3.1.4 \(\int (a g+b g x) (A+B \log (e (\frac {a+b x}{c+d x})^n)) \, dx\) [4]

3.1.4.1 Optimal result

Integrand size = 31, antiderivative size = 86 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {B (b c-a d) g n x}{2 d}+\frac {g (a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b}+\frac {B (b c-a d)^2 g n \log (c+d x)}{2 b d^2} \]

output

-1/2*B*(-a*d+b*c)*g*n*x/d+1/2*g*(b*x+a)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/ 
b+1/2*B*(-a*d+b*c)^2*g*n*ln(d*x+c)/b/d^2
 

3.1.4.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g \left ((a+b x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+\frac {B (-b c+a d) n (b d x+(-b c+a d) \log (c+d x))}{d^2}\right )}{2 b} \]

input

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

output

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

3.1.4.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2947, 27, 49, 2009}

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.

\(\displaystyle \int (a g+b g x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right ) \, dx\)

\(\Big \downarrow \) 2947

\(\displaystyle \frac {g (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b}-\frac {B n (b c-a d) \int \frac {g^2 (a+b x)}{c+d x}dx}{2 b g}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {g (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b}-\frac {B g n (b c-a d) \int \frac {a+b x}{c+d x}dx}{2 b}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {g (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b}-\frac {B g n (b c-a d) \int \left (\frac {b}{d}+\frac {a d-b c}{d (c+d x)}\right )dx}{2 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {g (a+b x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b}-\frac {B g n (b c-a d) \left (\frac {b x}{d}-\frac {(b c-a d) \log (c+d x)}{d^2}\right )}{2 b}\)

input

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

output

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

3.1.4.3.1 Defintions of rubi rules used

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 49

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] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009

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

rule 2947

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + 
 B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(m + 1))), x] - Simp[B*n*((b*c - a*d) 
/(g*(m + 1)))   Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; Free 
Q[{a, b, c, d, e, f, g, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] 
&& NeQ[m, -2]
 

3.1.4.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(275\) vs. \(2(80)=160\).

Time = 1.07 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.21

method	result	size
parallelrisch	\(\frac {B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} d^{2} g n +A \,x^{2} b^{2} d^{2} g n +B \ln \left (b x +a \right ) a^{2} d^{2} g \,n^{2}-2 B \ln \left (b x +a \right ) a b c d g \,n^{2}+B \ln \left (b x +a \right ) b^{2} c^{2} g \,n^{2}+2 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a b \,d^{2} g n +B x a b \,d^{2} g \,n^{2}-B x \,b^{2} c d g \,n^{2}+2 A x a b \,d^{2} g n +2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) a b c d g n -B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} c^{2} g n -B \,a^{2} d^{2} g \,n^{2}+B \,b^{2} c^{2} g \,n^{2}-2 A \,a^{2} d^{2} g n -3 A a b c d g n}{2 b \,d^{2} n}\)	\(276\)

input

int((b*g*x+a*g)*(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x,method=_RETURNVERBOSE)
 

output

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

3.1.4.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.86 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {A b^{2} d^{2} g x^{2} + B a^{2} d^{2} g n \log \left (b x + a\right ) + {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} g n \log \left (d x + c\right ) + {\left (2 \, A a b d^{2} g - {\left (B b^{2} c d - B a b d^{2}\right )} g n\right )} x + {\left (B b^{2} d^{2} g x^{2} + 2 \, B a b d^{2} g x\right )} \log \left (e\right ) + {\left (B b^{2} d^{2} g n x^{2} + 2 \, B a b d^{2} g n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{2 \, b d^{2}} \]

input

integrate((b*g*x+a*g)*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="frica 
s")
 

output

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

3.1.4.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (73) = 146\).

Time = 57.62 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.09 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\begin {cases} a g x \left (A + B \log {\left (e \left (\frac {a}{c}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \wedge d = 0 \\a g \left (A x + \frac {B c \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )}}{d} + B n x + B x \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \\A a g x + \frac {A b g x^{2}}{2} + \frac {B a^{2} g \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )}}{2 b} - \frac {B a g n x}{2} + B a g x \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )} - \frac {B b g n x^{2}}{4} + \frac {B b g x^{2} \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )}}{2} & \text {for}\: d = 0 \\A a g x + \frac {A b g x^{2}}{2} + \frac {B a^{2} g n \log {\left (\frac {c}{d} + x \right )}}{2 b} + \frac {B a^{2} g \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{2 b} - \frac {B a c g n \log {\left (\frac {c}{d} + x \right )}}{d} + \frac {B a g n x}{2} + B a g x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + \frac {B b c^{2} g n \log {\left (\frac {c}{d} + x \right )}}{2 d^{2}} - \frac {B b c g n x}{2 d} + \frac {B b g x^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{2} & \text {otherwise} \end {cases} \]

input

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

output

Piecewise((a*g*x*(A + B*log(e*(a/c)**n)), Eq(b, 0) & Eq(d, 0)), (a*g*(A*x 
+ B*c*log(e*(a/(c + d*x))**n)/d + B*n*x + B*x*log(e*(a/(c + d*x))**n)), Eq 
(b, 0)), (A*a*g*x + A*b*g*x**2/2 + B*a**2*g*log(e*(a/c + b*x/c)**n)/(2*b) 
- B*a*g*n*x/2 + B*a*g*x*log(e*(a/c + b*x/c)**n) - B*b*g*n*x**2/4 + B*b*g*x 
**2*log(e*(a/c + b*x/c)**n)/2, Eq(d, 0)), (A*a*g*x + A*b*g*x**2/2 + B*a**2 
*g*n*log(c/d + x)/(2*b) + B*a**2*g*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) 
/(2*b) - B*a*c*g*n*log(c/d + x)/d + B*a*g*n*x/2 + B*a*g*x*log(e*(a/(c + d* 
x) + b*x/(c + d*x))**n) + B*b*c**2*g*n*log(c/d + x)/(2*d**2) - B*b*c*g*n*x 
/(2*d) + B*b*g*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/2, True))
 

3.1.4.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.81 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{2} \, B b g x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{2} \, A b g x^{2} - \frac {1}{2} \, B b g n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B a g n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B a g x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A a g x \]

input

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

output

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

3.1.4.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 880 vs. \(2 (80) = 160\).

Time = 0.48 (sec) , antiderivative size = 880, normalized size of antiderivative = 10.23 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {1}{2} \, {\left (\frac {{\left (B b^{4} c^{3} g n - 3 \, B a b^{3} c^{2} d g n - \frac {2 \, {\left (b x + a\right )} B b^{3} c^{3} d g n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} g n + \frac {6 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g n}{d x + c} - B a^{3} b d^{3} g n - \frac {6 \, {\left (b x + a\right )} B a^{2} b c d^{3} g n}{d x + c} + \frac {2 \, {\left (b x + a\right )} B a^{3} d^{4} g n}{d x + c}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d^{2} - \frac {2 \, {\left (b x + a\right )} b d^{3}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{4}}{{\left (d x + c\right )}^{2}}} + \frac {B b^{4} c^{3} g n - 3 \, B a b^{3} c^{2} d g n - \frac {{\left (b x + a\right )} B b^{3} c^{3} d g n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} g n + \frac {3 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g n}{d x + c} - B a^{3} b d^{3} g n - \frac {3 \, {\left (b x + a\right )} B a^{2} b c d^{3} g n}{d x + c} + \frac {{\left (b x + a\right )} B a^{3} d^{4} g n}{d x + c} + B b^{4} c^{3} g \log \left (e\right ) - 3 \, B a b^{3} c^{2} d g \log \left (e\right ) - \frac {2 \, {\left (b x + a\right )} B b^{3} c^{3} d g \log \left (e\right )}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} g \log \left (e\right ) + \frac {6 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g \log \left (e\right )}{d x + c} - B a^{3} b d^{3} g \log \left (e\right ) - \frac {6 \, {\left (b x + a\right )} B a^{2} b c d^{3} g \log \left (e\right )}{d x + c} + \frac {2 \, {\left (b x + a\right )} B a^{3} d^{4} g \log \left (e\right )}{d x + c} + A b^{4} c^{3} g - 3 \, A a b^{3} c^{2} d g - \frac {2 \, {\left (b x + a\right )} A b^{3} c^{3} d g}{d x + c} + 3 \, A a^{2} b^{2} c d^{2} g + \frac {6 \, {\left (b x + a\right )} A a b^{2} c^{2} d^{2} g}{d x + c} - A a^{3} b d^{3} g - \frac {6 \, {\left (b x + a\right )} A a^{2} b c d^{3} g}{d x + c} + \frac {2 \, {\left (b x + a\right )} A a^{3} d^{4} g}{d x + c}}{b^{2} d^{2} - \frac {2 \, {\left (b x + a\right )} b d^{3}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{4}}{{\left (d x + c\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b d^{2}} - \frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b d^{2}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

input

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

output

-1/2*((B*b^4*c^3*g*n - 3*B*a*b^3*c^2*d*g*n - 2*(b*x + a)*B*b^3*c^3*d*g*n/( 
d*x + c) + 3*B*a^2*b^2*c*d^2*g*n + 6*(b*x + a)*B*a*b^2*c^2*d^2*g*n/(d*x + 
c) - B*a^3*b*d^3*g*n - 6*(b*x + a)*B*a^2*b*c*d^3*g*n/(d*x + c) + 2*(b*x + 
a)*B*a^3*d^4*g*n/(d*x + c))*log((b*x + a)/(d*x + c))/(b^2*d^2 - 2*(b*x + a 
)*b*d^3/(d*x + c) + (b*x + a)^2*d^4/(d*x + c)^2) + (B*b^4*c^3*g*n - 3*B*a* 
b^3*c^2*d*g*n - (b*x + a)*B*b^3*c^3*d*g*n/(d*x + c) + 3*B*a^2*b^2*c*d^2*g* 
n + 3*(b*x + a)*B*a*b^2*c^2*d^2*g*n/(d*x + c) - B*a^3*b*d^3*g*n - 3*(b*x + 
 a)*B*a^2*b*c*d^3*g*n/(d*x + c) + (b*x + a)*B*a^3*d^4*g*n/(d*x + c) + B*b^ 
4*c^3*g*log(e) - 3*B*a*b^3*c^2*d*g*log(e) - 2*(b*x + a)*B*b^3*c^3*d*g*log( 
e)/(d*x + c) + 3*B*a^2*b^2*c*d^2*g*log(e) + 6*(b*x + a)*B*a*b^2*c^2*d^2*g* 
log(e)/(d*x + c) - B*a^3*b*d^3*g*log(e) - 6*(b*x + a)*B*a^2*b*c*d^3*g*log( 
e)/(d*x + c) + 2*(b*x + a)*B*a^3*d^4*g*log(e)/(d*x + c) + A*b^4*c^3*g - 3* 
A*a*b^3*c^2*d*g - 2*(b*x + a)*A*b^3*c^3*d*g/(d*x + c) + 3*A*a^2*b^2*c*d^2* 
g + 6*(b*x + a)*A*a*b^2*c^2*d^2*g/(d*x + c) - A*a^3*b*d^3*g - 6*(b*x + a)* 
A*a^2*b*c*d^3*g/(d*x + c) + 2*(b*x + a)*A*a^3*d^4*g/(d*x + c))/(b^2*d^2 - 
2*(b*x + a)*b*d^3/(d*x + c) + (b*x + a)^2*d^4/(d*x + c)^2) + (B*b^3*c^3*g* 
n - 3*B*a*b^2*c^2*d*g*n + 3*B*a^2*b*c*d^2*g*n - B*a^3*d^3*g*n)*log(-b + (b 
*x + a)*d/(d*x + c))/(b*d^2) - (B*b^3*c^3*g*n - 3*B*a*b^2*c^2*d*g*n + 3*B* 
a^2*b*c*d^2*g*n - B*a^3*d^3*g*n)*log((b*x + a)/(d*x + c))/(b*d^2))*(b*c/(b 
*c - a*d)^2 - a*d/(b*c - a*d)^2)
 

3.1.4.9 Mupad [B] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int (a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=x\,\left (\frac {g\,\left (4\,A\,a\,d+2\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{2\,d}-\frac {A\,g\,\left (2\,a\,d+2\,b\,c\right )}{2\,d}\right )+\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {B\,b\,g\,x^2}{2}+B\,a\,g\,x\right )+\frac {\ln \left (c+d\,x\right )\,\left (B\,b\,c^2\,g\,n-2\,B\,a\,c\,d\,g\,n\right )}{2\,d^2}+\frac {A\,b\,g\,x^2}{2}+\frac {B\,a^2\,g\,n\,\ln \left (a+b\,x\right )}{2\,b} \]

input

int((a*g + b*g*x)*(A + B*log(e*((a + b*x)/(c + d*x))^n)),x)
 

output

x*((g*(4*A*a*d + 2*A*b*c + B*a*d*n - B*b*c*n))/(2*d) - (A*g*(2*a*d + 2*b*c 
))/(2*d)) + log(e*((a + b*x)/(c + d*x))^n)*((B*b*g*x^2)/2 + B*a*g*x) + (lo 
g(c + d*x)*(B*b*c^2*g*n - 2*B*a*c*d*g*n))/(2*d^2) + (A*b*g*x^2)/2 + (B*a^2 
*g*n*log(a + b*x))/(2*b)