\(\int \frac {(A+B \log (e (a+b x)^n (c+d x)^{-n}))^2}{a+b x} \, dx\) [159]

Optimal result

Integrand size = 33, antiderivative size = 131 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=-\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B^2 n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]

[Out]

-(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2*ln(1-b*(d*x+c)/d/(b*x+a))/b+2*B*n*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))*polyl
og(2,b*(d*x+c)/d/(b*x+a))/b+2*B^2*n^2*polylog(3,b*(d*x+c)/d/(b*x+a))/b

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2573, 2549, 2379, 2421, 6724} \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\frac {2 B n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{b}-\frac {\log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{b}+\frac {2 B^2 n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]

[In]

Int[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2/(a + b*x),x]

[Out]

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

Rule 2379

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] + Dist[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]

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c*x^n])^p/m), x] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2549

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/b)^m, Subst[Int[x^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + 2)), x]
, x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m,
p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rule 2573

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)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{a+b x} \, dx,e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\text {Subst}\left (\int \frac {\left (A+B \log \left (e x^n\right )\right )^2}{x (b-d x)} \, dx,x,\frac {a+b x}{c+d x}\right ),e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = -\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\text {Subst}\left (\frac {(2 B n) \text {Subst}\left (\int \frac {\log \left (1-\frac {b}{d x}\right ) \left (A+B \log \left (e x^n\right )\right )}{x} \, dx,x,\frac {a+b x}{c+d x}\right )}{b},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = -\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b}-\text {Subst}\left (\frac {\left (2 B^2 n^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b}{d x}\right )}{x} \, dx,x,\frac {a+b x}{c+d x}\right )}{b},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = -\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b}+\frac {2 B^2 n^2 \text {Li}_3\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(269\) vs. \(2(131)=262\).

Time = 0.14 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.05 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\frac {-A B n \log ^2\left (\frac {-b c+a d}{d (a+b x)}\right )+A^2 \log (a+b x)-2 A B n \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )-2 A B \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )-B^2 \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log ^2\left (e (a+b x)^n (c+d x)^{-n}\right )+2 A B n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )+2 B^2 n \log \left (e (a+b x)^n (c+d x)^{-n}\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )+2 B^2 n^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]

[In]

Integrate[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2/(a + b*x),x]

[Out]

(-(A*B*n*Log[(-(b*c) + a*d)/(d*(a + b*x))]^2) + A^2*Log[a + b*x] - 2*A*B*n*Log[(-(b*c) + a*d)/(d*(a + b*x))]*L
og[(b*(c + d*x))/(b*c - a*d)] - 2*A*B*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(e*(a + b*x)^n)/(c + d*x)^n] - B^2
*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(e*(a + b*x)^n)/(c + d*x)^n]^2 + 2*A*B*n*PolyLog[2, (d*(a + b*x))/(-(b*
c) + a*d)] + 2*B^2*n*Log[(e*(a + b*x)^n)/(c + d*x)^n]*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))] + 2*B^2*n^2*Poly
Log[3, (b*(c + d*x))/(d*(a + b*x))])/b

Maple [F]

\[\int \frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{2}}{b x +a}d x\]

[In]

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/(b*x+a),x)

[Out]

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/(b*x+a),x)

Fricas [F]

\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a} \,d x } \]

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2/(b*x+a),x, algorithm="fricas")

[Out]

integral((B^2*log((b*x + a)^n*e/(d*x + c)^n)^2 + 2*A*B*log((b*x + a)^n*e/(d*x + c)^n) + A^2)/(b*x + a), x)

Sympy [F]

\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int \frac {\left (A + B \log {\left (e \left (a + b x\right )^{n} \left (c + d x\right )^{- n} \right )}\right )^{2}}{a + b x}\, dx \]

[In]

integrate((A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**2/(b*x+a),x)

[Out]

Integral((A + B*log(e*(a + b*x)**n/(c + d*x)**n))**2/(a + b*x), x)

Maxima [F]

\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a} \,d x } \]

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2/(b*x+a),x, algorithm="maxima")

[Out]

B^2*log(b*x + a)*log((d*x + c)^n)^2/b + A^2*log(b*x + a)/b - integrate(-(B^2*b*c*log(e)^2 + 2*A*B*b*c*log(e) +
(B^2*b*d*x + B^2*b*c)*log((b*x + a)^n)^2 + (B^2*b*d*log(e)^2 + 2*A*B*b*d*log(e))*x + 2*(B^2*b*c*log(e) + A*B*
b*c + (B^2*b*d*log(e) + A*B*b*d)*x)*log((b*x + a)^n) - 2*(B^2*b*c*log(e) + A*B*b*c + (B^2*b*d*log(e) + A*B*b*d
)*x + (B^2*b*d*n*x + B^2*a*d*n)*log(b*x + a) + (B^2*b*d*x + B^2*b*c)*log((b*x + a)^n))*log((d*x + c)^n))/(b^2*
d*x^2 + a*b*c + (b^2*c + a*b*d)*x), x)

Giac [F]

\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a} \,d x } \]

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2/(b*x+a),x, algorithm="giac")

[Out]

integrate((B*log((b*x + a)^n*e/(d*x + c)^n) + A)^2/(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{a+b x} \, dx=\int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^2}{a+b\,x} \,d x \]

[In]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^2/(a + b*x),x)

[Out]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^2/(a + b*x), x)