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

Optimal. Leaf size=79 \[ \frac {B n \text {Li}_2\left (\frac {b c-a d}{d (a+b x)}+1\right )}{b}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 87, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {6742, 2488, 2411, 2343, 2333, 2315} \[ \frac {B n \text {PolyLog}\left (2,\frac {b c-a d}{d (a+b x)}+1\right )}{b}+\frac {A \log (a+b x)}{b}-\frac {B \log \left (-\frac {b c-a d}{d (a+b x)}\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*Log[a + b*x])/b - (B*Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[(e*(a + b*x)^n)/(c + d*x)^n])/b + (B*n*PolyLog[2
, 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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 129, normalized size = 1.63 \[ \frac {2 A \log (a+b x)-2 B \log \left (\frac {a d-b c}{d (a+b x)}\right ) \left (\log \left (e (a+b x)^n (c+d x)^{-n}\right )+n \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )+2 B n \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )-B n \log ^2\left (\frac {a d-b c}{d (a+b x)}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.24, size = 523, normalized size = 6.62 \[ -\frac {i \pi B \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \ln \left (b x +a \right )}{2 b}+\frac {i \pi B \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (b x +a \right )}{2 b}-\frac {i \pi B \,\mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \ln \left (b x +a \right )}{2 b}+\frac {i \pi B \,\mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (b x +a \right )}{2 b}+\frac {i \pi B \,\mathrm {csgn}\left (i \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (b x +a \right )}{2 b}-\frac {i \pi B \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3} \ln \left (b x +a \right )}{2 b}+\frac {i \pi B \,\mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \ln \left (b x +a \right )}{2 b}-\frac {i \pi B \mathrm {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3} \ln \left (b x +a \right )}{2 b}+\frac {B n \ln \left (\frac {-a d +b c +\left (b x +a \right ) d}{-a d +b c}\right ) \ln \left (b x +a \right )}{b}+\frac {B n \dilog \left (\frac {-a d +b c +\left (b x +a \right ) d}{-a d +b c}\right )}{b}+\frac {B \ln \relax (e ) \ln \left (b x +a \right )}{b}-\frac {B \ln \left (\left (d x +c \right )^{n}\right ) \ln \left (b x +a \right )}{b}+\frac {A \ln \left (b x +a \right )}{b}+\frac {B \ln \left (\left (b x +a \right )^{n}\right )^{2}}{2 b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B/b*ln(b*x+a)*ln((d*x+c)^n)+1/b*B*n*dilog((-a*d+b*c+d*(b*x+a))/(-a*d+b*c))+1/b*B*n*ln(b*x+a)*ln((-a*d+b*c+d*(
b*x+a))/(-a*d+b*c))+1/2*I/b*B*ln(b*x+a)*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I/b*B*ln(b*x+a)*Pi*
csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I/b*B*ln(b*x+a)*Pi*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/
((d*x+c)^n))^2+1/2*I/b*B*ln(b*x+a)*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+A*ln(b*x
+a)/b+1/b*B*ln(b*x+a)*ln(e)+1/2/b*B/n*ln((b*x+a)^n)^2-1/2*I/b*B*ln(b*x+a)*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1
/2*I/b*B*ln(b*x+a)*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/2*I/b*B*ln(b*x+a)*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/b*B*ln(b*x+a)*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(
b*x+a)^n/((d*x+c)^n))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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