3.629 \(\int \log (c (d+e (f+g x)^3)^q) \, dx\)

Optimal. Leaf size=169 \[ \frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}-\frac{\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac{\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac{\sqrt{3} \sqrt [3]{d} q \tan ^{-1}\left (\frac{\sqrt [3]{d}-2 \sqrt [3]{e} (f+g x)}{\sqrt{3} \sqrt [3]{d}}\right )}{\sqrt [3]{e} g}-3 q x \]

[Out]

-3*q*x - (Sqrt[3]*d^(1/3)*q*ArcTan[(d^(1/3) - 2*e^(1/3)*(f + g*x))/(Sqrt[3]*d^(1/3))])/(e^(1/3)*g) + (d^(1/3)*
q*Log[d^(1/3) + e^(1/3)*(f + g*x)])/(e^(1/3)*g) - (d^(1/3)*q*Log[d^(2/3) - d^(1/3)*e^(1/3)*(f + g*x) + e^(2/3)
*(f + g*x)^2])/(2*e^(1/3)*g) + ((f + g*x)*Log[c*(d + e*(f + g*x)^3)^q])/g

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Rubi [A]  time = 0.205736, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {2483, 2448, 321, 200, 31, 634, 617, 204, 628} \[ \frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}-\frac{\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac{\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac{\sqrt{3} \sqrt [3]{d} q \tan ^{-1}\left (\frac{\sqrt [3]{d}-2 \sqrt [3]{e} (f+g x)}{\sqrt{3} \sqrt [3]{d}}\right )}{\sqrt [3]{e} g}-3 q x \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e*(f + g*x)^3)^q],x]

[Out]

-3*q*x - (Sqrt[3]*d^(1/3)*q*ArcTan[(d^(1/3) - 2*e^(1/3)*(f + g*x))/(Sqrt[3]*d^(1/3))])/(e^(1/3)*g) + (d^(1/3)*
q*Log[d^(1/3) + e^(1/3)*(f + g*x)])/(e^(1/3)*g) - (d^(1/3)*q*Log[d^(2/3) - d^(1/3)*e^(1/3)*(f + g*x) + e^(2/3)
*(f + g*x)^2])/(2*e^(1/3)*g) + ((f + g*x)*Log[c*(d + e*(f + g*x)^3)^q])/g

Rule 2483

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*((f_.) + (g_.)*(x_))^(n_))^(p_.)]*(b_.))^(q_.), x_Symbol] :> Dist[1/g, Su
bst[Int[(a + b*Log[c*(d + e*x^n)^p])^q, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IGtQ[q
, 0] && (EqQ[q, 1] || IntegerQ[n])

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \log \left (c \left (d+e x^3\right )^q\right ) \, dx,x,f+g x\right )}{g}\\ &=\frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}-\frac{(3 e q) \operatorname{Subst}\left (\int \frac{x^3}{d+e x^3} \, dx,x,f+g x\right )}{g}\\ &=-3 q x+\frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac{(3 d q) \operatorname{Subst}\left (\int \frac{1}{d+e x^3} \, dx,x,f+g x\right )}{g}\\ &=-3 q x+\frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac{\left (\sqrt [3]{d} q\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{d}+\sqrt [3]{e} x} \, dx,x,f+g x\right )}{g}+\frac{\left (\sqrt [3]{d} q\right ) \operatorname{Subst}\left (\int \frac{2 \sqrt [3]{d}-\sqrt [3]{e} x}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx,x,f+g x\right )}{g}\\ &=-3 q x+\frac{\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}+\frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac{\left (3 d^{2/3} q\right ) \operatorname{Subst}\left (\int \frac{1}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx,x,f+g x\right )}{2 g}-\frac{\left (\sqrt [3]{d} q\right ) \operatorname{Subst}\left (\int \frac{-\sqrt [3]{d} \sqrt [3]{e}+2 e^{2/3} x}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx,x,f+g x\right )}{2 \sqrt [3]{e} g}\\ &=-3 q x+\frac{\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac{\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac{\left (3 \sqrt [3]{d} q\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{e} (f+g x)}{\sqrt [3]{d}}\right )}{\sqrt [3]{e} g}\\ &=-3 q x-\frac{\sqrt{3} \sqrt [3]{d} q \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{e} (f+g x)}{\sqrt [3]{d}}}{\sqrt{3}}\right )}{\sqrt [3]{e} g}+\frac{\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac{\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}\\ \end{align*}

Mathematica [A]  time = 0.0991185, size = 147, normalized size = 0.87 \[ \frac{(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac{\sqrt [3]{d} q \left (-\log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )+2 \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{e} (f+g x)-\sqrt [3]{d}}{\sqrt{3} \sqrt [3]{d}}\right )\right )}{2 \sqrt [3]{e} g}-3 q x \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*(f + g*x)^3)^q],x]

[Out]

-3*q*x + (d^(1/3)*q*(2*Sqrt[3]*ArcTan[(-d^(1/3) + 2*e^(1/3)*(f + g*x))/(Sqrt[3]*d^(1/3))] + 2*Log[d^(1/3) + e^
(1/3)*(f + g*x)] - Log[d^(2/3) - d^(1/3)*e^(1/3)*(f + g*x) + e^(2/3)*(f + g*x)^2]))/(2*e^(1/3)*g) + ((f + g*x)
*Log[c*(d + e*(f + g*x)^3)^q])/g

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Maple [C]  time = 0.569, size = 145, normalized size = 0.9 \begin{align*} \ln \left ( c \left ( e{g}^{3}{x}^{3}+3\,ef{g}^{2}{x}^{2}+3\,e{f}^{2}gx+e{f}^{3}+d \right ) ^{q} \right ) x-3\,qx-{\frac{q}{eg}\sum _{{\it \_R}={\it RootOf} \left ( e{g}^{3}{{\it \_Z}}^{3}+3\,ef{g}^{2}{{\it \_Z}}^{2}+3\,e{f}^{2}g{\it \_Z}+e{f}^{3}+d \right ) }{\frac{ \left ( -{{\it \_R}}^{2}ef{g}^{2}-2\,{\it \_R}\,e{f}^{2}g-e{f}^{3}-d \right ) \ln \left ( x-{\it \_R} \right ) }{{g}^{2}{{\it \_R}}^{2}+2\,fg{\it \_R}+{f}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(d+e*(g*x+f)^3)^q),x)

[Out]

ln(c*(e*g^3*x^3+3*e*f*g^2*x^2+3*e*f^2*g*x+e*f^3+d)^q)*x-3*q*x-1/g/e*q*sum((-_R^2*e*f*g^2-2*_R*e*f^2*g-e*f^3-d)
/(_R^2*g^2+2*_R*f*g+f^2)*ln(x-_R),_R=RootOf(_Z^3*e*g^3+3*_Z^2*e*f*g^2+3*_Z*e*f^2*g+e*f^3+d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(d+e*(g*x+f)^3)^q),x, algorithm="maxima")

[Out]

-(3*q - log(c))*x + 3*q*integrate((e*f*g^2*x^2 + 2*e*f^2*g*x + e*f^3 + d)/(e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2
*g*x + e*f^3 + d), x) + x*log((e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2*g*x + e*f^3 + d)^q)

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Fricas [C]  time = 9.78525, size = 3071, normalized size = 18.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(d+e*(g*x+f)^3)^q),x, algorithm="fricas")

[Out]

1/4*(4*g*q*x*log(e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2*g*x + e*f^3 + d) - 12*g*q*x - 4*sqrt(3)*g*sqrt((((-1/2*f^
3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 4*((
-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q
+ 4*f^2*q^2)/g^2)*arctan(-1/24*(2*sqrt(3)*sqrt(4*g^2*q^2*x^2 + 12*f*g*q^2*x + ((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/
(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 12*f^2*q^2 + 2*(g^2*q*x +
3*f*g*q)*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f
*q/g))*(((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*
q/g)*e*g^2 + 2*e*f*g*q)*sqrt((((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*
(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 4*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))
^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2) - sqrt(3)*(8*e*f*g^2*q^2*x + ((-1/2*f^3*q^3/g^3 + 1/
2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*e*g^3 + 12*e*f^2*g*q^2 +
4*(e*g^3*q*x + 2*e*f*g^2*q)*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(
I*sqrt(3) + 1) - 2*f*q/g))*sqrt((((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/
3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 4*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^
3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2))/(d*q^3)) - 2*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g
^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g*log(q*x - 1/2*(-1/2*f^3*q^3/g^3 + 1/
2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) + f*q/g) + 4*g*x*log(c) + (((-1/2*f^3
*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g + 6*f*q)*lo
g(4*g^2*q^2*x^2 + 12*f*g*q^2*x + ((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/
3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 12*f^2*q^2 + 2*(g^2*q*x + 3*f*g*q)*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3
) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)))/g

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Sympy [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(ln(c*(d+e*(g*x+f)**3)**q),x)

[Out]

Timed out

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Giac [B]  time = 1.42366, size = 402, normalized size = 2.38 \begin{align*} q x \log \left (g^{3} x^{3} e + 3 \, f g^{2} x^{2} e + 3 \, f^{2} g x e + f^{3} e + d\right ) + \sqrt{3} \left (\frac{d q^{3}}{g^{3}}\right )^{\frac{1}{3}} \arctan \left (-\frac{g^{4} x e^{2} + f g^{3} e^{2} + d^{\frac{1}{3}} g^{3} e^{\frac{5}{3}}}{\sqrt{3} g^{4} x e^{2} + \sqrt{3} f g^{3} e^{2} - \sqrt{3} d^{\frac{1}{3}} g^{3} e^{\frac{5}{3}}}\right ) e^{\left (-\frac{1}{3}\right )} - \frac{1}{2} \, \left (\frac{d q^{3}}{g^{3}}\right )^{\frac{1}{3}} e^{\left (-\frac{1}{3}\right )} \log \left (9 \,{\left (\sqrt{3} g^{4} x e^{2} + \sqrt{3} f g^{3} e^{2} - \sqrt{3} d^{\frac{1}{3}} g^{3} e^{\frac{5}{3}}\right )}^{2} + 9 \,{\left (g^{4} x e^{2} + f g^{3} e^{2} + d^{\frac{1}{3}} g^{3} e^{\frac{5}{3}}\right )}^{2}\right ) + \left (\frac{d q^{3}}{g^{3}}\right )^{\frac{1}{3}} e^{\left (-\frac{1}{3}\right )} \log \left ({\left | 3 \, g^{4} x e^{2} + 3 \, f g^{3} e^{2} + 3 \, d^{\frac{1}{3}} g^{3} e^{\frac{5}{3}} \right |}\right ) - 3 \, q x + x \log \left (c\right ) + \frac{f q \log \left ({\left | g^{3} x^{3} e + 3 \, f g^{2} x^{2} e + 3 \, f^{2} g x e + f^{3} e + d \right |}\right )}{g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(d+e*(g*x+f)^3)^q),x, algorithm="giac")

[Out]

q*x*log(g^3*x^3*e + 3*f*g^2*x^2*e + 3*f^2*g*x*e + f^3*e + d) + sqrt(3)*(d*q^3/g^3)^(1/3)*arctan(-(g^4*x*e^2 +
f*g^3*e^2 + d^(1/3)*g^3*e^(5/3))/(sqrt(3)*g^4*x*e^2 + sqrt(3)*f*g^3*e^2 - sqrt(3)*d^(1/3)*g^3*e^(5/3)))*e^(-1/
3) - 1/2*(d*q^3/g^3)^(1/3)*e^(-1/3)*log(9*(sqrt(3)*g^4*x*e^2 + sqrt(3)*f*g^3*e^2 - sqrt(3)*d^(1/3)*g^3*e^(5/3)
)^2 + 9*(g^4*x*e^2 + f*g^3*e^2 + d^(1/3)*g^3*e^(5/3))^2) + (d*q^3/g^3)^(1/3)*e^(-1/3)*log(abs(3*g^4*x*e^2 + 3*
f*g^3*e^2 + 3*d^(1/3)*g^3*e^(5/3))) - 3*q*x + x*log(c) + f*q*log(abs(g^3*x^3*e + 3*f*g^2*x^2*e + 3*f^2*g*x*e +
f^3*e + d))/g