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

Optimal. Leaf size=103 \[ \frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}} \]

[Out]

(f^3*(2*a + b*(d + e*x)^2))/(2*(b^2 - 4*a*c)*e*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) - (b*f^3*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 = 0.137605, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {1142, 1114, 638, 618, 206} \[ \frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^3*(2*a + b*(d + e*x)^2))/(2*(b^2 - 4*a*c)*e*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) - (b*f^3*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 1114

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

Rule 638

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

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)^3}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx &=\frac{f^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b x^2+c x^4\right )^2} \, dx,x,d+e x\right )}{e}\\ &=\frac{f^3 \operatorname{Subst}\left (\int \frac{x}{\left (a+b x+c x^2\right )^2} \, dx,x,(d+e x)^2\right )}{2 e}\\ &=\frac{f^3 \left (2 a+b (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{\left (b f^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,(d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e}\\ &=\frac{f^3 \left (2 a+b (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{\left (b f^3\right ) \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^3 \left (2 a+b (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{b f^3 \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 = 0.134716, size = 103, normalized size = 1. \[ \frac{f^3 \left (\frac{2 a+b (d+e x)^2}{\left (b^2-4 a c\right ) \left (a+(d+e x)^2 \left (b+c (d+e x)^2\right )\right )}-\frac{2 b \tan ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.017, size = 500, normalized size = 4.9 \begin{align*} -{\frac{{f}^{3}be{x}^{2}}{ \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 ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{f}^{3}bdx}{ \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{{f}^{3}b{d}^{2}}{ \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{{f}^{3}a}{ \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{{f}^{3}b}{ \left ( 8\,ac-2\,{b}^{2} \right ) e}\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)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -b f^{3} \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{b e^{2} f^{3} x^{2} + 2 \, b d e f^{3} x +{\left (b d^{2} + 2 \, a\right )} f^{3}}{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)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4)^2,x, algorithm="maxima")

[Out]

-b*f^3*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*(b*e^2*f^3*x^2 + 2*b*d*e*f^3*x + (b*d^2 + 2*a)*f^3)/((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 = 1.67859, size = 2290, normalized size = 22.23 \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)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4)^2,x, algorithm="fricas")

[Out]

[1/2*((b^3 - 4*a*b*c)*e^2*f^3*x^2 + 2*(b^3 - 4*a*b*c)*d*e*f^3*x + (2*a*b^2 - 8*a^2*c + (b^3 - 4*a*b*c)*d^2)*f^
3 - (b*c*e^4*f^3*x^4 + 4*b*c*d*e^3*f^3*x^3 + (6*b*c*d^2 + b^2)*e^2*f^3*x^2 + 2*(2*b*c*d^3 + b^2*d)*e*f^3*x + (
b*c*d^4 + b^2*d^2 + a*b)*f^3)*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^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 + 1
6*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*((b^3 - 4*a*b*c)*e^2*f^3*x^2 + 2*(b^
3 - 4*a*b*c)*d*e*f^3*x + (2*a*b^2 - 8*a^2*c + (b^3 - 4*a*b*c)*d^2)*f^3 - 2*(b*c*e^4*f^3*x^4 + 4*b*c*d*e^3*f^3*
x^3 + (6*b*c*d^2 + b^2)*e^2*f^3*x^2 + 2*(2*b*c*d^3 + b^2*d)*e*f^3*x + (b*c*d^4 + b^2*d^2 + a*b)*f^3)*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^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 = 39.0261, size = 554, normalized size = 5.38 \begin{align*} \frac{b f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 16 a^{2} b c^{2} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2} f^{3} + 2 b c d^{2} f^{3}}{2 b c e^{2} f^{3}} \right )}}{2 e} - \frac{b f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{16 a^{2} b c^{2} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2} f^{3} + 2 b c d^{2} f^{3}}{2 b c e^{2} f^{3}} \right )}}{2 e} - \frac{2 a f^{3} + b d^{2} f^{3} + 2 b d e f^{3} x + b e^{2} f^{3} 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)**3/(a+b*(e*x+d)**2+c*(e*x+d)**4)**2,x)

[Out]

b*f**3*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (-16*a**2*b*c**2*f**3*sqrt(-1/(4*a*c - b**2)**3) + 8*a*
b**3*c*f**3*sqrt(-1/(4*a*c - b**2)**3) - b**5*f**3*sqrt(-1/(4*a*c - b**2)**3) + b**2*f**3 + 2*b*c*d**2*f**3)/(
2*b*c*e**2*f**3))/(2*e) - b*f**3*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (16*a**2*b*c**2*f**3*sqrt(-1/
(4*a*c - b**2)**3) - 8*a*b**3*c*f**3*sqrt(-1/(4*a*c - b**2)**3) + b**5*f**3*sqrt(-1/(4*a*c - b**2)**3) + b**2*
f**3 + 2*b*c*d**2*f**3)/(2*b*c*e**2*f**3))/(2*e) - (2*a*f**3 + b*d**2*f**3 + 2*b*d*e*f**3*x + b*e**2*f**3*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 = 2.37564, size = 522, normalized size = 5.07 \begin{align*} \frac{{\left (b^{3} f^{3} e - 4 \, a b c f^{3} 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 )}{2 \,{\left (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}\right )}} - \frac{{\left (b^{3} f^{3} e - 4 \, a b c f^{3} 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 )}{2 \,{\left (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}\right )}} + \frac{b f^{3} x^{2} e^{2} + 2 \, b d f^{3} x e + b d^{2} f^{3} + 2 \, a f^{3}}{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)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4)^2,x, algorithm="giac")

[Out]

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