3.180 \(\int \frac{\log (c (a+b x)^p)}{d+e x} \, dx\)

Optimal. Leaf size=58 \[ \frac{p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{e}+\frac{\log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.0504011, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2394, 2393, 2391} \[ \frac{p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{e}+\frac{\log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x)^p]/(d + e*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0035091, size = 57, normalized size = 0.98 \[ \frac{p \text{PolyLog}\left (2,\frac{e (a+b x)}{a e-b d}\right )}{e}+\frac{\log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.633, size = 242, normalized size = 4.2 \begin{align*}{\frac{\ln \left ( ex+d \right ) \ln \left ( \left ( bx+a \right ) ^{p} \right ) }{e}}-{\frac{p}{e}{\it dilog} \left ({\frac{b \left ( ex+d \right ) +ae-bd}{ae-bd}} \right ) }-{\frac{p\ln \left ( ex+d \right ) }{e}\ln \left ({\frac{b \left ( ex+d \right ) +ae-bd}{ae-bd}} \right ) }-{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( bx+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) }{e}}+{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{2}}{e}}+{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{2}}{e}}-{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{3}}{e}}+{\frac{\ln \left ( ex+d \right ) \ln \left ( c \right ) }{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.06048, size = 159, normalized size = 2.74 \begin{align*} \frac{b p{\left (\frac{\log \left (b x + a\right ) \log \left (e x + d\right )}{b} - \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}\right )}}{e} - \frac{p \log \left (b x + a\right ) \log \left (e x + d\right )}{e} + \frac{\log \left ({\left (b x + a\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{p} c\right )}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x + a)^p*c)/(e*x + d), 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 )^{p} \right )}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(a + b*x)**p)/(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 )}^{p} c\right )}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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