3.2.86 \(\int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx\) [186]

3.2.86.1 Optimal result
3.2.86.2 Mathematica [A] (verified)
3.2.86.3 Rubi [A] (verified)
3.2.86.4 Maple [A] (verified)
3.2.86.5 Fricas [A] (verification not implemented)
3.2.86.6 Sympy [A] (verification not implemented)
3.2.86.7 Maxima [B] (verification not implemented)
3.2.86.8 Giac [A] (verification not implemented)
3.2.86.9 Mupad [B] (verification not implemented)

3.2.86.1 Optimal result

Integrand size = 30, antiderivative size = 113 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=-\frac {2 a b i x}{d f}+\frac {2 b^2 i x}{d f}-\frac {2 b^2 i (e+f x) \log (c (e+f x))}{d f^2}+\frac {i (e+f x) (a+b \log (c (e+f x)))^2}{d f^2}+\frac {(f h-e i) (a+b \log (c (e+f x)))^3}{3 b d f^2} \]

output
-2*a*b*i*x/d/f+2*b^2*i*x/d/f-2*b^2*i*(f*x+e)*ln(c*(f*x+e))/d/f^2+i*(f*x+e) 
*(a+b*ln(c*(f*x+e)))^2/d/f^2+1/3*(-e*i+f*h)*(a+b*ln(c*(f*x+e)))^3/b/d/f^2
 
3.2.86.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {-6 (a-b) b f i x-6 b^2 i (e+f x) \log (c (e+f x))+3 i (e+f x) (a+b \log (c (e+f x)))^2+\frac {(f h-e i) (a+b \log (c (e+f x)))^3}{b}}{3 d f^2} \]

input
Integrate[((h + i*x)*(a + b*Log[c*(e + f*x)])^2)/(d*e + d*f*x),x]
 
output
(-6*(a - b)*b*f*i*x - 6*b^2*i*(e + f*x)*Log[c*(e + f*x)] + 3*i*(e + f*x)*( 
a + b*Log[c*(e + f*x)])^2 + ((f*h - e*i)*(a + b*Log[c*(e + f*x)])^3)/b)/(3 
*d*f^2)
 
3.2.86.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2858, 27, 2788, 2733, 2009, 2739, 15}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx\)

\(\Big \downarrow \) 2858

\(\displaystyle \frac {\int \frac {\left (f \left (h-\frac {e i}{f}\right )+i (e+f x)\right ) (a+b \log (c (e+f x)))^2}{d f (e+f x)}d(e+f x)}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(f h-e i+i (e+f x)) (a+b \log (c (e+f x)))^2}{e+f x}d(e+f x)}{d f^2}\)

\(\Big \downarrow \) 2788

\(\displaystyle \frac {(f h-e i) \int \frac {(a+b \log (c (e+f x)))^2}{e+f x}d(e+f x)+i \int (a+b \log (c (e+f x)))^2d(e+f x)}{d f^2}\)

\(\Big \downarrow \) 2733

\(\displaystyle \frac {(f h-e i) \int \frac {(a+b \log (c (e+f x)))^2}{e+f x}d(e+f x)+i \left ((e+f x) (a+b \log (c (e+f x)))^2-2 b \int (a+b \log (c (e+f x)))d(e+f x)\right )}{d f^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(f h-e i) \int \frac {(a+b \log (c (e+f x)))^2}{e+f x}d(e+f x)+i \left ((e+f x) (a+b \log (c (e+f x)))^2-2 b (a (e+f x)+b (e+f x) \log (c (e+f x))-b (e+f x))\right )}{d f^2}\)

\(\Big \downarrow \) 2739

\(\displaystyle \frac {\frac {(f h-e i) \int (a+b \log (c (e+f x)))^2d(a+b \log (c (e+f x)))}{b}+i \left ((e+f x) (a+b \log (c (e+f x)))^2-2 b (a (e+f x)+b (e+f x) \log (c (e+f x))-b (e+f x))\right )}{d f^2}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {\frac {(f h-e i) (a+b \log (c (e+f x)))^3}{3 b}+i \left ((e+f x) (a+b \log (c (e+f x)))^2-2 b (a (e+f x)+b (e+f x) \log (c (e+f x))-b (e+f x))\right )}{d f^2}\)

input
Int[((h + i*x)*(a + b*Log[c*(e + f*x)])^2)/(d*e + d*f*x),x]
 
output
(((f*h - e*i)*(a + b*Log[c*(e + f*x)])^3)/(3*b) + i*((e + f*x)*(a + b*Log[ 
c*(e + f*x)])^2 - 2*b*(a*(e + f*x) - b*(e + f*x) + b*(e + f*x)*Log[c*(e + 
f*x)])))/(d*f^2)
 

3.2.86.3.1 Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2733
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b 
*Log[c*x^n])^p, x] - Simp[b*n*p   Int[(a + b*Log[c*x^n])^(p - 1), x], x] /; 
 FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]
 

rule 2739
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1/( 
b*n)   Subst[Int[x^p, x], x, a + b*Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p} 
, x]
 

rule 2788
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.)) 
/(x_), x_Symbol] :> Simp[d   Int[(d + e*x)^(q - 1)*((a + b*Log[c*x^n])^p/x) 
, x], x] + Simp[e   Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /; F 
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]
 

rule 2858
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ 
.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e   Subst[In 
t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + 
e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - 
d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
 
3.2.86.4 Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56

method result size
norman \(\frac {i \left (a^{2}-2 a b +2 b^{2}\right ) x}{d f}+\frac {b^{2} i x \ln \left (c \left (f x +e \right )\right )^{2}}{d f}-\frac {\left (a^{2} e i -a^{2} f h -2 a b e i +2 b^{2} e i \right ) \ln \left (c \left (f x +e \right )\right )}{d \,f^{2}}-\frac {b \left (a e i -a f h -b e i \right ) \ln \left (c \left (f x +e \right )\right )^{2}}{d \,f^{2}}-\frac {b^{2} \left (e i -f h \right ) \ln \left (c \left (f x +e \right )\right )^{3}}{3 d \,f^{2}}+\frac {2 b i \left (a -b \right ) x \ln \left (c \left (f x +e \right )\right )}{d f}\) \(176\)
parts \(\frac {a^{2} \left (\frac {x i}{f}+\frac {\left (-e i +f h \right ) \ln \left (f x +e \right )}{f^{2}}\right )}{d}+\frac {b^{2} \left (-\frac {c e i \ln \left (c f x +c e \right )^{3}}{3 f}+\frac {c h \ln \left (c f x +c e \right )^{3}}{3}+\frac {i \left (\left (c f x +c e \right ) \ln \left (c f x +c e \right )^{2}-2 \left (c f x +c e \right ) \ln \left (c f x +c e \right )+2 c f x +2 c e \right )}{f}\right )}{d c f}+\frac {2 a b \left (-\frac {c e i \ln \left (c f x +c e \right )^{2}}{2 f}+\frac {c h \ln \left (c f x +c e \right )^{2}}{2}+\frac {i \left (\left (c f x +c e \right ) \ln \left (c f x +c e \right )-c f x -c e \right )}{f}\right )}{d c f}\) \(217\)
risch \(-\frac {b^{2} \ln \left (c \left (f x +e \right )\right )^{3} e i}{3 d \,f^{2}}+\frac {b^{2} \ln \left (c \left (f x +e \right )\right )^{3} h}{3 d f}-\frac {b \left (-b f i x +a e i -a f h -b e i \right ) \ln \left (c \left (f x +e \right )\right )^{2}}{d \,f^{2}}+\frac {2 b i \left (a -b \right ) x \ln \left (c \left (f x +e \right )\right )}{d f}-\frac {\ln \left (f x +e \right ) a^{2} e i}{d \,f^{2}}+\frac {\ln \left (f x +e \right ) a^{2} h}{d f}+\frac {2 \ln \left (f x +e \right ) a b e i}{d \,f^{2}}-\frac {2 \ln \left (f x +e \right ) b^{2} e i}{d \,f^{2}}+\frac {a^{2} i x}{d f}-\frac {2 a b i x}{d f}+\frac {2 b^{2} i x}{d f}\) \(221\)
parallelrisch \(\frac {3 x \ln \left (c \left (f x +e \right )\right )^{2} b^{2} f i -\ln \left (c \left (f x +e \right )\right )^{3} b^{2} e i +\ln \left (c \left (f x +e \right )\right )^{3} b^{2} f h +6 x \ln \left (c \left (f x +e \right )\right ) a b f i -6 x \ln \left (c \left (f x +e \right )\right ) b^{2} f i -3 \ln \left (c \left (f x +e \right )\right )^{2} a b e i +3 \ln \left (c \left (f x +e \right )\right )^{2} a b f h +3 \ln \left (c \left (f x +e \right )\right )^{2} b^{2} e i +3 a^{2} f i x -6 a b f i x +6 b^{2} i x f -3 \ln \left (c \left (f x +e \right )\right ) a^{2} e i +3 \ln \left (c \left (f x +e \right )\right ) a^{2} f h +6 \ln \left (c \left (f x +e \right )\right ) a b e i -6 \ln \left (c \left (f x +e \right )\right ) b^{2} e i -6 a^{2} e i +12 a b e i -12 b^{2} e i}{3 d \,f^{2}}\) \(243\)
derivativedivides \(\frac {-\frac {a^{2} c e i \ln \left (c f x +c e \right )}{f d}+\frac {a^{2} h c \ln \left (c f x +c e \right )}{d}+\frac {a^{2} i \left (c f x +c e \right )}{f d}-\frac {a b c e i \ln \left (c f x +c e \right )^{2}}{f d}+\frac {a b h c \ln \left (c f x +c e \right )^{2}}{d}+\frac {2 a b i \left (\left (c f x +c e \right ) \ln \left (c f x +c e \right )-c f x -c e \right )}{f d}-\frac {b^{2} c e i \ln \left (c f x +c e \right )^{3}}{3 f d}+\frac {b^{2} h c \ln \left (c f x +c e \right )^{3}}{3 d}+\frac {b^{2} i \left (\left (c f x +c e \right ) \ln \left (c f x +c e \right )^{2}-2 \left (c f x +c e \right ) \ln \left (c f x +c e \right )+2 c f x +2 c e \right )}{f d}}{c f}\) \(257\)
default \(\frac {-\frac {a^{2} c e i \ln \left (c f x +c e \right )}{f d}+\frac {a^{2} h c \ln \left (c f x +c e \right )}{d}+\frac {a^{2} i \left (c f x +c e \right )}{f d}-\frac {a b c e i \ln \left (c f x +c e \right )^{2}}{f d}+\frac {a b h c \ln \left (c f x +c e \right )^{2}}{d}+\frac {2 a b i \left (\left (c f x +c e \right ) \ln \left (c f x +c e \right )-c f x -c e \right )}{f d}-\frac {b^{2} c e i \ln \left (c f x +c e \right )^{3}}{3 f d}+\frac {b^{2} h c \ln \left (c f x +c e \right )^{3}}{3 d}+\frac {b^{2} i \left (\left (c f x +c e \right ) \ln \left (c f x +c e \right )^{2}-2 \left (c f x +c e \right ) \ln \left (c f x +c e \right )+2 c f x +2 c e \right )}{f d}}{c f}\) \(257\)

input
int((i*x+h)*(a+b*ln(c*(f*x+e)))^2/(d*f*x+d*e),x,method=_RETURNVERBOSE)
 
output
i*(a^2-2*a*b+2*b^2)/d/f*x+b^2*i*x/d/f*ln(c*(f*x+e))^2-(a^2*e*i-a^2*f*h-2*a 
*b*e*i+2*b^2*e*i)/d/f^2*ln(c*(f*x+e))-b*(a*e*i-a*f*h-b*e*i)/d/f^2*ln(c*(f* 
x+e))^2-1/3*b^2*(e*i-f*h)/d/f^2*ln(c*(f*x+e))^3+2*b*i*(a-b)/d/f*x*ln(c*(f* 
x+e))
 
3.2.86.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.25 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {3 \, {\left (a^{2} - 2 \, a b + 2 \, b^{2}\right )} f i x + {\left (b^{2} f h - b^{2} e i\right )} \log \left (c f x + c e\right )^{3} + 3 \, {\left (b^{2} f i x + a b f h - {\left (a b - b^{2}\right )} e i\right )} \log \left (c f x + c e\right )^{2} + 3 \, {\left (a^{2} f h + 2 \, {\left (a b - b^{2}\right )} f i x - {\left (a^{2} - 2 \, a b + 2 \, b^{2}\right )} e i\right )} \log \left (c f x + c e\right )}{3 \, d f^{2}} \]

input
integrate((i*x+h)*(a+b*log(c*(f*x+e)))^2/(d*f*x+d*e),x, algorithm="fricas" 
)
 
output
1/3*(3*(a^2 - 2*a*b + 2*b^2)*f*i*x + (b^2*f*h - b^2*e*i)*log(c*f*x + c*e)^ 
3 + 3*(b^2*f*i*x + a*b*f*h - (a*b - b^2)*e*i)*log(c*f*x + c*e)^2 + 3*(a^2* 
f*h + 2*(a*b - b^2)*f*i*x - (a^2 - 2*a*b + 2*b^2)*e*i)*log(c*f*x + c*e))/( 
d*f^2)
 
3.2.86.6 Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.55 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=x \left (\frac {a^{2} i}{d f} - \frac {2 a b i}{d f} + \frac {2 b^{2} i}{d f}\right ) + \frac {\left (2 a b i x - 2 b^{2} i x\right ) \log {\left (c \left (e + f x\right ) \right )}}{d f} + \frac {\left (- b^{2} e i + b^{2} f h\right ) \log {\left (c \left (e + f x\right ) \right )}^{3}}{3 d f^{2}} - \frac {\left (a^{2} e i - a^{2} f h - 2 a b e i + 2 b^{2} e i\right ) \log {\left (e + f x \right )}}{d f^{2}} + \frac {\left (- a b e i + a b f h + b^{2} e i + b^{2} f i x\right ) \log {\left (c \left (e + f x\right ) \right )}^{2}}{d f^{2}} \]

input
integrate((i*x+h)*(a+b*ln(c*(f*x+e)))**2/(d*f*x+d*e),x)
 
output
x*(a**2*i/(d*f) - 2*a*b*i/(d*f) + 2*b**2*i/(d*f)) + (2*a*b*i*x - 2*b**2*i* 
x)*log(c*(e + f*x))/(d*f) + (-b**2*e*i + b**2*f*h)*log(c*(e + f*x))**3/(3* 
d*f**2) - (a**2*e*i - a**2*f*h - 2*a*b*e*i + 2*b**2*e*i)*log(e + f*x)/(d*f 
**2) + (-a*b*e*i + a*b*f*h + b**2*e*i + b**2*f*i*x)*log(c*(e + f*x))**2/(d 
*f**2)
 
3.2.86.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (111) = 222\).

Time = 0.23 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.69 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=2 \, a b i {\left (\frac {x}{d f} - \frac {e \log \left (f x + e\right )}{d f^{2}}\right )} \log \left (c f x + c e\right ) - a b h {\left (\frac {2 \, \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} - \frac {\log \left (f x + e\right )^{2} + 2 \, \log \left (f x + e\right ) \log \left (c\right )}{d f}\right )} + a^{2} i {\left (\frac {x}{d f} - \frac {e \log \left (f x + e\right )}{d f^{2}}\right )} + \frac {b^{2} h \log \left (c f x + c e\right )^{3}}{3 \, d f} + \frac {2 \, a b h \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} + \frac {a^{2} h \log \left (d f x + d e\right )}{d f} + \frac {{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} a b i}{d f^{2}} - \frac {{\left (c^{2} e \log \left (c f x + c e\right )^{3} - 3 \, {\left (c f x + c e\right )} {\left (c \log \left (c f x + c e\right )^{2} - 2 \, c \log \left (c f x + c e\right ) + 2 \, c\right )}\right )} b^{2} i}{3 \, c^{2} d f^{2}} \]

input
integrate((i*x+h)*(a+b*log(c*(f*x+e)))^2/(d*f*x+d*e),x, algorithm="maxima" 
)
 
output
2*a*b*i*(x/(d*f) - e*log(f*x + e)/(d*f^2))*log(c*f*x + c*e) - a*b*h*(2*log 
(c*f*x + c*e)*log(d*f*x + d*e)/(d*f) - (log(f*x + e)^2 + 2*log(f*x + e)*lo 
g(c))/(d*f)) + a^2*i*(x/(d*f) - e*log(f*x + e)/(d*f^2)) + 1/3*b^2*h*log(c* 
f*x + c*e)^3/(d*f) + 2*a*b*h*log(c*f*x + c*e)*log(d*f*x + d*e)/(d*f) + a^2 
*h*log(d*f*x + d*e)/(d*f) + (e*log(f*x + e)^2 - 2*f*x + 2*e*log(f*x + e))* 
a*b*i/(d*f^2) - 1/3*(c^2*e*log(c*f*x + c*e)^3 - 3*(c*f*x + c*e)*(c*log(c*f 
*x + c*e)^2 - 2*c*log(c*f*x + c*e) + 2*c))*b^2*i/(c^2*d*f^2)
 
3.2.86.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.58 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx={\left (\frac {b^{2} i x}{d f} + \frac {a b f h - a b e i + b^{2} e i}{d f^{2}}\right )} \log \left (c f x + c e\right )^{2} + \frac {2 \, {\left (a b i - b^{2} i\right )} x \log \left (c f x + c e\right )}{d f} + \frac {{\left (b^{2} f h - b^{2} e i\right )} \log \left (c f x + c e\right )^{3}}{3 \, d f^{2}} + \frac {{\left (a^{2} i - 2 \, a b i + 2 \, b^{2} i\right )} x}{d f} + \frac {{\left (a^{2} f h - a^{2} e i + 2 \, a b e i - 2 \, b^{2} e i\right )} \log \left (f x + e\right )}{d f^{2}} \]

input
integrate((i*x+h)*(a+b*log(c*(f*x+e)))^2/(d*f*x+d*e),x, algorithm="giac")
 
output
(b^2*i*x/(d*f) + (a*b*f*h - a*b*e*i + b^2*e*i)/(d*f^2))*log(c*f*x + c*e)^2 
 + 2*(a*b*i - b^2*i)*x*log(c*f*x + c*e)/(d*f) + 1/3*(b^2*f*h - b^2*e*i)*lo 
g(c*f*x + c*e)^3/(d*f^2) + (a^2*i - 2*a*b*i + 2*b^2*i)*x/(d*f) + (a^2*f*h 
- a^2*e*i + 2*a*b*e*i - 2*b^2*e*i)*log(f*x + e)/(d*f^2)
 
3.2.86.9 Mupad [B] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.44 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^2}{d e+d f x} \, dx={\ln \left (c\,\left (e+f\,x\right )\right )}^2\,\left (\frac {b\,\left (a\,f\,h-a\,e\,i+b\,e\,i\right )}{d\,f^2}+\frac {b^2\,i\,x}{d\,f}\right )-\frac {\ln \left (e+f\,x\right )\,\left (a^2\,e\,i-a^2\,f\,h+2\,b^2\,e\,i-2\,a\,b\,e\,i\right )}{d\,f^2}+\frac {i\,x\,\left (a^2-2\,a\,b+2\,b^2\right )}{d\,f}-\frac {b^2\,{\ln \left (c\,\left (e+f\,x\right )\right )}^3\,\left (e\,i-f\,h\right )}{3\,d\,f^2}+\frac {2\,b\,i\,x\,\ln \left (c\,\left (e+f\,x\right )\right )\,\left (a-b\right )}{d\,f} \]

input
int(((h + i*x)*(a + b*log(c*(e + f*x)))^2)/(d*e + d*f*x),x)
 
output
log(c*(e + f*x))^2*((b*(a*f*h - a*e*i + b*e*i))/(d*f^2) + (b^2*i*x)/(d*f)) 
 - (log(e + f*x)*(a^2*e*i - a^2*f*h + 2*b^2*e*i - 2*a*b*e*i))/(d*f^2) + (i 
*x*(a^2 - 2*a*b + 2*b^2))/(d*f) - (b^2*log(c*(e + f*x))^3*(e*i - f*h))/(3* 
d*f^2) + (2*b*i*x*log(c*(e + f*x))*(a - b))/(d*f)