3.2.51 \(\int \frac {(a g+b g x)^3 (A+B \log (e (\frac {a+b x}{c+d x})^n))}{(c i+d i x)^3} \, dx\) [151]

3.2.51.1 Optimal result
3.2.51.2 Mathematica [A] (verified)
3.2.51.3 Rubi [A] (verified)
3.2.51.4 Maple [F]
3.2.51.5 Fricas [F]
3.2.51.6 Sympy [F(-1)]
3.2.51.7 Maxima [B] (verification not implemented)
3.2.51.8 Giac [F]
3.2.51.9 Mupad [F(-1)]

3.2.51.1 Optimal result

Integrand size = 43, antiderivative size = 382 \[ \int \frac {(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^3} \, dx=-\frac {3 B (b c-a d) g^3 n (a+b x)^2}{4 d^2 i^3 (c+d x)^2}-\frac {3 b B (b c-a d) g^3 n (a+b x)}{d^3 i^3 (c+d x)}+\frac {b (b c-a d) g^3 (3 A+B n) (a+b x)}{d^3 i^3 (c+d x)}+\frac {3 b B (b c-a d) g^3 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d^3 i^3 (c+d x)}+\frac {g^3 (a+b x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d i^3 (c+d x)^2}+\frac {(b c-a d) g^3 (a+b x)^2 \left (3 A+B n+3 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d^2 i^3 (c+d x)^2}+\frac {b^2 (b c-a d) g^3 \left (3 A+B n+3 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{d^4 i^3}+\frac {3 b^2 B (b c-a d) g^3 n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^4 i^3} \]

output
-3/4*B*(-a*d+b*c)*g^3*n*(b*x+a)^2/d^2/i^3/(d*x+c)^2-3*b*B*(-a*d+b*c)*g^3*n 
*(b*x+a)/d^3/i^3/(d*x+c)+b*(-a*d+b*c)*g^3*(B*n+3*A)*(b*x+a)/d^3/i^3/(d*x+c 
)+3*b*B*(-a*d+b*c)*g^3*(b*x+a)*ln(e*((b*x+a)/(d*x+c))^n)/d^3/i^3/(d*x+c)+g 
^3*(b*x+a)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/d/i^3/(d*x+c)^2+1/2*(-a*d+b*c 
)*g^3*(b*x+a)^2*(3*A+B*n+3*B*ln(e*((b*x+a)/(d*x+c))^n))/d^2/i^3/(d*x+c)^2+ 
b^2*(-a*d+b*c)*g^3*(3*A+B*n+3*B*ln(e*((b*x+a)/(d*x+c))^n))*ln((-a*d+b*c)/b 
/(d*x+c))/d^4/i^3+3*b^2*B*(-a*d+b*c)*g^3*n*polylog(2,d*(b*x+a)/b/(d*x+c))/ 
d^4/i^3
 
3.2.51.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.87 \[ \int \frac {(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^3} \, dx=\frac {g^3 \left (4 A b^3 d x-\frac {B (b c-a d)^3 n}{(c+d x)^2}+\frac {10 b B (b c-a d)^2 n}{c+d x}+10 b^2 B (b c-a d) n \log (a+b x)+4 b^2 B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+\frac {2 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c+d x)^2}-\frac {12 b (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{c+d x}-14 b^2 B (b c-a d) n \log (c+d x)-12 b^2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+6 b^2 B (b c-a d) n \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{4 d^4 i^3} \]

input
Integrate[((a*g + b*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(c*i + 
d*i*x)^3,x]
 
output
(g^3*(4*A*b^3*d*x - (B*(b*c - a*d)^3*n)/(c + d*x)^2 + (10*b*B*(b*c - a*d)^ 
2*n)/(c + d*x) + 10*b^2*B*(b*c - a*d)*n*Log[a + b*x] + 4*b^2*B*d*(a + b*x) 
*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]))/(c + d*x)^2 - (12*b*(b*c - a*d)^2*(A + B*Log[e*((a + b*x) 
/(c + d*x))^n]))/(c + d*x) - 14*b^2*B*(b*c - a*d)*n*Log[c + d*x] - 12*b^2* 
(b*c - a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[c + d*x] + 6*b^2*B* 
(b*c - a*d)*n*((2*Log[(d*(a + b*x))/(-(b*c) + a*d)] - Log[c + d*x])*Log[c 
+ d*x] + 2*PolyLog[2, (b*(c + d*x))/(b*c - a*d)])))/(4*d^4*i^3)
 
3.2.51.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {2961, 2784, 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.

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

\(\Big \downarrow \) 2961

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

\(\Big \downarrow \) 2784

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

\(\Big \downarrow \) 2793

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

\(\Big \downarrow \) 2009

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

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

3.2.51.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2784
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* 
(x_))^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] 
)/(e*(q + 1))), x] - Simp[f/(e*(q + 1))   Int[(f*x)^(m - 1)*(d + e*x)^(q + 
1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, 
x] && ILtQ[q, -1] && GtQ[m, 0]
 

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]
 
3.2.51.4 Maple [F]

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

input
int((b*g*x+a*g)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x)
 
output
int((b*g*x+a*g)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x)
 
3.2.51.5 Fricas [F]

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

input
integrate((b*g*x+a*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x, 
algorithm="fricas")
 
output
integral((A*b^3*g^3*x^3 + 3*A*a*b^2*g^3*x^2 + 3*A*a^2*b*g^3*x + A*a^3*g^3 
+ (B*b^3*g^3*x^3 + 3*B*a*b^2*g^3*x^2 + 3*B*a^2*b*g^3*x + B*a^3*g^3)*log(e* 
((b*x + a)/(d*x + c))^n))/(d^3*i^3*x^3 + 3*c*d^2*i^3*x^2 + 3*c^2*d*i^3*x + 
 c^3*i^3), x)
 
3.2.51.6 Sympy [F(-1)]

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

input
integrate((b*g*x+a*g)**3*(A+B*ln(e*((b*x+a)/(d*x+c))**n))/(d*i*x+c*i)**3,x 
)
 
output
Timed out
 
3.2.51.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2894 vs. \(2 (377) = 754\).

Time = 0.55 (sec) , antiderivative size = 2894, normalized size of antiderivative = 7.58 \[ \int \frac {(a g+b g x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(c i+d i x)^3} \, dx=\text {Too large to display} \]

input
integrate((b*g*x+a*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x, 
algorithm="maxima")
 
output
3/4*B*a^2*b*g^3*n*((b*c^2 - 3*a*c*d + 2*(b*c*d - 2*a*d^2)*x)/((b*c*d^4 - a 
*d^5)*i^3*x^2 + 2*(b*c^2*d^3 - a*c*d^4)*i^3*x + (b*c^3*d^2 - a*c^2*d^3)*i^ 
3) + 2*(b^2*c - 2*a*b*d)*log(b*x + a)/((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^ 
4)*i^3) - 2*(b^2*c - 2*a*b*d)*log(d*x + c)/((b^2*c^2*d^2 - 2*a*b*c*d^3 + a 
^2*d^4)*i^3)) + 1/4*B*a^3*g^3*n*((2*b*d*x + 3*b*c - a*d)/((b*c*d^3 - a*d^4 
)*i^3*x^2 + 2*(b*c^2*d^2 - a*c*d^3)*i^3*x + (b*c^3*d - a*c^2*d^2)*i^3) + 2 
*b^2*log(b*x + a)/((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*i^3) - 2*b^2*log(d* 
x + c)/((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*i^3)) - 1/2*A*b^3*g^3*((6*c^2* 
d*x + 5*c^3)/(d^6*i^3*x^2 + 2*c*d^5*i^3*x + c^2*d^4*i^3) - 2*x/(d^3*i^3) + 
 6*c*log(d*x + c)/(d^4*i^3)) + 3/2*A*a*b^2*g^3*((4*c*d*x + 3*c^2)/(d^5*i^3 
*x^2 + 2*c*d^4*i^3*x + c^2*d^3*i^3) + 2*log(d*x + c)/(d^3*i^3)) - 3/2*(2*d 
*x + c)*B*a^2*b*g^3*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(d^4*i^3*x^2 + 
2*c*d^3*i^3*x + c^2*d^2*i^3) - 3/2*(2*d*x + c)*A*a^2*b*g^3/(d^4*i^3*x^2 + 
2*c*d^3*i^3*x + c^2*d^2*i^3) - 1/2*B*a^3*g^3*log(e*(b*x/(d*x + c) + a/(d*x 
 + c))^n)/(d^3*i^3*x^2 + 2*c*d^2*i^3*x + c^2*d*i^3) - 1/2*A*a^3*g^3/(d^3*i 
^3*x^2 + 2*c*d^2*i^3*x + c^2*d*i^3) + 1/2*(6*a^3*b^2*d^3*g^3*log(e) - (7*g 
^3*n + 6*g^3*log(e))*b^5*c^3 + (19*g^3*n + 18*g^3*log(e))*a*b^4*c^2*d - 2* 
(7*g^3*n + 9*g^3*log(e))*a^2*b^3*c*d^2)*B*log(d*x + c)/(b^2*c^2*d^4*i^3 - 
2*a*b*c*d^5*i^3 + a^2*d^6*i^3) + 1/4*(4*(b^5*c^2*d^3*g^3*log(e) - 2*a*b^4* 
c*d^4*g^3*log(e) + a^2*b^3*d^5*g^3*log(e))*B*x^3 + 8*(b^5*c^3*d^2*g^3*l...
 
3.2.51.8 Giac [F]

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

input
integrate((b*g*x+a*g)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x, 
algorithm="giac")
 
output
integrate((b*g*x + a*g)^3*(B*log(e*((b*x + a)/(d*x + c))^n) + A)/(d*i*x + 
c*i)^3, x)
 
3.2.51.9 Mupad [F(-1)]

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

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