\(\int (g x)^{-1+2 m} (a+b \log (c x^n)) \log (d (e+f x^m)^k) \, dx\) [157]

Optimal result

Integrand size = 32, antiderivative size = 363 \[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {b k n (g x)^{2 m}}{4 g m^2}-\frac {3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \operatorname {PolyLog}\left (2,1+\frac {f x^m}{e}\right )}{2 f^2 g m^2} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.70 (sec) , antiderivative size = 352, normalized size of antiderivative = 0.97 \[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {x^{-2 m} (g x)^{2 m} \left (2 a e f k m x^m-3 b e f k n x^m-a f^2 k m x^{2 m}+b f^2 k n x^{2 m}+2 b e^2 k m^2 n \log ^2(x)+2 b e f k m x^m \log \left (c x^n\right )-b f^2 k m x^{2 m} \log \left (c x^n\right )-2 a e^2 k m \log \left (e-e x^m\right )+b e^2 k n \log \left (e-e x^m\right )-2 b e^2 k m \log \left (c x^n\right ) \log \left (e-e x^m\right )+2 b e^2 k n \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )+e^2 k m \log (x) \left (-2 a m+b n-2 b m \log \left (c x^n\right )+2 b n \log \left (e-e x^m\right )-2 b n \log \left (e+f x^m\right )\right )+2 a f^2 m x^{2 m} \log \left (d \left (e+f x^m\right )^k\right )-b f^2 n x^{2 m} \log \left (d \left (e+f x^m\right )^k\right )+2 b f^2 m x^{2 m} \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )+2 b e^2 k n \operatorname {PolyLog}\left (2,1+\frac {f x^m}{e}\right )\right )}{4 f^2 g m^2} \] Input:

Integrate[(g*x)^(-1 + 2*m)*(a + b*Log[c*x^n])*Log[d*(e + f*x^m)^k],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, 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.

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

\(\Big \downarrow \) 2823

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

\(\Big \downarrow \) 2009

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

Input:

Int[(g*x)^(-1 + 2*m)*(a + b*Log[c*x^n])*Log[d*(e + f*x^m)^k],x]
 

Output:

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

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 [F]

\[\int \left (g x \right )^{-1+2 m} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \,x^{m}\right )^{k}\right )d x\]

Input:

int((g*x)^(-1+2*m)*(a+b*ln(c*x^n))*ln(d*(e+f*x^m)^k),x)
 

Output:

int((g*x)^(-1+2*m)*(a+b*ln(c*x^n))*ln(d*(e+f*x^m)^k),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.83 \[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=-\frac {2 \, b e^{2} g^{2 \, m - 1} k m n \log \left (x\right ) \log \left (\frac {f x^{m} + e}{e}\right ) + 2 \, b e^{2} g^{2 \, m - 1} k n {\rm Li}_2\left (-\frac {f x^{m} + e}{e} + 1\right ) + {\left (b f^{2} k m \log \left (c\right ) + a f^{2} k m - b f^{2} k n - {\left (2 \, b f^{2} m \log \left (c\right ) + 2 \, a f^{2} m - b f^{2} n\right )} \log \left (d\right ) + {\left (b f^{2} k m n - 2 \, b f^{2} m n \log \left (d\right )\right )} \log \left (x\right )\right )} g^{2 \, m - 1} x^{2 \, m} - {\left (2 \, b e f k m n \log \left (x\right ) + 2 \, b e f k m \log \left (c\right ) + 2 \, a e f k m - 3 \, b e f k n\right )} g^{2 \, m - 1} x^{m} - {\left ({\left (2 \, b f^{2} k m n \log \left (x\right ) + 2 \, b f^{2} k m \log \left (c\right ) + 2 \, a f^{2} k m - b f^{2} k n\right )} g^{2 \, m - 1} x^{2 \, m} - {\left (2 \, b e^{2} k m \log \left (c\right ) + 2 \, a e^{2} k m - b e^{2} k n\right )} g^{2 \, m - 1}\right )} \log \left (f x^{m} + e\right )}{4 \, f^{2} m^{2}} \] Input:

integrate((g*x)^(-1+2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\text {Timed out} \] Input:

integrate((g*x)**(-1+2*m)*(a+b*ln(c*x**n))*ln(d*(e+f*x**m)**k),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right ) \,d x } \] Input:

integrate((g*x)^(-1+2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x, algorithm= 
"maxima")
 

Output:

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

Giac [F]

\[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right ) \,d x } \] Input:

integrate((g*x)^(-1+2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x, algorithm= 
"giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int \ln \left (d\,{\left (e+f\,x^m\right )}^k\right )\,{\left (g\,x\right )}^{2\,m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:

int(log(d*(e + f*x^m)^k)*(g*x)^(2*m - 1)*(a + b*log(c*x^n)),x)
 

Output:

int(log(d*(e + f*x^m)^k)*(g*x)^(2*m - 1)*(a + b*log(c*x^n)), x)
 

Reduce [F]

\[ \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {g^{2 m} \left (2 x^{2 m} \mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,f^{2} m +2 x^{2 m} \mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) a \,f^{2} m -x^{2 m} \mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) b \,f^{2} n -x^{2 m} \mathrm {log}\left (x^{n} c \right ) b \,f^{2} k m -x^{2 m} a \,f^{2} k m +x^{2 m} b \,f^{2} k n +2 x^{m} \mathrm {log}\left (x^{n} c \right ) b e f k m +2 x^{m} a e f k m -3 x^{m} b e f k n -2 \left (\int \frac {x^{m} \mathrm {log}\left (x^{n} c \right )}{x^{m} f x +e x}d x \right ) b \,e^{2} f k \,m^{2}-2 \,\mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) a \,e^{2} m +\mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) b \,e^{2} n \right )}{4 f^{2} g \,m^{2}} \] Input:

int((g*x)^(-1+2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x)
 

Output:

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