3.3.0 ∫ (ag + bgx)−2−m(ci + dix)m(A + B log(e(a+bx c+dx)n))3 dx [218]

Integrand size = 49, antiderivative size = 309 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=-\frac {6 B^3 n^3 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m}}{(b c-a d) i^2 (1+m)^4 (c+d x)}-\frac {6 B^2 n^2 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i^2 (1+m)^3 (c+d x)}-\frac {3 B n (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (1+m)^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)} \]

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

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

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

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

Rubi [A] (veriﬁed)

Time = 0.21 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2563, 2342, 2341} \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=-\frac {6 B^2 n^2 (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{i^2 (m+1)^3 (c+d x) (b c-a d)}-\frac {(a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{i^2 (m+1) (c+d x) (b c-a d)}-\frac {3 B n (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{i^2 (m+1)^2 (c+d x) (b c-a d)}-\frac {6 B^3 n^3 (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2}}{i^2 (m+1)^4 (c+d x) (b c-a d)} \]

Int[(a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^3,x]

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

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

- a*d)*i^2*(1 + m)^3*(c + d*x)) - (3*B*n*(a + b*x)*(g*(a + b*x))^(-2 - m)*(i*(c + d*x))^(2 + m)*(A + B*Log[e*(

(a + b*x)/(c + d*x))^n])^2)/((b*c - a*d)*i^2*(1 + m)^2*(c + d*x)) - ((a + b*x)*(g*(a + b*x))^(-2 - m)*(i*(c +

d*x))^(2 + m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^3)/((b*c - a*d)*i^2*(1 + m)*(c + d*x))

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^

n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo

g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{

a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m

_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[d^2*((g*((a + b*x)/b))^m/(i^2*(b*c - a*d)*(i*((c + d*x)/d))^

m*((a + b*x)/(c + d*x))^m)), Subst[Int[x^m*(A + B*Log[e*x^n])^p, x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a,

b, c, d, e, f, g, h, i, A, B, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] &

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((g (a+b x))^{-2-m} \left (\frac {a+b x}{c+d x}\right )^{2+m} (i (c+d x))^{2+m}\right ) \text {Subst}\left (\int x^{-2-m} \left (A+B \log \left (e x^n\right )\right )^3 \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) i^2} \\ & = -\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)}+\frac {\left (3 B n (g (a+b x))^{-2-m} \left (\frac {a+b x}{c+d x}\right )^{2+m} (i (c+d x))^{2+m}\right ) \text {Subst}\left (\int x^{-2-m} \left (A+B \log \left (e x^n\right )\right )^2 \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) i^2 (1+m)} \\ & = -\frac {3 B n (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (1+m)^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)}+\frac {\left (6 B^2 n^2 (g (a+b x))^{-2-m} \left (\frac {a+b x}{c+d x}\right )^{2+m} (i (c+d x))^{2+m}\right ) \text {Subst}\left (\int x^{-2-m} \left (A+B \log \left (e x^n\right )\right ) \, dx,x,\frac {a+b x}{c+d x}\right )}{(b c-a d) i^2 (1+m)^2} \\ & = -\frac {6 B^3 n^3 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m}}{(b c-a d) i^2 (1+m)^4 (c+d x)}-\frac {6 B^2 n^2 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i^2 (1+m)^3 (c+d x)}-\frac {3 B n (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (1+m)^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 4.82 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.67 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=-\frac {(g (a+b x))^{-1-m} (c+d x) (i (c+d x))^m \left (A^3 (1+m)^3+3 A^2 B (1+m)^2 n+6 A B^2 (1+m) n^2+6 B^3 n^3+3 B (1+m) \left (A^2 (1+m)^2+2 A B (1+m) n+2 B^2 n^2\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+3 B^2 (1+m)^2 (A+A m+B n) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B^3 (1+m)^3 \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) g (1+m)^4} \]

Integrate[(a*g + b*g*x)^(-2 - m)*(c*i + d*i*x)^m*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^3,x]

-(((g*(a + b*x))^(-1 - m)*(c + d*x)*(i*(c + d*x))^m*(A^3*(1 + m)^3 + 3*A^2*B*(1 + m)^2*n + 6*A*B^2*(1 + m)*n^2

+ 6*B^3*n^3 + 3*B*(1 + m)*(A^2*(1 + m)^2 + 2*A*B*(1 + m)*n + 2*B^2*n^2)*Log[e*((a + b*x)/(c + d*x))^n] + 3*B^

2*(1 + m)^2*(A + A*m + B*n)*Log[e*((a + b*x)/(c + d*x))^n]^2 + B^3*(1 + m)^3*Log[e*((a + b*x)/(c + d*x))^n]^3)

Maple [F]

\[\int \left (b g x +a g \right )^{-2-m} \left (d i x +c i \right )^{m} {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{3}d x\]

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

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2669 vs. \(2 (309) = 618\).

Time = 0.44 (sec) , antiderivative size = 2669, normalized size of antiderivative = 8.64 \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\text {Too large to display} \]

integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*log(e*((b*x+a)/(d*x+c))^n))^3,x, algorithm="fricas")

-(A^3*a*c*m^3 + 6*B^3*a*c*n^3 + 3*A^3*a*c*m^2 + 3*A^3*a*c*m + A^3*a*c + (B^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a

*c*m + B^3*a*c + (B^3*b*d*m^3 + 3*B^3*b*d*m^2 + 3*B^3*b*d*m + B^3*b*d)*x^2 + (B^3*b*c + B^3*a*d + (B^3*b*c + B

^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*(B^3*b*c + B^3*a*d)*m)*x)*log(e)^3 + ((B^3*b*d*m^3 + 3*B^3*b*d*m^2

+ 3*B^3*b*d*m + B^3*b*d)*n^3*x^2 + (B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 +

3*(B^3*b*c + B^3*a*d)*m)*n^3*x + (B^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c)*n^3)*log((b*x + a)/(d*

x + c))^3 + 6*(A*B^2*a*c*m + A*B^2*a*c)*n^2 + (A^3*b*d*m^3 + 6*B^3*b*d*n^3 + 3*A^3*b*d*m^2 + 3*A^3*b*d*m + A^3

*b*d + 6*(A*B^2*b*d*m + A*B^2*b*d)*n^2 + 3*(A^2*B*b*d*m^2 + 2*A^2*B*b*d*m + A^2*B*b*d)*n)*x^2 + 3*(A*B^2*a*c*m

^3 + 3*A*B^2*a*c*m^2 + 3*A*B^2*a*c*m + A*B^2*a*c + (A*B^2*b*d*m^3 + 3*A*B^2*b*d*m^2 + 3*A*B^2*b*d*m + A*B^2*b*

d + (B^3*b*d*m^2 + 2*B^3*b*d*m + B^3*b*d)*n)*x^2 + (B^3*a*c*m^2 + 2*B^3*a*c*m + B^3*a*c)*n + (A*B^2*b*c + A*B^

2*a*d + (A*B^2*b*c + A*B^2*a*d)*m^3 + 3*(A*B^2*b*c + A*B^2*a*d)*m^2 + 3*(A*B^2*b*c + A*B^2*a*d)*m + (B^3*b*c +

B^3*a*d + (B^3*b*c + B^3*a*d)*m^2 + 2*(B^3*b*c + B^3*a*d)*m)*n)*x + ((B^3*b*d*m^3 + 3*B^3*b*d*m^2 + 3*B^3*b*d

*m + B^3*b*d)*n*x^2 + (B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*(B^3*b*c +

B^3*a*d)*m)*n*x + (B^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c)*n)*log((b*x + a)/(d*x + c)))*log(e)^2

+ 3*((B^3*a*c*m^2 + 2*B^3*a*c*m + B^3*a*c)*n^3 + (A*B^2*a*c*m^3 + 3*A*B^2*a*c*m^2 + 3*A*B^2*a*c*m + A*B^2*a*c)

*n^2 + ((B^3*b*d*m^2 + 2*B^3*b*d*m + B^3*b*d)*n^3 + (A*B^2*b*d*m^3 + 3*A*B^2*b*d*m^2 + 3*A*B^2*b*d*m + A*B^2*b

*d)*n^2)*x^2 + ((B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^2 + 2*(B^3*b*c + B^3*a*d)*m)*n^3 + (A*B^2*b*c + A*B

^2*a*d + (A*B^2*b*c + A*B^2*a*d)*m^3 + 3*(A*B^2*b*c + A*B^2*a*d)*m^2 + 3*(A*B^2*b*c + A*B^2*a*d)*m)*n^2)*x)*lo

g((b*x + a)/(d*x + c))^2 + 3*(A^2*B*a*c*m^2 + 2*A^2*B*a*c*m + A^2*B*a*c)*n + (A^3*b*c + A^3*a*d + (A^3*b*c + A

^3*a*d)*m^3 + 6*(B^3*b*c + B^3*a*d)*n^3 + 3*(A^3*b*c + A^3*a*d)*m^2 + 6*(A*B^2*b*c + A*B^2*a*d + (A*B^2*b*c +

A*B^2*a*d)*m)*n^2 + 3*(A^3*b*c + A^3*a*d)*m + 3*(A^2*B*b*c + A^2*B*a*d + (A^2*B*b*c + A^2*B*a*d)*m^2 + 2*(A^2*

B*b*c + A^2*B*a*d)*m)*n)*x + 3*(A^2*B*a*c*m^3 + 3*A^2*B*a*c*m^2 + 3*A^2*B*a*c*m + A^2*B*a*c + 2*(B^3*a*c*m + B

^3*a*c)*n^2 + (A^2*B*b*d*m^3 + 3*A^2*B*b*d*m^2 + 3*A^2*B*b*d*m + A^2*B*b*d + 2*(B^3*b*d*m + B^3*b*d)*n^2 + 2*(

A*B^2*b*d*m^2 + 2*A*B^2*b*d*m + A*B^2*b*d)*n)*x^2 + ((B^3*b*d*m^3 + 3*B^3*b*d*m^2 + 3*B^3*b*d*m + B^3*b*d)*n^2

*x^2 + (B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*(B^3*b*c + B^3*a*d)*m)*n^2

*x + (B^3*a*c*m^3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c)*n^2)*log((b*x + a)/(d*x + c))^2 + 2*(A*B^2*a*c*m^2

+ 2*A*B^2*a*c*m + A*B^2*a*c)*n + (A^2*B*b*c + A^2*B*a*d + (A^2*B*b*c + A^2*B*a*d)*m^3 + 3*(A^2*B*b*c + A^2*B*a

*d)*m^2 + 2*(B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m)*n^2 + 3*(A^2*B*b*c + A^2*B*a*d)*m + 2*(A*B^2*b*c + A*B

^2*a*d + (A*B^2*b*c + A*B^2*a*d)*m^2 + 2*(A*B^2*b*c + A*B^2*a*d)*m)*n)*x + 2*((B^3*a*c*m^2 + 2*B^3*a*c*m + B^3

*a*c)*n^2 + ((B^3*b*d*m^2 + 2*B^3*b*d*m + B^3*b*d)*n^2 + (A*B^2*b*d*m^3 + 3*A*B^2*b*d*m^2 + 3*A*B^2*b*d*m + A*

B^2*b*d)*n)*x^2 + (A*B^2*a*c*m^3 + 3*A*B^2*a*c*m^2 + 3*A*B^2*a*c*m + A*B^2*a*c)*n + ((B^3*b*c + B^3*a*d + (B^3

*b*c + B^3*a*d)*m^2 + 2*(B^3*b*c + B^3*a*d)*m)*n^2 + (A*B^2*b*c + A*B^2*a*d + (A*B^2*b*c + A*B^2*a*d)*m^3 + 3*

(A*B^2*b*c + A*B^2*a*d)*m^2 + 3*(A*B^2*b*c + A*B^2*a*d)*m)*n)*x)*log((b*x + a)/(d*x + c)))*log(e) + 3*(2*(B^3*

a*c*m + B^3*a*c)*n^3 + 2*(A*B^2*a*c*m^2 + 2*A*B^2*a*c*m + A*B^2*a*c)*n^2 + (2*(B^3*b*d*m + B^3*b*d)*n^3 + 2*(A

*B^2*b*d*m^2 + 2*A*B^2*b*d*m + A*B^2*b*d)*n^2 + (A^2*B*b*d*m^3 + 3*A^2*B*b*d*m^2 + 3*A^2*B*b*d*m + A^2*B*b*d)*

n)*x^2 + (A^2*B*a*c*m^3 + 3*A^2*B*a*c*m^2 + 3*A^2*B*a*c*m + A^2*B*a*c)*n + (2*(B^3*b*c + B^3*a*d + (B^3*b*c +

B^3*a*d)*m)*n^3 + 2*(A*B^2*b*c + A*B^2*a*d + (A*B^2*b*c + A*B^2*a*d)*m^2 + 2*(A*B^2*b*c + A*B^2*a*d)*m)*n^2 +

(A^2*B*b*c + A^2*B*a*d + (A^2*B*b*c + A^2*B*a*d)*m^3 + 3*(A^2*B*b*c + A^2*B*a*d)*m^2 + 3*(A^2*B*b*c + A^2*B*a*

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

) + m*log(i/g))/((b*c - a*d)*m^4 + 4*(b*c - a*d)*m^3 + 6*(b*c - a*d)*m^2 + b*c - a*d + 4*(b*c - a*d)*m)

Sympy [F(-2)]

Exception generated. \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]

integrate((b*g*x+a*g)**(-2-m)*(d*i*x+c*i)**m*(A+B*ln(e*((b*x+a)/(d*x+c))**n))**3,x)

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int { {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{3} {\left (b g x + a g\right )}^{-m - 2} {\left (d i x + c i\right )}^{m} \,d x } \]

integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*log(e*((b*x+a)/(d*x+c))^n))^3,x, algorithm="maxima")

integrate((B*log(e*((b*x + a)/(d*x + c))^n) + A)^3*(b*g*x + a*g)^(-m - 2)*(d*i*x + c*i)^m, x)

Giac [F]

integrate((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*log(e*((b*x+a)/(d*x+c))^n))^3,x, algorithm="giac")

integrate((B*log(e*((b*x + a)/(d*x + c))^n) + A)^3*(b*g*x + a*g)^(-m - 2)*(d*i*x + c*i)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (a g+b g x)^{-2-m} (c i+d i x)^m \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^m\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^3}{{\left (a\,g+b\,g\,x\right )}^{m+2}} \,d x \]

int(((c*i + d*i*x)^m*(A + B*log(e*((a + b*x)/(c + d*x))^n))^3)/(a*g + b*g*x)^(m + 2),x)

int(((c*i + d*i*x)^m*(A + B*log(e*((a + b*x)/(c + d*x))^n))^3)/(a*g + b*g*x)^(m + 2), x)

\(\int (a g+b g x)^{-2-m} (c i+d i x)^m (A+B \log (e (\frac {a+b x}{c+d x})^n))^3 \, dx\) [218]

Optimal result