\(\int \frac {(c i+d i x) (A+B \log (e (\frac {a+b x}{c+d x})^n))}{(a g+b g x)^3} \, dx\) [114]

Optimal result

Integrand size = 41, antiderivative size = 89 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=-\frac {B i n (c+d x)^2}{4 (b c-a d) g^3 (a+b x)^2}-\frac {i (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 (b c-a d) g^3 (a+b x)^2} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(216\) vs. \(2(89)=178\).

Time = 0.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.43 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\frac {i \left (-\frac {(b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 (a+b x)^2}-\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 (a+b x)}-\frac {B d n \left (\frac {1}{a+b x}+\frac {d \log (a+b x)}{b c-a d}-\frac {d \log (c+d x)}{b c-a d}\right )}{b^2}-\frac {B n \left (\frac {b c-a d}{(a+b x)^2}-\frac {2 d}{a+b x}-\frac {2 d^2 \log (a+b x)}{b c-a d}+\frac {2 d^2 \log (c+d x)}{b c-a d}\right )}{4 b^2}\right )}{g^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {2961, 2741}

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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2741

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2741

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]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(236\) vs. \(2(85)=170\).

Time = 4.73 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.66

method	result	size
parallelrisch	\(-\frac {-4 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c \,d^{2} i n +B \,a^{2} b^{2} d^{3} i \,n^{2}-B \,b^{4} c^{2} d i \,n^{2}+2 A \,a^{2} b^{2} d^{3} i n -2 A \,b^{4} c^{2} d i n -2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} c^{2} d i n -2 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{4} d^{3} i n +2 B x a \,b^{3} d^{3} i \,n^{2}-2 B x \,b^{4} c \,d^{2} i \,n^{2}+4 A x a \,b^{3} d^{3} i n -4 A x \,b^{4} c \,d^{2} i n}{4 g^{3} \left (b x +a \right )^{2} b^{4} d n \left (d a -b c \right )}\)	\(237\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (85) = 170\).

Time = 0.09 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.81 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=-\frac {{\left (B b^{2} c^{2} - B a^{2} d^{2}\right )} i n + 2 \, {\left (A b^{2} c^{2} - A a^{2} d^{2}\right )} i + 2 \, {\left ({\left (B b^{2} c d - B a b d^{2}\right )} i n + 2 \, {\left (A b^{2} c d - A a b d^{2}\right )} i\right )} x + 2 \, {\left (2 \, {\left (B b^{2} c d - B a b d^{2}\right )} i x + {\left (B b^{2} c^{2} - B a^{2} d^{2}\right )} i\right )} \log \left (e\right ) + 2 \, {\left (B b^{2} d^{2} i n x^{2} + 2 \, B b^{2} c d i n x + B b^{2} c^{2} i n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{5} c - a b^{4} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c - a^{2} b^{3} d\right )} g^{3} x + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} g^{3}\right )}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 582 vs. \(2 (85) = 170\).

Time = 0.06 (sec) , antiderivative size = 582, normalized size of antiderivative = 6.54 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=-\frac {1}{4} \, B d i n {\left (\frac {3 \, a b c - a^{2} d + 2 \, {\left (2 \, b^{2} c - a b d\right )} x}{{\left (b^{5} c - a b^{4} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c - a^{2} b^{3} d\right )} g^{3} x + {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} g^{3}} + \frac {2 \, {\left (2 \, b c d - a d^{2}\right )} \log \left (b x + a\right )}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} g^{3}} - \frac {2 \, {\left (2 \, b c d - a d^{2}\right )} \log \left (d x + c\right )}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} g^{3}}\right )} + \frac {1}{4} \, B c i n {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {{\left (2 \, b x + a\right )} B d i \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{2 \, {\left (b^{4} g^{3} x^{2} + 2 \, a b^{3} g^{3} x + a^{2} b^{2} g^{3}\right )}} - \frac {{\left (2 \, b x + a\right )} A d i}{2 \, {\left (b^{4} g^{3} x^{2} + 2 \, a b^{3} g^{3} x + a^{2} b^{2} g^{3}\right )}} - \frac {B c i \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} - \frac {A c i}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.12 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, {\left (d x + c\right )}^{2} B i n \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b x + a\right )}^{2} g^{3}} + \frac {{\left (B i n + 2 \, B i \log \left (e\right ) + 2 \, A i\right )} {\left (d x + c\right )}^{2}}{{\left (b x + a\right )}^{2} g^{3}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 26.43 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.29 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=-\frac {x\,\left (2\,A\,b\,d\,i+B\,b\,d\,i\,n\right )+A\,a\,d\,i+A\,b\,c\,i+\frac {B\,a\,d\,i\,n}{2}+\frac {B\,b\,c\,i\,n}{2}}{2\,a^2\,b^2\,g^3+4\,a\,b^3\,g^3\,x+2\,b^4\,g^3\,x^2}-\frac {\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {B\,c\,i}{2\,b}+\frac {B\,a\,d\,i}{2\,b^2}+\frac {B\,d\,i\,x}{b}\right )}{a^2\,g^3+2\,a\,b\,g^3\,x+b^2\,g^3\,x^2}-\frac {B\,d^2\,i\,n\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^2\,g^3\,\left (a\,d-b\,c\right )} \] Input:

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

Output:

- (x*(2*A*b*d*i + B*b*d*i*n) + A*a*d*i + A*b*c*i + (B*a*d*i*n)/2 + (B*b*c* 
i*n)/2)/(2*a^2*b^2*g^3 + 2*b^4*g^3*x^2 + 4*a*b^3*g^3*x) - (log(e*((a + b*x 
)/(c + d*x))^n)*((B*c*i)/(2*b) + (B*a*d*i)/(2*b^2) + (B*d*i*x)/b))/(a^2*g^ 
3 + b^2*g^3*x^2 + 2*a*b*g^3*x) - (B*d^2*i*n*atan((b*c*2i + b*d*x*2i)/(a*d 
- b*c) + 1i)*1i)/(b^2*g^3*(a*d - b*c))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 353, normalized size of antiderivative = 3.97 \[ \int \frac {(c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\frac {i \left (2 \,\mathrm {log}\left (b x +a \right ) a^{2} b c d n +4 \,\mathrm {log}\left (b x +a \right ) a \,b^{2} c d n x +2 \,\mathrm {log}\left (b x +a \right ) b^{3} c d n \,x^{2}-2 \,\mathrm {log}\left (d x +c \right ) a^{2} b c d n -4 \,\mathrm {log}\left (d x +c \right ) a \,b^{2} c d n x -2 \,\mathrm {log}\left (d x +c \right ) b^{3} c d n \,x^{2}-2 \,\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a^{2} b c d +2 \,\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a \,b^{2} c^{2}+2 \,\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a \,b^{2} d^{2} x^{2}-2 \,\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b^{3} c d \,x^{2}-2 a^{3} c d +2 a^{2} b \,c^{2}-a^{2} b c d n +2 a^{2} b \,d^{2} x^{2}+a \,b^{2} c^{2} n -2 a \,b^{2} c d \,x^{2}+a \,b^{2} d^{2} n \,x^{2}-b^{3} c d n \,x^{2}\right )}{4 a b \,g^{3} \left (a \,b^{2} d \,x^{2}-b^{3} c \,x^{2}+2 a^{2} b d x -2 a \,b^{2} c x +a^{3} d -a^{2} b c \right )} \] Input:

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

Output:

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