\(\int \frac {(a g+b g x) (A+B \log (\frac {e (a+b x)}{c+d x}))^2}{c i+d i x} \, dx\) [86]

Optimal result

Integrand size = 40, antiderivative size = 283 \[ \int \frac {(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{c i+d i x} \, dx=\frac {2 B (b c-a d) g \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{d^2 i}+\frac {g (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d i}+\frac {(b c-a d) g \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d^2 i}+\frac {2 B^2 (b c-a d) g \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i}+\frac {2 B (b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i}-\frac {2 B^2 (b c-a d) g \operatorname {PolyLog}\left (3,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1227\) vs. \(2(283)=566\).

Time = 1.17 (sec) , antiderivative size = 1227, normalized size of antiderivative = 4.34 \[ \int \frac {(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{c i+d i x} \, dx =\text {Too large to display} \] Input:

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

Output:

(g*(3*A^2*b*d*x - 3*A^2*(b*c - a*d)*Log[c + d*x] - 3*a*A*B*d*(Log[c/d + x] 
^2 + 2*(Log[a/b + x] - Log[c/d + x] - Log[(e*(a + b*x))/(c + d*x)])*Log[c 
+ d*x] - 2*(Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] + PolyLog[2, (d*(a 
 + b*x))/(-(b*c) + a*d)])) - 3*A*B*(-2*d*(a + b*x)*(-1 + Log[a/b + x]) + 2 
*b*(c + d*x)*(-1 + Log[c/d + x]) - b*c*Log[c/d + x]^2 + 2*b*(Log[a/b + x] 
- Log[c/d + x] - Log[(e*(a + b*x))/(c + d*x)])*(d*x - c*Log[c + d*x]) + 2* 
b*c*(Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] + PolyLog[2, (d*(a + b*x) 
)/(-(b*c) + a*d)])) + a*B^2*d*(Log[c/d + x]^3 + 3*Log[c/d + x]^2*(-Log[a/b 
 + x] + Log[(d*(a + b*x))/(-(b*c) + a*d)]) + 3*(-Log[a/b + x] + Log[c/d + 
x] + Log[(e*(a + b*x))/(c + d*x)])^2*Log[c + d*x] + 3*Log[a/b + x]^2*Log[( 
b*(c + d*x))/(b*c - a*d)] + 6*Log[a/b + x]*PolyLog[2, (d*(a + b*x))/(-(b*c 
) + a*d)] + 3*(Log[a/b + x] - Log[c/d + x] - Log[(e*(a + b*x))/(c + d*x)]) 
*(Log[c/d + x]^2 - 2*(Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] + PolyLo 
g[2, (d*(a + b*x))/(-(b*c) + a*d)])) + 6*Log[c/d + x]*PolyLog[2, (b*(c + d 
*x))/(b*c - a*d)] - 6*PolyLog[3, (d*(a + b*x))/(-(b*c) + a*d)] - 6*PolyLog 
[3, (b*(c + d*x))/(b*c - a*d)]) + B^2*(3*d*(2*b*x - 2*(a + b*x)*Log[a/b + 
x] + (a + b*x)*Log[a/b + x]^2) - b*c*Log[c/d + x]^3 + 3*b*(2*d*x - 2*(c + 
d*x)*Log[c/d + x] + (c + d*x)*Log[c/d + x]^2) + 3*b*(-Log[a/b + x] + Log[c 
/d + x] + Log[(e*(a + b*x))/(c + d*x)])^2*(d*x - c*Log[c + d*x]) - 6*(a*d 
+ 2*b*d*x - b*d*x*Log[c/d + x] - b*c*Log[c + d*x] + Log[a/b + x]*(-(d*(...
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2962, 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.

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

\(\Big \downarrow \) 2962

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

\(\Big \downarrow \) 2795

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 2962

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ 
)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Sy 
mbol] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + 
 B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; 
 FreeQ[{a, b, c, d, e, f, g, h, i, A, B, n, p}, x] && EqQ[n + mn, 0] && IGt 
Q[n, 0] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && I 
ntegersQ[m, q]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

g*(Integral(A**2*a/(c + d*x), x) + Integral(A**2*b*x/(c + d*x), x) + Integ 
ral(B**2*a*log(a*e/(c + d*x) + b*e*x/(c + d*x))**2/(c + d*x), x) + Integra 
l(2*A*B*a*log(a*e/(c + d*x) + b*e*x/(c + d*x))/(c + d*x), x) + Integral(B* 
*2*b*x*log(a*e/(c + d*x) + b*e*x/(c + d*x))**2/(c + d*x), x) + Integral(2* 
A*B*b*x*log(a*e/(c + d*x) + b*e*x/(c + d*x))/(c + d*x), x))/i
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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