Integrand size = 29, antiderivative size = 129 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x^m}{d}\right )}{e m}+\frac {2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{e m^2}-\frac {2 b^2 n^2 x^{1-m} (f x)^{-1+m} \operatorname {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{e m^3} \]
x^(1-m)*(f*x)^(-1+m)*(a+b*ln(c*x^n))^2*ln(1+e*x^m/d)/e/m+2*b*n*x^(1-m)*(f* x)^(-1+m)*(a+b*ln(c*x^n))*polylog(2,-e*x^m/d)/e/m^2-2*b^2*n^2*x^(1-m)*(f*x )^(-1+m)*polylog(3,-e*x^m/d)/e/m^3
Leaf count is larger than twice the leaf count of optimal. \(502\) vs. \(2(129)=258\).
Time = 0.40 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.89 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\frac {x^{-m} (f x)^m \left (3 a^2 m^3 \log (x)-6 a b m^3 n \log ^2(x)+4 b^2 m^3 n^2 \log ^3(x)+6 a b m^3 \log (x) \log \left (c x^n\right )-6 b^2 m^3 n \log ^2(x) \log \left (c x^n\right )+3 b^2 m^3 \log (x) \log ^2\left (c x^n\right )+3 b^2 m^2 n^2 \log ^2(x) \log \left (1+\frac {d x^{-m}}{e}\right )+3 a^2 m^2 \log \left (d-d x^m\right )-6 a b m^2 n \log (x) \log \left (d-d x^m\right )+3 b^2 m^2 n^2 \log ^2(x) \log \left (d-d x^m\right )+6 a b m^2 \log \left (c x^n\right ) \log \left (d-d x^m\right )-6 b^2 m^2 n \log (x) \log \left (c x^n\right ) \log \left (d-d x^m\right )+3 b^2 m^2 \log ^2\left (c x^n\right ) \log \left (d-d x^m\right )+6 a b m^2 n \log (x) \log \left (d+e x^m\right )-6 b^2 m^2 n^2 \log ^2(x) \log \left (d+e x^m\right )-6 a b m n \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+6 b^2 m n^2 \log (x) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+6 b^2 m^2 n \log (x) \log \left (c x^n\right ) \log \left (d+e x^m\right )-6 b^2 m n \log \left (-\frac {e x^m}{d}\right ) \log \left (c x^n\right ) \log \left (d+e x^m\right )-6 b^2 m n^2 \log (x) \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )-6 b m n \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,1+\frac {e x^m}{d}\right )-6 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {d x^{-m}}{e}\right )\right )}{3 e f m^3} \]
((f*x)^m*(3*a^2*m^3*Log[x] - 6*a*b*m^3*n*Log[x]^2 + 4*b^2*m^3*n^2*Log[x]^3 + 6*a*b*m^3*Log[x]*Log[c*x^n] - 6*b^2*m^3*n*Log[x]^2*Log[c*x^n] + 3*b^2*m ^3*Log[x]*Log[c*x^n]^2 + 3*b^2*m^2*n^2*Log[x]^2*Log[1 + d/(e*x^m)] + 3*a^2 *m^2*Log[d - d*x^m] - 6*a*b*m^2*n*Log[x]*Log[d - d*x^m] + 3*b^2*m^2*n^2*Lo g[x]^2*Log[d - d*x^m] + 6*a*b*m^2*Log[c*x^n]*Log[d - d*x^m] - 6*b^2*m^2*n* Log[x]*Log[c*x^n]*Log[d - d*x^m] + 3*b^2*m^2*Log[c*x^n]^2*Log[d - d*x^m] + 6*a*b*m^2*n*Log[x]*Log[d + e*x^m] - 6*b^2*m^2*n^2*Log[x]^2*Log[d + e*x^m] - 6*a*b*m*n*Log[-((e*x^m)/d)]*Log[d + e*x^m] + 6*b^2*m*n^2*Log[x]*Log[-(( e*x^m)/d)]*Log[d + e*x^m] + 6*b^2*m^2*n*Log[x]*Log[c*x^n]*Log[d + e*x^m] - 6*b^2*m*n*Log[-((e*x^m)/d)]*Log[c*x^n]*Log[d + e*x^m] - 6*b^2*m*n^2*Log[x ]*PolyLog[2, -(d/(e*x^m))] - 6*b*m*n*(a - b*n*Log[x] + b*Log[c*x^n])*PolyL og[2, 1 + (e*x^m)/d] - 6*b^2*n^2*PolyLog[3, -(d/(e*x^m))]))/(3*e*f*m^3*x^m )
Time = 0.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.78, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2777, 2775, 2821, 7143}
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.
| \(\displaystyle \int \frac {(f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx\) |
| \(\Big \downarrow \) 2777 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \int \frac {x^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{e x^m+d}dx\) |
| \(\Big \downarrow \) 2775 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e m}-\frac {2 b n \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (\frac {e x^m}{d}+1\right )}{x}dx}{e m}\right )\) |
| \(\Big \downarrow \) 2821 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e m}-\frac {2 b n \left (\frac {b n \int \frac {\operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{x}dx}{m}-\frac {\operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{m}\right )}{e m}\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e m}-\frac {2 b n \left (\frac {b n \operatorname {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{m^2}-\frac {\operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{m}\right )}{e m}\right )\) |
x^(1 - m)*(f*x)^(-1 + m)*(((a + b*Log[c*x^n])^2*Log[1 + (e*x^m)/d])/(e*m) - (2*b*n*(-(((a + b*Log[c*x^n])*PolyLog[2, -((e*x^m)/d)])/m) + (b*n*PolyLo g[3, -((e*x^m)/d)])/m^2))/(e*m))
3.4.63.3.1 Defintions of rubi rules used
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Simp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c* x^n])^p/(e*r)), x] - Simp[b*f^m*n*(p/(e*r)) Int[Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] & & EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + ( e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[(f*x)^m/x^m Int[x^m*(d + e*x^r)^ q*(a + b*Log[c*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && !(IntegerQ[m] || GtQ[f, 0])
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b _.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c *x^n])^p/m), x] + Simp[b*n*(p/m) Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c *x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
\[\int \frac {\left (f x \right )^{m -1} {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{d +e \,x^{m}}d x\]
Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.38 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=-\frac {2 \, b^{2} f^{m - 1} n^{2} {\rm polylog}\left (3, -\frac {e x^{m}}{d}\right ) - 2 \, {\left (b^{2} m n^{2} \log \left (x\right ) + b^{2} m n \log \left (c\right ) + a b m n\right )} f^{m - 1} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - {\left (b^{2} m^{2} \log \left (c\right )^{2} + 2 \, a b m^{2} \log \left (c\right ) + a^{2} m^{2}\right )} f^{m - 1} \log \left (e x^{m} + d\right ) - {\left (b^{2} m^{2} n^{2} \log \left (x\right )^{2} + 2 \, {\left (b^{2} m^{2} n \log \left (c\right ) + a b m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} \log \left (\frac {e x^{m} + d}{d}\right )}{e m^{3}} \]
-(2*b^2*f^(m - 1)*n^2*polylog(3, -e*x^m/d) - 2*(b^2*m*n^2*log(x) + b^2*m*n *log(c) + a*b*m*n)*f^(m - 1)*dilog(-(e*x^m + d)/d + 1) - (b^2*m^2*log(c)^2 + 2*a*b*m^2*log(c) + a^2*m^2)*f^(m - 1)*log(e*x^m + d) - (b^2*m^2*n^2*log (x)^2 + 2*(b^2*m^2*n*log(c) + a*b*m^2*n)*log(x))*f^(m - 1)*log((e*x^m + d) /d))/(e*m^3)
\[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\int \frac {\left (f x\right )^{m - 1} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{d + e x^{m}}\, dx \]
\[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \left (f x\right )^{m - 1}}{e x^{m} + d} \,d x } \]
a^2*f^(m - 1)*log((e*x^m + d)/e)/(e*m) + integrate((b^2*f^m*x^m*log(x^n)^2 + 2*(b^2*f^m*log(c) + a*b*f^m)*x^m*log(x^n) + (b^2*f^m*log(c)^2 + 2*a*b*f ^m*log(c))*x^m)/(e*f*x*x^m + d*f*x), x)
\[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \left (f x\right )^{m - 1}}{e x^{m} + d} \,d x } \]
Timed out. \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\int \frac {{\left (f\,x\right )}^{m-1}\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{d+e\,x^m} \,d x \]