\(\int x (a+b \log (c x^n)) \log (d (e+f x)^m) \, dx\) [78]

Optimal result

Integrand size = 22, antiderivative size = 203 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=-\frac {3 b e m n x}{4 f}+\frac {1}{4} b m n x^2+\frac {e m x \left (a+b \log \left (c x^n\right )\right )}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b e^2 m n \log (e+f x)}{4 f^2}+\frac {b e^2 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{2 f^2}-\frac {1}{4} b n x^2 \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \operatorname {PolyLog}\left (2,1+\frac {f x}{e}\right )}{2 f^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.02 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\frac {2 a e f m x-3 b e f m n x-a f^2 m x^2+b f^2 m n x^2-2 a e^2 m \log (e+f x)+b e^2 m n \log (e+f x)+2 b e^2 m n \log (x) \log (e+f x)+2 a f^2 x^2 \log \left (d (e+f x)^m\right )-b f^2 n x^2 \log \left (d (e+f x)^m\right )+b \log \left (c x^n\right ) \left (-2 e^2 m \log (e+f x)+f x \left (2 e m-f m x+2 f x \log \left (d (e+f x)^m\right )\right )\right )-2 b e^2 m n \log (x) \log \left (1+\frac {f x}{e}\right )-2 b e^2 m n \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{4 f^2} \] Input:

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

Output:

(2*a*e*f*m*x - 3*b*e*f*m*n*x - a*f^2*m*x^2 + b*f^2*m*n*x^2 - 2*a*e^2*m*Log 
[e + f*x] + b*e^2*m*n*Log[e + f*x] + 2*b*e^2*m*n*Log[x]*Log[e + f*x] + 2*a 
*f^2*x^2*Log[d*(e + f*x)^m] - b*f^2*n*x^2*Log[d*(e + f*x)^m] + b*Log[c*x^n 
]*(-2*e^2*m*Log[e + f*x] + f*x*(2*e*m - f*m*x + 2*f*x*Log[d*(e + f*x)^m])) 
 - 2*b*e^2*m*n*Log[x]*Log[1 + (f*x)/e] - 2*b*e^2*m*n*PolyLog[2, -((f*x)/e) 
])/(4*f^2)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2823, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx\)

\(\Big \downarrow \) 2823

\(\displaystyle -b n \int \left (-\frac {m \log (e+f x) e^2}{2 f^2 x}+\frac {m e}{2 f}-\frac {m x}{4}+\frac {1}{2} x \log \left (d (e+f x)^m\right )\right )dx+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-\frac {e^2 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{2 f^2}+\frac {e m x \left (a+b \log \left (c x^n\right )\right )}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )-\frac {e^2 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{2 f^2}+\frac {e m x \left (a+b \log \left (c x^n\right )\right )}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {1}{4} x^2 \log \left (d (e+f x)^m\right )-\frac {e^2 m \operatorname {PolyLog}\left (2,\frac {f x}{e}+1\right )}{2 f^2}-\frac {e^2 m \log (e+f x)}{4 f^2}-\frac {e^2 m \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{2 f^2}+\frac {3 e m x}{4 f}-\frac {m x^2}{4}\right )\)

Input:

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

Output:

(e*m*x*(a + b*Log[c*x^n]))/(2*f) - (m*x^2*(a + b*Log[c*x^n]))/4 - (e^2*m*( 
a + b*Log[c*x^n])*Log[e + f*x])/(2*f^2) + (x^2*(a + b*Log[c*x^n])*Log[d*(e 
 + f*x)^m])/2 - b*n*((3*e*m*x)/(4*f) - (m*x^2)/4 - (e^2*m*Log[e + f*x])/(4 
*f^2) - (e^2*m*Log[-((f*x)/e)]*Log[e + f*x])/(2*f^2) + (x^2*Log[d*(e + f*x 
)^m])/4 - (e^2*m*PolyLog[2, 1 + (f*x)/e])/(2*f^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2823

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* 
(e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n])   u, x] - Simp[b*n   Int[1/x 
 u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q 
+ 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
 

Maple [C] (warning: unable to verify)

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

Time = 28.78 (sec) , antiderivative size = 885, normalized size of antiderivative = 4.36

method	result	size
risch	\(\text {Expression too large to display}\)	\(885\)

Input:

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

Output:

-1/2*m*a*e^2/f^2*ln(f*x+e)+(-1/4*I*Pi*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d 
*(f*x+e)^m)+1/4*I*Pi*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2+1/4*I*Pi*csgn(I*(f*x+ 
e)^m)*csgn(I*d*(f*x+e)^m)^2-1/4*I*Pi*csgn(I*d*(f*x+e)^m)^3+1/2*ln(d))*(1/2 
*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn 
(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)+2* 
a)*x^2+b*x^2*ln(x^n)-1/2*b*n*x^2)+(1/2*b*x^2*ln(x^n)+1/4*x^2*(I*Pi*b*csgn( 
I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*c 
sgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)-n*b+2*a))*ln((f* 
x+e)^m)+1/2*m/f*b*ln(x^n)*e*x-5/8*m/f^2*b*n*e^2-1/4*I*m/f*e*x*Pi*b*csgn(I* 
x^n)*csgn(I*c*x^n)*csgn(I*c)+1/4*I*m/f^2*e^2*ln(f*x+e)*Pi*b*csgn(I*x^n)*cs 
gn(I*c*x^n)*csgn(I*c)-1/2*m/f^2*b*ln(x^n)*e^2*ln(f*x+e)-1/2*m/f^2*e^2*ln(f 
*x+e)*b*ln(c)+1/8*I*m*x^2*Pi*b*csgn(I*c*x^n)^3+1/2*n*b/f^2*e^2*m*dilog(-f* 
x/e)+1/2*m/f*e*x*a-1/4*m*b*ln(x^n)*x^2-1/8*I*m*x^2*Pi*b*csgn(I*x^n)*csgn(I 
*c*x^n)^2-1/8*I*m*x^2*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)-3/4*b*e*m*n*x/f+1/4*b 
*e^2*m*n*ln(f*x+e)/f^2+1/4*I*m/f^2*e^2*ln(f*x+e)*Pi*b*csgn(I*c*x^n)^3-1/4* 
I*m/f*e*x*Pi*b*csgn(I*c*x^n)^3+1/8*I*m*x^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)* 
csgn(I*c)-1/4*x^2*ln(c)*b*m+1/4*b*m*n*x^2+1/4*I*m/f*e*x*Pi*b*csgn(I*c*x^n) 
^2*csgn(I*c)-1/4*I*m/f^2*e^2*ln(f*x+e)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/ 
4*I*m/f^2*e^2*ln(f*x+e)*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+1/4*I*m/f*e*x*Pi*b* 
csgn(I*x^n)*csgn(I*c*x^n)^2+1/2*m/f*e*x*b*ln(c)-1/4*x^2*a*m+1/2*b*e^2*m...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.33 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=-\frac {{\left (\log \left (\frac {f x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {f x}{e}\right )\right )} b e^{2} m n}{2 \, f^{2}} - \frac {{\left (2 \, a e^{2} m - {\left (e^{2} m n - 2 \, e^{2} m \log \left (c\right )\right )} b\right )} \log \left (f x + e\right )}{4 \, f^{2}} + \frac {2 \, b e^{2} m n \log \left (f x + e\right ) \log \left (x\right ) - {\left ({\left (f^{2} m - 2 \, f^{2} \log \left (d\right )\right )} a - {\left (f^{2} m n - f^{2} n \log \left (d\right ) - {\left (f^{2} m - 2 \, f^{2} \log \left (d\right )\right )} \log \left (c\right )\right )} b\right )} x^{2} + {\left (2 \, a e f m - {\left (3 \, e f m n - 2 \, e f m \log \left (c\right )\right )} b\right )} x + {\left (2 \, b f^{2} x^{2} \log \left (x^{n}\right ) + {\left (2 \, a f^{2} - {\left (f^{2} n - 2 \, f^{2} \log \left (c\right )\right )} b\right )} x^{2}\right )} \log \left ({\left (f x + e\right )}^{m}\right ) + {\left (2 \, b e f m x - 2 \, b e^{2} m \log \left (f x + e\right ) - {\left (f^{2} m - 2 \, f^{2} \log \left (d\right )\right )} b x^{2}\right )} \log \left (x^{n}\right )}{4 \, f^{2}} \] Input:

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

Output:

-1/2*(log(f*x/e + 1)*log(x) + dilog(-f*x/e))*b*e^2*m*n/f^2 - 1/4*(2*a*e^2* 
m - (e^2*m*n - 2*e^2*m*log(c))*b)*log(f*x + e)/f^2 + 1/4*(2*b*e^2*m*n*log( 
f*x + e)*log(x) - ((f^2*m - 2*f^2*log(d))*a - (f^2*m*n - f^2*n*log(d) - (f 
^2*m - 2*f^2*log(d))*log(c))*b)*x^2 + (2*a*e*f*m - (3*e*f*m*n - 2*e*f*m*lo 
g(c))*b)*x + (2*b*f^2*x^2*log(x^n) + (2*a*f^2 - (f^2*n - 2*f^2*log(c))*b)* 
x^2)*log((f*x + e)^m) + (2*b*e*f*m*x - 2*b*e^2*m*log(f*x + e) - (f^2*m - 2 
*f^2*log(d))*b*x^2)*log(x^n))/f^2
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right ) \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{2}+e x}d x \right ) b \,e^{3} m n +2 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,f^{2} n \,x^{2}-2 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) a \,e^{2} n +2 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) a \,f^{2} n \,x^{2}+\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) b \,e^{2} n^{2}-\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) b \,f^{2} n^{2} x^{2}-\mathrm {log}\left (x^{n} c \right )^{2} b \,e^{2} m +2 \,\mathrm {log}\left (x^{n} c \right ) b e f m n x -\mathrm {log}\left (x^{n} c \right ) b \,f^{2} m n \,x^{2}+2 a e f m n x -a \,f^{2} m n \,x^{2}-3 b e f m \,n^{2} x +b \,f^{2} m \,n^{2} x^{2}}{4 f^{2} n} \] Input:

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

Output:

(2*int(log(x**n*c)/(e*x + f*x**2),x)*b*e**3*m*n + 2*log((e + f*x)**m*d)*lo 
g(x**n*c)*b*f**2*n*x**2 - 2*log((e + f*x)**m*d)*a*e**2*n + 2*log((e + f*x) 
**m*d)*a*f**2*n*x**2 + log((e + f*x)**m*d)*b*e**2*n**2 - log((e + f*x)**m* 
d)*b*f**2*n**2*x**2 - log(x**n*c)**2*b*e**2*m + 2*log(x**n*c)*b*e*f*m*n*x 
- log(x**n*c)*b*f**2*m*n*x**2 + 2*a*e*f*m*n*x - a*f**2*m*n*x**2 - 3*b*e*f* 
m*n**2*x + b*f**2*m*n**2*x**2)/(4*f**2*n)