\(\int \frac {(a g+b g x) (A+B \log (e (\frac {a+b x}{c+d x})^n))^2}{(c i+d i x)^2} \, dx\) [196]

Optimal result

Integrand size = 43, antiderivative size = 282 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx=\frac {2 A B g n (a+b x)}{d i^2 (c+d x)}-\frac {2 B^2 g n^2 (a+b x)}{d i^2 (c+d x)}+\frac {2 B^2 g n (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d i^2 (c+d x)}-\frac {g (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{d i^2 (c+d x)}-\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \log \left (\frac {b c-a d}{b (c+d x)}\right )}{d^2 i^2}-\frac {2 b B g n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i^2}+\frac {2 b B^2 g n^2 \operatorname {PolyLog}\left (3,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i^2} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1570\) vs. \(2(282)=564\).

Time = 1.84 (sec) , antiderivative size = 1570, normalized size of antiderivative = 5.57 \[ \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(c i+d i x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2961, 2795, 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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2795

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2795

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(g*( - int((log(((a + b*x)**n*e)/(c + d*x)**n)**2*x)/(c**2 + 2*c*d*x + d** 
2*x**2),x)*a*b**3*c**2*d**3 - int((log(((a + b*x)**n*e)/(c + d*x)**n)**2*x 
)/(c**2 + 2*c*d*x + d**2*x**2),x)*a*b**3*c*d**4*x + int((log(((a + b*x)**n 
*e)/(c + d*x)**n)**2*x)/(c**2 + 2*c*d*x + d**2*x**2),x)*b**4*c**3*d**2 + i 
nt((log(((a + b*x)**n*e)/(c + d*x)**n)**2*x)/(c**2 + 2*c*d*x + d**2*x**2), 
x)*b**4*c**2*d**3*x - 2*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x)/(c**2 + 
 2*c*d*x + d**2*x**2),x)*a**2*b**2*c**2*d**3 - 2*int((log(((a + b*x)**n*e) 
/(c + d*x)**n)*x)/(c**2 + 2*c*d*x + d**2*x**2),x)*a**2*b**2*c*d**4*x + 2*i 
nt((log(((a + b*x)**n*e)/(c + d*x)**n)*x)/(c**2 + 2*c*d*x + d**2*x**2),x)* 
a*b**3*c**3*d**2 + 2*int((log(((a + b*x)**n*e)/(c + d*x)**n)*x)/(c**2 + 2* 
c*d*x + d**2*x**2),x)*a*b**3*c**2*d**3*x + 2*log(a + b*x)*a**3*b*c*d**2*n 
+ 2*log(a + b*x)*a**3*b*d**3*n*x - 2*log(a + b*x)*a**2*b**2*c*d**2*n**2 - 
2*log(a + b*x)*a**2*b**2*d**3*n**2*x - log(c + d*x)*a**3*b*c**2*d - 2*log( 
c + d*x)*a**3*b*c*d**2*n - log(c + d*x)*a**3*b*c*d**2*x - 2*log(c + d*x)*a 
**3*b*d**3*n*x + log(c + d*x)*a**2*b**2*c**3 + log(c + d*x)*a**2*b**2*c**2 
*d*x + 2*log(c + d*x)*a**2*b**2*c*d**2*n**2 + 2*log(c + d*x)*a**2*b**2*d** 
3*n**2*x + log(((a + b*x)**n*e)/(c + d*x)**n)**2*a**2*b**2*c*d**2 + log((( 
a + b*x)**n*e)/(c + d*x)**n)**2*a*b**3*c*d**2*x - 2*log(((a + b*x)**n*e)/( 
c + d*x)**n)*a**3*b*d**3*x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b** 
2*c*d**2*x + 2*log(((a + b*x)**n*e)/(c + d*x)**n)*a**2*b**2*d**3*n*x - ...