\(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))}{d+e x^m} \, dx\) [355]

Optimal result

Integrand size = 27, antiderivative size = 77 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d+e x^m} \, dx=\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^m}{d}\right )}{e m}+\frac {b n x^{1-m} (f x)^{-1+m} \operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{e m^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2377, 2375, 2438} \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d+e x^m} \, dx=\frac {x^{1-m} (f x)^{m-1} \log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e m}+\frac {b n x^{1-m} (f x)^{m-1} \operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{e m^2} \]

[In]

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

[Out]

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

Rule 2375

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^p/(e*r)), x] - Dist[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]

Rule 2377

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :>
Dist[(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])

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps \begin{align*} \text {integral}& = \left (x^{1-m} (f x)^{-1+m}\right ) \int \frac {x^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d+e x^m} \, dx \\ & = \frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^m}{d}\right )}{e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \int \frac {\log \left (1+\frac {e x^m}{d}\right )}{x} \, dx}{e m} \\ & = \frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^m}{d}\right )}{e m}+\frac {b n x^{1-m} (f x)^{-1+m} \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.83 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d+e x^m} \, dx=\frac {x^{-m} (f x)^m \left (-b m^2 n \log ^2(x)+a m \log \left (d-d x^m\right )+b m \log \left (c x^n\right ) \log \left (d-d x^m\right )-b n \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+m \log (x) \left (a m+b m \log \left (c x^n\right )-b n \log \left (d-d x^m\right )+b n \log \left (d+e x^m\right )\right )-b n \operatorname {PolyLog}\left (2,1+\frac {e x^m}{d}\right )\right )}{e f m^2} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{d+e x^m} \, dx=\frac {b f^{m - 1} m n \log \left (x\right ) \log \left (\frac {e x^{m} + d}{d}\right ) + b f^{m - 1} n {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) + {\left (b m \log \left (c\right ) + a m\right )} f^{m - 1} \log \left (e x^{m} + d\right )}{e m^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((f*x)**(m - 1)*(a + b*log(c*x**n))/(d + e*x**m), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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