\(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))^2}{(d+e x^m)^2} \, dx\) [362]

Optimal result

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

-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
 

Mathematica [A] (warning: unable to verify)

Time = 0.94 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\frac {x^{-m} (f x)^m \left (-\frac {m^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m}-\frac {2 a b m n \log \left (d-d x^m\right )}{d}+\frac {2 b^2 m n \left (n \log (x)-\log \left (c x^n\right )\right ) \log \left (d-d x^m\right )}{d}+\frac {2 b^2 n^2 \left (\frac {1}{2} m^2 \log ^2(x)+\left (-m \log (x)+\log \left (-\frac {e x^m}{d}\right )\right ) \log \left (d+e x^m\right )+\operatorname {PolyLog}\left (2,1+\frac {e x^m}{d}\right )\right )}{d}\right )}{e f m^3} \] Input:

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

Output:

((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] - Log[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)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2777, 2776, 2779, 2838}

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.

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

\(\Big \downarrow \) 2777

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

\(\Big \downarrow \) 2776

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

\(\Big \downarrow \) 2779

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

\(\Big \downarrow \) 2838

\(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {2 b n \left (\frac {b n \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )}{d m^2}-\frac {\log \left (\frac {d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d m}\right )}{e m}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2776

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*L 
og[c*x^n])^p/(e*r*(q + 1))), x] - Simp[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] || G 
tQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]
 

rule 2777

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])
 

rule 2779

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] + Simp[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 2838

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.93 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\frac {{\left (b^{2} e m^{2} n^{2} \log \left (x\right )^{2} + 2 \, {\left (b^{2} e m^{2} n \log \left (c\right ) + a b e m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} x^{m} - {\left (b^{2} d m^{2} \log \left (c\right )^{2} + 2 \, a b d m^{2} \log \left (c\right ) + a^{2} d m^{2}\right )} f^{m - 1} - 2 \, {\left (b^{2} e f^{m - 1} n^{2} x^{m} + b^{2} d f^{m - 1} n^{2}\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - 2 \, {\left ({\left (b^{2} e m n \log \left (c\right ) + a b e m n\right )} f^{m - 1} x^{m} + {\left (b^{2} d m n \log \left (c\right ) + a b d m n\right )} f^{m - 1}\right )} \log \left (e x^{m} + d\right ) - 2 \, {\left (b^{2} e f^{m - 1} m n^{2} x^{m} \log \left (x\right ) + b^{2} d f^{m - 1} m n^{2} \log \left (x\right )\right )} \log \left (\frac {e x^{m} + d}{d}\right )}{d e^{2} m^{3} x^{m} + d^{2} e m^{3}} \] Input:

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

Output:

((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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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) - integrate((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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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