3.2.51 \(\int \frac {A+B \log (e (a+b x)^n (c+d x)^{-n})}{a+b x} \, dx\) [151]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2542, 2458, 2378, 2370, 2352} \begin {gather*} \frac {B n \text {PolyLog}\left (2,\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Log[-((b*c - a*d)/(d*(a + b*x)))]*(A + B*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 2352

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

Rule 2370

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 2378

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 2458

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 2542

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))/((f_.) + (g_.)*(x_)), x_S
ymbol] :> Simp[(-Log[-(b*c - a*d)/(d*(a + b*x))])*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/g), x] + Dist[B*n*
((b*c - a*d)/g), Int[Log[-(b*c - a*d)/(d*(a + b*x))]/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f,
g, A, B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0]

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) \text {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) \text {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) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 95, normalized size = 1.20 \begin {gather*} \frac {\log (a+b x) \left (-B n \log (a+b x)+2 \left (A+B n \log \left (\frac {b (c+d x)}{b c-a d}\right )+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )\right )+2 B n \text {Li}_2\left (\frac {d (a+b x)}{-b c+a d}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.50, size = 523, normalized size = 6.62

method	result	size
risch	\(-\frac {B \ln \left (b x +a \right ) \ln \left (\left (d x +c \right )^{n}\right )}{b}+\frac {B n \dilog \left (\frac {-a d +c b +\left (b x +a \right ) d}{-a d +c b}\right )}{b}+\frac {B n \ln \left (b x +a \right ) \ln \left (\frac {-a d +c b +\left (b x +a \right ) d}{-a d +c b}\right )}{b}+\frac {i B \ln \left (b x +a \right ) \pi \,\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}}{2 b}+\frac {i B \ln \left (b x +a \right ) \pi \,\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}}{2 b}+\frac {i B \ln \left (b x +a \right ) \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}}{2 b}+\frac {i B \ln \left (b x +a \right ) \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}}{2 b}+\frac {A \ln \left (b x +a \right )}{b}+\frac {B \ln \left (b x +a \right ) \ln \left (e \right )}{b}+\frac {B \ln \left (\left (b x +a \right )^{n}\right )^{2}}{2 b n}-\frac {i B \ln \left (b x +a \right ) \pi \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}}{2 b}-\frac {i B \ln \left (b x +a \right ) \pi \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{3}}{2 b}-\frac {i B \ln \left (b x +a \right ) \pi \,\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 )}{2 b}-\frac {i B \ln \left (b x +a \right ) \pi \,\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 (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )}{2 b}\)	\(523\)

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,method=_RETURNVERBOSE)

[Out]

-B/b*ln(b*x+a)*ln((d*x+c)^n)+1/b*B*n*dilog((-a*d+c*b+(b*x+a)*d)/(-a*d+b*c))+1/b*B*n*ln(b*x+a)*ln((-a*d+c*b+(b*
x+a)*d)/(-a*d+b*c))+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*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/((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*(b*x+a)^n)*csgn(I/((d*x+
c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 + b*c - (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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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)

________________________________________________________________________________________