\(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))^2}{d+e x^m} \, dx\) [363]

Optimal result

Integrand size = 29, antiderivative size = 129 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x^m}{d}\right )}{e m}+\frac {2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{e m^2}-\frac {2 b^2 n^2 x^{1-m} (f x)^{-1+m} \operatorname {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{e m^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 129, 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, 2375, 2421, 6724} \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\frac {2 b n x^{1-m} (f x)^{m-1} \operatorname {PolyLog}\left (2,-\frac {e x^m}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e m^2}+\frac {x^{1-m} (f x)^{m-1} \log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e m}-\frac {2 b^2 n^2 x^{1-m} (f x)^{m-1} \operatorname {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{e m^3} \]

[In]

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

[Out]

(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*x^(1 - m)*(f*x)^(-1 + m)*(a
+ b*Log[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)

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 2421

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] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[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]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

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}{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 {\left (2 b n x^{1-m} (f x)^{-1+m}\right ) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \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 )^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 ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2}-\frac {\left (2 b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac {\text {Li}_2\left (-\frac {e x^m}{d}\right )}{x} \, dx}{e m^2} \\ & = \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 ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2}-\frac {2 b^2 n^2 x^{1-m} (f x)^{-1+m} \text {Li}_3\left (-\frac {e x^m}{d}\right )}{e m^3} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

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 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=\frac {x^{-m} (f x)^m \left (3 a^2 m^3 \log (x)-6 a b m^3 n \log ^2(x)+4 b^2 m^3 n^2 \log ^3(x)+6 a b m^3 \log (x) \log \left (c x^n\right )-6 b^2 m^3 n \log ^2(x) \log \left (c x^n\right )+3 b^2 m^3 \log (x) \log ^2\left (c x^n\right )+3 b^2 m^2 n^2 \log ^2(x) \log \left (1+\frac {d x^{-m}}{e}\right )+3 a^2 m^2 \log \left (d-d x^m\right )-6 a b m^2 n \log (x) \log \left (d-d x^m\right )+3 b^2 m^2 n^2 \log ^2(x) \log \left (d-d x^m\right )+6 a b m^2 \log \left (c x^n\right ) \log \left (d-d x^m\right )-6 b^2 m^2 n \log (x) \log \left (c x^n\right ) \log \left (d-d x^m\right )+3 b^2 m^2 \log ^2\left (c x^n\right ) \log \left (d-d x^m\right )+6 a b m^2 n \log (x) \log \left (d+e x^m\right )-6 b^2 m^2 n^2 \log ^2(x) \log \left (d+e x^m\right )-6 a b m n \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+6 b^2 m n^2 \log (x) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+6 b^2 m^2 n \log (x) \log \left (c x^n\right ) \log \left (d+e x^m\right )-6 b^2 m n \log \left (-\frac {e x^m}{d}\right ) \log \left (c x^n\right ) \log \left (d+e x^m\right )-6 b^2 m n^2 \log (x) \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )-6 b m n \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,1+\frac {e x^m}{d}\right )-6 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {d x^{-m}}{e}\right )\right )}{3 e f m^3} \]

[In]

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

[Out]

((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*Log[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)]*L
og[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])*PolyLog[2, 1 + (e*x^m)/d] - 6*b^2*n^2*PolyLog[3, -(d/(e*x^m))]))/(3*e*f
*m^3*x^m)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.38 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m} \, dx=-\frac {2 \, b^{2} f^{m - 1} n^{2} {\rm polylog}\left (3, -\frac {e x^{m}}{d}\right ) - 2 \, {\left (b^{2} m n^{2} \log \left (x\right ) + b^{2} m n \log \left (c\right ) + a b m n\right )} f^{m - 1} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - {\left (b^{2} m^{2} \log \left (c\right )^{2} + 2 \, a b m^{2} \log \left (c\right ) + a^{2} m^{2}\right )} f^{m - 1} \log \left (e x^{m} + d\right ) - {\left (b^{2} m^{2} n^{2} \log \left (x\right )^{2} + 2 \, {\left (b^{2} m^{2} n \log \left (c\right ) + a b m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} \log \left (\frac {e x^{m} + d}{d}\right )}{e m^{3}} \]

[In]

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

[Out]

-(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*l
og(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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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