\(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{a+b x} \, dx\) [11]

Optimal result

Integrand size = 29, antiderivative size = 107 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{a+b x} \, dx=-\frac {p r \log ^2(a+b x)}{2 b}-\frac {q r \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b}+\frac {\log (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b}-\frac {q r \operatorname {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2580, 2437, 2338, 2441, 2440, 2438} \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{a+b x} \, dx=\frac {\log (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b}-\frac {q r \operatorname {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b}-\frac {q r \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b}-\frac {p r \log ^2(a+b x)}{2 b} \]

[In]

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

[Out]

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

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 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 2440

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 2441

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{a+b x} \, dx=-\frac {\log (a+b x) \left (p r \log (a+b x)+2 q r \log \left (\frac {b (c+d x)}{b c-a d}\right )-2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )\right )+2 q r \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )}{2 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 9.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.17

method	result	size
parts	\(\frac {\ln \left (b x +a \right ) \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )}{b}-\frac {r \left (\frac {b p \ln \left (b x +a \right )^{2}}{2}+b d q \left (\frac {\operatorname {dilog}\left (\frac {-a d +c b +d \left (b x +a \right )}{-a d +c b}\right )}{d}+\frac {\ln \left (b x +a \right ) \ln \left (\frac {-a d +c b +d \left (b x +a \right )}{-a d +c b}\right )}{d}\right )\right )}{b^{2}}\)	\(125\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{a+b x} \, dx=-\frac {{\left (\frac {2 \, {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} f q}{b} - \frac {f p \log \left (b x + a\right )^{2} + 2 \, f q \log \left (b x + a\right ) \log \left (d x + c\right )}{b}\right )} r}{2 \, f} - \frac {{\left (f p \log \left (b x + a\right ) + f q \log \left (d x + c\right )\right )} r \log \left (b x + a\right )}{b f} + \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

-1/2*(2*(log(b*x + a)*log((b*d*x + a*d)/(b*c - a*d) + 1) + dilog(-(b*d*x + a*d)/(b*c - a*d)))*f*q/b - (f*p*log
(b*x + a)^2 + 2*f*q*log(b*x + a)*log(d*x + c))/b)*r/f - (f*p*log(b*x + a) + f*q*log(d*x + c))*r*log(b*x + a)/(
b*f) + log(((b*x + a)^p*(d*x + c)^q*f)^r*e)*log(b*x + a)/b

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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