Integrand size = 45, antiderivative size = 472 \[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=-\frac {2 B^2 (b c-a d) i^2 n^2 (c+d x)}{b^2 g^2 (a+b x)}-\frac {2 B (b c-a d) i^2 n (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^2 (a+b x)}+\frac {d^2 i^2 (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{b^3 g^2}-\frac {(b c-a d) i^2 (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{b^2 g^2 (a+b x)}+\frac {2 B d (b c-a d) i^2 n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b^3 g^2}-\frac {2 d (b c-a d) i^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b^3 g^2}+\frac {2 B^2 d (b c-a d) i^2 n^2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b^3 g^2}+\frac {4 B d (b c-a d) i^2 n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^3 g^2}+\frac {4 B^2 d (b c-a d) i^2 n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b^3 g^2} \]
-2*B^2*(-a*d+b*c)*i^2*n^2*(d*x+c)/b^2/g^2/(b*x+a)-2*B*(-a*d+b*c)*i^2*n*(d* x+c)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b^2/g^2/(b*x+a)+d^2*i^2*(b*x+a)*(A+B* ln(e*((b*x+a)/(d*x+c))^n))^2/b^3/g^2-(-a*d+b*c)*i^2*(d*x+c)*(A+B*ln(e*((b* x+a)/(d*x+c))^n))^2/b^2/g^2/(b*x+a)+2*B*d*(-a*d+b*c)*i^2*n*(A+B*ln(e*((b*x +a)/(d*x+c))^n))*ln((-a*d+b*c)/b/(d*x+c))/b^3/g^2-2*d*(-a*d+b*c)*i^2*(A+B* ln(e*((b*x+a)/(d*x+c))^n))^2*ln(1-b*(d*x+c)/d/(b*x+a))/b^3/g^2+2*B^2*d*(-a *d+b*c)*i^2*n^2*polylog(2,d*(b*x+a)/b/(d*x+c))/b^3/g^2+4*B*d*(-a*d+b*c)*i^ 2*n*(A+B*ln(e*((b*x+a)/(d*x+c))^n))*polylog(2,b*(d*x+c)/d/(b*x+a))/b^3/g^2 +4*B^2*d*(-a*d+b*c)*i^2*n^2*polylog(3,b*(d*x+c)/d/(b*x+a))/b^3/g^2
Leaf count is larger than twice the leaf count of optimal. \(2885\) vs. \(2(472)=944\).
Time = 2.78 (sec) , antiderivative size = 2885, normalized size of antiderivative = 6.11 \[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\text {Result too large to show} \]
(i^2*(3*b*d^2*x*(A + B*Log[e*((a + b*x)/(c + d*x))^n] - B*n*Log[(a + b*x)/ (c + d*x)])^2 - (3*(b*c - a*d)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n] - B *n*Log[(a + b*x)/(c + d*x)])^2)/(a + b*x) + 6*d*(b*c - a*d)*Log[a + b*x]*( A + B*Log[e*((a + b*x)/(c + d*x))^n] - B*n*Log[(a + b*x)/(c + d*x)])^2 + ( 6*b^2*B*c^2*n*(-A - B*Log[e*((a + b*x)/(c + d*x))^n] + B*n*Log[(a + b*x)/( c + d*x)])*(-(d*(a + b*x)*Log[c/d + x]) + d*(a + b*x)*Log[(d*(a + b*x))/(- (b*c) + a*d)] + (b*c - a*d)*(1 + Log[(a + b*x)/(c + d*x)])))/((b*c - a*d)* (a + b*x)) + (3*b^2*B^2*c^2*n^2*(-2*b*c + 2*a*d - 2*d*(a + b*x)*Log[a + b* x] - 2*(b*c - a*d)*Log[(a + b*x)/(c + d*x)] - 2*d*(a + b*x)*Log[a + b*x]*L og[(a + b*x)/(c + d*x)] - (b*c - a*d)*Log[(a + b*x)/(c + d*x)]^2 + 2*d*(a + b*x)*Log[c + d*x] - 2*d*(a + b*x)*Log[(a + b*x)/(c + d*x)]*Log[(b*c - a* d)/(b*c + b*d*x)] + d*(a + b*x)*(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)]) + d*(a + b*x)*(Log[(b*c - a*d)/(b*c + b*d*x)]*(2*Log[(d*(a + b*x))/(-(b*c) + a*d )] + Log[(b*c - a*d)/(b*c + b*d*x)]) - 2*PolyLog[2, (b*(c + d*x))/(b*c - a *d)])))/((b*c - a*d)*(a + b*x)) + 6*b*B*c*d*n*(A + B*Log[e*((a + b*x)/(c + d*x))^n] - B*n*Log[(a + b*x)/(c + d*x)])*(Log[a/b + x]^2 - 2*Log[a/b + x] *Log[a + b*x] - 2*Log[c/d + x]*Log[(d*(a + b*x))/(-(b*c) + a*d)] + 2*Log[a + b*x]*((a*d)/(b*c - a*d) + Log[c/d + x] + Log[(a + b*x)/(c + d*x)]) + 2* a*((a + b*x)^(-1) + Log[(a + b*x)/(c + d*x)]/(a + b*x) + (d*Log[c + d*x...
Time = 0.72 (sec) , antiderivative size = 396, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2961, 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.
\(\displaystyle \int \frac {(c i+d i x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{(a g+b g x)^2} \, dx\) |
\(\Big \downarrow \) 2961 |
\(\displaystyle \frac {i^2 (b c-a d) \int \frac {(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a+b x)^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}d\frac {a+b x}{c+d x}}{g^2}\) |
\(\Big \downarrow \) 2795 |
\(\displaystyle \frac {i^2 (b c-a d) \int \left (\frac {(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{b^2 (a+b x)^2}+\frac {2 d (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{b^2 (a+b x) \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {d^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{b^2 \left (b-\frac {d (a+b x)}{c+d x}\right )^2}\right )d\frac {a+b x}{c+d x}}{g^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i^2 (b c-a d) \left (\frac {d^2 (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b^3 (c+d x) \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {4 B d n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^3}+\frac {2 B d n \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^3}-\frac {2 d \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b^3}-\frac {2 B n (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^2 (a+b x)}-\frac {(c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{b^2 (a+b x)}+\frac {2 B^2 d n^2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b^3}+\frac {4 B^2 d n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b^3}-\frac {2 B^2 n^2 (c+d x)}{b^2 (a+b x)}\right )}{g^2}\) |
((b*c - a*d)*i^2*((-2*B^2*n^2*(c + d*x))/(b^2*(a + b*x)) - (2*B*n*(c + d*x )*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(b^2*(a + b*x)) - ((c + d*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(b^2*(a + b*x)) + (d^2*(a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^2)/(b^3*(c + d*x)*(b - (d*(a + b*x))/ (c + d*x))) + (2*B*d*n*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[1 - (d*( a + b*x))/(b*(c + d*x))])/b^3 - (2*d*(A + B*Log[e*((a + b*x)/(c + d*x))^n] )^2*Log[1 - (b*(c + d*x))/(d*(a + b*x))])/b^3 + (2*B^2*d*n^2*PolyLog[2, (d *(a + b*x))/(b*(c + d*x))])/b^3 + (4*B*d*n*(A + B*Log[e*((a + b*x)/(c + d* x))^n])*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))])/b^3 + (4*B^2*d*n^2*PolyLo g[3, (b*(c + d*x))/(d*(a + b*x))])/b^3))/g^2
3.2.73.3.1 Defintions of rubi rules used
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]))
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]
\[\int \frac {\left (d i x +c i \right )^{2} {\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{2}}{\left (b g x +a g \right )^{2}}d x\]
\[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\int { \frac {{\left (d i x + c i\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}{{\left (b g x + a g\right )}^{2}} \,d x } \]
integrate((d*i*x+c*i)^2*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^2,x , algorithm="fricas")
integral((A^2*d^2*i^2*x^2 + 2*A^2*c*d*i^2*x + A^2*c^2*i^2 + (B^2*d^2*i^2*x ^2 + 2*B^2*c*d*i^2*x + B^2*c^2*i^2)*log(e*((b*x + a)/(d*x + c))^n)^2 + 2*( A*B*d^2*i^2*x^2 + 2*A*B*c*d*i^2*x + A*B*c^2*i^2)*log(e*((b*x + a)/(d*x + c ))^n))/(b^2*g^2*x^2 + 2*a*b*g^2*x + a^2*g^2), x)
Timed out. \[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\text {Timed out} \]
\[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\int { \frac {{\left (d i x + c i\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}{{\left (b g x + a g\right )}^{2}} \,d x } \]
integrate((d*i*x+c*i)^2*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^2,x , algorithm="maxima")
-2*A*B*c^2*i^2*n*(1/(b^2*g^2*x + a*b*g^2) + d*log(b*x + a)/((b^2*c - a*b*d )*g^2) - d*log(d*x + c)/((b^2*c - a*b*d)*g^2)) - A^2*(a^2/(b^4*g^2*x + a*b ^3*g^2) - x/(b^2*g^2) + 2*a*log(b*x + a)/(b^3*g^2))*d^2*i^2 + 2*A^2*c*d*i^ 2*(a/(b^3*g^2*x + a*b^2*g^2) + log(b*x + a)/(b^2*g^2)) - 2*A*B*c^2*i^2*log (e*(b*x/(d*x + c) + a/(d*x + c))^n)/(b^2*g^2*x + a*b*g^2) - A^2*c^2*i^2/(b ^2*g^2*x + a*b*g^2) + (B^2*b^2*d^2*i^2*x^2 + B^2*a*b*d^2*i^2*x - (b^2*c^2* i^2 - 2*a*b*c*d*i^2 + a^2*d^2*i^2)*B^2 + 2*((b^2*c*d*i^2 - a*b*d^2*i^2)*B^ 2*x + (a*b*c*d*i^2 - a^2*d^2*i^2)*B^2)*log(b*x + a))*log((d*x + c)^n)^2/(b ^4*g^2*x + a*b^3*g^2) - integrate(-(B^2*b^3*c^3*i^2*log(e)^2 + (B^2*b^3*d^ 3*i^2*log(e)^2 + 2*A*B*b^3*d^3*i^2*log(e))*x^3 + 3*(B^2*b^3*c*d^2*i^2*log( e)^2 + 2*A*B*b^3*c*d^2*i^2*log(e))*x^2 + (B^2*b^3*d^3*i^2*x^3 + 3*B^2*b^3* c*d^2*i^2*x^2 + 3*B^2*b^3*c^2*d*i^2*x + B^2*b^3*c^3*i^2)*log((b*x + a)^n)^ 2 + (3*B^2*b^3*c^2*d*i^2*log(e)^2 + 4*A*B*b^3*c^2*d*i^2*log(e))*x + 2*(B^2 *b^3*c^3*i^2*log(e) + (B^2*b^3*d^3*i^2*log(e) + A*B*b^3*d^3*i^2)*x^3 + 3*( B^2*b^3*c*d^2*i^2*log(e) + A*B*b^3*c*d^2*i^2)*x^2 + (3*B^2*b^3*c^2*d*i^2*l og(e) + 2*A*B*b^3*c^2*d*i^2)*x)*log((b*x + a)^n) - 2*((A*B*b^3*d^3*i^2 + ( i^2*n + i^2*log(e))*B^2*b^3*d^3)*x^3 - (a*b^2*c^2*d*i^2*n - 2*a^2*b*c*d^2* i^2*n + a^3*d^3*i^2*n - b^3*c^3*i^2*log(e))*B^2 + (3*A*B*b^3*c*d^2*i^2 + ( 2*a*b^2*d^3*i^2*n + 3*b^3*c*d^2*i^2*log(e))*B^2)*x^2 + (2*A*B*b^3*c^2*d*i^ 2 + (2*a*b^2*c*d^2*i^2*n - (i^2*n - 3*i^2*log(e))*b^3*c^2*d)*B^2)*x + 2...
\[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\int { \frac {{\left (d i x + c i\right )}^{2} {\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}{{\left (b g x + a g\right )}^{2}} \,d x } \]
integrate((d*i*x+c*i)^2*(A+B*log(e*((b*x+a)/(d*x+c))^n))^2/(b*g*x+a*g)^2,x , algorithm="giac")
Timed out. \[ \int \frac {(c i+d i x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(a g+b g x)^2} \, dx=\int \frac {{\left (c\,i+d\,i\,x\right )}^2\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^2}{{\left (a\,g+b\,g\,x\right )}^2} \,d x \]