\(\int \frac {\log (c (a+b x)^p)}{x (d+e x)} \, dx\) [223]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x (d+e x)} \, dx=\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d}-\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}-\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {36, 29, 31, 2463, 2441, 2352, 2440, 2438} \[ \int \frac {\log \left (c (a+b x)^p\right )}{x (d+e x)} \, dx=-\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d}-\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {b x}{a}+1\right )}{d} \]

[In]

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

[Out]

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

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 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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 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 2463

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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x (d+e x)} \, dx=\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d}-\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {a+b x}{a}\right )}{d}-\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.61

method	result	size
parts	\(-\frac {\ln \left (c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{d}+\frac {\ln \left (c \left (b x +a \right )^{p}\right ) \ln \left (x \right )}{d}-p b \left (\frac {\operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d b}+\frac {\ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d b}-\frac {\operatorname {dilog}\left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d b}-\frac {\ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d b}\right )\)	\(156\)
risch	\(-\frac {\ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{d}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) \ln \left (x \right )}{d}-\frac {p \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d}-\frac {p \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d}+\frac {p \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d}+\frac {p \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-\frac {\ln \left (e x +d \right )}{d}+\frac {\ln \left (x \right )}{d}\right )\)	\(263\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.27 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x (d+e x)} \, dx=-b p {\left (\frac {\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )}{b d} - \frac {\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )}{b d}\right )} - {\left (\frac {\log \left (e x + d\right )}{d} - \frac {\log \left (x\right )}{d}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

[In]

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

[Out]

-b*p*((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)^p*c)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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