\(\int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx\) [269]

Optimal result

Integrand size = 33, antiderivative size = 170 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=-\frac {\sqrt {b-\sqrt {b^2-4 a c}} f^2 \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2-4 a c} e}+\frac {\sqrt {b+\sqrt {b^2-4 a c}} f^2 \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2-4 a c} e} \] Output:

-1/2*(b-(-4*a*c+b^2)^(1/2))^(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)/c^(1/2)/(-4*a*c+b^2)^(1/2)/e+1/2*(b+(-4*a 
*c+b^2)^(1/2))^(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)/c^(1/2)/(-4*a*c+b^2)^(1/2)/e
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\frac {f^2 \left (\left (-b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )+\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {b+\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}} e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1462, 1450, 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.

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

\(\Big \downarrow \) 1462

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

\(\Big \downarrow \) 1450

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(f^2*(((1 - b/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[c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (( 
1 + b/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[c]*Sqrt[b + Sqrt[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 1450

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Wi 
th[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(d^2/2)*(b/q + 1)   Int[(d*x)^(m - 2)/(b/ 
2 + q/2 + c*x^2), x], x] - Simp[(d^2/2)*(b/q - 1)   Int[(d*x)^(m - 2)/(b/2 
- q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && 
 GeQ[m, 2]
 

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]
 

Maple [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {f^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 b d e \right ) \textit {\_Z} +c \,d^{4}+b \,d^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{2} e^{2}+2 \textit {\_R} d e +d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{2 c \,e^{3} \textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 c \,d^{3}+e b \textit {\_R} +b d}\right )}{2 e}\)	\(143\)
risch	\(\frac {f^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 b d e \right ) \textit {\_Z} +c \,d^{4}+b \,d^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{2} e^{2}+2 \textit {\_R} d e +d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{2 c \,e^{3} \textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 c \,d^{3}+e b \textit {\_R} +b d}\right )}{2 e}\)	\(143\)

Input:

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

Output:

1/2*f^2/e*sum((_R^2*e^2+2*_R*d*e+d^2)/(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. 799 vs. \(2 (135) = 270\).

Time = 0.12 (sec) , antiderivative size = 799, normalized size of antiderivative = 4.70 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*sqrt(1/2)*sqrt(-(b*f^4 + (b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^ 
3)*e^4))*e^2)/((b^2*c - 4*a*c^2)*e^2))*log(e*f^6*x + d*f^6 + sqrt(1/2)*(b^ 
2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^3*sqrt(-(b*f^4 + (b^2 
*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^2)/((b^2*c - 4*a*c^2)* 
e^2))) - 1/2*sqrt(1/2)*sqrt(-(b*f^4 + (b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 
 - 4*a*c^3)*e^4))*e^2)/((b^2*c - 4*a*c^2)*e^2))*log(e*f^6*x + d*f^6 - sqrt 
(1/2)*(b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^3*sqrt(-(b*f 
^4 + (b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^2)/((b^2*c - 
4*a*c^2)*e^2))) - 1/2*sqrt(1/2)*sqrt(-(b*f^4 - (b^2*c - 4*a*c^2)*sqrt(f^8/ 
((b^2*c^2 - 4*a*c^3)*e^4))*e^2)/((b^2*c - 4*a*c^2)*e^2))*log(e*f^6*x + d*f 
^6 + sqrt(1/2)*(b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^3*s 
qrt(-(b*f^4 - (b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^2)/( 
(b^2*c - 4*a*c^2)*e^2))) + 1/2*sqrt(1/2)*sqrt(-(b*f^4 - (b^2*c - 4*a*c^2)* 
sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4))*e^2)/((b^2*c - 4*a*c^2)*e^2))*log(e*f^ 
6*x + d*f^6 - sqrt(1/2)*(b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^ 
4))*e^3*sqrt(-(b*f^4 - (b^2*c - 4*a*c^2)*sqrt(f^8/((b^2*c^2 - 4*a*c^3)*e^4 
))*e^2)/((b^2*c - 4*a*c^2)*e^2)))
 

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (256 a^{2} c^{3} e^{4} - 128 a b^{2} c^{2} e^{4} + 16 b^{4} c e^{4}\right ) + t^{2} \left (- 16 a b c e^{2} f^{4} + 4 b^{3} e^{2} f^{4}\right ) + a f^{8}, \left ( t \mapsto t \log {\left (x + \frac {64 t^{3} a c^{2} e^{3} - 16 t^{3} b^{2} c e^{3} - 2 t b e f^{4} + d f^{6}}{e f^{6}} \right )} \right )\right )} \] Input:

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

Output:

RootSum(_t**4*(256*a**2*c**3*e**4 - 128*a*b**2*c**2*e**4 + 16*b**4*c*e**4) 
 + _t**2*(-16*a*b*c*e**2*f**4 + 4*b**3*e**2*f**4) + a*f**8, Lambda(_t, _t* 
log(x + (64*_t**3*a*c**2*e**3 - 16*_t**3*b**2*c*e**3 - 2*_t*b*e*f**4 + d*f 
**6)/(e*f**6))))
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1443 vs. \(2 (135) = 270\).

Time = 0.14 (sec) , antiderivative size = 1443, normalized size of antiderivative = 8.49 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(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 - 2*d*e*f^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4 
)) + d/e) + d^2*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 + 6*c*d^2*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 
 - 4*a*c)*e^2)/(c*e^4)) + d/e) - 2*c*d^3*e + b*e^2*(sqrt(1/2)*sqrt(-(b*e^2 
 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e) - b*d*e) + 1/2*(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 + 2*d*e*f^2*(sq 
rt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) - d/e) + d^2*f^2)*l 
og(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 + 6*c*d^2*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e^2)/(c*e^4) 
) - d/e) + 2*c*d^3*e + b*e^2*(sqrt(1/2)*sqrt(-(b*e^2 + sqrt(b^2 - 4*a*c)*e 
^2)/(c*e^4)) - d/e) + b*d*e) - 1/2*(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 - 2*d*e*f^2*(sqrt(1/2)*sqrt(-(b*e^2 - 
 sqrt(b^2 - 4*a*c)*e^2)/(c*e^4)) + d/e) + d^2*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/...
 

Mupad [B] (verification not implemented)

Time = 11.30 (sec) , antiderivative size = 683, normalized size of antiderivative = 4.02 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=-2\,\mathrm {atanh}\left (\frac {\sqrt {-\frac {b^3\,f^4+f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}}\,\left (x\,\left (4\,a\,c^2\,e^{12}\,f^4-2\,b^2\,c\,e^{12}\,f^4\right )+\frac {\left (x\,\left (8\,b^3\,c^2\,e^{14}-32\,a\,b\,c^3\,e^{14}\right )+8\,b^3\,c^2\,d\,e^{13}-32\,a\,b\,c^3\,d\,e^{13}\right )\,\left (b^3\,f^4+f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-4\,a\,b\,c\,f^4\right )}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}+4\,a\,c^2\,d\,e^{11}\,f^4-2\,b^2\,c\,d\,e^{11}\,f^4\right )}{a\,c\,e^{10}\,f^6}\right )\,\sqrt {-\frac {b^3\,f^4+f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}}-2\,\mathrm {atanh}\left (\frac {\sqrt {\frac {f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,f^4+4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}}\,\left (x\,\left (4\,a\,c^2\,e^{12}\,f^4-2\,b^2\,c\,e^{12}\,f^4\right )-\frac {\left (x\,\left (8\,b^3\,c^2\,e^{14}-32\,a\,b\,c^3\,e^{14}\right )+8\,b^3\,c^2\,d\,e^{13}-32\,a\,b\,c^3\,d\,e^{13}\right )\,\left (f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,f^4+4\,a\,b\,c\,f^4\right )}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}+4\,a\,c^2\,d\,e^{11}\,f^4-2\,b^2\,c\,d\,e^{11}\,f^4\right )}{a\,c\,e^{10}\,f^6}\right )\,\sqrt {\frac {f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,f^4+4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}} \] Input:

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

Output:

- 2*atanh(((-(b^3*f^4 + f^4*(-(4*a*c - b^2)^3)^(1/2) - 4*a*b*c*f^4)/(8*(b^ 
4*c*e^2 + 16*a^2*c^3*e^2 - 8*a*b^2*c^2*e^2)))^(1/2)*(x*(4*a*c^2*e^12*f^4 - 
 2*b^2*c*e^12*f^4) + ((x*(8*b^3*c^2*e^14 - 32*a*b*c^3*e^14) + 8*b^3*c^2*d* 
e^13 - 32*a*b*c^3*d*e^13)*(b^3*f^4 + f^4*(-(4*a*c - b^2)^3)^(1/2) - 4*a*b* 
c*f^4))/(8*(b^4*c*e^2 + 16*a^2*c^3*e^2 - 8*a*b^2*c^2*e^2)) + 4*a*c^2*d*e^1 
1*f^4 - 2*b^2*c*d*e^11*f^4))/(a*c*e^10*f^6))*(-(b^3*f^4 + f^4*(-(4*a*c - b 
^2)^3)^(1/2) - 4*a*b*c*f^4)/(8*(b^4*c*e^2 + 16*a^2*c^3*e^2 - 8*a*b^2*c^2*e 
^2)))^(1/2) - 2*atanh((((f^4*(-(4*a*c - b^2)^3)^(1/2) - b^3*f^4 + 4*a*b*c* 
f^4)/(8*(b^4*c*e^2 + 16*a^2*c^3*e^2 - 8*a*b^2*c^2*e^2)))^(1/2)*(x*(4*a*c^2 
*e^12*f^4 - 2*b^2*c*e^12*f^4) - ((x*(8*b^3*c^2*e^14 - 32*a*b*c^3*e^14) + 8 
*b^3*c^2*d*e^13 - 32*a*b*c^3*d*e^13)*(f^4*(-(4*a*c - b^2)^3)^(1/2) - b^3*f 
^4 + 4*a*b*c*f^4))/(8*(b^4*c*e^2 + 16*a^2*c^3*e^2 - 8*a*b^2*c^2*e^2)) + 4* 
a*c^2*d*e^11*f^4 - 2*b^2*c*d*e^11*f^4))/(a*c*e^10*f^6))*((f^4*(-(4*a*c - b 
^2)^3)^(1/2) - b^3*f^4 + 4*a*b*c*f^4)/(8*(b^4*c*e^2 + 16*a^2*c^3*e^2 - 8*a 
*b^2*c^2*e^2)))^(1/2)
 

Reduce [F]

\[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=f^{2} \left (\left (\int \frac {x^{2}}{c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a}d x \right ) e^{2}+2 \left (\int \frac {x}{c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a}d x \right ) d e +\left (\int \frac {1}{c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a}d x \right ) d^{2}\right ) \] Input:

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

Output:

f**2*(int(x**2/(a + b*d**2 + 2*b*d*e*x + b*e**2*x**2 + c*d**4 + 4*c*d**3*e 
*x + 6*c*d**2*e**2*x**2 + 4*c*d*e**3*x**3 + c*e**4*x**4),x)*e**2 + 2*int(x 
/(a + b*d**2 + 2*b*d*e*x + b*e**2*x**2 + c*d**4 + 4*c*d**3*e*x + 6*c*d**2* 
e**2*x**2 + 4*c*d*e**3*x**3 + c*e**4*x**4),x)*d*e + int(1/(a + b*d**2 + 2* 
b*d*e*x + b*e**2*x**2 + c*d**4 + 4*c*d**3*e*x + 6*c*d**2*e**2*x**2 + 4*c*d 
*e**3*x**3 + c*e**4*x**4),x)*d**2)