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

Optimal. Leaf size=79 \[ \frac{B n \text{PolyLog}\left (2,\frac{b c-a d}{d (a+b x)}+1\right )}{b}-\frac{\log \left (-\frac{b c-a d}{d (a+b x)}\right ) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{b} \]

[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

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Rubi [A]  time = 0.26782, antiderivative size = 87, normalized size of antiderivative = 1.1, 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 6742

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

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.0972473, size = 129, normalized size = 1.63 \[ \frac{2 B n \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )+2 A \log (a+b x)-2 B \log \left (\frac{a d-b c}{d (a+b x)}\right ) \left (\log \left (e (a+b x)^n (c+d x)^{-n}\right )+n \log \left (\frac{b (c+d x)}{b c-a d}\right )\right )-B n \log ^2\left (\frac{a d-b c}{d (a+b x)}\right )}{2 b} \]

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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Maple [C]  time = 1.534, size = 523, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

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/((d*x+c)^n))*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+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*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/((d*x+c)^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., size = 0, normalized size = 0. \begin{align*} 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 \left (e\right ) + b c \log \left (e\right ) -{\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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)