\(\int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx\) [640]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1156, 1144, 211} \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\frac {f^2 \sqrt {\sqrt {b^2-4 a c}+b} \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} e \sqrt {b^2-4 a c}}-\frac {f^2 \sqrt {b-\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} e \sqrt {b^2-4 a c}} \]

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(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 1156

Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {f^2 \text {Subst}\left (\int \frac {x^2}{a+b x^2+c x^4} \, dx,x,d+e x\right )}{e} \\ & = \frac {\left (\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) f^2\right ) \text {Subst}\left (\int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx,x,d+e x\right )}{2 e}+\frac {\left (\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) f^2\right ) \text {Subst}\left (\int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx,x,d+e x\right )}{2 e} \\ & = -\frac {\sqrt {b-\sqrt {b^2-4 a c}} f^2 \tan ^{-1}\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 \tan ^{-1}\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\frac {f^2 \left (\left (-b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )+\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {b+\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} (d+e x)}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}} e} \]

[In]

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

[Out]

(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]]]))/(
Sqrt[2]*Sqrt[c]*Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*e)

Maple [C] (verified)

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

Time = 0.59 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.84

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

[In]

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

[Out]

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+d^4*c+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 = 0.28 (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=\frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b f^{4} + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}} \log \left (e f^{6} x + d f^{6} + \sqrt {\frac {1}{2}} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{3} \sqrt {-\frac {b f^{4} + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b f^{4} + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}} \log \left (e f^{6} x + d f^{6} - \sqrt {\frac {1}{2}} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{3} \sqrt {-\frac {b f^{4} + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b f^{4} - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}} \log \left (e f^{6} x + d f^{6} + \sqrt {\frac {1}{2}} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{3} \sqrt {-\frac {b f^{4} - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}}\right ) + \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b f^{4} - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}} \log \left (e f^{6} x + d f^{6} - \sqrt {\frac {1}{2}} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{3} \sqrt {-\frac {b f^{4} - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {f^{8}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{4}}} e^{2}}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}}}\right ) \]

[In]

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

[Out]

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)*sqr
t(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)))

Sympy [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (256 a^{2} c^{3} e^{4} - 128 a b^{2} c^{2} e^{4} + 16 b^{4} c e^{4}\right ) + t^{2} \left (- 16 a b c e^{2} f^{4} + 4 b^{3} e^{2} f^{4}\right ) + a f^{8}, \left ( t \mapsto t \log {\left (x + \frac {64 t^{3} a c^{2} e^{3} - 16 t^{3} b^{2} c e^{3} - 2 t b e f^{4} + d f^{6}}{e f^{6}} \right )} \right )\right )} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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 = 0.33 (sec) , antiderivative size = 1443, normalized size of antiderivative = 8.49 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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*(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*(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 + sqr
t(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(1/2)*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)

Mupad [B] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 683, normalized size of antiderivative = 4.02 \[ \int \frac {(d f+e f x)^2}{a+b (d+e x)^2+c (d+e x)^4} \, dx=-2\,\mathrm {atanh}\left (\frac {\sqrt {-\frac {b^3\,f^4+f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}}\,\left (x\,\left (4\,a\,c^2\,e^{12}\,f^4-2\,b^2\,c\,e^{12}\,f^4\right )+\frac {\left (x\,\left (8\,b^3\,c^2\,e^{14}-32\,a\,b\,c^3\,e^{14}\right )+8\,b^3\,c^2\,d\,e^{13}-32\,a\,b\,c^3\,d\,e^{13}\right )\,\left (b^3\,f^4+f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-4\,a\,b\,c\,f^4\right )}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}+4\,a\,c^2\,d\,e^{11}\,f^4-2\,b^2\,c\,d\,e^{11}\,f^4\right )}{a\,c\,e^{10}\,f^6}\right )\,\sqrt {-\frac {b^3\,f^4+f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}}-2\,\mathrm {atanh}\left (\frac {\sqrt {\frac {f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,f^4+4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}}\,\left (x\,\left (4\,a\,c^2\,e^{12}\,f^4-2\,b^2\,c\,e^{12}\,f^4\right )-\frac {\left (x\,\left (8\,b^3\,c^2\,e^{14}-32\,a\,b\,c^3\,e^{14}\right )+8\,b^3\,c^2\,d\,e^{13}-32\,a\,b\,c^3\,d\,e^{13}\right )\,\left (f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,f^4+4\,a\,b\,c\,f^4\right )}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}+4\,a\,c^2\,d\,e^{11}\,f^4-2\,b^2\,c\,d\,e^{11}\,f^4\right )}{a\,c\,e^{10}\,f^6}\right )\,\sqrt {\frac {f^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,f^4+4\,a\,b\,c\,f^4}{8\,\left (16\,a^2\,c^3\,e^2-8\,a\,b^2\,c^2\,e^2+b^4\,c\,e^2\right )}} \]

[In]

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

[Out]

- 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^11*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)