\(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))}{x} \, dx\) [362]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b n \log ^2\left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 m}-b n \log \left (f x^m\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+b m n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right ) \]

[Out]

1/2*ln(f*x^m)^2*(a+b*ln(c*(e*x+d)^n))/m-1/2*b*n*ln(f*x^m)^2*ln(1+e*x/d)/m-b*n*ln(f*x^m)*polylog(2,-e*x/d)+b*m*
n*polylog(3,-e*x/d)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2472, 2354, 2421, 6724} \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \log \left (f x^m\right )-\frac {b n \log \left (\frac {e x}{d}+1\right ) \log ^2\left (f x^m\right )}{2 m}+b m n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right ) \]

[In]

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

[Out]

(Log[f*x^m]^2*(a + b*Log[c*(d + e*x)^n]))/(2*m) - (b*n*Log[f*x^m]^2*Log[1 + (e*x)/d])/(2*m) - b*n*Log[f*x^m]*P
olyLog[2, -((e*x)/d)] + b*m*n*PolyLog[3, -((e*x)/d)]

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 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c*x^n])^p/m), x] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2472

Int[(Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)))/(x_), x_Symbol] :> Simp[Log[f
*x^m]^2*((a + b*Log[c*(d + e*x)^n])/(2*m)), x] - Dist[b*e*(n/(2*m)), Int[Log[f*x^m]^2/(d + e*x), x], x] /; Fre
eQ[{a, b, c, d, e, f, m, n}, x]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {(b e n) \int \frac {\log ^2\left (f x^m\right )}{d+e x} \, dx}{2 m} \\ & = \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b n \log ^2\left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 m}+(b n) \int \frac {\log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{x} \, dx \\ & = \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b n \log ^2\left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 m}-b n \log \left (f x^m\right ) \text {Li}_2\left (-\frac {e x}{d}\right )+(b m n) \int \frac {\text {Li}_2\left (-\frac {e x}{d}\right )}{x} \, dx \\ & = \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b n \log ^2\left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 m}-b n \log \left (f x^m\right ) \text {Li}_2\left (-\frac {e x}{d}\right )+b m n \text {Li}_3\left (-\frac {e x}{d}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.45 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\frac {1}{2} \left (\frac {a \log ^2\left (f x^m\right )}{m}-b m \log ^2(x) \log \left (c (d+e x)^n\right )+2 b \log (x) \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b m n \log ^2(x) \log \left (1+\frac {e x}{d}\right )-2 b n \log (x) \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )-2 b n \log \left (f x^m\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+2 b m n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )\right ) \]

[In]

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

[Out]

((a*Log[f*x^m]^2)/m - b*m*Log[x]^2*Log[c*(d + e*x)^n] + 2*b*Log[x]*Log[f*x^m]*Log[c*(d + e*x)^n] + b*m*n*Log[x
]^2*Log[1 + (e*x)/d] - 2*b*n*Log[x]*Log[f*x^m]*Log[1 + (e*x)/d] - 2*b*n*Log[f*x^m]*PolyLog[2, -((e*x)/d)] + 2*
b*m*n*PolyLog[3, -((e*x)/d)])/2

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.60 (sec) , antiderivative size = 756, normalized size of antiderivative = 8.59

method result size
risch \(\text {Expression too large to display}\) \(756\)

[In]

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

[Out]

(b*ln(x)*ln(x^m)-1/2*b*m*ln(x)^2-1/2*I*Pi*ln(x)*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/2*I*Pi*ln(x)*b*csgn(I*
f)*csgn(I*f*x^m)^2+1/2*I*Pi*ln(x)*b*csgn(I*x^m)*csgn(I*f*x^m)^2-1/2*I*Pi*ln(x)*b*csgn(I*f*x^m)^3+ln(f)*ln(x)*b
)*ln((e*x+d)^n)+(-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*
x+d)^n)^2+1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*b*ln(c)+1/2*
a)*(I*Pi*csgn(I*f)*csgn(I*f*x^m)^2*ln(x)+I*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2*ln(x)+2*ln(x)*ln(f)-I*Pi*csgn(I*f*x^
m)^3*ln(x)-I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)*ln(x)+1/m*ln(x^m)^2)-1/2*I*n*b*ln(x)*ln((e*x+d)/d)*Pi*csgn
(I*x^m)*csgn(I*f*x^m)^2-1/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/2*I*n*b*dilog((e*x+d)/d)*Pi*
csgn(I*x^m)*csgn(I*f*x^m)^2+1/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/2*I*n*b*ln(x)*
ln((e*x+d)/d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/2*I*n*b*ln(x)*ln((e*x+d)/d)*Pi*csgn(I*f)*csgn(I*f*x^m)^
2+1/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*f*x^m)^3+1/2*I*n*b*ln(x)*ln((e*x+d)/d)*Pi*csgn(I*f*x^m)^3-1/2*n*b*m*ln(
x)^2*ln(1+e*x/d)+n*b*ln(x)^2*ln((e*x+d)/d)*m-n*b*m*ln(x)*polylog(2,-e*x/d)+n*b*dilog((e*x+d)/d)*m*ln(x)-n*b*ln
(x)*ln((e*x+d)/d)*ln(x^m)-n*b*ln(x)*ln((e*x+d)/d)*ln(f)+b*m*n*polylog(3,-e*x/d)-n*b*dilog((e*x+d)/d)*ln(x^m)-n
*b*dilog((e*x+d)/d)*ln(f)

Fricas [F]

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

[In]

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

[Out]

integral((b*log((e*x + d)^n*c)*log(f*x^m) + a*log(f*x^m))/x, x)

Sympy [F(-1)]

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

[In]

integrate(ln(f*x**m)*(a+b*ln(c*(e*x+d)**n))/x,x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

-1/2*(b*m*log(x)^2 - 2*b*log(f)*log(x) - 2*b*log(x)*log(x^m))*log((e*x + d)^n) - integrate(-1/2*(b*e*m*n*x*log
(x)^2 - 2*b*e*n*x*log(f)*log(x) + 2*b*d*log(c)*log(f) + 2*a*d*log(f) + 2*(b*e*log(c)*log(f) + a*e*log(f))*x -
2*(b*e*n*x*log(x) - b*d*log(c) - a*d - (b*e*log(c) + a*e)*x)*log(x^m))/(e*x^2 + d*x), x)

Giac [F]

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

[In]

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

[Out]

integrate((b*log((e*x + d)^n*c) + a)*log(f*x^m)/x, x)

Mupad [F(-1)]

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

[In]

int((log(f*x^m)*(a + b*log(c*(d + e*x)^n)))/x,x)

[Out]

int((log(f*x^m)*(a + b*log(c*(d + e*x)^n)))/x, x)