Integrand size = 49, antiderivative size = 123 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=-\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \log \left (1-\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{(b f-a g) h}+\frac {B n \operatorname {PolyLog}\left (2,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{(b f-a g) h} \]
-(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))*ln(1-(-a*g+b*f)*(d*x+c)/(-c*g+d*f)/(b*x +a))/(-a*g+b*f)/h+B*n*polylog(2,(-a*g+b*f)*(d*x+c)/(-c*g+d*f)/(b*x+a))/(-a *g+b*f)/h
Leaf count is larger than twice the leaf count of optimal. \(303\) vs. \(2(123)=246\).
Time = 0.18 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.46 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=-\frac {-2 A \log (a+b x)+B n \log ^2(a+b x)-2 B n \log (a+b x) \log (c+d x)+2 B n \log \left (\frac {d (a+b x)}{-b c+a d}\right ) \log (c+d x)-2 B \log (a+b x) \log \left (e (a+b x)^n (c+d x)^{-n}\right )+2 A \log (f+g x)-2 B n \log (a+b x) \log (f+g x)+2 B n \log (c+d x) \log (f+g x)+2 B \log \left (e (a+b x)^n (c+d x)^{-n}\right ) \log (f+g x)+2 B n \log (a+b x) \log \left (\frac {b (f+g x)}{b f-a g}\right )-2 B n \log (c+d x) \log \left (\frac {d (f+g x)}{d f-c g}\right )+2 B n \operatorname {PolyLog}\left (2,\frac {g (a+b x)}{-b f+a g}\right )+2 B n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )-2 B n \operatorname {PolyLog}\left (2,\frac {g (c+d x)}{-d f+c g}\right )}{(2 b f-2 a g) h} \]
Integrate[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(a*f*h + b*g*h*x^2 + h* (b*f*x + a*g*x)),x]
-((-2*A*Log[a + b*x] + B*n*Log[a + b*x]^2 - 2*B*n*Log[a + b*x]*Log[c + d*x ] + 2*B*n*Log[(d*(a + b*x))/(-(b*c) + a*d)]*Log[c + d*x] - 2*B*Log[a + b*x ]*Log[(e*(a + b*x)^n)/(c + d*x)^n] + 2*A*Log[f + g*x] - 2*B*n*Log[a + b*x] *Log[f + g*x] + 2*B*n*Log[c + d*x]*Log[f + g*x] + 2*B*Log[(e*(a + b*x)^n)/ (c + d*x)^n]*Log[f + g*x] + 2*B*n*Log[a + b*x]*Log[(b*(f + g*x))/(b*f - a* g)] - 2*B*n*Log[c + d*x]*Log[(d*(f + g*x))/(d*f - c*g)] + 2*B*n*PolyLog[2, (g*(a + b*x))/(-(b*f) + a*g)] + 2*B*n*PolyLog[2, (b*(c + d*x))/(b*c - a*d )] - 2*B*n*PolyLog[2, (g*(c + d*x))/(-(d*f) + c*g)])/((2*b*f - 2*a*g)*h))
Time = 0.62 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {2973, 2976, 2026, 2779, 2838}
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.
\(\displaystyle \int \frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{h (a g x+b f x)+a f h+b g h x^2} \, dx\) |
\(\Big \downarrow \) 2973 |
\(\displaystyle \int \frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{h (a g x+b f x)+a f h+b g h x^2}dx\) |
\(\Big \downarrow \) 2976 |
\(\displaystyle (b c-a d) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{\frac {(b c-a d) (b f-a g) h (a+b x)}{c+d x}-\frac {(b c-a d) (d f-c g) h (a+b x)^2}{(c+d x)^2}}d\frac {a+b x}{c+d x}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle (b c-a d) \int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(a+b x) \left ((b c-a d) (b f-a g) h-\frac {(b c-a d) (d f-c g) h (a+b x)}{c+d x}\right )}d\frac {a+b x}{c+d x}\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle (b c-a d) \left (\frac {B n \int \frac {(c+d x) \log \left (1-\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{a+b x}d\frac {a+b x}{c+d x}}{h (b c-a d) (b f-a g)}-\frac {\log \left (1-\frac {(c+d x) (b f-a g)}{(a+b x) (d f-c g)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{h (b c-a d) (b f-a g)}\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle (b c-a d) \left (\frac {B n \operatorname {PolyLog}\left (2,\frac {(b f-a g) (c+d x)}{(d f-c g) (a+b x)}\right )}{h (b c-a d) (b f-a g)}-\frac {\log \left (1-\frac {(c+d x) (b f-a g)}{(a+b x) (d f-c g)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{h (b c-a d) (b f-a g)}\right )\) |
(b*c - a*d)*(-(((A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[1 - ((b*f - a*g )*(c + d*x))/((d*f - c*g)*(a + b*x))])/((b*c - a*d)*(b*f - a*g)*h)) + (B*n *PolyLog[2, ((b*f - a*g)*(c + d*x))/((d*f - c*g)*(a + b*x))])/((b*c - a*d) *(b*f - a*g)*h))
3.3.61.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; Fr eeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] && !Intege rQ[n]
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))^(p_.)*(P2x_)^(m_.), x_Symbol] :> With[{f = Coeff[P2x, x, 0], g = Coef f[P2x, x, 1], h = Coeff[P2x, x, 2]}, Simp[(b*c - a*d) Subst[Int[(b^2*f - a*b*g + a^2*h - (2*b*d*f - b*c*g - a*d*g + 2*a*c*h)*x + (d^2*f - c*d*g + c^ 2*h)*x^2)^m*((A + B*Log[e*x^n])^p/(b - d*x)^(2*(m + 1))), x], x, (a + b*x)/ (c + d*x)], x]] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && PolyQ[P2x, x, 2] & & NeQ[b*c - a*d, 0] && IntegerQ[m] && IGtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 25.90 (sec) , antiderivative size = 1447, normalized size of antiderivative = 11.76
int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/(a*f*h+b*g*h*x^2+h*(a*g*x+b*f*x)),x, method=_RETURNVERBOSE)
1/2*I/h/(a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csg n(I*(b*x+a)^n/((d*x+c)^n))-1/2*I/h/(a*g-b*f)*ln(g*x+f)*B*Pi*csgn(I*e)*csgn (I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-1/2*I/h/(a*g-b*f )*ln(g*x+f)*B*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/(( d*x+c)^n))+1/2*I/h/(a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I*e)*csgn(I*(b*x+a)^n/((d *x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-1/h/(a*g-b*f)*ln(b*x+a)*A+1/h/(a *g-b*f)*ln(g*x+f)*A+1/2*I/h/(a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I*(b*x+a)^n/((d* x+c)^n))^3+1/2*I/h/(a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n )^3-1/2*I/h/(a*g-b*f)*ln(g*x+f)*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/2*I /h/(a*g-b*f)*ln(g*x+f)*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/2*I/h/(a*g -b*f)*ln(b*x+a)*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/2*I/h/( a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2- 1/2*I/h/(a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d* x+c)^n))^2+1/2*I/h/(a*g-b*f)*ln(g*x+f)*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))* csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/2*I/h/(a*g-b*f)*ln(b*x+a)*B*Pi*csgn(I* (b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I/h/(a*g-b*f) *ln(g*x+f)*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I/h/(a*g-b *f)*ln(g*x+f)*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I /h/(a*g-b*f)*ln(g*x+f)*B*Pi*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^ n))^2-1/h*B*n/(a*g-b*f)*ln(g*x+f)*ln((b*(g*x+f)+a*g-b*f)/(a*g-b*f))+1/h...
\[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=\int { \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A}{b g h x^{2} + a f h + {\left (b f x + a g x\right )} h} \,d x } \]
integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(a*f*h+b*g*h*x^2+h*(a*g*x+b*f *x)),x, algorithm="fricas")
Timed out. \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=\int { \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A}{b g h x^{2} + a f h + {\left (b f x + a g x\right )} h} \,d x } \]
integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(a*f*h+b*g*h*x^2+h*(a*g*x+b*f *x)),x, algorithm="maxima")
A*(log(b*x + a)/((b*f - a*g)*h) - log(g*x + f)/((b*f - a*g)*h)) - B*integr ate(-(log((b*x + a)^n) - log((d*x + c)^n) + log(e))/(b*g*h*x^2 + a*f*h + ( b*f*h + a*g*h)*x), x)
\[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=\int { \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A}{b g h x^{2} + a f h + {\left (b f x + a g x\right )} h} \,d x } \]
integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(a*f*h+b*g*h*x^2+h*(a*g*x+b*f *x)),x, algorithm="giac")
integrate((B*log((b*x + a)^n*e/(d*x + c)^n) + A)/(b*g*h*x^2 + a*f*h + (b*f *x + a*g*x)*h), x)
Timed out. \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{a f h+b g h x^2+h (b f x+a g x)} \, dx=\int \frac {A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}{h\,\left (a\,g\,x+b\,f\,x\right )+a\,f\,h+b\,g\,h\,x^2} \,d x \]