\(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{x} \, dx\) [59]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=-p r \log (x) \log \left (1+\frac {b x}{a}\right )+\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-q r \log (x) \log \left (1+\frac {d x}{c}\right )-p r \operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )-q r \operatorname {PolyLog}\left (2,-\frac {d x}{c}\right ) \]

[Out]

-p*r*ln(x)*ln(1+b*x/a)+ln(x)*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)-q*r*ln(x)*ln(1+d*x/c)-p*r*polylog(2,-b*x/a)-q*r*p
olylog(2,-d*x/c)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2580, 2354, 2438} \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=\log (x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-p r \operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )-p r \log (x) \log \left (\frac {b x}{a}+1\right )-q r \operatorname {PolyLog}\left (2,-\frac {d x}{c}\right )-q r \log (x) \log \left (\frac {d x}{c}+1\right ) \]

[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 2354

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 2438

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]

Rule 2580

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=\log (x) \left (-p r \log \left (1+\frac {b x}{a}\right )+\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )-q r \log \left (1+\frac {d x}{c}\right )\right )-p r \operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )-q r \operatorname {PolyLog}\left (2,-\frac {d x}{c}\right ) \]

[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)]

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.27

method	result	size
parts	\(\ln \left (x \right ) \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )-\frac {r \left (b f p \left (\frac {\operatorname {dilog}\left (\frac {b x +a}{a}\right )}{b}+\frac {\ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{b}\right )+d f q \left (\frac {\operatorname {dilog}\left (\frac {d x +c}{c}\right )}{d}+\frac {\ln \left (x \right ) \ln \left (\frac {d x +c}{c}\right )}{d}\right )\right )}{f}\)	\(103\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=\int { \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{x} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=\int \frac {\log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{x}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=-\frac {{\left (f p \log \left (b x + a\right ) + f q \log \left (d x + c\right )\right )} r \log \left (x\right )}{f} + \log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) \log \left (x\right ) + \frac {{\left ({\left (\log \left (b x + a\right ) \log \left (-\frac {b x + a}{a} + 1\right ) + {\rm Li}_2\left (\frac {b x + a}{a}\right )\right )} f p + {\left (\log \left (d x + c\right ) \log \left (-\frac {d x + c}{c} + 1\right ) + {\rm Li}_2\left (\frac {d x + c}{c}\right )\right )} f q\right )} r}{f} \]

[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

Giac [F]

\[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=\int { \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{x} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x} \, dx=\int \frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )}{x} \,d x \]

[In]

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

[Out]

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