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

Optimal. Leaf size=131 \[ \frac {2 B n \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{b}-\frac {\log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{b}+\frac {2 B^2 n^2 \text {Li}_3\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]

[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

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Rubi [A]  time = 0.50, antiderivative size = 227, normalized size of antiderivative = 1.73, number of steps used = 10, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {6742, 2488, 2411, 2343, 2333, 2315, 2506, 6610} \[ \frac {2 A B n \text {PolyLog}\left (2,\frac {b c-a d}{d (a+b x)}+1\right )}{b}+\frac {2 B^2 n \text {PolyLog}\left (2,\frac {b c-a d}{d (a+b x)}+1\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{b}+\frac {2 B^2 n^2 \text {PolyLog}\left (3,\frac {b c-a d}{d (a+b x)}+1\right )}{b}+\frac {A^2 \log (a+b x)}{b}-\frac {2 A B \log \left (-\frac {b c-a d}{d (a+b x)}\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{b}-\frac {B^2 \log \left (-\frac {b c-a d}{d (a+b x)}\right ) \log ^2\left (e (a+b x)^n (c+d x)^{-n}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2333

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

Rule 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Dist[1/n, Subst[Int[(a
+ b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2488

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)/((g_.) + (h_.)*(x_)),
x_Symbol] :> -Simp[(Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/h, x] + Dist[(p
*r*s*(b*c - a*d))/h, Int[(Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a
+ b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q,
0] && EqQ[b*g - a*h, 0] && IGtQ[s, 0]

Rule 2506

Int[Log[v_]*Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_), x_Symbo
l] :> With[{g = Simplify[((v - 1)*(c + d*x))/(a + b*x)], h = Simplify[u*(a + b*x)*(c + d*x)]}, -Simp[(h*PolyLo
g[2, 1 - v]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(b*c - a*d), x] + Dist[h*p*r*s, Int[(PolyLog[2, 1 - v]*Log
[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{g, h}, x]] /; FreeQ[{a, b,
c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && EqQ[p + q, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [B]  time = 0.19, size = 269, normalized size = 2.05 \[ \frac {A^2 \log (a+b x)-2 A B \log \left (\frac {a d-b c}{d (a+b x)}\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )+2 A B n \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )-A B n \log ^2\left (\frac {a d-b c}{d (a+b x)}\right )-2 A B n \log \left (\frac {a d-b c}{d (a+b x)}\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )+2 B^2 n \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )-B^2 \log \left (\frac {a d-b c}{d (a+b x)}\right ) \log ^2\left (e (a+b x)^n (c+d x)^{-n}\right )+2 B^2 n^2 \text {Li}_3\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[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

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B^{2} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )^{2} + 2 \, A B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A^{2}}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{2}}{b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 2.57, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )+A \right )^{2}}{b x +a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {B^{2} \log \left (b x + a\right ) \log \left ({\left (d x + c\right )}^{n}\right )^{2}}{b} + \frac {A^{2} \log \left (b x + a\right )}{b} - \int -\frac {B^{2} b c \log \relax (e)^{2} + 2 \, A B b c \log \relax (e) + {\left (B^{2} b d x + B^{2} b c\right )} \log \left ({\left (b x + a\right )}^{n}\right )^{2} + {\left (B^{2} b d \log \relax (e)^{2} + 2 \, A B b d \log \relax (e)\right )} x + 2 \, {\left (B^{2} b c \log \relax (e) + A B b c + {\left (B^{2} b d \log \relax (e) + A B b d\right )} x\right )} \log \left ({\left (b x + a\right )}^{n}\right ) - 2 \, {\left (B^{2} b c \log \relax (e) + A B b c + {\left (B^{2} b d \log \relax (e) + A B b d\right )} x + {\left (B^{2} b d n x + B^{2} a d n\right )} \log \left (b x + a\right ) + {\left (B^{2} b d x + B^{2} b c\right )} \log \left ({\left (b x + a\right )}^{n}\right )\right )} \log \left ({\left (d x + c\right )}^{n}\right )}{b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^2}{a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \log {\left (e \left (a + b x\right )^{n} \left (c + d x\right )^{- n} \right )}\right )^{2}}{a + b x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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