\(\int \frac {(a+b \log (c x^n)) \log (d (e+f x)^m)}{x^4} \, dx\) [83]

Optimal result

Integrand size = 24, antiderivative size = 274 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x^4} \, dx=-\frac {5 b f m n}{36 e x^2}+\frac {4 b f^2 m n}{9 e^2 x}+\frac {b f^3 m n \log (x)}{9 e^3}-\frac {b f^3 m n \log ^2(x)}{6 e^3}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {b f^3 m n \log (e+f x)}{9 e^3}+\frac {b f^3 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {b n \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {b f^3 m n \operatorname {PolyLog}\left (2,1+\frac {f x}{e}\right )}{3 e^3} \] Output:

-5/36*b*f*m*n/e/x^2+4/9*b*f^2*m*n/e^2/x+1/9*b*f^3*m*n*ln(x)/e^3-1/6*b*f^3* 
m*n*ln(x)^2/e^3-1/6*f*m*(a+b*ln(c*x^n))/e/x^2+1/3*f^2*m*(a+b*ln(c*x^n))/e^ 
2/x+1/3*f^3*m*ln(x)*(a+b*ln(c*x^n))/e^3-1/9*b*f^3*m*n*ln(f*x+e)/e^3+1/3*b* 
f^3*m*n*ln(-f*x/e)*ln(f*x+e)/e^3-1/3*f^3*m*(a+b*ln(c*x^n))*ln(f*x+e)/e^3-1 
/9*b*n*ln(d*(f*x+e)^m)/x^3-1/3*(a+b*ln(c*x^n))*ln(d*(f*x+e)^m)/x^3+1/3*b*f 
^3*m*n*polylog(2,1+f*x/e)/e^3
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x^4} \, dx=-\frac {6 a e^2 f m x+5 b e^2 f m n x-12 a e f^2 m x^2-16 b e f^2 m n x^2+6 b f^3 m n x^3 \log ^2(x)+6 b e^2 f m x \log \left (c x^n\right )-12 b e f^2 m x^2 \log \left (c x^n\right )+12 a f^3 m x^3 \log (e+f x)+4 b f^3 m n x^3 \log (e+f x)+12 b f^3 m x^3 \log \left (c x^n\right ) \log (e+f x)+12 a e^3 \log \left (d (e+f x)^m\right )+4 b e^3 n \log \left (d (e+f x)^m\right )+12 b e^3 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )-4 f^3 m x^3 \log (x) \left (3 a+b n+3 b \log \left (c x^n\right )+3 b n \log (e+f x)-3 b n \log \left (1+\frac {f x}{e}\right )\right )+12 b f^3 m n x^3 \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )}{36 e^3 x^3} \] Input:

Integrate[((a + b*Log[c*x^n])*Log[d*(e + f*x)^m])/x^4,x]
 

Output:

-1/36*(6*a*e^2*f*m*x + 5*b*e^2*f*m*n*x - 12*a*e*f^2*m*x^2 - 16*b*e*f^2*m*n 
*x^2 + 6*b*f^3*m*n*x^3*Log[x]^2 + 6*b*e^2*f*m*x*Log[c*x^n] - 12*b*e*f^2*m* 
x^2*Log[c*x^n] + 12*a*f^3*m*x^3*Log[e + f*x] + 4*b*f^3*m*n*x^3*Log[e + f*x 
] + 12*b*f^3*m*x^3*Log[c*x^n]*Log[e + f*x] + 12*a*e^3*Log[d*(e + f*x)^m] + 
 4*b*e^3*n*Log[d*(e + f*x)^m] + 12*b*e^3*Log[c*x^n]*Log[d*(e + f*x)^m] - 4 
*f^3*m*x^3*Log[x]*(3*a + b*n + 3*b*Log[c*x^n] + 3*b*n*Log[e + f*x] - 3*b*n 
*Log[1 + (f*x)/e]) + 12*b*f^3*m*n*x^3*PolyLog[2, -((f*x)/e)])/(e^3*x^3)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

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

\(\Big \downarrow \) 2823

\(\displaystyle -b n \int \left (\frac {m \log (x) f^3}{3 e^3 x}-\frac {m \log (e+f x) f^3}{3 e^3 x}+\frac {m f^2}{3 e^2 x^2}-\frac {m f}{6 e x^3}-\frac {\log \left (d (e+f x)^m\right )}{3 x^4}\right )dx-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {f^3 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {f^3 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}-b n \left (\frac {\log \left (d (e+f x)^m\right )}{9 x^3}-\frac {f^3 m \operatorname {PolyLog}\left (2,\frac {f x}{e}+1\right )}{3 e^3}+\frac {f^3 m \log ^2(x)}{6 e^3}-\frac {f^3 m \log (x)}{9 e^3}+\frac {f^3 m \log (e+f x)}{9 e^3}-\frac {f^3 m \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}-\frac {4 f^2 m}{9 e^2 x}+\frac {5 f m}{36 e x^2}\right )\)

Input:

Int[((a + b*Log[c*x^n])*Log[d*(e + f*x)^m])/x^4,x]
 

Output:

-1/6*(f*m*(a + b*Log[c*x^n]))/(e*x^2) + (f^2*m*(a + b*Log[c*x^n]))/(3*e^2* 
x) + (f^3*m*Log[x]*(a + b*Log[c*x^n]))/(3*e^3) - (f^3*m*(a + b*Log[c*x^n]) 
*Log[e + f*x])/(3*e^3) - ((a + b*Log[c*x^n])*Log[d*(e + f*x)^m])/(3*x^3) - 
 b*n*((5*f*m)/(36*e*x^2) - (4*f^2*m)/(9*e^2*x) - (f^3*m*Log[x])/(9*e^3) + 
(f^3*m*Log[x]^2)/(6*e^3) + (f^3*m*Log[e + f*x])/(9*e^3) - (f^3*m*Log[-((f* 
x)/e)]*Log[e + f*x])/(3*e^3) + Log[d*(e + f*x)^m]/(9*x^3) - (f^3*m*PolyLog 
[2, 1 + (f*x)/e])/(3*e^3))
 

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 = 30.99 (sec) , antiderivative size = 1127, normalized size of antiderivative = 4.11

method	result	size
risch	\(\text {Expression too large to display}\)	\(1127\)

Input:

int((a+b*ln(c*x^n))*ln(d*(f*x+e)^m)/x^4,x,method=_RETURNVERBOSE)
 

Output:

(-1/3*b/x^3*ln(x^n)-1/18*(3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-3*I*b*Pi*cs 
gn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3*I*b*Pi*csgn(I*c*x^n)^3+3*I*b*Pi*csgn(I 
*c*x^n)^2*csgn(I*c)+6*b*ln(c)+2*n*b+6*a)/x^3)*ln((f*x+e)^m)+(-1/4*I*Pi*csg 
n(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))*(-1/3*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*P 
i*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^3-2/3*b/x^3*ln(x^n)-2/9*b/x^3*n)-1/ 
3*m*f^3/e^3*ln(f*x+e)*a+1/3*m*f^3/e^3*ln(x)*a-1/3*m*f^3*b*ln(x^n)/e^3*ln(f 
*x+e)+1/3*m*f^3*b*ln(x^n)/e^3*ln(x)+1/3*m*f^3*b*n/e^3*dilog(-f*x/e)-1/3*m* 
f^3/e^3*ln(f*x+e)*b*ln(c)+1/3*m*f^3/e^3*ln(x)*b*ln(c)+1/3*m*f^2/e^2/x*a-1/ 
6*m*f/e/x^2*a-5/36*b*f*m*n/e/x^2+4/9*b*f^2*m*n/e^2/x+1/9*b*f^3*m*n*ln(x)/e 
^3-1/6*b*f^3*m*n*ln(x)^2/e^3-1/9*b*f^3*m*n*ln(f*x+e)/e^3+1/3*m*f^2/e^2/x*b 
*ln(c)-1/6*m*f/e/x^2*b*ln(c)+1/3*m*f^2*b*ln(x^n)/e^2/x-1/6*m*f*b*ln(x^n)/e 
/x^2+1/12*I*m*f/e/x^2*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/6*I*m*f^3 
/e^3*ln(f*x+e)*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/6*I*m*f^3/e^3*ln 
(x)*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/6*I*m*f^2/e^2/x*b*Pi*csgn(I 
*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/12*I*m*f/e/x^2*b*Pi*csgn(I*c*x^n)^3+1/6*I* 
m*f^3/e^3*ln(f*x+e)*b*Pi*csgn(I*c*x^n)^3-1/6*I*m*f^3/e^3*ln(x)*b*Pi*csgn(I 
*c*x^n)^3-1/6*I*m*f^2/e^2/x*b*Pi*csgn(I*c*x^n)^3+1/6*I*m*f^3/e^3*ln(x)*...
 

Fricas [F]

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

integrate((a+b*log(c*x^n))*log(d*(f*x+e)^m)/x^4,x, algorithm="fricas")
 

Output:

integral((b*log(c*x^n) + a)*log((f*x + e)^m*d)/x^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x^4} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*x**n))*ln(d*(f*x+e)**m)/x**4,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x^4} \, dx=-\frac {{\left (\log \left (\frac {f x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {f x}{e}\right )\right )} b f^{3} m n}{3 \, e^{3}} - \frac {{\left (3 \, a f^{3} m + {\left (f^{3} m n + 3 \, f^{3} m \log \left (c\right )\right )} b\right )} \log \left (f x + e\right )}{9 \, e^{3}} + \frac {12 \, b f^{3} m n x^{3} \log \left (f x + e\right ) \log \left (x\right ) - 6 \, b f^{3} m n x^{3} \log \left (x\right )^{2} - 12 \, a e^{3} \log \left (d\right ) + 4 \, {\left (3 \, a f^{3} m + {\left (f^{3} m n + 3 \, f^{3} m \log \left (c\right )\right )} b\right )} x^{3} \log \left (x\right ) + 4 \, {\left (3 \, a e f^{2} m + {\left (4 \, e f^{2} m n + 3 \, e f^{2} m \log \left (c\right )\right )} b\right )} x^{2} - 4 \, {\left (e^{3} n \log \left (d\right ) + 3 \, e^{3} \log \left (c\right ) \log \left (d\right )\right )} b - {\left (6 \, a e^{2} f m + {\left (5 \, e^{2} f m n + 6 \, e^{2} f m \log \left (c\right )\right )} b\right )} x - 4 \, {\left (3 \, b e^{3} \log \left (x^{n}\right ) + 3 \, a e^{3} + {\left (e^{3} n + 3 \, e^{3} \log \left (c\right )\right )} b\right )} \log \left ({\left (f x + e\right )}^{m}\right ) - 6 \, {\left (2 \, b f^{3} m x^{3} \log \left (f x + e\right ) - 2 \, b f^{3} m x^{3} \log \left (x\right ) - 2 \, b e f^{2} m x^{2} + b e^{2} f m x + 2 \, b e^{3} \log \left (d\right )\right )} \log \left (x^{n}\right )}{36 \, e^{3} x^{3}} \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x+e)^m)/x^4,x, algorithm="maxima")
 

Output:

-1/3*(log(f*x/e + 1)*log(x) + dilog(-f*x/e))*b*f^3*m*n/e^3 - 1/9*(3*a*f^3* 
m + (f^3*m*n + 3*f^3*m*log(c))*b)*log(f*x + e)/e^3 + 1/36*(12*b*f^3*m*n*x^ 
3*log(f*x + e)*log(x) - 6*b*f^3*m*n*x^3*log(x)^2 - 12*a*e^3*log(d) + 4*(3* 
a*f^3*m + (f^3*m*n + 3*f^3*m*log(c))*b)*x^3*log(x) + 4*(3*a*e*f^2*m + (4*e 
*f^2*m*n + 3*e*f^2*m*log(c))*b)*x^2 - 4*(e^3*n*log(d) + 3*e^3*log(c)*log(d 
))*b - (6*a*e^2*f*m + (5*e^2*f*m*n + 6*e^2*f*m*log(c))*b)*x - 4*(3*b*e^3*l 
og(x^n) + 3*a*e^3 + (e^3*n + 3*e^3*log(c))*b)*log((f*x + e)^m) - 6*(2*b*f^ 
3*m*x^3*log(f*x + e) - 2*b*f^3*m*x^3*log(x) - 2*b*e*f^2*m*x^2 + b*e^2*f*m* 
x + 2*b*e^3*log(d))*log(x^n))/(e^3*x^3)
 

Giac [F]

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

integrate((a+b*log(c*x^n))*log(d*(f*x+e)^m)/x^4,x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*log((f*x + e)^m*d)/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x^4} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \] Input:

int((log(d*(e + f*x)^m)*(a + b*log(c*x^n)))/x^4,x)
 

Output:

int((log(d*(e + f*x)^m)*(a + b*log(c*x^n)))/x^4, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x^4} \, dx=\frac {-18 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{5}+e \,x^{4}}d x \right ) b \,e^{4} m \,x^{3}-18 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{3}-18 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) a \,e^{3}-18 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) a \,f^{3} x^{3}-6 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) b \,e^{3} n -6 \,\mathrm {log}\left (\left (f x +e \right )^{m} d \right ) b \,f^{3} n \,x^{3}-6 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} m +18 \,\mathrm {log}\left (x \right ) a \,f^{3} m \,x^{3}+6 \,\mathrm {log}\left (x \right ) b \,f^{3} m n \,x^{3}-9 a \,e^{2} f m x +18 a e \,f^{2} m \,x^{2}-2 b \,e^{3} m n -3 b \,e^{2} f m n x +6 b e \,f^{2} m n \,x^{2}}{54 e^{3} x^{3}} \] Input:

int((a+b*log(c*x^n))*log(d*(f*x+e)^m)/x^4,x)
 

Output:

( - 18*int(log(x**n*c)/(e*x**4 + f*x**5),x)*b*e**4*m*x**3 - 18*log((e + f* 
x)**m*d)*log(x**n*c)*b*e**3 - 18*log((e + f*x)**m*d)*a*e**3 - 18*log((e + 
f*x)**m*d)*a*f**3*x**3 - 6*log((e + f*x)**m*d)*b*e**3*n - 6*log((e + f*x)* 
*m*d)*b*f**3*n*x**3 - 6*log(x**n*c)*b*e**3*m + 18*log(x)*a*f**3*m*x**3 + 6 
*log(x)*b*f**3*m*n*x**3 - 9*a*e**2*f*m*x + 18*a*e*f**2*m*x**2 - 2*b*e**3*m 
*n - 3*b*e**2*f*m*n*x + 6*b*e*f**2*m*n*x**2)/(54*e**3*x**3)