3.264 \(\int (f+g x)^2 (A+B \log (\frac{e (a+b x)^2}{(c+d x)^2})) \, dx\)

Optimal. Leaf size=152 \[ \frac{(f+g x)^3 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{3 g}-\frac{2 B g x (b c-a d) (-a d g-b c g+3 b d f)}{3 b^2 d^2}-\frac{2 B (b f-a g)^3 \log (a+b x)}{3 b^3 g}-\frac{B g^2 x^2 (b c-a d)}{3 b d}+\frac{2 B (d f-c g)^3 \log (c+d x)}{3 d^3 g} \]

[Out]

(-2*B*(b*c - a*d)*g*(3*b*d*f - b*c*g - a*d*g)*x)/(3*b^2*d^2) - (B*(b*c - a*d)*g^2*x^2)/(3*b*d) - (2*B*(b*f - a
*g)^3*Log[a + b*x])/(3*b^3*g) + ((f + g*x)^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(3*g) + (2*B*(d*f - c*g
)^3*Log[c + d*x])/(3*d^3*g)

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Rubi [A]  time = 0.160903, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2525, 12, 72} \[ \frac{(f+g x)^3 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{3 g}-\frac{2 B g x (b c-a d) (-a d g-b c g+3 b d f)}{3 b^2 d^2}-\frac{2 B (b f-a g)^3 \log (a+b x)}{3 b^3 g}-\frac{B g^2 x^2 (b c-a d)}{3 b d}+\frac{2 B (d f-c g)^3 \log (c+d x)}{3 d^3 g} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]

[Out]

(-2*B*(b*c - a*d)*g*(3*b*d*f - b*c*g - a*d*g)*x)/(3*b^2*d^2) - (B*(b*c - a*d)*g^2*x^2)/(3*b*d) - (2*B*(b*f - a
*g)^3*Log[a + b*x])/(3*b^3*g) + ((f + g*x)^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(3*g) + (2*B*(d*f - c*g
)^3*Log[c + d*x])/(3*d^3*g)

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int (f+g x)^2 \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx &=\frac{(f+g x)^3 \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 g}-\frac{B \int \frac{2 (b c-a d) (f+g x)^3}{(a+b x) (c+d x)} \, dx}{3 g}\\ &=\frac{(f+g x)^3 \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 g}-\frac{(2 B (b c-a d)) \int \frac{(f+g x)^3}{(a+b x) (c+d x)} \, dx}{3 g}\\ &=\frac{(f+g x)^3 \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 g}-\frac{(2 B (b c-a d)) \int \left (\frac{g^2 (3 b d f-b c g-a d g)}{b^2 d^2}+\frac{g^3 x}{b d}+\frac{(b f-a g)^3}{b^2 (b c-a d) (a+b x)}+\frac{(d f-c g)^3}{d^2 (-b c+a d) (c+d x)}\right ) \, dx}{3 g}\\ &=-\frac{2 B (b c-a d) g (3 b d f-b c g-a d g) x}{3 b^2 d^2}-\frac{B (b c-a d) g^2 x^2}{3 b d}-\frac{2 B (b f-a g)^3 \log (a+b x)}{3 b^3 g}+\frac{(f+g x)^3 \left (A+B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )\right )}{3 g}+\frac{2 B (d f-c g)^3 \log (c+d x)}{3 d^3 g}\\ \end{align*}

Mathematica [A]  time = 0.131152, size = 142, normalized size = 0.93 \[ \frac{(f+g x)^3 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )-\frac{B \left (b^2 d^2 g^3 x^2 (b c-a d)+2 b d g^2 x (b c-a d) (-a d g-b c g+3 b d f)+2 d^3 (b f-a g)^3 \log (a+b x)-2 b^3 (d f-c g)^3 \log (c+d x)\right )}{b^3 d^3}}{3 g} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]

[Out]

((f + g*x)^3*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) - (B*(2*b*d*(b*c - a*d)*g^2*(3*b*d*f - b*c*g - a*d*g)*x
+ b^2*d^2*(b*c - a*d)*g^3*x^2 + 2*d^3*(b*f - a*g)^3*Log[a + b*x] - 2*b^3*(d*f - c*g)^3*Log[c + d*x]))/(b^3*d^3
))/(3*g)

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Maple [B]  time = 0.237, size = 1188, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x)

[Out]

-2/3/d*B*g^2*a^2/b^2*c-2/d*B*g*c*f*x-1/d^2*B*g*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*f*c^2-2/d^2*B*g*ln(1/
(d*x+c)*a*d-b*c/(d*x+c)+b)*c^2*f-2*B*g/b^2*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a^2*f-2/d^2*B*ln(1/(d*x+c))*c^2*f*g
-4*B/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a*c*f^2+2*B*g/b^2*ln(1/(d*x+c))*a^2*f+2*B*g/b*a*f*x+A*x^2*f*g+B
*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*x*f^2+1/3/d^3*A*c^3*g^2+1/d*A*c*f^2+1/d^3*B*c^3*g^2+1/3*B*g^2*ln(e*
(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*x^3-2/3*B*g^2*a^3/b^3*ln(1/(d*x+c))+2/3*B*g^2*a^3/b^3*ln(1/(d*x+c)*a*d-b*
c/(d*x+c)+b)+B*g*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*f*x^2+1/d*B*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^
2)*c*f^2+2/d*B*ln(1/(d*x+c))*c*f^2-1/3/d*B*c*g^2*x^2+2/3/d^2*B*c^2*g^2*x+1/3*B*g^2*a/b*x^2+1/3/d^3*B*g^2*ln(e*
(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*c^3+2/3/d^3*B*g^2*c^3*ln(1/(d*x+c))+4/3/d^3*B*g^2*c^3*ln(1/(d*x+c)*a*d-b*
c/(d*x+c)+b)-2*B/b*ln(1/(d*x+c))*a*f^2+2/d*B*g/b*a*f*c+1/3*A*x^3*g^2+A*x*f^2+2/d*B/b/(a*d-b*c)*ln(1/(d*x+c)*a*
d-b*c/(d*x+c)+b)*a^2*c^2*g^2-4*B/b/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a^2*c*f*g+8/d*B/(a*d-b*c)*ln(1/(d
*x+c)*a*d-b*c/(d*x+c)+b)*a*c^2*f*g-4/d^2*B/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*c^3*b*f*g+4/d*B*g/b*ln(1/
(d*x+c)*a*d-b*c/(d*x+c)+b)*a*c*f-2/3*B*g^2*a^2/b^2*x-2/d^2*B*g*c^2*f-1/d^2*A*c^2*f*g-1/3/d^2*B*g^2*a/b*c^2+2/d
^3*B/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*c^4*b*g^2+2*d*B/b/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a^2
*f^2+2/d*B/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*c^2*b*f^2-4/d^2*B/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+
b)*a*c^3*g^2-2/d^2*B*g^2*a/b*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*c^2

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Maxima [B]  time = 1.32247, size = 566, normalized size = 3.72 \begin{align*} \frac{1}{3} \, A g^{2} x^{3} + A f g x^{2} +{\left (x \log \left (\frac{b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac{2 \, a \log \left (b x + a\right )}{b} - \frac{2 \, c \log \left (d x + c\right )}{d}\right )} B f^{2} +{\left (x^{2} \log \left (\frac{b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - \frac{2 \, a^{2} \log \left (b x + a\right )}{b^{2}} + \frac{2 \, c^{2} \log \left (d x + c\right )}{d^{2}} - \frac{2 \,{\left (b c - a d\right )} x}{b d}\right )} B f g + \frac{1}{3} \,{\left (x^{3} \log \left (\frac{b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac{2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac{2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac{{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \,{\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} B g^{2} + A f^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*A*g^2*x^3 + A*f*g*x^2 + (x*log(b^2*e*x^2/(d^2*x^2 + 2*c*d*x + c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) +
a^2*e/(d^2*x^2 + 2*c*d*x + c^2)) + 2*a*log(b*x + a)/b - 2*c*log(d*x + c)/d)*B*f^2 + (x^2*log(b^2*e*x^2/(d^2*x
^2 + 2*c*d*x + c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c*d*x + c^2)) - 2*a^2*log(b*x +
a)/b^2 + 2*c^2*log(d*x + c)/d^2 - 2*(b*c - a*d)*x/(b*d))*B*f*g + 1/3*(x^3*log(b^2*e*x^2/(d^2*x^2 + 2*c*d*x +
c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c*d*x + c^2)) + 2*a^3*log(b*x + a)/b^3 - 2*c^3
*log(d*x + c)/d^3 - ((b^2*c*d - a*b*d^2)*x^2 - 2*(b^2*c^2 - a^2*d^2)*x)/(b^2*d^2))*B*g^2 + A*f^2*x

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Fricas [B]  time = 1.33308, size = 621, normalized size = 4.09 \begin{align*} \frac{A b^{3} d^{3} g^{2} x^{3} +{\left (3 \, A b^{3} d^{3} f g -{\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} g^{2}\right )} x^{2} +{\left (3 \, A b^{3} d^{3} f^{2} - 6 \,{\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} f g + 2 \,{\left (B b^{3} c^{2} d - B a^{2} b d^{3}\right )} g^{2}\right )} x + 2 \,{\left (3 \, B a b^{2} d^{3} f^{2} - 3 \, B a^{2} b d^{3} f g + B a^{3} d^{3} g^{2}\right )} \log \left (b x + a\right ) - 2 \,{\left (3 \, B b^{3} c d^{2} f^{2} - 3 \, B b^{3} c^{2} d f g + B b^{3} c^{3} g^{2}\right )} \log \left (d x + c\right ) +{\left (B b^{3} d^{3} g^{2} x^{3} + 3 \, B b^{3} d^{3} f g x^{2} + 3 \, B b^{3} d^{3} f^{2} x\right )} \log \left (\frac{b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{3 \, b^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(A*b^3*d^3*g^2*x^3 + (3*A*b^3*d^3*f*g - (B*b^3*c*d^2 - B*a*b^2*d^3)*g^2)*x^2 + (3*A*b^3*d^3*f^2 - 6*(B*b^3
*c*d^2 - B*a*b^2*d^3)*f*g + 2*(B*b^3*c^2*d - B*a^2*b*d^3)*g^2)*x + 2*(3*B*a*b^2*d^3*f^2 - 3*B*a^2*b*d^3*f*g +
B*a^3*d^3*g^2)*log(b*x + a) - 2*(3*B*b^3*c*d^2*f^2 - 3*B*b^3*c^2*d*f*g + B*b^3*c^3*g^2)*log(d*x + c) + (B*b^3*
d^3*g^2*x^3 + 3*B*b^3*d^3*f*g*x^2 + 3*B*b^3*d^3*f^2*x)*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x
+ c^2)))/(b^3*d^3)

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Sympy [B]  time = 10.3726, size = 719, normalized size = 4.73 \begin{align*} \frac{A g^{2} x^{3}}{3} + \frac{2 B a \left (a^{2} g^{2} - 3 a b f g + 3 b^{2} f^{2}\right ) \log{\left (x + \frac{2 B a^{3} c d^{2} g^{2} - 6 B a^{2} b c d^{2} f g + \frac{2 B a^{2} d^{3} \left (a^{2} g^{2} - 3 a b f g + 3 b^{2} f^{2}\right )}{b} + 2 B a b^{2} c^{3} g^{2} - 6 B a b^{2} c^{2} d f g + 12 B a b^{2} c d^{2} f^{2} - 2 B a c d^{2} \left (a^{2} g^{2} - 3 a b f g + 3 b^{2} f^{2}\right )}{2 B a^{3} d^{3} g^{2} - 6 B a^{2} b d^{3} f g + 6 B a b^{2} d^{3} f^{2} + 2 B b^{3} c^{3} g^{2} - 6 B b^{3} c^{2} d f g + 6 B b^{3} c d^{2} f^{2}} \right )}}{3 b^{3}} - \frac{2 B c \left (c^{2} g^{2} - 3 c d f g + 3 d^{2} f^{2}\right ) \log{\left (x + \frac{2 B a^{3} c d^{2} g^{2} - 6 B a^{2} b c d^{2} f g + 2 B a b^{2} c^{3} g^{2} - 6 B a b^{2} c^{2} d f g + 12 B a b^{2} c d^{2} f^{2} - 2 B a b^{2} c \left (c^{2} g^{2} - 3 c d f g + 3 d^{2} f^{2}\right ) + \frac{2 B b^{3} c^{2} \left (c^{2} g^{2} - 3 c d f g + 3 d^{2} f^{2}\right )}{d}}{2 B a^{3} d^{3} g^{2} - 6 B a^{2} b d^{3} f g + 6 B a b^{2} d^{3} f^{2} + 2 B b^{3} c^{3} g^{2} - 6 B b^{3} c^{2} d f g + 6 B b^{3} c d^{2} f^{2}} \right )}}{3 d^{3}} + \left (B f^{2} x + B f g x^{2} + \frac{B g^{2} x^{3}}{3}\right ) \log{\left (\frac{e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )} + \frac{x^{2} \left (3 A b d f g + B a d g^{2} - B b c g^{2}\right )}{3 b d} - \frac{x \left (- 3 A b^{2} d^{2} f^{2} + 2 B a^{2} d^{2} g^{2} - 6 B a b d^{2} f g - 2 B b^{2} c^{2} g^{2} + 6 B b^{2} c d f g\right )}{3 b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2*(A+B*ln(e*(b*x+a)**2/(d*x+c)**2)),x)

[Out]

A*g**2*x**3/3 + 2*B*a*(a**2*g**2 - 3*a*b*f*g + 3*b**2*f**2)*log(x + (2*B*a**3*c*d**2*g**2 - 6*B*a**2*b*c*d**2*
f*g + 2*B*a**2*d**3*(a**2*g**2 - 3*a*b*f*g + 3*b**2*f**2)/b + 2*B*a*b**2*c**3*g**2 - 6*B*a*b**2*c**2*d*f*g + 1
2*B*a*b**2*c*d**2*f**2 - 2*B*a*c*d**2*(a**2*g**2 - 3*a*b*f*g + 3*b**2*f**2))/(2*B*a**3*d**3*g**2 - 6*B*a**2*b*
d**3*f*g + 6*B*a*b**2*d**3*f**2 + 2*B*b**3*c**3*g**2 - 6*B*b**3*c**2*d*f*g + 6*B*b**3*c*d**2*f**2))/(3*b**3) -
2*B*c*(c**2*g**2 - 3*c*d*f*g + 3*d**2*f**2)*log(x + (2*B*a**3*c*d**2*g**2 - 6*B*a**2*b*c*d**2*f*g + 2*B*a*b**
2*c**3*g**2 - 6*B*a*b**2*c**2*d*f*g + 12*B*a*b**2*c*d**2*f**2 - 2*B*a*b**2*c*(c**2*g**2 - 3*c*d*f*g + 3*d**2*f
**2) + 2*B*b**3*c**2*(c**2*g**2 - 3*c*d*f*g + 3*d**2*f**2)/d)/(2*B*a**3*d**3*g**2 - 6*B*a**2*b*d**3*f*g + 6*B*
a*b**2*d**3*f**2 + 2*B*b**3*c**3*g**2 - 6*B*b**3*c**2*d*f*g + 6*B*b**3*c*d**2*f**2))/(3*d**3) + (B*f**2*x + B*
f*g*x**2 + B*g**2*x**3/3)*log(e*(a + b*x)**2/(c + d*x)**2) + x**2*(3*A*b*d*f*g + B*a*d*g**2 - B*b*c*g**2)/(3*b
*d) - x*(-3*A*b**2*d**2*f**2 + 2*B*a**2*d**2*g**2 - 6*B*a*b*d**2*f*g - 2*B*b**2*c**2*g**2 + 6*B*b**2*c*d*f*g)/
(3*b**2*d**2)

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Giac [A]  time = 14.6037, size = 377, normalized size = 2.48 \begin{align*} \frac{1}{3} \,{\left (A g^{2} + B g^{2}\right )} x^{3} + \frac{1}{3} \,{\left (B g^{2} x^{3} + 3 \, B f g x^{2} + 3 \, B f^{2} x\right )} \log \left (\frac{b^{2} x^{2} + 2 \, a b x + a^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac{{\left (3 \, A b d f g + 3 \, B b d f g - B b c g^{2} + B a d g^{2}\right )} x^{2}}{3 \, b d} + \frac{2 \,{\left (3 \, B a b^{2} f^{2} - 3 \, B a^{2} b f g + B a^{3} g^{2}\right )} \log \left (b x + a\right )}{3 \, b^{3}} - \frac{2 \,{\left (3 \, B c d^{2} f^{2} - 3 \, B c^{2} d f g + B c^{3} g^{2}\right )} \log \left (-d x - c\right )}{3 \, d^{3}} + \frac{{\left (3 \, A b^{2} d^{2} f^{2} + 3 \, B b^{2} d^{2} f^{2} - 6 \, B b^{2} c d f g + 6 \, B a b d^{2} f g + 2 \, B b^{2} c^{2} g^{2} - 2 \, B a^{2} d^{2} g^{2}\right )} x}{3 \, b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(A*g^2 + B*g^2)*x^3 + 1/3*(B*g^2*x^3 + 3*B*f*g*x^2 + 3*B*f^2*x)*log((b^2*x^2 + 2*a*b*x + a^2)/(d^2*x^2 + 2
*c*d*x + c^2)) + 1/3*(3*A*b*d*f*g + 3*B*b*d*f*g - B*b*c*g^2 + B*a*d*g^2)*x^2/(b*d) + 2/3*(3*B*a*b^2*f^2 - 3*B*
a^2*b*f*g + B*a^3*g^2)*log(b*x + a)/b^3 - 2/3*(3*B*c*d^2*f^2 - 3*B*c^2*d*f*g + B*c^3*g^2)*log(-d*x - c)/d^3 +
1/3*(3*A*b^2*d^2*f^2 + 3*B*b^2*d^2*f^2 - 6*B*b^2*c*d*f*g + 6*B*a*b*d^2*f*g + 2*B*b^2*c^2*g^2 - 2*B*a^2*d^2*g^2
)*x/(b^2*d^2)