(16*a 
*b*c*d*e**2 + 32*a*c**2*d**3*e**2 - 4*b**3*d*e**2 - 8*b**2*c*d**3*e**2)) + 
 RootSum(_t**4*(1048576*a**7*c**6*e**4 - 1572864*a**6*b**2*c**5*e**4 + 983 
040*a**5*b**4*c**4*e**4 - 327680*a**4*b**6*c**3*e**4 + 61440*a**3*b**8*c** 
2*e**4 - 6144*a**2*b**10*c*e**4 + 256*a*b**12*e**4) + _t**2*(-12288*a**4*b 
*c**4*e**2*f**4 + 8192*a**3*b**3*c**3*e**2*f**4 - 1536*a**2*b**5*c**2*e**2 
*f**4 + 16*b**9*e**2*f**4) + 16*a**2*c**3*f**8 + 24*a*b**2*c**2*f**8 + 9*b 
**4*c*f**8, Lambda(_t, _t*log(x + (16384*_t**3*a**5*c**4*e**3 - 8192*_t**3 
*a**4*b**2*c**3*e**3 + 512*_t**3*a**2*b**6*c*e**3 - 64*_t**3*a*b**8*e**3 - 
 128*_t*a**2*b*c**2*e*f**4 - 16*_t*a*b**3*c*e*f**4 - 4*_t*b**5*e*f**4 + 4* 
a*c**2*d*f**6 + 3*b**2*c*d*f**6)/(4*a*c**2*e*f**6 + 3*b**2*c*e*f**6))))
 

Maxima [F]

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

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

Output:

1/2*f^2*integrate(-(2*c*e^2*x^2 + 4*c*d*e*x + 2*c*d^2 - b)/((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) - 1 
/2*(2*c*e^3*f^2*x^3 + 6*c*d*e^2*f^2*x^2 + (6*c*d^2 + b)*e*f^2*x + (2*c*d^3 
 + b*d)*f^2)/((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^4*x^3 + 
(b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2) 
*d^3 + (b^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 - 4*a^2*c 
 + (b^3 - 4*a*b*c)*d^2)*e)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1482 vs. \(2 (222) = 444\).

Time = 0.14 (sec) , antiderivative size = 1482, normalized size of antiderivative = 5.63 \[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((2*c*e^2*f^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4) 
) + d/e)^2 - 4*c*d*e*f^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/ 
(c*e^4)) + d/e) + 2*c*d^2*f^2 - b*f^2)*log(x + sqrt(1/2)*sqrt(-(b*e^2 + sq 
rt(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e)/(2*c*e^4*(sqrt(1/2)*sqrt(-(b*e^2 + sq 
rt(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)) - (2*c*e^2*f^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e 
^4)) - d/e)^2 + 4*c*d*e*f^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^ 
2)/(c*e^4)) - d/e) + 2*c*d^2*f^2 - b*f^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)) + (2*c*e^2*f^2*(sqrt(1/2)*sqrt(-(b*e^2 - sqrt(b^2 - 4*a*c)*e^2)/( 
c*e^4)) + d/e)^2 - 4*c*d*e*f^2*(sqrt(1/2)*sqrt(-(b*e^2 - sqrt(b^2 - 4*a*c) 
*e^2)/(c*e^4)) + d/e) + 2*c*d^2*f^2 - b*f^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 +...
 

Mupad [B] (verification not implemented)

Time = 12.64 (sec) , antiderivative size = 7835, normalized size of antiderivative = 29.79 \[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((x*(b*f^2 + 6*c*d^2*f^2))/(2*(4*a*c - b^2)) + (2*c*d^3*f^2 + b*d*f^2)/(2* 
e*(4*a*c - b^2)) + (c*e^2*f^2*x^3)/(4*a*c - b^2) + (3*c*d*e*f^2*x^2)/(4*a* 
c - b^2))/(a + x^2*(b*e^2 + 6*c*d^2*e^2) + b*d^2 + c*d^4 + x*(2*b*d*e + 4* 
c*d^3*e) + c*e^4*x^4 + 4*c*d*e^3*x^3) + atan(((((f^4*(-(4*a*c - b^2)^9)^(1 
/2))/32 - (b^9*f^4)/32 + 24*a^4*b*c^4*f^4 + 3*a^2*b^5*c^2*f^4 - 16*a^3*b^3 
*c^3*f^4)/(a*b^12*e^2 + 4096*a^7*c^6*e^2 - 24*a^2*b^10*c*e^2 + 240*a^3*b^8 
*c^2*e^2 - 1280*a^4*b^6*c^3*e^2 + 3840*a^5*b^4*c^4*e^2 - 6144*a^6*b^2*c^5* 
e^2))^(1/2)*((((f^4*(-(4*a*c - b^2)^9)^(1/2))/32 - (b^9*f^4)/32 + 24*a^4*b 
*c^4*f^4 + 3*a^2*b^5*c^2*f^4 - 16*a^3*b^3*c^3*f^4)/(a*b^12*e^2 + 4096*a^7* 
c^6*e^2 - 24*a^2*b^10*c*e^2 + 240*a^3*b^8*c^2*e^2 - 1280*a^4*b^6*c^3*e^2 + 
 3840*a^5*b^4*c^4*e^2 - 6144*a^6*b^2*c^5*e^2))^(1/2)*(((8*b^9*c^2*d*e^13 - 
 128*a*b^7*c^3*d*e^13 + 2048*a^4*b*c^6*d*e^13 + 768*a^2*b^5*c^4*d*e^13 - 2 
048*a^3*b^3*c^5*d*e^13)/(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c) + 
 (x*(8*b^7*c^2*e^14 - 96*a*b^5*c^3*e^14 - 512*a^3*b*c^5*e^14 + 384*a^2*b^3 
*c^4*e^14))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c))*(((f^4*(-(4*a*c - b^2)^9)^(1/2 
))/32 - (b^9*f^4)/32 + 24*a^4*b*c^4*f^4 + 3*a^2*b^5*c^2*f^4 - 16*a^3*b^3*c 
^3*f^4)/(a*b^12*e^2 + 4096*a^7*c^6*e^2 - 24*a^2*b^10*c*e^2 + 240*a^3*b^8*c 
^2*e^2 - 1280*a^4*b^6*c^3*e^2 + 3840*a^5*b^4*c^4*e^2 - 6144*a^6*b^2*c^5*e^ 
2))^(1/2) + (2*b^7*c^2*e^12*f^2 + 96*a^2*b^3*c^4*e^12*f^2 - 24*a*b^5*c^3*e 
^12*f^2 - 128*a^3*b*c^5*e^12*f^2)/(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - ...
 

Reduce [F]

\[ \int \frac {(d f+e f x)^2}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx=\text {too large to display} \] Input:

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

Output:

(f**2*( - 4*int(x**3/(a**2*b + 2*a**2*c*d**2 + 2*a*b**2*d**2 + 4*a*b**2*d* 
e*x + 2*a*b**2*e**2*x**2 + 6*a*b*c*d**4 + 16*a*b*c*d**3*e*x + 16*a*b*c*d** 
2*e**2*x**2 + 8*a*b*c*d*e**3*x**3 + 2*a*b*c*e**4*x**4 + 4*a*c**2*d**6 + 16 
*a*c**2*d**5*e*x + 24*a*c**2*d**4*e**2*x**2 + 16*a*c**2*d**3*e**3*x**3 + 4 
*a*c**2*d**2*e**4*x**4 + b**3*d**4 + 4*b**3*d**3*e*x + 6*b**3*d**2*e**2*x* 
*2 + 4*b**3*d*e**3*x**3 + b**3*e**4*x**4 + 4*b**2*c*d**6 + 20*b**2*c*d**5* 
e*x + 42*b**2*c*d**4*e**2*x**2 + 48*b**2*c*d**3*e**3*x**3 + 32*b**2*c*d**2 
*e**4*x**4 + 12*b**2*c*d*e**5*x**5 + 2*b**2*c*e**6*x**6 + 5*b*c**2*d**8 + 
32*b*c**2*d**7*e*x + 88*b*c**2*d**6*e**2*x**2 + 136*b*c**2*d**5*e**3*x**3 
+ 130*b*c**2*d**4*e**4*x**4 + 80*b*c**2*d**3*e**5*x**5 + 32*b*c**2*d**2*e* 
*6*x**6 + 8*b*c**2*d*e**7*x**7 + b*c**2*e**8*x**8 + 2*c**3*d**10 + 16*c**3 
*d**9*e*x + 56*c**3*d**8*e**2*x**2 + 112*c**3*d**7*e**3*x**3 + 140*c**3*d* 
*6*e**4*x**4 + 112*c**3*d**5*e**5*x**5 + 56*c**3*d**4*e**6*x**6 + 16*c**3* 
d**3*e**7*x**7 + 2*c**3*d**2*e**8*x**8),x)*a*b*c*d*e**4 - 8*int(x**3/(a**2 
*b + 2*a**2*c*d**2 + 2*a*b**2*d**2 + 4*a*b**2*d*e*x + 2*a*b**2*e**2*x**2 + 
 6*a*b*c*d**4 + 16*a*b*c*d**3*e*x + 16*a*b*c*d**2*e**2*x**2 + 8*a*b*c*d*e* 
*3*x**3 + 2*a*b*c*e**4*x**4 + 4*a*c**2*d**6 + 16*a*c**2*d**5*e*x + 24*a*c* 
*2*d**4*e**2*x**2 + 16*a*c**2*d**3*e**3*x**3 + 4*a*c**2*d**2*e**4*x**4 + b 
**3*d**4 + 4*b**3*d**3*e*x + 6*b**3*d**2*e**2*x**2 + 4*b**3*d*e**3*x**3 + 
b**3*e**4*x**4 + 4*b**2*c*d**6 + 20*b**2*c*d**5*e*x + 42*b**2*c*d**4*e*...