3.59 \(\int \frac{\log (e (f (a+b x)^p (c+d x)^q)^r)}{x} \, dx\)

Optimal. Leaf size=81 \[ -p r \text{PolyLog}\left (2,-\frac{b x}{a}\right )-q r \text{PolyLog}\left (2,-\frac{d x}{c}\right )+\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-p r \log (x) \log \left (\frac{b x}{a}+1\right )-q r \log (x) \log \left (\frac{d x}{c}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.0657538, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2494, 2317, 2391} \[ -p r \text{PolyLog}\left (2,-\frac{b x}{a}\right )-q r \text{PolyLog}\left (2,-\frac{d x}{c}\right )+\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-p r \log (x) \log \left (\frac{b x}{a}+1\right )-q r \log (x) \log \left (\frac{d x}{c}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/x,x]

[Out]

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

Rule 2494

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[(Log[g + h*x]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/h, x] + (-Dist[(b*p*r)/h, Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[(d*q*r)/h, Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 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 (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx &=\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-(b p r) \int \frac{\log (x)}{a+b x} \, dx-(d q r) \int \frac{\log (x)}{c+d x} \, dx\\ &=-p r \log (x) \log \left (1+\frac{b x}{a}\right )+\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-q r \log (x) \log \left (1+\frac{d x}{c}\right )+(p r) \int \frac{\log \left (1+\frac{b x}{a}\right )}{x} \, dx+(q r) \int \frac{\log \left (1+\frac{d x}{c}\right )}{x} \, dx\\ &=-p r \log (x) \log \left (1+\frac{b x}{a}\right )+\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-q r \log (x) \log \left (1+\frac{d x}{c}\right )-p r \text{Li}_2\left (-\frac{b x}{a}\right )-q r \text{Li}_2\left (-\frac{d x}{c}\right )\\ \end{align*}

Mathematica [A]  time = 0.0630908, size = 78, normalized size = 0.96 \[ -p r \text{PolyLog}\left (2,-\frac{b x}{a}\right )-q r \text{PolyLog}\left (2,-\frac{d x}{c}\right )+\log (x) \left (\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-p r \log \left (\frac{b x}{a}+1\right )-q r \log \left (\frac{d x}{c}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.123, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( e \left ( f \left ( bx+a \right ) ^{p} \left ( dx+c \right ) ^{q} \right ) ^{r} \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20753, size = 170, normalized size = 2.1 \begin{align*} -\frac{{\left (f p \log \left (b x + a\right ) + f q \log \left (d x + c\right )\right )} r \log \left (x\right )}{f} + \log \left (\left ({\left (b x + a\right )}^{p}{\left (d x + c\right )}^{q} f\right )^{r} e\right ) \log \left (x\right ) + \frac{{\left ({\left (\log \left (b x + a\right ) \log \left (-\frac{b x + a}{a} + 1\right ) +{\rm Li}_2\left (\frac{b x + a}{a}\right )\right )} f p +{\left (\log \left (d x + c\right ) \log \left (-\frac{d x + c}{c} + 1\right ) +{\rm Li}_2\left (\frac{d x + c}{c}\right )\right )} f q\right )} r}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(((b*x + a)^p*(d*x + c)^q*f)^r*e)/x, 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(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/x,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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