\(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))^2}{(d+e x^m)^2} \, dx\) [364]

Optimal result

Integrand size = 29, antiderivative size = 138 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=-\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}-\frac {2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x^{-m}}{e}\right )}{d e m^2}+\frac {2 b^2 n^2 x^{1-m} (f x)^{-1+m} \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )}{d e m^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2377, 2376, 2379, 2438} \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=-\frac {2 b n x^{1-m} (f x)^{m-1} \log \left (\frac {d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d e m^2}-\frac {x^{1-m} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}+\frac {2 b^2 n^2 x^{1-m} (f x)^{m-1} \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )}{d e m^3} \]

[In]

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

[Out]

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

Rule 2376

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

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 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 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 )^2}{\left (d+e x^m\right )^2} \, dx \\ & = -\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}+\frac {\left (2 b n x^{1-m} (f x)^{-1+m}\right ) \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^m\right )} \, dx}{e m} \\ & = -\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}-\frac {2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x^{-m}}{e}\right )}{d e m^2}+\frac {\left (2 b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac {\log \left (1+\frac {d x^{-m}}{e}\right )}{x} \, dx}{d e m^2} \\ & = -\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}-\frac {2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x^{-m}}{e}\right )}{d e m^2}+\frac {2 b^2 n^2 x^{1-m} (f x)^{-1+m} \text {Li}_2\left (-\frac {d x^{-m}}{e}\right )}{d e m^3} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\frac {x^{-m} (f x)^m \left (-\frac {m^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m}-\frac {2 a b m n \log \left (d-d x^m\right )}{d}+\frac {2 b^2 m n \left (n \log (x)-\log \left (c x^n\right )\right ) \log \left (d-d x^m\right )}{d}+\frac {2 b^2 n^2 \left (\frac {1}{2} m^2 \log ^2(x)+\left (-m \log (x)+\log \left (-\frac {e x^m}{d}\right )\right ) \log \left (d+e x^m\right )+\operatorname {PolyLog}\left (2,1+\frac {e x^m}{d}\right )\right )}{d}\right )}{e f m^3} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.93 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\frac {{\left (b^{2} e m^{2} n^{2} \log \left (x\right )^{2} + 2 \, {\left (b^{2} e m^{2} n \log \left (c\right ) + a b e m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} x^{m} - {\left (b^{2} d m^{2} \log \left (c\right )^{2} + 2 \, a b d m^{2} \log \left (c\right ) + a^{2} d m^{2}\right )} f^{m - 1} - 2 \, {\left (b^{2} e f^{m - 1} n^{2} x^{m} + b^{2} d f^{m - 1} n^{2}\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - 2 \, {\left ({\left (b^{2} e m n \log \left (c\right ) + a b e m n\right )} f^{m - 1} x^{m} + {\left (b^{2} d m n \log \left (c\right ) + a b d m n\right )} f^{m - 1}\right )} \log \left (e x^{m} + d\right ) - 2 \, {\left (b^{2} e f^{m - 1} m n^{2} x^{m} \log \left (x\right ) + b^{2} d f^{m - 1} m n^{2} \log \left (x\right )\right )} \log \left (\frac {e x^{m} + d}{d}\right )}{d e^{2} m^{3} x^{m} + d^{2} e m^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

2*a*b*f^m*n*(log(x)/(d*e*f*m) - log(e*x^m + d)/(d*e*f*m^2)) - (f^m*log(x^n)^2/(e^2*f*m*x^m + d*e*f*m) - integr
ate((e*f^m*m*x^m*log(c)^2 + 2*(d*f^m*n + (e*f^m*m*log(c) + e*f^m*n)*x^m)*log(x^n))/(e^3*f*m*x*x^(2*m) + 2*d*e^
2*f*m*x*x^m + d^2*e*f*m*x), x))*b^2 - 2*a*b*f^m*log(c*x^n)/(e^2*f*m*x^m + d*e*f*m) - a^2*f^m/(e^2*f*m*x^m + d*
e*f*m)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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