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

Optimal. Leaf size=97 \[ -\frac{n \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d}+\frac{n \text{PolyLog}\left (2,\frac{b x}{a}+1\right )}{d}-\frac{\log \left (c (a+b x)^n\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d}+\frac{\log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^n\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.123538, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {1593, 36, 29, 31, 2416, 2394, 2315, 2393, 2391} \[ -\frac{n \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d}+\frac{n \text{PolyLog}\left (2,\frac{b x}{a}+1\right )}{d}-\frac{\log \left (c (a+b x)^n\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d}+\frac{\log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^n\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

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

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 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

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

Rubi steps

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

Mathematica [A]  time = 0.0217651, size = 98, normalized size = 1.01 \[ -\frac{n \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d}+\frac{n \text{PolyLog}\left (2,\frac{a+b x}{a}\right )}{d}-\frac{\log \left (c (a+b x)^n\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d}+\frac{\log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^n\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.519, size = 420, normalized size = 4.3 \begin{align*} -{\frac{\ln \left ( ex+d \right ) \ln \left ( \left ( bx+a \right ) ^{n} \right ) }{d}}+{\frac{\ln \left ( \left ( bx+a \right ) ^{n} \right ) \ln \left ( x \right ) }{d}}-{\frac{n}{d}{\it dilog} \left ({\frac{bx+a}{a}} \right ) }-{\frac{n\ln \left ( x \right ) }{d}\ln \left ({\frac{bx+a}{a}} \right ) }+{\frac{n}{d}{\it dilog} \left ({\frac{b \left ( ex+d \right ) +ae-bd}{ae-bd}} \right ) }+{\frac{n\ln \left ( ex+d \right ) }{d}\ln \left ({\frac{b \left ( ex+d \right ) +ae-bd}{ae-bd}} \right ) }-{\frac{{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ) \right ) ^{3}\ln \left ( x \right ) }{d}}-{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{n} \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ) \right ) ^{2}\ln \left ( ex+d \right ) }{d}}+{\frac{{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ) \right ) ^{3}\ln \left ( ex+d \right ) }{d}}+{\frac{{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) \ln \left ( x \right ) }{d}}-{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{n} \right ){\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ){\it csgn} \left ( ic \right ) \ln \left ( x \right ) }{d}}+{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{n} \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ) \right ) ^{2}\ln \left ( x \right ) }{d}}+{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{n} \right ){\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ){\it csgn} \left ( ic \right ) \ln \left ( ex+d \right ) }{d}}-{\frac{{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) \ln \left ( ex+d \right ) }{d}}-{\frac{\ln \left ( c \right ) \ln \left ( ex+d \right ) }{d}}+{\frac{\ln \left ( c \right ) \ln \left ( x \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.33211, size = 166, normalized size = 1.71 \begin{align*} -b n{\left (\frac{\log \left (\frac{b x}{a} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a}\right )}{b d} - \frac{\log \left (e x + d\right ) \log \left (-\frac{b e x + b d}{b d - a e} + 1\right ) +{\rm Li}_2\left (\frac{b e x + b d}{b d - a e}\right )}{b d}\right )} -{\left (\frac{\log \left (e x + d\right )}{d} - \frac{\log \left (x\right )}{d}\right )} \log \left ({\left (b x + a\right )}^{n} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*n*((log(b*x/a + 1)*log(x) + dilog(-b*x/a))/(b*d) - (log(e*x + d)*log(-(b*e*x + b*d)/(b*d - a*e) + 1) + dilo
g((b*e*x + b*d)/(b*d - a*e)))/(b*d)) - (log(e*x + d)/d - log(x)/d)*log((b*x + a)^n*c)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (b x + a\right )}^{n} c\right )}{e x^{2} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (b x + a\right )}^{n} c\right )}{e x^{2} + d x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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