\(\int \frac {(c i+d i x)^3 (A+B \log (e (\frac {a+b x}{c+d x})^n))}{(a g+b g x)^2} \, dx\) [132]

Optimal result

Integrand size = 43, antiderivative size = 390 \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^2} \, dx=-\frac {B d^2 (b c-a d) i^3 n x}{2 b^3 g^2}-\frac {B (b c-a d)^2 i^3 n (c+d x)}{b^3 g^2 (a+b x)}+\frac {2 d^2 (b c-a d) i^3 (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^4 g^2}-\frac {(b c-a d)^2 i^3 (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^3 g^2 (a+b x)}+\frac {d i^3 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g^2}-\frac {B d (b c-a d)^2 i^3 n \log \left (\frac {a+b x}{c+d x}\right )}{2 b^4 g^2}-\frac {5 B d (b c-a d)^2 i^3 n \log (c+d x)}{2 b^4 g^2}-\frac {3 d (b c-a d)^2 i^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b^4 g^2}+\frac {3 B d (b c-a d)^2 i^3 n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^4 g^2} \] Output:

-1/2*B*d^2*(-a*d+b*c)*i^3*n*x/b^3/g^2-B*(-a*d+b*c)^2*i^3*n*(d*x+c)/b^3/g^2 
/(b*x+a)+2*d^2*(-a*d+b*c)*i^3*(b*x+a)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^4/ 
g^2-(-a*d+b*c)^2*i^3*(d*x+c)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^3/g^2/(b*x+ 
a)+1/2*d*i^3*(d*x+c)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^2/g^2-1/2*B*d*(-a 
*d+b*c)^2*i^3*n*ln((b*x+a)/(d*x+c))/b^4/g^2-5/2*B*d*(-a*d+b*c)^2*i^3*n*ln( 
d*x+c)/b^4/g^2-3*d*(-a*d+b*c)^2*i^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))*ln(1-b 
*(d*x+c)/d/(b*x+a))/b^4/g^2+3*B*d*(-a*d+b*c)^2*i^3*n*polylog(2,b*(d*x+c)/d 
/(b*x+a))/b^4/g^2
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.01 \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^2} \, dx=\frac {i^3 \left (2 A b d^2 (3 b c-2 a d) x-b B d^2 (b c-a d) n x-\frac {2 B (b c-a d)^3 n}{a+b x}-a^2 B d^3 n \log (a+b x)-2 B d (b c-a d)^2 n \log (a+b x)+2 B d^2 (3 b c-2 a d) (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+b^2 d^3 x^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-\frac {2 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{a+b x}+6 d (b c-a d)^2 \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+b^2 B c^2 d n \log (c+d x)+2 B d (b c-a d)^2 n \log (c+d x)-2 B d (-b c+a d) (-3 b c+2 a d) n \log (c+d x)-3 B d (b c-a d)^2 n \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{2 b^4 g^2} \] Input:

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

Output:

(i^3*(2*A*b*d^2*(3*b*c - 2*a*d)*x - b*B*d^2*(b*c - a*d)*n*x - (2*B*(b*c - 
a*d)^3*n)/(a + b*x) - a^2*B*d^3*n*Log[a + b*x] - 2*B*d*(b*c - a*d)^2*n*Log 
[a + b*x] + 2*B*d^2*(3*b*c - 2*a*d)*(a + b*x)*Log[e*((a + b*x)/(c + d*x))^ 
n] + b^2*d^3*x^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) - (2*(b*c - a*d)^3 
*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(a + b*x) + 6*d*(b*c - a*d)^2*Log 
[a + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + b^2*B*c^2*d*n*Log[c + d 
*x] + 2*B*d*(b*c - a*d)^2*n*Log[c + d*x] - 2*B*d*(-(b*c) + a*d)*(-3*b*c + 
2*a*d)*n*Log[c + d*x] - 3*B*d*(b*c - a*d)^2*n*(Log[a + b*x]*(Log[a + b*x] 
- 2*Log[(b*(c + d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + 
 a*d)])))/(2*b^4*g^2)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 343, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2961, 2793, 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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2793

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i^3 (b c-a d)^2 \left (\frac {2 d^2 (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4 (c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )}-\frac {3 d \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4}-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^3 (a+b x)}+\frac {d \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}+\frac {3 B d n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^4}-\frac {B d n \log \left (\frac {a+b x}{c+d x}\right )}{2 b^4}+\frac {5 B d n \log \left (b-\frac {d (a+b x)}{c+d x}\right )}{2 b^4}-\frac {B d n}{2 b^3 \left (b-\frac {d (a+b x)}{c+d x}\right )}-\frac {B n (c+d x)}{b^3 (a+b x)}\right )}{g^2}\)

Input:

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

Output:

((b*c - a*d)^2*i^3*(-((B*n*(c + d*x))/(b^3*(a + b*x))) - (B*d*n)/(2*b^3*(b 
 - (d*(a + b*x))/(c + d*x))) - ((c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x 
))^n]))/(b^3*(a + b*x)) + (d*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*b^ 
2*(b - (d*(a + b*x))/(c + d*x))^2) + (2*d^2*(a + b*x)*(A + B*Log[e*((a + b 
*x)/(c + d*x))^n]))/(b^4*(c + d*x)*(b - (d*(a + b*x))/(c + d*x))) - (B*d*n 
*Log[(a + b*x)/(c + d*x)])/(2*b^4) + (5*B*d*n*Log[b - (d*(a + b*x))/(c + d 
*x)])/(2*b^4) - (3*d*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[1 - (b*(c 
+ d*x))/(d*(a + b*x))])/b^4 + (3*B*d*n*PolyLog[2, (b*(c + d*x))/(d*(a + b* 
x))])/b^4))/g^2
 

Defintions of rubi rules used

rule 2009

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

rule 2793

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* 
(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], 
 (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, 
 f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer 
Q[r]))
 

rule 2961

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + B*L 
og[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1785 vs. \(2 (381) = 762\).

Time = 0.38 (sec) , antiderivative size = 1785, normalized size of antiderivative = 4.58 \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-B*c^3*i^3*n*(1/(b^2*g^2*x + a*b*g^2) + d*log(b*x + a)/((b^2*c - a*b*d)*g^ 
2) - d*log(d*x + c)/((b^2*c - a*b*d)*g^2)) - 3*A*(a^2/(b^4*g^2*x + a*b^3*g 
^2) - x/(b^2*g^2) + 2*a*log(b*x + a)/(b^3*g^2))*c*d^2*i^3 + 1/2*(2*a^3/(b^ 
5*g^2*x + a*b^4*g^2) + 6*a^2*log(b*x + a)/(b^4*g^2) + (b*x^2 - 4*a*x)/(b^3 
*g^2))*A*d^3*i^3 + 3*A*c^2*d*i^3*(a/(b^3*g^2*x + a*b^2*g^2) + log(b*x + a) 
/(b^2*g^2)) - B*c^3*i^3*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(b^2*g^2*x 
+ a*b*g^2) - A*c^3*i^3/(b^2*g^2*x + a*b*g^2) - 1/2*(5*b^3*c^3*d*i^3*n - 3* 
a*b^2*c^2*d^2*i^3*n - 2*a^2*b*c*d^3*i^3*n + 2*a^3*d^4*i^3*n)*B*log(d*x + c 
)/(b^5*c*g^2 - a*b^4*d*g^2) + 1/2*((b^4*c*d^3*i^3*log(e) - a*b^3*d^4*i^3*l 
og(e))*B*x^3 - ((i^3*n - 6*i^3*log(e))*b^4*c^2*d^2 - (2*i^3*n - 9*i^3*log( 
e))*a*b^3*c*d^3 + (i^3*n - 3*i^3*log(e))*a^2*b^2*d^4)*B*x^2 - ((i^3*n - 6* 
i^3*log(e))*a*b^3*c^2*d^2 - 2*(i^3*n - 5*i^3*log(e))*a^2*b^2*c*d^3 + (i^3* 
n - 4*i^3*log(e))*a^3*b*d^4)*B*x - 3*((b^4*c^3*d*i^3*n - 3*a*b^3*c^2*d^2*i 
^3*n + 3*a^2*b^2*c*d^3*i^3*n - a^3*b*d^4*i^3*n)*B*x + (a*b^3*c^3*d*i^3*n - 
 3*a^2*b^2*c^2*d^2*i^3*n + 3*a^3*b*c*d^3*i^3*n - a^4*d^4*i^3*n)*B)*log(b*x 
 + a)^2 + 2*(3*(i^3*n + i^3*log(e))*a*b^3*c^3*d - 6*(i^3*n + i^3*log(e))*a 
^2*b^2*c^2*d^2 + 4*(i^3*n + i^3*log(e))*a^3*b*c*d^3 - (i^3*n + i^3*log(e)) 
*a^4*d^4)*B + ((6*b^4*c^3*d*i^3*log(e) + 6*(2*i^3*n - 3*i^3*log(e))*a*b^3* 
c^2*d^2 - (17*i^3*n - 18*i^3*log(e))*a^2*b^2*c*d^3 + (7*i^3*n - 6*i^3*log( 
e))*a^3*b*d^4)*B*x + (6*a*b^3*c^3*d*i^3*log(e) + 6*(2*i^3*n - 3*i^3*log...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(i*( - 2*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x**3)/(a**2 + 2*a*b*x + b 
**2*x**2),x)*a**3*b**5*d**4 + 2*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x* 
*3)/(a**2 + 2*a*b*x + b**2*x**2),x)*a**2*b**6*c*d**3 - 2*int((log(((a + b* 
x)**n*e)/(c + d*x)**n)*x**3)/(a**2 + 2*a*b*x + b**2*x**2),x)*a**2*b**6*d** 
4*x + 2*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x**3)/(a**2 + 2*a*b*x + b* 
*2*x**2),x)*a*b**7*c*d**3*x - 6*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x* 
*2)/(a**2 + 2*a*b*x + b**2*x**2),x)*a**3*b**5*c*d**3 + 6*int((log(((a + b* 
x)**n*e)/(c + d*x)**n)*x**2)/(a**2 + 2*a*b*x + b**2*x**2),x)*a**2*b**6*c** 
2*d**2 - 6*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x**2)/(a**2 + 2*a*b*x + 
 b**2*x**2),x)*a**2*b**6*c*d**3*x + 6*int((log(((a + b*x)**n*e)/(c + d*x)* 
*n)*x**2)/(a**2 + 2*a*b*x + b**2*x**2),x)*a*b**7*c**2*d**2*x - 6*int((log( 
((a + b*x)**n*e)/(c + d*x)**n)*x)/(a**2 + 2*a*b*x + b**2*x**2),x)*a**3*b** 
5*c**2*d**2 + 6*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x)/(a**2 + 2*a*b*x 
 + b**2*x**2),x)*a**2*b**6*c**3*d - 6*int((log(((a + b*x)**n*e)/(c + d*x)* 
*n)*x)/(a**2 + 2*a*b*x + b**2*x**2),x)*a**2*b**6*c**2*d**2*x + 6*int((log( 
((a + b*x)**n*e)/(c + d*x)**n)*x)/(a**2 + 2*a*b*x + b**2*x**2),x)*a*b**7*c 
**3*d*x - 6*log(a + b*x)*a**6*d**4 + 18*log(a + b*x)*a**5*b*c*d**3 - 6*log 
(a + b*x)*a**5*b*d**4*x - 18*log(a + b*x)*a**4*b**2*c**2*d**2 + 18*log(a + 
 b*x)*a**4*b**2*c*d**3*x + 6*log(a + b*x)*a**3*b**3*c**3*d - 18*log(a + b* 
x)*a**3*b**3*c**2*d**2*x + 6*log(a + b*x)*a**2*b**4*c**3*d*x - 2*log(a ...