Integrand size = 32, antiderivative size = 414 \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=-\frac {3 b f k n x^m (g x)^{-2 m}}{4 e g m^2}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log (x)}{4 e^2 g m}+\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log ^2(x)}{4 e^2 g}-\frac {f k x^m (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e g m}-\frac {f^2 k x^{2 m} (g x)^{-2 m} \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g}+\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right )}{4 e^2 g m^2}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 e^2 g m^2}+\frac {f^2 k x^{2 m} (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 e^2 g m}-\frac {b n (g x)^{-2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}-\frac {(g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \operatorname {PolyLog}\left (2,1+\frac {f x^m}{e}\right )}{2 e^2 g m^2} \]
-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))
Time = 0.39 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.73 \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {(g x)^{-2 m} \left (-2 a e f k m x^m-3 b e f k n x^m+b f^2 k m^2 n x^{2 m} \log ^2(x)-2 b e f k m x^m \log \left (c x^n\right )+2 a f^2 k m x^{2 m} \log \left (f-f x^{-m}\right )+b f^2 k n x^{2 m} \log \left (f-f x^{-m}\right )+2 b f^2 k m x^{2 m} \log \left (c x^n\right ) \log \left (f-f x^{-m}\right )-2 a e^2 m \log \left (d \left (e+f x^m\right )^k\right )-b e^2 n \log \left (d \left (e+f x^m\right )^k\right )-2 b e^2 m \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )-f^2 k m x^{2 m} \log (x) \left (2 a m+b n+2 b m \log \left (c x^n\right )+2 b n \log \left (f-f x^{-m}\right )-2 b n \log \left (1+\frac {f x^m}{e}\right )\right )+2 b f^2 k n x^{2 m} \operatorname {PolyLog}\left (2,-\frac {f x^m}{e}\right )\right )}{4 e^2 g m^2} \]
(-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))
Time = 0.71 (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 |
\(\displaystyle -\frac {(g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {f^2 k x^{2 m} \log (x) (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g}+\frac {f^2 k x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g m}-\frac {f k x^m (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e g m}-b n \left (\frac {(g x)^{-2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {f^2 k x^{2 m} (g x)^{-2 m} \operatorname {PolyLog}\left (2,\frac {f x^m}{e}+1\right )}{2 e^2 g m^2}-\frac {f^2 k x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right )}{4 e^2 g m^2}+\frac {f^2 k x^{2 m} (g x)^{-2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 e^2 g m^2}-\frac {f^2 k x^{2 m} \log ^2(x) (g x)^{-2 m}}{4 e^2 g}+\frac {f^2 k x^{2 m} \log (x) (g x)^{-2 m}}{4 e^2 g m}+\frac {3 f k x^m (g x)^{-2 m}}{4 e g m^2}\right )\) |
-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)))
3.2.54.3.1 Defintions of rubi rules used
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]
\[\int \left (g x \right )^{-1-2 m} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \,x^{m}\right )^{k}\right )d x\]
Time = 0.31 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.82 \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {2 \, b f^{2} g^{-2 \, m - 1} k m n x^{2 \, m} \log \left (x\right ) \log \left (\frac {f x^{m} + e}{e}\right ) + 2 \, b f^{2} g^{-2 \, m - 1} k n x^{2 \, m} {\rm Li}_2\left (-\frac {f x^{m} + e}{e} + 1\right ) - {\left (b f^{2} k m^{2} n \log \left (x\right )^{2} + {\left (2 \, b f^{2} k m^{2} \log \left (c\right ) + 2 \, a f^{2} k m^{2} + b f^{2} k m n\right )} \log \left (x\right )\right )} g^{-2 \, m - 1} x^{2 \, m} - {\left (2 \, b e f k m n \log \left (x\right ) + 2 \, b e f k m \log \left (c\right ) + 2 \, a e f k m + 3 \, b e f k n\right )} g^{-2 \, m - 1} x^{m} - {\left (2 \, b e^{2} m n \log \left (d\right ) \log \left (x\right ) + {\left (2 \, b e^{2} m \log \left (c\right ) + 2 \, a e^{2} m + b e^{2} n\right )} \log \left (d\right )\right )} g^{-2 \, m - 1} + {\left ({\left (2 \, b f^{2} k m \log \left (c\right ) + 2 \, a f^{2} k m + b f^{2} k n\right )} g^{-2 \, m - 1} x^{2 \, m} - {\left (2 \, b e^{2} k m n \log \left (x\right ) + 2 \, b e^{2} k m \log \left (c\right ) + 2 \, a e^{2} k m + b e^{2} k n\right )} g^{-2 \, m - 1}\right )} \log \left (f x^{m} + e\right )}{4 \, e^{2} m^{2} x^{2 \, m}} \]
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))
Timed out. \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\text {Timed out} \]
\[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{-2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right ) \,d x } \]
-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)
\[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{-2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right ) \,d x } \]
Timed out. \[ \int (g x)^{-1-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int \frac {\ln \left (d\,{\left (e+f\,x^m\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (g\,x\right )}^{2\,m+1}} \,d x \]