3.644 \(\int \frac{1}{(d f+e f x)^3 (a+b (d+e x)^2+c (d+e x)^4)} \, dx\)

Optimal. Leaf size=133 \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 e f^3 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 a^2 e f^3}-\frac{b \log (d+e x)}{a^2 e f^3}-\frac{1}{2 a e f^3 (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.19531, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {1142, 1114, 709, 800, 634, 618, 206, 628} \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 e f^3 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 a^2 e f^3}-\frac{b \log (d+e x)}{a^2 e f^3}-\frac{1}{2 a e f^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

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

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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.133494, size = 157, normalized size = 1.18 \[ \frac{\frac{\left (b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \log \left (-\sqrt{b^2-4 a c}+b+2 c (d+e x)^2\right )}{\sqrt{b^2-4 a c}}+\frac{\left (b \sqrt{b^2-4 a c}+2 a c-b^2\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c (d+e x)^2\right )}{\sqrt{b^2-4 a c}}-\frac{2 a}{(d+e x)^2}-4 b \log (d+e x)}{4 a^2 e f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.007, size = 222, normalized size = 1.7 \begin{align*} -{\frac{1}{2\,ae{f}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{b\ln \left ( ex+d \right ) }{{a}^{2}e{f}^{3}}}+{\frac{1}{2\,{a}^{2}e{f}^{3}}\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}}^{3}bc{e}^{3}+3\,{{\it \_R}}^{2}bcd{e}^{2}+e \left ( 3\,c{d}^{2}b-ac+{b}^{2} \right ){\it \_R}+bc{d}^{3}-acd+{b}^{2}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(1/(e*f*x+d*f)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4),x)

[Out]

-1/2/a/e/f^3/(e*x+d)^2-b*ln(e*x+d)/a^2/e/f^3+1/2/f^3/a^2/e*sum((_R^3*b*c*e^3+3*_R^2*b*c*d*e^2+e*(3*b*c*d^2-a*c
+b^2)*_R+b*c*d^3-a*c*d+b^2*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*f*x+d*f)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.41604, size = 1814, normalized size = 13.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*f*x+d*f)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 17.9185, size = 532, normalized size = 4. \begin{align*} \left (\frac{b}{4 a^{2} e f^{3}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} e f^{3} \left (4 a c - b^{2}\right )}\right ) \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 8 a^{3} c e f^{3} \left (\frac{b}{4 a^{2} e f^{3}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} e f^{3} \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} e f^{3} \left (\frac{b}{4 a^{2} e f^{3}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} e f^{3} \left (4 a c - b^{2}\right )}\right ) + 3 a b c + 2 a c^{2} d^{2} - b^{3} - b^{2} c d^{2}}{2 a c^{2} e^{2} - b^{2} c e^{2}} \right )} + \left (\frac{b}{4 a^{2} e f^{3}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} e f^{3} \left (4 a c - b^{2}\right )}\right ) \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 8 a^{3} c e f^{3} \left (\frac{b}{4 a^{2} e f^{3}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} e f^{3} \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} e f^{3} \left (\frac{b}{4 a^{2} e f^{3}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} e f^{3} \left (4 a c - b^{2}\right )}\right ) + 3 a b c + 2 a c^{2} d^{2} - b^{3} - b^{2} c d^{2}}{2 a c^{2} e^{2} - b^{2} c e^{2}} \right )} - \frac{1}{2 a d^{2} e f^{3} + 4 a d e^{2} f^{3} x + 2 a e^{3} f^{3} x^{2}} - \frac{b \log{\left (\frac{d}{e} + x \right )}}{a^{2} e f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.46316, size = 618, normalized size = 4.65 \begin{align*} \frac{{\left (a^{2} b^{2} f^{3} e - 2 \, a^{3} c f^{3} e\right )} \sqrt{b^{2} - 4 \, a c} \log \left ({\left | 4 \, a^{3} c f^{3} e^{4} + 2 \,{\left (a^{2} b c + \sqrt{b^{2} - 4 \, a c} a^{2} c\right )} f^{3} x^{2} e^{6} + 4 \,{\left (a^{2} b c + \sqrt{b^{2} - 4 \, a c} a^{2} c\right )} d f^{3} x e^{5} + 2 \,{\left (a^{2} b c + \sqrt{b^{2} - 4 \, a c} a^{2} c\right )} d^{2} f^{3} e^{4} \right |}\right )}{4 \,{\left (a^{4} b^{2} f^{6} e^{2} - 4 \, a^{5} c f^{6} e^{2}\right )}} - \frac{{\left (a^{2} b^{2} f^{3} e - 2 \, a^{3} c f^{3} e\right )} \sqrt{b^{2} - 4 \, a c} \log \left ({\left | -4 \, a^{3} c f^{3} e^{4} - 2 \,{\left (a^{2} b c - \sqrt{b^{2} - 4 \, a c} a^{2} c\right )} f^{3} x^{2} e^{6} - 4 \,{\left (a^{2} b c - \sqrt{b^{2} - 4 \, a c} a^{2} c\right )} d f^{3} x e^{5} - 2 \,{\left (a^{2} b c - \sqrt{b^{2} - 4 \, a c} a^{2} c\right )} d^{2} f^{3} e^{4} \right |}\right )}{4 \,{\left (a^{4} b^{2} f^{6} e^{2} - 4 \, a^{5} c f^{6} e^{2}\right )}} + \frac{b e^{\left (-1\right )} \log \left ({\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 |}\right )}{4 \, a^{2} f^{3}} - \frac{b e^{\left (-1\right )} \log \left ({\left | x e + d \right |}\right )}{a^{2} f^{3}} - \frac{e^{\left (-1\right )}}{2 \,{\left (x e + d\right )}^{2} a f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*f*x+d*f)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4),x, algorithm="giac")

[Out]

1/4*(a^2*b^2*f^3*e - 2*a^3*c*f^3*e)*sqrt(b^2 - 4*a*c)*log(abs(4*a^3*c*f^3*e^4 + 2*(a^2*b*c + sqrt(b^2 - 4*a*c)
*a^2*c)*f^3*x^2*e^6 + 4*(a^2*b*c + sqrt(b^2 - 4*a*c)*a^2*c)*d*f^3*x*e^5 + 2*(a^2*b*c + sqrt(b^2 - 4*a*c)*a^2*c
)*d^2*f^3*e^4))/(a^4*b^2*f^6*e^2 - 4*a^5*c*f^6*e^2) - 1/4*(a^2*b^2*f^3*e - 2*a^3*c*f^3*e)*sqrt(b^2 - 4*a*c)*lo
g(abs(-4*a^3*c*f^3*e^4 - 2*(a^2*b*c - sqrt(b^2 - 4*a*c)*a^2*c)*f^3*x^2*e^6 - 4*(a^2*b*c - sqrt(b^2 - 4*a*c)*a^
2*c)*d*f^3*x*e^5 - 2*(a^2*b*c - sqrt(b^2 - 4*a*c)*a^2*c)*d^2*f^3*e^4))/(a^4*b^2*f^6*e^2 - 4*a^5*c*f^6*e^2) + 1
/4*b*e^(-1)*log(abs(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))/(a^2*f^3) - b*e^(-1)*log(abs(x*e + d))/(a^2*f^3) - 1/2*e^(-1)/((x*e + d)^2*a*f^3)