\(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^3} \, dx\) [426]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=\frac {b f p q}{2 h (f g-e h) (g+h x)}+\frac {b f^2 p q \log (e+f x)}{2 h (f g-e h)^2}-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}-\frac {b f^2 p q \log (g+h x)}{2 h (f g-e h)^2} \]

[Out]

1/2*b*f*p*q/h/(-e*h+f*g)/(h*x+g)+1/2*b*f^2*p*q*ln(f*x+e)/h/(-e*h+f*g)^2+1/2*(-a-b*ln(c*(d*(f*x+e)^p)^q))/h/(h*
x+g)^2-1/2*b*f^2*p*q*ln(h*x+g)/h/(-e*h+f*g)^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2442, 46, 2495} \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}+\frac {b f^2 p q \log (e+f x)}{2 h (f g-e h)^2}-\frac {b f^2 p q \log (g+h x)}{2 h (f g-e h)^2}+\frac {b f p q}{2 h (g+h x) (f g-e h)} \]

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^3,x]

[Out]

(b*f*p*q)/(2*h*(f*g - e*h)*(g + h*x)) + (b*f^2*p*q*Log[e + f*x])/(2*h*(f*g - e*h)^2) - (a + b*Log[c*(d*(e + f*
x)^p)^q])/(2*h*(g + h*x)^2) - (b*f^2*p*q*Log[g + h*x])/(2*h*(f*g - e*h)^2)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {a+b \log \left (c d^q (e+f x)^{p q}\right )}{(g+h x)^3} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}+\text {Subst}\left (\frac {(b f p q) \int \frac {1}{(e+f x) (g+h x)^2} \, dx}{2 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}+\text {Subst}\left (\frac {(b f p q) \int \left (\frac {f^2}{(f g-e h)^2 (e+f x)}-\frac {h}{(f g-e h) (g+h x)^2}-\frac {f h}{(f g-e h)^2 (g+h x)}\right ) \, dx}{2 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = \frac {b f p q}{2 h (f g-e h) (g+h x)}+\frac {b f^2 p q \log (e+f x)}{2 h (f g-e h)^2}-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}-\frac {b f^2 p q \log (g+h x)}{2 h (f g-e h)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-\frac {b f p q (g+h x) (f g-e h+f (g+h x) \log (e+f x)-f (g+h x) \log (g+h x))}{(f g-e h)^2}}{2 h (g+h x)^2} \]

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^3,x]

[Out]

-1/2*(a + b*Log[c*(d*(e + f*x)^p)^q] - (b*f*p*q*(g + h*x)*(f*g - e*h + f*(g + h*x)*Log[e + f*x] - f*(g + h*x)*
Log[g + h*x]))/(f*g - e*h)^2)/(h*(g + h*x)^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs. \(2(114)=228\).

Time = 3.76 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.56

method	result	size
parallelrisch	\(\frac {2 \ln \left (f x +e \right ) x b \,f^{3} g \,h^{2} p q -a \,e^{2} f \,h^{3}-a \,f^{3} g^{2} h +\ln \left (f x +e \right ) x^{2} b \,f^{3} h^{3} p q -\ln \left (h x +g \right ) x^{2} b \,f^{3} h^{3} p q +\ln \left (f x +e \right ) b \,f^{3} g^{2} h p q -\ln \left (h x +g \right ) b \,f^{3} g^{2} h p q -x b e \,f^{2} h^{3} p q +x b \,f^{3} g \,h^{2} p q -2 \ln \left (h x +g \right ) x b \,f^{3} g \,h^{2} p q -b e \,f^{2} g \,h^{2} p q +b \,f^{3} g^{2} h p q +2 a e \,f^{2} g \,h^{2}+2 \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b e \,f^{2} g \,h^{2}-\ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b \,e^{2} f \,h^{3}-\ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b \,f^{3} g^{2} h}{2 \left (e^{2} h^{2}-2 e f g h +f^{2} g^{2}\right ) \left (h x +g \right )^{2} f \,h^{2}}\)	\(305\)

[In]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^3,x,method=_RETURNVERBOSE)

[Out]

1/2*(2*ln(f*x+e)*x*b*f^3*g*h^2*p*q-a*e^2*f*h^3-a*f^3*g^2*h+ln(f*x+e)*x^2*b*f^3*h^3*p*q-ln(h*x+g)*x^2*b*f^3*h^3
*p*q+ln(f*x+e)*b*f^3*g^2*h*p*q-ln(h*x+g)*b*f^3*g^2*h*p*q-x*b*e*f^2*h^3*p*q+x*b*f^3*g*h^2*p*q-2*ln(h*x+g)*x*b*f
^3*g*h^2*p*q-b*e*f^2*g*h^2*p*q+b*f^3*g^2*h*p*q+2*a*e*f^2*g*h^2+2*ln(c*(d*(f*x+e)^p)^q)*b*e*f^2*g*h^2-ln(c*(d*(
f*x+e)^p)^q)*b*e^2*f*h^3-ln(c*(d*(f*x+e)^p)^q)*b*f^3*g^2*h)/(e^2*h^2-2*e*f*g*h+f^2*g^2)/(h*x+g)^2/f/h^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (111) = 222\).

Time = 0.31 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.61 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=-\frac {a f^{2} g^{2} - 2 \, a e f g h + a e^{2} h^{2} - {\left (b f^{2} g h - b e f h^{2}\right )} p q x - {\left (b f^{2} g^{2} - b e f g h\right )} p q + {\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} q \log \left (d\right ) - {\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + {\left (2 \, b e f g h - b e^{2} h^{2}\right )} p q\right )} \log \left (f x + e\right ) + {\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (h x + g\right ) + {\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} \log \left (c\right )}{2 \, {\left (f^{2} g^{4} h - 2 \, e f g^{3} h^{2} + e^{2} g^{2} h^{3} + {\left (f^{2} g^{2} h^{3} - 2 \, e f g h^{4} + e^{2} h^{5}\right )} x^{2} + 2 \, {\left (f^{2} g^{3} h^{2} - 2 \, e f g^{2} h^{3} + e^{2} g h^{4}\right )} x\right )}} \]

[In]

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

[Out]

-1/2*(a*f^2*g^2 - 2*a*e*f*g*h + a*e^2*h^2 - (b*f^2*g*h - b*e*f*h^2)*p*q*x - (b*f^2*g^2 - b*e*f*g*h)*p*q + (b*f
^2*g^2 - 2*b*e*f*g*h + b*e^2*h^2)*q*log(d) - (b*f^2*h^2*p*q*x^2 + 2*b*f^2*g*h*p*q*x + (2*b*e*f*g*h - b*e^2*h^2
)*p*q)*log(f*x + e) + (b*f^2*h^2*p*q*x^2 + 2*b*f^2*g*h*p*q*x + b*f^2*g^2*p*q)*log(h*x + g) + (b*f^2*g^2 - 2*b*
e*f*g*h + b*e^2*h^2)*log(c))/(f^2*g^4*h - 2*e*f*g^3*h^2 + e^2*g^2*h^3 + (f^2*g^2*h^3 - 2*e*f*g*h^4 + e^2*h^5)*
x^2 + 2*(f^2*g^3*h^2 - 2*e*f*g^2*h^3 + e^2*g*h^4)*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1984 vs. \(2 (105) = 210\).

Time = 15.25 (sec) , antiderivative size = 1984, normalized size of antiderivative = 16.67 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))/(h*x+g)**3,x)

[Out]

Piecewise(((a*x + b*e*log(c*(d*(e + f*x)**p)**q)/f - b*p*q*x + b*x*log(c*(d*(e + f*x)**p)**q))/g**3, Eq(h, 0))
, (-2*a/(4*g**2*h + 8*g*h**2*x + 4*h**3*x**2) - b*p*q/(4*g**2*h + 8*g*h**2*x + 4*h**3*x**2) - 2*b*log(c*(d*(f*
g/h + f*x)**p)**q)/(4*g**2*h + 8*g*h**2*x + 4*h**3*x**2), Eq(e, f*g/h)), (-a*e**2*h**2/(2*e**2*g**2*h**3 + 4*e
**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*
f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) + 2*a*e*f*g*h/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2
- 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*
h**3*x**2) - a*f**2*g**2/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2
*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) - b*e**2*h**2*log(c*
(d*(e + f*x)**p)**q)/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**
3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) - b*e*f*g*h*p*q/(2*e**2*
g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f
**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) + 2*b*e*f*g*h*log(c*(d*(e + f*x)**p)**q)/(2*e**2*g**2
*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*
g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) - b*e*f*h**2*p*q*x/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x +
2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*
x + 2*f**2*g**2*h**3*x**2) - b*f**2*g**2*p*q*log(g/h + x)/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x*
*2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**
2*h**3*x**2) + b*f**2*g**2*p*q/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*
f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) - 2*b*f**2*g*h
*p*q*x*log(g/h + x)/(2*e**2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3
*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) + b*f**2*g*h*p*q*x/(2*e**
2*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2
*f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) + 2*b*f**2*g*h*x*log(c*(d*(e + f*x)**p)**q)/(2*e**2
*g**2*h**3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*
f**2*g**4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) - b*f**2*h**2*p*q*x**2*log(g/h + x)/(2*e**2*g**2*h**
3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**4
*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2) + b*f**2*h**2*x**2*log(c*(d*(e + f*x)**p)**q)/(2*e**2*g**2*h*
*3 + 4*e**2*g*h**4*x + 2*e**2*h**5*x**2 - 4*e*f*g**3*h**2 - 8*e*f*g**2*h**3*x - 4*e*f*g*h**4*x**2 + 2*f**2*g**
4*h + 4*f**2*g**3*h**2*x + 2*f**2*g**2*h**3*x**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.45 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=\frac {1}{2} \, b f p q {\left (\frac {f \log \left (f x + e\right )}{f^{2} g^{2} h - 2 \, e f g h^{2} + e^{2} h^{3}} - \frac {f \log \left (h x + g\right )}{f^{2} g^{2} h - 2 \, e f g h^{2} + e^{2} h^{3}} + \frac {1}{f g^{2} h - e g h^{2} + {\left (f g h^{2} - e h^{3}\right )} x}\right )} - \frac {b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{2 \, {\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} - \frac {a}{2 \, {\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} \]

[In]

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

[Out]

1/2*b*f*p*q*(f*log(f*x + e)/(f^2*g^2*h - 2*e*f*g*h^2 + e^2*h^3) - f*log(h*x + g)/(f^2*g^2*h - 2*e*f*g*h^2 + e^
2*h^3) + 1/(f*g^2*h - e*g*h^2 + (f*g*h^2 - e*h^3)*x)) - 1/2*b*log(((f*x + e)^p*d)^q*c)/(h^3*x^2 + 2*g*h^2*x +
g^2*h) - 1/2*a/(h^3*x^2 + 2*g*h^2*x + g^2*h)

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.86 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=\frac {b f^{2} p q \log \left (f x + e\right )}{2 \, {\left (f^{2} g^{2} h - 2 \, e f g h^{2} + e^{2} h^{3}\right )}} - \frac {b f^{2} p q \log \left (h x + g\right )}{2 \, {\left (f^{2} g^{2} h - 2 \, e f g h^{2} + e^{2} h^{3}\right )}} - \frac {b p q \log \left (f x + e\right )}{2 \, {\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} + \frac {b f h p q x + b f g p q - b f g q \log \left (d\right ) + b e h q \log \left (d\right ) - b f g \log \left (c\right ) + b e h \log \left (c\right ) - a f g + a e h}{2 \, {\left (f g h^{3} x^{2} - e h^{4} x^{2} + 2 \, f g^{2} h^{2} x - 2 \, e g h^{3} x + f g^{3} h - e g^{2} h^{2}\right )}} \]

[In]

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

[Out]

1/2*b*f^2*p*q*log(f*x + e)/(f^2*g^2*h - 2*e*f*g*h^2 + e^2*h^3) - 1/2*b*f^2*p*q*log(h*x + g)/(f^2*g^2*h - 2*e*f
*g*h^2 + e^2*h^3) - 1/2*b*p*q*log(f*x + e)/(h^3*x^2 + 2*g*h^2*x + g^2*h) + 1/2*(b*f*h*p*q*x + b*f*g*p*q - b*f*
g*q*log(d) + b*e*h*q*log(d) - b*f*g*log(c) + b*e*h*log(c) - a*f*g + a*e*h)/(f*g*h^3*x^2 - e*h^4*x^2 + 2*f*g^2*
h^2*x - 2*e*g*h^3*x + f*g^3*h - e*g^2*h^2)

Mupad [B] (verification not implemented)

Time = 3.49 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.51 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^3} \, dx=\frac {b\,f^2\,p\,q\,\mathrm {atanh}\left (\frac {2\,e^2\,h^3-2\,f^2\,g^2\,h}{2\,h\,{\left (e\,h-f\,g\right )}^2}+\frac {2\,f\,h\,x}{e\,h-f\,g}\right )}{h\,{\left (e\,h-f\,g\right )}^2}-\frac {b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{2\,h\,\left (g^2+2\,g\,h\,x+h^2\,x^2\right )}-\frac {\frac {a\,e\,h-a\,f\,g+b\,f\,g\,p\,q}{e\,h-f\,g}+\frac {b\,f\,h\,p\,q\,x}{e\,h-f\,g}}{2\,g^2\,h+4\,g\,h^2\,x+2\,h^3\,x^2} \]

[In]

int((a + b*log(c*(d*(e + f*x)^p)^q))/(g + h*x)^3,x)

[Out]

(b*f^2*p*q*atanh((2*e^2*h^3 - 2*f^2*g^2*h)/(2*h*(e*h - f*g)^2) + (2*f*h*x)/(e*h - f*g)))/(h*(e*h - f*g)^2) - (
b*log(c*(d*(e + f*x)^p)^q))/(2*h*(g^2 + h^2*x^2 + 2*g*h*x)) - ((a*e*h - a*f*g + b*f*g*p*q)/(e*h - f*g) + (b*f*
h*p*q*x)/(e*h - f*g))/(2*g^2*h + 2*h^3*x^2 + 4*g*h^2*x)