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

Optimal. Leaf size=98 \[ \frac{2 c f \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}}-\frac{f \left (b+2 c (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.127071, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {1142, 1107, 614, 618, 206} \[ \frac{2 c f \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}}-\frac{f \left (b+2 c (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1142

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]

Rule 1107

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

Rule 614

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{d f+e f x}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx &=\frac{f \operatorname{Subst}\left (\int \frac{x}{\left (a+b x^2+c x^4\right )^2} \, dx,x,d+e x\right )}{e}\\ &=\frac{f \operatorname{Subst}\left (\int \frac{1}{\left (a+b x+c x^2\right )^2} \, dx,x,(d+e x)^2\right )}{2 e}\\ &=-\frac{f \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{(c f) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,(d+e x)^2\right )}{\left (b^2-4 a c\right ) e}\\ &=-\frac{f \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{(2 c f) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c (d+e x)^2\right )}{\left (b^2-4 a c\right ) e}\\ &=-\frac{f \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{2 c f \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} e}\\ \end{align*}

Mathematica [A]  time = 0.128798, size = 99, normalized size = 1.01 \[ -\frac{f \left (\frac{4 c \tan ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\frac{b+2 c (d+e x)^2}{a+b (d+e x)^2+c (d+e x)^4}\right )}{2 e \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.019, size = 484, normalized size = 4.9 \begin{align*}{\frac{fce{x}^{2}}{ \left ( c{e}^{4}{x}^{4}+4\,cd{e}^{3}{x}^{3}+6\,c{d}^{2}{e}^{2}{x}^{2}+4\,c{d}^{3}ex+b{e}^{2}{x}^{2}+c{d}^{4}+2\,bdex+b{d}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+2\,{\frac{fcdx}{ \left ( c{e}^{4}{x}^{4}+4\,cd{e}^{3}{x}^{3}+6\,c{d}^{2}{e}^{2}{x}^{2}+4\,c{d}^{3}ex+b{e}^{2}{x}^{2}+c{d}^{4}+2\,bdex+b{d}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{fc{d}^{2}}{ \left ( c{e}^{4}{x}^{4}+4\,cd{e}^{3}{x}^{3}+6\,c{d}^{2}{e}^{2}{x}^{2}+4\,c{d}^{3}ex+b{e}^{2}{x}^{2}+c{d}^{4}+2\,bdex+b{d}^{2}+a \right ) e \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{fb}{ \left ( 2\,c{e}^{4}{x}^{4}+8\,cd{e}^{3}{x}^{3}+12\,c{d}^{2}{e}^{2}{x}^{2}+8\,c{d}^{3}ex+2\,b{e}^{2}{x}^{2}+2\,c{d}^{4}+4\,bdex+2\,b{d}^{2}+2\,a \right ) e \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{fc}{e \left ( 4\,ac-{b}^{2} \right ) }\sum _{{\it \_R}={\it RootOf} \left ( c{e}^{4}{{\it \_Z}}^{4}+4\,cd{e}^{3}{{\it \_Z}}^{3}+ \left ( 6\,c{d}^{2}{e}^{2}+b{e}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,c{d}^{3}e+2\,bde \right ){\it \_Z}+c{d}^{4}+b{d}^{2}+a \right ) }{\frac{ \left ({\it \_R}\,e+d \right ) \ln \left ( x-{\it \_R} \right ) }{2\,c{e}^{3}{{\it \_R}}^{3}+6\,cd{e}^{2}{{\it \_R}}^{2}+6\,c{d}^{2}e{\it \_R}+2\,c{d}^{3}+be{\it \_R}+bd}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f/(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)*c*e/(4*a*c-b^2)*x^2+
2*f/(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)*c*d/(4*a*c-b^2)*x+
f/(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)/e/(4*a*c-b^2)*c*d^2+
1/2*f/(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)/e/(4*a*c-b^2)*b+
f*c/(4*a*c-b^2)/e*sum((_R*e+d)/(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=RootO
f(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, c f \int -\frac{e x + d}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{4} x^{4} + 4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e^{3} x^{3} +{\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} +{\left (b^{3} - 4 \, a b c + 6 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{2} x^{2} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} d^{2} + 2 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} +{\left (b^{3} - 4 \, a b c\right )} d\right )} e x}\,{d x} - \frac{2 \, c e^{2} f x^{2} + 4 \, c d e f x +{\left (2 \, c d^{2} + b\right )} f}{2 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{5} x^{4} + 4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e^{4} x^{3} +{\left (b^{3} - 4 \, a b c + 6 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{3} x^{2} + 2 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} +{\left (b^{3} - 4 \, a b c\right )} d\right )} e^{2} x +{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.75624, size = 2268, normalized size = 23.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(b^2*c - 4*a*c^2)*e^2*f*x^2 + 4*(b^2*c - 4*a*c^2)*d*e*f*x + 2*(c^2*e^4*f*x^4 + 4*c^2*d*e^3*f*x^3 + (6
*c^2*d^2 + b*c)*e^2*f*x^2 + 2*(2*c^2*d^3 + b*c*d)*e*f*x + (c^2*d^4 + b*c*d^2 + a*c)*f)*sqrt(b^2 - 4*a*c)*log((
2*c^2*e^4*x^4 + 8*c^2*d*e^3*x^3 + 2*c^2*d^4 + 2*(6*c^2*d^2 + b*c)*e^2*x^2 + 2*b*c*d^2 + 4*(2*c^2*d^3 + b*c*d)*
e*x + b^2 - 2*a*c - (2*c*e^2*x^2 + 4*c*d*e*x + 2*c*d^2 + b)*sqrt(b^2 - 4*a*c))/(c*e^4*x^4 + 4*c*d*e^3*x^3 + c*
d^4 + (6*c*d^2 + b)*e^2*x^2 + b*d^2 + 2*(2*c*d^3 + b*d)*e*x + a)) + (b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*d^2)*
f)/((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e^5*x^4 + 4*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*e^4*x^3 + (b^5 - 8*a*b
^3*c + 16*a^2*b*c^2 + 6*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^2)*e^3*x^2 + 2*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c
^3)*d^3 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d)*e^2*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2
+ 16*a^2*c^3)*d^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d^2)*e), -1/2*(2*(b^2*c - 4*a*c^2)*e^2*f*x^2 + 4*(b^2*c
- 4*a*c^2)*d*e*f*x - 4*(c^2*e^4*f*x^4 + 4*c^2*d*e^3*f*x^3 + (6*c^2*d^2 + b*c)*e^2*f*x^2 + 2*(2*c^2*d^3 + b*c*d
)*e*f*x + (c^2*d^4 + b*c*d^2 + a*c)*f)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*e^2*x^2 + 4*c*d*e*x + 2*c*d^2 + b)*sqrt
(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + (b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*d^2)*f)/((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^
3)*e^5*x^4 + 4*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*e^4*x^3 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2 + 6*(b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3)*d^2)*e^3*x^2 + 2*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^3 + (b^5 - 8*a*b^3*c + 16*a^2*
b*c^2)*d)*e^2*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^4 + (b^5 - 8*a*b^3*
c + 16*a^2*b*c^2)*d^2)*e)]

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Sympy [B]  time = 26.7481, size = 525, normalized size = 5.36 \begin{align*} - \frac{c f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 16 a^{2} c^{3} f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{2} c^{2} f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - b^{4} c f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b c f + 2 c^{2} d^{2} f}{2 c^{2} e^{2} f} \right )}}{e} + \frac{c f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{16 a^{2} c^{3} f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{2} c^{2} f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{4} c f \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b c f + 2 c^{2} d^{2} f}{2 c^{2} e^{2} f} \right )}}{e} + \frac{b f + 2 c d^{2} f + 4 c d e f x + 2 c e^{2} f x^{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} \left (8 a c^{2} e^{5} - 2 b^{2} c e^{5}\right ) + x^{3} \left (32 a c^{2} d e^{4} - 8 b^{2} c d e^{4}\right ) + x^{2} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*f*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (-16*a**2*c**3*f*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**2*c*
*2*f*sqrt(-1/(4*a*c - b**2)**3) - b**4*c*f*sqrt(-1/(4*a*c - b**2)**3) + b*c*f + 2*c**2*d**2*f)/(2*c**2*e**2*f)
)/e + c*f*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (16*a**2*c**3*f*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**
2*c**2*f*sqrt(-1/(4*a*c - b**2)**3) + b**4*c*f*sqrt(-1/(4*a*c - b**2)**3) + b*c*f + 2*c**2*d**2*f)/(2*c**2*e**
2*f))/e + (b*f + 2*c*d**2*f + 4*c*d*e*f*x + 2*c*e**2*f*x**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))

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Giac [B]  time = 2.40469, size = 506, normalized size = 5.16 \begin{align*} -\frac{{\left (b^{2} c f e - 4 \, a c^{2} f e\right )} \sqrt{b^{2} - 4 \, a c} \log \left ({\left |{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{2} + 2 \,{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d x e +{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d^{2} + 2 \, a \right |}\right )}{b^{6} e^{2} - 12 \, a b^{4} c e^{2} + 48 \, a^{2} b^{2} c^{2} e^{2} - 64 \, a^{3} c^{3} e^{2}} + \frac{{\left (b^{2} c f e - 4 \, a c^{2} f e\right )} \sqrt{b^{2} - 4 \, a c} \log \left ({\left | -{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{2} - 2 \,{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d x e -{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d^{2} - 2 \, a \right |}\right )}{b^{6} e^{2} - 12 \, a b^{4} c e^{2} + 48 \, a^{2} b^{2} c^{2} e^{2} - 64 \, a^{3} c^{3} e^{2}} - \frac{2 \, c f x^{2} e^{2} + 4 \, c d f x e + 2 \, c d^{2} f + b f}{2 \,{\left (c x^{4} e^{4} + 4 \, c d x^{3} e^{3} + 6 \, c d^{2} x^{2} e^{2} + 4 \, c d^{3} x e + c d^{4} + b x^{2} e^{2} + 2 \, b d x e + b d^{2} + a\right )}{\left (b^{2} e - 4 \, a c e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*c*f*e - 4*a*c^2*f*e)*sqrt(b^2 - 4*a*c)*log(abs((b + sqrt(b^2 - 4*a*c))*x^2*e^2 + 2*(b + sqrt(b^2 - 4*a*c
))*d*x*e + (b + sqrt(b^2 - 4*a*c))*d^2 + 2*a))/(b^6*e^2 - 12*a*b^4*c*e^2 + 48*a^2*b^2*c^2*e^2 - 64*a^3*c^3*e^2
) + (b^2*c*f*e - 4*a*c^2*f*e)*sqrt(b^2 - 4*a*c)*log(abs(-(b - sqrt(b^2 - 4*a*c))*x^2*e^2 - 2*(b - sqrt(b^2 - 4
*a*c))*d*x*e - (b - sqrt(b^2 - 4*a*c))*d^2 - 2*a))/(b^6*e^2 - 12*a*b^4*c*e^2 + 48*a^2*b^2*c^2*e^2 - 64*a^3*c^3
*e^2) - 1/2*(2*c*f*x^2*e^2 + 4*c*d*f*x*e + 2*c*d^2*f + b*f)/((c*x^4*e^4 + 4*c*d*x^3*e^3 + 6*c*d^2*x^2*e^2 + 4*
c*d^3*x*e + c*d^4 + b*x^2*e^2 + 2*b*d*x*e + b*d^2 + a)*(b^2*e - 4*a*c*e))