\(\int (g x)^{-1-2 m} (a+b \log (c x^n)) \log (d (e+f x^m)^k) \, dx\) [160]

Optimal result

Integrand size = 32, antiderivative size = 414 \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=-\frac {3 b f k n x^m (g x)^{-2 m}}{4 e g m^2}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log (x)}{4 e^2 g m}+\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log ^2(x)}{4 e^2 g}-\frac {f k x^m (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e g m}-\frac {f^2 k x^{2 m} (g x)^{-2 m} \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g}+\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right )}{4 e^2 g m^2}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 e^2 g m^2}+\frac {f^2 k x^{2 m} (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 e^2 g m}-\frac {b n (g x)^{-2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}-\frac {(g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \operatorname {PolyLog}\left (2,1+\frac {f x^m}{e}\right )}{2 e^2 g m^2} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.73 \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {(g x)^{-2 m} \left (-2 a e f k m x^m-3 b e f k n x^m+b f^2 k m^2 n x^{2 m} \log ^2(x)-2 b e f k m x^m \log \left (c x^n\right )+2 a f^2 k m x^{2 m} \log \left (f-f x^{-m}\right )+b f^2 k n x^{2 m} \log \left (f-f x^{-m}\right )+2 b f^2 k m x^{2 m} \log \left (c x^n\right ) \log \left (f-f x^{-m}\right )-2 a e^2 m \log \left (d \left (e+f x^m\right )^k\right )-b e^2 n \log \left (d \left (e+f x^m\right )^k\right )-2 b e^2 m \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )-f^2 k m x^{2 m} \log (x) \left (2 a m+b n+2 b m \log \left (c x^n\right )+2 b n \log \left (f-f x^{-m}\right )-2 b n \log \left (1+\frac {f x^m}{e}\right )\right )+2 b f^2 k n x^{2 m} \operatorname {PolyLog}\left (2,-\frac {f x^m}{e}\right )\right )}{4 e^2 g m^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 405, 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

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(f*k*x^m*(a + b*Log[c*x^n]))/(e*g*m*(g*x)^(2*m)) - (f^2*k*x^(2*m)*Log 
[x]*(a + b*Log[c*x^n]))/(2*e^2*g*(g*x)^(2*m)) + (f^2*k*x^(2*m)*(a + b*Log[ 
c*x^n])*Log[e + f*x^m])/(2*e^2*g*m*(g*x)^(2*m)) - ((a + b*Log[c*x^n])*Log[ 
d*(e + f*x^m)^k])/(2*g*m*(g*x)^(2*m)) - b*n*((3*f*k*x^m)/(4*e*g*m^2*(g*x)^ 
(2*m)) + (f^2*k*x^(2*m)*Log[x])/(4*e^2*g*m*(g*x)^(2*m)) - (f^2*k*x^(2*m)*L 
og[x]^2)/(4*e^2*g*(g*x)^(2*m)) - (f^2*k*x^(2*m)*Log[e + f*x^m])/(4*e^2*g*m 
^2*(g*x)^(2*m)) + (f^2*k*x^(2*m)*Log[-((f*x^m)/e)]*Log[e + f*x^m])/(2*e^2* 
g*m^2*(g*x)^(2*m)) + Log[d*(e + f*x^m)^k]/(4*g*m^2*(g*x)^(2*m)) + (f^2*k*x 
^(2*m)*PolyLog[2, 1 + (f*x^m)/e])/(2*e^2*g*m^2*(g*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 = 0.10 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.82 \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {2 \, b f^{2} g^{-2 \, m - 1} k m n x^{2 \, m} \log \left (x\right ) \log \left (\frac {f x^{m} + e}{e}\right ) + 2 \, b f^{2} g^{-2 \, m - 1} k n x^{2 \, m} {\rm Li}_2\left (-\frac {f x^{m} + e}{e} + 1\right ) - {\left (b f^{2} k m^{2} n \log \left (x\right )^{2} + {\left (2 \, b f^{2} k m^{2} \log \left (c\right ) + 2 \, a f^{2} k m^{2} + b f^{2} k m n\right )} \log \left (x\right )\right )} g^{-2 \, m - 1} x^{2 \, m} - {\left (2 \, b e f k m n \log \left (x\right ) + 2 \, b e f k m \log \left (c\right ) + 2 \, a e f k m + 3 \, b e f k n\right )} g^{-2 \, m - 1} x^{m} - {\left (2 \, b e^{2} m n \log \left (d\right ) \log \left (x\right ) + {\left (2 \, b e^{2} m \log \left (c\right ) + 2 \, a e^{2} m + b e^{2} n\right )} \log \left (d\right )\right )} g^{-2 \, m - 1} + {\left ({\left (2 \, b f^{2} k m \log \left (c\right ) + 2 \, a f^{2} k m + b f^{2} k n\right )} g^{-2 \, m - 1} x^{2 \, m} - {\left (2 \, b e^{2} k m n \log \left (x\right ) + 2 \, b e^{2} k m \log \left (c\right ) + 2 \, a e^{2} k m + b e^{2} k n\right )} g^{-2 \, m - 1}\right )} \log \left (f x^{m} + e\right )}{4 \, e^{2} m^{2} x^{2 \, m}} \] 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*f^2*g^(-2*m - 1)*k*m*n*x^(2*m)*log(x)*log((f*x^m + e)/e) + 2*b*f^ 
2*g^(-2*m - 1)*k*n*x^(2*m)*dilog(-(f*x^m + e)/e + 1) - (b*f^2*k*m^2*n*log( 
x)^2 + (2*b*f^2*k*m^2*log(c) + 2*a*f^2*k*m^2 + b*f^2*k*m*n)*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*e^2*m*n*log(d)*log(x) + (2*b*e^2*m* 
log(c) + 2*a*e^2*m + b*e^2*n)*log(d))*g^(-2*m - 1) + ((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*n*log(x) + 
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) 
)/(e^2*m^2*x^(2*m))
 

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

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

Output:

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

Reduce [F]

\[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {-4 x^{2 m} \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{x^{3 m} f x +x^{2 m} e x}d x \right ) b \,e^{3} k \,m^{2}+4 x^{2 m} \mathrm {log}\left (x^{m} f +e \right ) a \,f^{2} k m +2 x^{2 m} \mathrm {log}\left (x^{m} f +e \right ) b \,f^{2} k n -4 x^{2 m} \mathrm {log}\left (x \right ) a \,f^{2} k \,m^{2}-2 x^{2 m} \mathrm {log}\left (x \right ) b \,f^{2} k m n -4 x^{m} a e f k m -2 x^{m} b e f k n -4 \,\mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{2} m -4 \,\mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) a \,e^{2} m -2 \,\mathrm {log}\left (\left (x^{m} f +e \right )^{k} d \right ) b \,e^{2} n -2 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} k m -b \,e^{2} k n}{8 x^{2 m} g^{2 m} e^{2} g \,m^{2}} \] Input:

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

Output:

( - 4*x**(2*m)*int(log(x**n*c)/(x**(3*m)*f*x + x**(2*m)*e*x),x)*b*e**3*k*m 
**2 + 4*x**(2*m)*log(x**m*f + e)*a*f**2*k*m + 2*x**(2*m)*log(x**m*f + e)*b 
*f**2*k*n - 4*x**(2*m)*log(x)*a*f**2*k*m**2 - 2*x**(2*m)*log(x)*b*f**2*k*m 
*n - 4*x**m*a*e*f*k*m - 2*x**m*b*e*f*k*n - 4*log((x**m*f + e)**k*d)*log(x* 
*n*c)*b*e**2*m - 4*log((x**m*f + e)**k*d)*a*e**2*m - 2*log((x**m*f + e)**k 
*d)*b*e**2*n - 2*log(x**n*c)*b*e**2*k*m - b*e**2*k*n)/(8*x**(2*m)*g**(2*m) 
*e**2*g*m**2)