\(\int (c i+d i x)^2 (A+B \log (\frac {e (a+b x)}{c+d x})) \, dx\) [13]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 118 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=-\frac {B (b c-a d)^2 i^2 x}{3 b^2}-\frac {B (b c-a d) i^2 (c+d x)^2}{6 b d}-\frac {B (b c-a d)^3 i^2 \log (a+b x)}{3 b^3 d}+\frac {i^2 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 d} \] Output:

-1/3*B*(-a*d+b*c)^2*i^2*x/b^2-1/6*B*(-a*d+b*c)*i^2*(d*x+c)^2/b/d-1/3*B*(-a 
*d+b*c)^3*i^2*ln(b*x+a)/b^3/d+1/3*i^2*(d*x+c)^3*(A+B*ln(e*(b*x+a)/(d*x+c)) 
)/d
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {i^2 \left (-\frac {B (b c-a d) \left (2 b d (b c-a d) x+b^2 (c+d x)^2+2 (b c-a d)^2 \log (a+b x)\right )}{2 b^3}+(c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )\right )}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2948, 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.

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

\(\Big \downarrow \) 2948

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 2948
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ 
)]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*( 
(A + B*Log[e*((a + b*x)^n/(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] / 
; FreeQ[{a, b, c, d, e, f, g, A, B, m, n}, x] && EqQ[n + mn, 0] && NeQ[b*c 
- a*d, 0] && NeQ[m, -1] &&  !(EqQ[m, -2] && IntegerQ[n])
 
Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.75

method result size
risch \(\frac {i^{2} \left (d x +c \right )^{3} B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )}{3 d}+\frac {i^{2} d^{2} A \,x^{3}}{3}+i^{2} d A c \,x^{2}+\frac {i^{2} d^{2} B a \,x^{2}}{6 b}-\frac {i^{2} d B c \,x^{2}}{6}+i^{2} A \,c^{2} x +\frac {i^{2} d^{2} B \ln \left (b x +a \right ) a^{3}}{3 b^{3}}-\frac {i^{2} d B \ln \left (b x +a \right ) a^{2} c}{b^{2}}+\frac {i^{2} B \ln \left (b x +a \right ) a \,c^{2}}{b}-\frac {i^{2} B \ln \left (b x +a \right ) c^{3}}{3 d}-\frac {i^{2} d^{2} B \,a^{2} x}{3 b^{2}}+\frac {i^{2} d B a c x}{b}-\frac {2 i^{2} B \,c^{2} x}{3}\) \(206\)
parts \(\frac {A \,i^{2} \left (d x +c \right )^{3}}{3 d}-B \,i^{2} \left (d a -b c \right )^{3} e^{3} \left (\frac {1}{3 b^{2} e^{2} d \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )}-\frac {1}{6 b e d \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )^{2}}+\frac {\ln \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )}{3 b^{3} e^{3} d}-\frac {\ln \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (3 b^{2} e^{2}-3 \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d b e +d^{2} \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right )^{2}\right )}{3 b^{3} e^{3} \left (\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d -b e \right )^{3}}\right )\) \(351\)
parallelrisch \(\frac {-6 A \,b^{3} c^{3} i^{2}-6 B \ln \left (b x +a \right ) a^{2} b c \,d^{2} i^{2}+6 B \ln \left (b x +a \right ) a \,b^{2} c^{2} d \,i^{2}+6 B \,x^{2} \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} c \,d^{2} i^{2}+6 B x \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} c^{2} d \,i^{2}+6 B x a \,b^{2} c \,d^{2} i^{2}+2 B \,a^{3} d^{3} i^{2}+4 B \,b^{3} c^{3} i^{2}-5 B \,a^{2} b c \,d^{2} i^{2}-B a \,b^{2} c^{2} d \,i^{2}-B \,x^{2} b^{3} c \,d^{2} i^{2}+6 A x \,b^{3} c^{2} d \,i^{2}-12 A a \,b^{2} c^{2} d \,i^{2}+2 A \,x^{3} b^{3} d^{3} i^{2}+2 B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} c^{3} i^{2}+2 B \ln \left (b x +a \right ) a^{3} d^{3} i^{2}-2 B \ln \left (b x +a \right ) b^{3} c^{3} i^{2}-4 B x \,b^{3} c^{2} d \,i^{2}-2 B x \,a^{2} b \,d^{3} i^{2}+2 B \,x^{3} \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) b^{3} d^{3} i^{2}+6 A \,x^{2} b^{3} c \,d^{2} i^{2}+B \,x^{2} a \,b^{2} d^{3} i^{2}}{6 b^{3} d}\) \(400\)
derivativedivides \(-\frac {e \left (d a -b c \right ) \left (\frac {A d \,e^{2} i^{2} \left (a^{2} d^{2}-2 a c d b +c^{2} b^{2}\right )}{3 \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )^{3}}+B \,d^{2} e^{2} i^{2} \left (a^{2} d^{2}-2 a c d b +c^{2} b^{2}\right ) \left (-\frac {1}{6 b e d \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )^{2}}-\frac {1}{3 b^{2} e^{2} d \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )}+\frac {\ln \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )}{3 b^{3} e^{3} d}+\frac {\ln \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (3 b^{2} e^{2}-3 \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d b e +d^{2} \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right )^{2}\right )}{3 b^{3} e^{3} \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )^{3}}\right )\right )}{d^{2}}\) \(429\)
default \(-\frac {e \left (d a -b c \right ) \left (\frac {A d \,e^{2} i^{2} \left (a^{2} d^{2}-2 a c d b +c^{2} b^{2}\right )}{3 \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )^{3}}+B \,d^{2} e^{2} i^{2} \left (a^{2} d^{2}-2 a c d b +c^{2} b^{2}\right ) \left (-\frac {1}{6 b e d \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )^{2}}-\frac {1}{3 b^{2} e^{2} d \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )}+\frac {\ln \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )}{3 b^{3} e^{3} d}+\frac {\ln \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) \left (3 b^{2} e^{2}-3 \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d b e +d^{2} \left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right )^{2}\right )}{3 b^{3} e^{3} \left (b e -\left (\frac {b e}{d}+\frac {\left (d a -b c \right ) e}{d \left (d x +c \right )}\right ) d \right )^{3}}\right )\right )}{d^{2}}\) \(429\)

Input:

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

Output:

1/3*i^2*(d*x+c)^3*B/d*ln(e*(b*x+a)/(d*x+c))+1/3*i^2*d^2*A*x^3+i^2*d*A*c*x^ 
2+1/6*i^2/b*d^2*B*a*x^2-1/6*i^2*d*B*c*x^2+i^2*A*c^2*x+1/3*i^2/b^3*d^2*B*ln 
(b*x+a)*a^3-i^2/b^2*d*B*ln(b*x+a)*a^2*c+i^2/b*B*ln(b*x+a)*a*c^2-1/3*i^2/d* 
B*ln(b*x+a)*c^3-1/3*i^2/b^2*d^2*B*a^2*x+i^2/b*d*B*a*c*x-2/3*i^2*B*c^2*x
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (110) = 220\).

Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.89 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {2 \, A b^{3} d^{3} i^{2} x^{3} - 2 \, B b^{3} c^{3} i^{2} \log \left (d x + c\right ) + {\left ({\left (6 \, A - B\right )} b^{3} c d^{2} + B a b^{2} d^{3}\right )} i^{2} x^{2} + 2 \, {\left ({\left (3 \, A - 2 \, B\right )} b^{3} c^{2} d + 3 \, B a b^{2} c d^{2} - B a^{2} b d^{3}\right )} i^{2} x + 2 \, {\left (3 \, B a b^{2} c^{2} d - 3 \, B a^{2} b c d^{2} + B a^{3} d^{3}\right )} i^{2} \log \left (b x + a\right ) + 2 \, {\left (B b^{3} d^{3} i^{2} x^{3} + 3 \, B b^{3} c d^{2} i^{2} x^{2} + 3 \, B b^{3} c^{2} d i^{2} x\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{6 \, b^{3} d} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (100) = 200\).

Time = 1.34 (sec) , antiderivative size = 491, normalized size of antiderivative = 4.16 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {A d^{2} i^{2} x^{3}}{3} + \frac {B a i^{2} \left (a^{2} d^{2} - 3 a b c d + 3 b^{2} c^{2}\right ) \log {\left (x + \frac {B a^{3} c d^{2} i^{2} - 3 B a^{2} b c^{2} d i^{2} + \frac {B a^{2} d i^{2} \left (a^{2} d^{2} - 3 a b c d + 3 b^{2} c^{2}\right )}{b} + 4 B a b^{2} c^{3} i^{2} - B a c i^{2} \left (a^{2} d^{2} - 3 a b c d + 3 b^{2} c^{2}\right )}{B a^{3} d^{3} i^{2} - 3 B a^{2} b c d^{2} i^{2} + 3 B a b^{2} c^{2} d i^{2} + B b^{3} c^{3} i^{2}} \right )}}{3 b^{3}} - \frac {B c^{3} i^{2} \log {\left (x + \frac {B a^{3} c d^{2} i^{2} - 3 B a^{2} b c^{2} d i^{2} + 3 B a b^{2} c^{3} i^{2} + \frac {B b^{3} c^{4} i^{2}}{d}}{B a^{3} d^{3} i^{2} - 3 B a^{2} b c d^{2} i^{2} + 3 B a b^{2} c^{2} d i^{2} + B b^{3} c^{3} i^{2}} \right )}}{3 d} + x^{2} \left (A c d i^{2} + \frac {B a d^{2} i^{2}}{6 b} - \frac {B c d i^{2}}{6}\right ) + x \left (A c^{2} i^{2} - \frac {B a^{2} d^{2} i^{2}}{3 b^{2}} + \frac {B a c d i^{2}}{b} - \frac {2 B c^{2} i^{2}}{3}\right ) + \left (B c^{2} i^{2} x + B c d i^{2} x^{2} + \frac {B d^{2} i^{2} x^{3}}{3}\right ) \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )} \] Input:

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

Output:

A*d**2*i**2*x**3/3 + B*a*i**2*(a**2*d**2 - 3*a*b*c*d + 3*b**2*c**2)*log(x 
+ (B*a**3*c*d**2*i**2 - 3*B*a**2*b*c**2*d*i**2 + B*a**2*d*i**2*(a**2*d**2 
- 3*a*b*c*d + 3*b**2*c**2)/b + 4*B*a*b**2*c**3*i**2 - B*a*c*i**2*(a**2*d** 
2 - 3*a*b*c*d + 3*b**2*c**2))/(B*a**3*d**3*i**2 - 3*B*a**2*b*c*d**2*i**2 + 
 3*B*a*b**2*c**2*d*i**2 + B*b**3*c**3*i**2))/(3*b**3) - B*c**3*i**2*log(x 
+ (B*a**3*c*d**2*i**2 - 3*B*a**2*b*c**2*d*i**2 + 3*B*a*b**2*c**3*i**2 + B* 
b**3*c**4*i**2/d)/(B*a**3*d**3*i**2 - 3*B*a**2*b*c*d**2*i**2 + 3*B*a*b**2* 
c**2*d*i**2 + B*b**3*c**3*i**2))/(3*d) + x**2*(A*c*d*i**2 + B*a*d**2*i**2/ 
(6*b) - B*c*d*i**2/6) + x*(A*c**2*i**2 - B*a**2*d**2*i**2/(3*b**2) + B*a*c 
*d*i**2/b - 2*B*c**2*i**2/3) + (B*c**2*i**2*x + B*c*d*i**2*x**2 + B*d**2*i 
**2*x**3/3)*log(e*(a + b*x)/(c + d*x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (110) = 220\).

Time = 0.04 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.37 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {1}{3} \, A d^{2} i^{2} x^{3} + A c d i^{2} x^{2} + {\left (x \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} B c^{2} i^{2} + {\left (x^{2} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) - \frac {a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {{\left (b c - a d\right )} x}{b d}\right )} B c d i^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} B d^{2} i^{2} + A c^{2} i^{2} x \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1056 vs. \(2 (110) = 220\).

Time = 0.24 (sec) , antiderivative size = 1056, normalized size of antiderivative = 8.95 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(2*(B*b^4*c^4*e^4*i^2 - 4*B*a*b^3*c^3*d*e^4*i^2 + 6*B*a^2*b^2*c^2*d^2* 
e^4*i^2 - 4*B*a^3*b*c*d^3*e^4*i^2 + B*a^4*d^4*e^4*i^2)*log((b*e*x + a*e)/( 
d*x + c))/(b^3*d*e^3 - 3*(b*e*x + a*e)*b^2*d^2*e^2/(d*x + c) + 3*(b*e*x + 
a*e)^2*b*d^3*e/(d*x + c)^2 - (b*e*x + a*e)^3*d^4/(d*x + c)^3) + (2*A*b^6*c 
^4*e^4*i^2 - 3*B*b^6*c^4*e^4*i^2 - 8*A*a*b^5*c^3*d*e^4*i^2 + 12*B*a*b^5*c^ 
3*d*e^4*i^2 + 12*A*a^2*b^4*c^2*d^2*e^4*i^2 - 18*B*a^2*b^4*c^2*d^2*e^4*i^2 
- 8*A*a^3*b^3*c*d^3*e^4*i^2 + 12*B*a^3*b^3*c*d^3*e^4*i^2 + 2*A*a^4*b^2*d^4 
*e^4*i^2 - 3*B*a^4*b^2*d^4*e^4*i^2 + 5*(b*e*x + a*e)*B*b^5*c^4*d*e^3*i^2/( 
d*x + c) - 20*(b*e*x + a*e)*B*a*b^4*c^3*d^2*e^3*i^2/(d*x + c) + 30*(b*e*x 
+ a*e)*B*a^2*b^3*c^2*d^3*e^3*i^2/(d*x + c) - 20*(b*e*x + a*e)*B*a^3*b^2*c* 
d^4*e^3*i^2/(d*x + c) + 5*(b*e*x + a*e)*B*a^4*b*d^5*e^3*i^2/(d*x + c) - 2* 
(b*e*x + a*e)^2*B*b^4*c^4*d^2*e^2*i^2/(d*x + c)^2 + 8*(b*e*x + a*e)^2*B*a* 
b^3*c^3*d^3*e^2*i^2/(d*x + c)^2 - 12*(b*e*x + a*e)^2*B*a^2*b^2*c^2*d^4*e^2 
*i^2/(d*x + c)^2 + 8*(b*e*x + a*e)^2*B*a^3*b*c*d^5*e^2*i^2/(d*x + c)^2 - 2 
*(b*e*x + a*e)^2*B*a^4*d^6*e^2*i^2/(d*x + c)^2)/(b^5*d*e^3 - 3*(b*e*x + a* 
e)*b^4*d^2*e^2/(d*x + c) + 3*(b*e*x + a*e)^2*b^3*d^3*e/(d*x + c)^2 - (b*e* 
x + a*e)^3*b^2*d^4/(d*x + c)^3) + 2*(B*b^4*c^4*e*i^2 - 4*B*a*b^3*c^3*d*e*i 
^2 + 6*B*a^2*b^2*c^2*d^2*e*i^2 - 4*B*a^3*b*c*d^3*e*i^2 + B*a^4*d^4*e*i^2)* 
log(-b*e + (b*e*x + a*e)*d/(d*x + c))/(b^3*d) - 2*(B*b^4*c^4*e*i^2 - 4*B*a 
*b^3*c^3*d*e*i^2 + 6*B*a^2*b^2*c^2*d^2*e*i^2 - 4*B*a^3*b*c*d^3*e*i^2 + ...
 

Mupad [B] (verification not implemented)

Time = 26.44 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.46 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=x^2\,\left (\frac {d\,i^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{6\,b}-\frac {A\,d\,i^2\,\left (3\,a\,d+3\,b\,c\right )}{6\,b}\right )-x\,\left (\frac {\left (3\,a\,d+3\,b\,c\right )\,\left (\frac {d\,i^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{3\,b}-\frac {A\,d\,i^2\,\left (3\,a\,d+3\,b\,c\right )}{3\,b}\right )}{3\,b\,d}-\frac {c\,i^2\,\left (3\,A\,a\,d+3\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{b}+\frac {A\,a\,c\,d\,i^2}{b}\right )+\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\,\left (B\,c^2\,i^2\,x+B\,c\,d\,i^2\,x^2+\frac {B\,d^2\,i^2\,x^3}{3}\right )+\frac {\ln \left (a+b\,x\right )\,\left (B\,a^3\,d^2\,i^2-3\,B\,a^2\,b\,c\,d\,i^2+3\,B\,a\,b^2\,c^2\,i^2\right )}{3\,b^3}+\frac {A\,d^2\,i^2\,x^3}{3}-\frac {B\,c^3\,i^2\,\ln \left (c+d\,x\right )}{3\,d} \] Input:

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

Output:

x^2*((d*i^2*(3*A*a*d + 9*A*b*c + B*a*d - B*b*c))/(6*b) - (A*d*i^2*(3*a*d + 
 3*b*c))/(6*b)) - x*(((3*a*d + 3*b*c)*((d*i^2*(3*A*a*d + 9*A*b*c + B*a*d - 
 B*b*c))/(3*b) - (A*d*i^2*(3*a*d + 3*b*c))/(3*b)))/(3*b*d) - (c*i^2*(3*A*a 
*d + 3*A*b*c + B*a*d - B*b*c))/b + (A*a*c*d*i^2)/b) + log((e*(a + b*x))/(c 
 + d*x))*((B*d^2*i^2*x^3)/3 + B*c^2*i^2*x + B*c*d*i^2*x^2) + (log(a + b*x) 
*(B*a^3*d^2*i^2 + 3*B*a*b^2*c^2*i^2 - 3*B*a^2*b*c*d*i^2))/(3*b^3) + (A*d^2 
*i^2*x^3)/3 - (B*c^3*i^2*log(c + d*x))/(3*d)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.27 \[ \int (c i+d i x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx=\frac {-2 \,\mathrm {log}\left (b x +a \right ) a^{3} d^{3}+6 \,\mathrm {log}\left (b x +a \right ) a^{2} b c \,d^{2}-6 \,\mathrm {log}\left (b x +a \right ) a \,b^{2} c^{2} d +2 \,\mathrm {log}\left (b x +a \right ) b^{3} c^{3}-2 \,\mathrm {log}\left (\frac {b e x +a e}{d x +c}\right ) b^{3} c^{3}-6 \,\mathrm {log}\left (\frac {b e x +a e}{d x +c}\right ) b^{3} c^{2} d x -6 \,\mathrm {log}\left (\frac {b e x +a e}{d x +c}\right ) b^{3} c \,d^{2} x^{2}-2 \,\mathrm {log}\left (\frac {b e x +a e}{d x +c}\right ) b^{3} d^{3} x^{3}+2 a^{2} b \,d^{3} x -6 a \,b^{2} c^{2} d x -6 a \,b^{2} c \,d^{2} x^{2}-6 a \,b^{2} c \,d^{2} x -2 a \,b^{2} d^{3} x^{3}-a \,b^{2} d^{3} x^{2}+4 b^{3} c^{2} d x +b^{3} c \,d^{2} x^{2}}{6 b^{2} d} \] Input:

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

Output:

( - 2*log(a + b*x)*a**3*d**3 + 6*log(a + b*x)*a**2*b*c*d**2 - 6*log(a + b* 
x)*a*b**2*c**2*d + 2*log(a + b*x)*b**3*c**3 - 2*log((a*e + b*e*x)/(c + d*x 
))*b**3*c**3 - 6*log((a*e + b*e*x)/(c + d*x))*b**3*c**2*d*x - 6*log((a*e + 
 b*e*x)/(c + d*x))*b**3*c*d**2*x**2 - 2*log((a*e + b*e*x)/(c + d*x))*b**3* 
d**3*x**3 + 2*a**2*b*d**3*x - 6*a*b**2*c**2*d*x - 6*a*b**2*c*d**2*x**2 - 6 
*a*b**2*c*d**2*x - 2*a*b**2*d**3*x**3 - a*b**2*d**3*x**2 + 4*b**3*c**2*d*x 
 + b**3*c*d**2*x**2)/(6*b**2*d)