Integrand size = 43, antiderivative size = 361 \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=-\frac {2 B d (b c-a d) i^3 n (c+d x)}{b^3 g^3 (a+b x)}-\frac {B (b c-a d) i^3 n (c+d x)^2}{4 b^2 g^3 (a+b x)^2}+\frac {d^3 i^3 (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^4 g^3}-\frac {2 d (b c-a d) i^3 (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^3 g^3 (a+b x)}-\frac {(b c-a d) i^3 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 b^2 g^3 (a+b x)^2}-\frac {B d^2 (b c-a d) i^3 n \log (c+d x)}{b^4 g^3}-\frac {3 d^2 (b c-a d) i^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b^4 g^3}+\frac {3 B d^2 (b c-a d) i^3 n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^4 g^3} \]
-2*B*d*(-a*d+b*c)*i^3*n*(d*x+c)/b^3/g^3/(b*x+a)-1/4*B*(-a*d+b*c)*i^3*n*(d* x+c)^2/b^2/g^3/(b*x+a)^2+d^3*i^3*(b*x+a)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b ^4/g^3-2*d*(-a*d+b*c)*i^3*(d*x+c)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^3/g^3/ (b*x+a)-1/2*(-a*d+b*c)*i^3*(d*x+c)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^2/g ^3/(b*x+a)^2-B*d^2*(-a*d+b*c)*i^3*n*ln(d*x+c)/b^4/g^3-3*d^2*(-a*d+b*c)*i^3 *(A+B*ln(e*((b*x+a)/(d*x+c))^n))*ln(1-b*(d*x+c)/d/(b*x+a))/b^4/g^3+3*B*d^2 *(-a*d+b*c)*i^3*n*polylog(2,b*(d*x+c)/d/(b*x+a))/b^4/g^3
Time = 0.23 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.92 \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\frac {i^3 \left (4 A b d^3 x-\frac {B (b c-a d)^3 n}{(a+b x)^2}-\frac {10 B d (b c-a d)^2 n}{a+b x}+10 B d^2 (-b c+a d) n \log (a+b x)+4 B d^3 (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-\frac {2 (b c-a d)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x)^2}-\frac {12 d (b c-a d)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{a+b x}+12 d^2 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+6 B d^2 (b c-a d) n \log (c+d x)+6 B d^2 (-b c+a d) n \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{4 b^4 g^3} \]
(i^3*(4*A*b*d^3*x - (B*(b*c - a*d)^3*n)/(a + b*x)^2 - (10*B*d*(b*c - a*d)^ 2*n)/(a + b*x) + 10*B*d^2*(-(b*c) + a*d)*n*Log[a + b*x] + 4*B*d^3*(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]))/(a + b*x)^2 - (12*d*(b*c - a*d)^2*(A + B*Log[e*((a + b*x )/(c + d*x))^n]))/(a + b*x) + 12*d^2*(b*c - a*d)*Log[a + b*x]*(A + B*Log[e *((a + b*x)/(c + d*x))^n]) + 6*B*d^2*(b*c - a*d)*n*Log[c + d*x] + 6*B*d^2* (-(b*c) + a*d)*n*(Log[a + b*x]*(Log[a + b*x] - 2*Log[(b*(c + d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)])))/(4*b^4*g^3)
Time = 0.56 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2961, 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.
| \(\displaystyle \int \frac {(c i+d i x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{(a g+b g x)^3} \, dx\) |
| \(\Big \downarrow \) 2961 |
| \(\displaystyle \frac {i^3 (b c-a d) \int \frac {(c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x)^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}d\frac {a+b x}{c+d x}}{g^3}\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \frac {i^3 (b c-a d) \int \left (\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) d^3}{b^3 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}+\frac {3 (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) d^2}{b^3 (a+b x) \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {2 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) d}{b^3 (a+b x)^2}+\frac {(c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 (a+b x)^3}\right )d\frac {a+b x}{c+d x}}{g^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {i^3 (b c-a d) \left (\frac {d^3 (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4 (c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )}-\frac {3 d^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^4}-\frac {2 d (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^3 (a+b x)}-\frac {(c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^2 (a+b x)^2}+\frac {3 B d^2 n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^4}+\frac {B d^2 n \log \left (b-\frac {d (a+b x)}{c+d x}\right )}{b^4}-\frac {2 B d n (c+d x)}{b^3 (a+b x)}-\frac {B n (c+d x)^2}{4 b^2 (a+b x)^2}\right )}{g^3}\) |
((b*c - a*d)*i^3*((-2*B*d*n*(c + d*x))/(b^3*(a + b*x)) - (B*n*(c + d*x)^2) /(4*b^2*(a + b*x)^2) - (2*d*(c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n ]))/(b^3*(a + b*x)) - ((c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])) /(2*b^2*(a + b*x)^2) + (d^3*(a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n ]))/(b^4*(c + d*x)*(b - (d*(a + b*x))/(c + d*x))) + (B*d^2*n*Log[b - (d*(a + b*x))/(c + d*x)])/b^4 - (3*d^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*L og[1 - (b*(c + d*x))/(d*(a + b*x))])/b^4 + (3*B*d^2*n*PolyLog[2, (b*(c + d *x))/(d*(a + b*x))])/b^4))/g^3
3.2.33.3.1 Defintions of rubi rules used
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]))
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]
\[\int \frac {\left (d i x +c i \right )^{3} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}{\left (b g x +a g \right )^{3}}d x\]
\[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\int { \frac {{\left (d i x + c i\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}}{{\left (b g x + a g\right )}^{3}} \,d x } \]
integral((A*d^3*i^3*x^3 + 3*A*c*d^2*i^3*x^2 + 3*A*c^2*d*i^3*x + A*c^3*i^3 + (B*d^3*i^3*x^3 + 3*B*c*d^2*i^3*x^2 + 3*B*c^2*d*i^3*x + B*c^3*i^3)*log(e* ((b*x + a)/(d*x + c))^n))/(b^3*g^3*x^3 + 3*a*b^2*g^3*x^2 + 3*a^2*b*g^3*x + a^3*g^3), x)
Timed out. \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2746 vs. \(2 (356) = 712\).
Time = 0.57 (sec) , antiderivative size = 2746, normalized size of antiderivative = 7.61 \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\text {Too large to display} \]
-3/4*B*c^2*d*i^3*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^3*i^3*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*A*d^3*i^3*((6*a^2 *b*x + 5*a^3)/(b^6*g^3*x^2 + 2*a*b^5*g^3*x + a^2*b^4*g^3) - 2*x/(b^3*g^3) + 6*a*log(b*x + a)/(b^4*g^3)) + 3/2*A*c*d^2*i^3*((4*a*b*x + 3*a^2)/(b^5*g^ 3*x^2 + 2*a*b^4*g^3*x + a^2*b^3*g^3) + 2*log(b*x + a)/(b^3*g^3)) - 3/2*(2* b*x + a)*B*c^2*d*i^3*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) - 3/2*(2*b*x + a)*A*c^2*d*i^3/(b^4*g^3*x^2 + 2*a*b^3*g^3*x + a^2*b^2*g^3) - 1/2*B*c^3*i^3*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^3*i^3/(b^3* g^3*x^2 + 2*a*b^2*g^3*x + a^2*b*g^3) - 1/2*(2*b^3*c^3*d^2*i^3*n + 8*a*b^2* c^2*d^3*i^3*n - 13*a^2*b*c*d^4*i^3*n + 5*a^3*d^5*i^3*n)*B*log(d*x + c)/(b^ 6*c^2*g^3 - 2*a*b^5*c*d*g^3 + a^2*b^4*d^2*g^3) + 1/4*(4*(b^5*c^2*d^3*i^3*l og(e) - 2*a*b^4*c*d^4*i^3*log(e) + a^2*b^3*d^5*i^3*log(e))*B*x^3 + 8*(a*b^ 4*c^2*d^3*i^3*log(e) - 2*a^2*b^3*c*d^4*i^3*log(e) + a^3*b^2*d^5*i^3*log...
\[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\int { \frac {{\left (d i x + c i\right )}^{3} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}}{{\left (b g x + a g\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(c i+d i x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^3\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}{{\left (a\,g+b\,g\,x\right )}^3} \,d x \]