Integrand size = 27, antiderivative size = 226 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2 b^2 d n^2 x (f x)^{-1+m}}{m^3}+\frac {b^2 e n^2 x^{1+m} (f x)^{-1+m}}{4 m^3}+\frac {b^2 d^2 n^2 x^{1-m} (f x)^{-1+m} \log ^2(x)}{2 e m}-\frac {2 b d n x (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{m^2}-\frac {b e n x^{1+m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )}{2 m^2}-\frac {b d^2 n x^{1-m} (f x)^{-1+m} \log (x) \left (a+b \log \left (c x^n\right )\right )}{e m}+\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m} \]
2*b^2*d*n^2*x*(f*x)^(-1+m)/m^3+1/4*b^2*e*n^2*x^(1+m)*(f*x)^(-1+m)/m^3+1/2* b^2*d^2*n^2*x^(1-m)*(f*x)^(-1+m)*ln(x)^2/e/m-2*b*d*n*x*(f*x)^(-1+m)*(a+b*l n(c*x^n))/m^2-1/2*b*e*n*x^(1+m)*(f*x)^(-1+m)*(a+b*ln(c*x^n))/m^2-b*d^2*n*x ^(1-m)*(f*x)^(-1+m)*ln(x)*(a+b*ln(c*x^n))/e/m+1/2*x^(1-m)*(f*x)^(-1+m)*(d+ e*x^m)^2*(a+b*ln(c*x^n))^2/e/m
Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.55 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {(f x)^m \left (2 a^2 m^2 \left (2 d+e x^m\right )-2 a b m n \left (4 d+e x^m\right )+b^2 n^2 \left (8 d+e x^m\right )-2 b m \left (-2 a m \left (2 d+e x^m\right )+b n \left (4 d+e x^m\right )\right ) \log \left (c x^n\right )+2 b^2 m^2 \left (2 d+e x^m\right ) \log ^2\left (c x^n\right )\right )}{4 f m^3} \]
((f*x)^m*(2*a^2*m^2*(2*d + e*x^m) - 2*a*b*m*n*(4*d + e*x^m) + b^2*n^2*(8*d + e*x^m) - 2*b*m*(-2*a*m*(2*d + e*x^m) + b*n*(4*d + e*x^m))*Log[c*x^n] + 2*b^2*m^2*(2*d + e*x^m)*Log[c*x^n]^2))/(4*f*m^3)
Time = 0.59 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2777, 2776, 2772, 27, 2010, 2009}
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.
| \(\displaystyle \int (f x)^{m-1} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 2777 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \int x^{m-1} \left (e x^m+d\right ) \left (a+b \log \left (c x^n\right )\right )^2dx\) |
| \(\Big \downarrow \) 2776 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}-\frac {b n \int \frac {\left (e x^m+d\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x}dx}{e m}\right )\) |
| \(\Big \downarrow \) 2772 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}-\frac {b n \left (-b n \int \frac {e \left (e x^m+4 d\right ) x^m+2 d^2 m \log (x)}{2 m x}dx+d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {2 d e x^m \left (a+b \log \left (c x^n\right )\right )}{m}+\frac {e^2 x^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 m}\right )}{e m}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}-\frac {b n \left (-\frac {b n \int \frac {e \left (e x^m+4 d\right ) x^m+2 d^2 m \log (x)}{x}dx}{2 m}+d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {2 d e x^m \left (a+b \log \left (c x^n\right )\right )}{m}+\frac {e^2 x^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 m}\right )}{e m}\right )\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}-\frac {b n \left (-\frac {b n \int \left (4 d e x^{m-1}+e^2 x^{2 m-1}+\frac {2 d^2 m \log (x)}{x}\right )dx}{2 m}+d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {2 d e x^m \left (a+b \log \left (c x^n\right )\right )}{m}+\frac {e^2 x^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 m}\right )}{e m}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}-\frac {b n \left (d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {2 d e x^m \left (a+b \log \left (c x^n\right )\right )}{m}+\frac {e^2 x^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 m}-\frac {b n \left (d^2 m \log ^2(x)+\frac {4 d e x^m}{m}+\frac {e^2 x^{2 m}}{2 m}\right )}{2 m}\right )}{e m}\right )\) |
x^(1 - m)*(f*x)^(-1 + m)*(((d + e*x^m)^2*(a + b*Log[c*x^n])^2)/(2*e*m) - ( b*n*(-1/2*(b*n*((4*d*e*x^m)/m + (e^2*x^(2*m))/(2*m) + d^2*m*Log[x]^2))/m + (2*d*e*x^m*(a + b*Log[c*x^n]))/m + (e^2*x^(2*m)*(a + b*Log[c*x^n]))/(2*m) + d^2*Log[x]*(a + b*Log[c*x^n])))/(e*m))
3.4.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ .))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^r)^q, x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] && !(EqQ[q , 1] && EqQ[m, -1])
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]
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])
Time = 10.75 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.15
| method | result | size |
| parallelrisch | \(-\frac {-2 e \,b^{2} \left (f x \right )^{m -1} \ln \left (c \,x^{n}\right )^{2} x^{m} x \,m^{2}-4 x \,x^{m} \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} a b e \,m^{2}+2 x \,x^{m} \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} b^{2} e m n -4 b^{2} d \left (f x \right )^{m -1} \ln \left (c \,x^{n}\right )^{2} x \,m^{2}-2 x \,x^{m} \left (f x \right )^{m -1} a^{2} e \,m^{2}+2 x \,x^{m} \left (f x \right )^{m -1} a b e m n -x \,x^{m} \left (f x \right )^{m -1} b^{2} e \,n^{2}-8 x \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} a b d \,m^{2}+8 x \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} b^{2} d m n -4 x \left (f x \right )^{m -1} a^{2} d \,m^{2}+8 x \left (f x \right )^{m -1} a b d m n -8 x \left (f x \right )^{m -1} b^{2} d \,n^{2}}{4 m^{3}}\) | \(261\) |
| risch | \(\text {Expression too large to display}\) | \(1919\) |
-1/4*(-2*e*b^2*(f*x)^(m-1)*ln(c*x^n)^2*x^m*x*m^2-4*x*x^m*ln(c*x^n)*(f*x)^( m-1)*a*b*e*m^2+2*x*x^m*ln(c*x^n)*(f*x)^(m-1)*b^2*e*m*n-4*b^2*d*(f*x)^(m-1) *ln(c*x^n)^2*x*m^2-2*x*x^m*(f*x)^(m-1)*a^2*e*m^2+2*x*x^m*(f*x)^(m-1)*a*b*e *m*n-x*x^m*(f*x)^(m-1)*b^2*e*n^2-8*x*ln(c*x^n)*(f*x)^(m-1)*a*b*d*m^2+8*x*l n(c*x^n)*(f*x)^(m-1)*b^2*d*m*n-4*x*(f*x)^(m-1)*a^2*d*m^2+8*x*(f*x)^(m-1)*a *b*d*m*n-8*x*(f*x)^(m-1)*b^2*d*n^2)/m^3
Time = 0.32 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.08 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {{\left (2 \, b^{2} e m^{2} n^{2} \log \left (x\right )^{2} + 2 \, b^{2} e m^{2} \log \left (c\right )^{2} + 2 \, a^{2} e m^{2} - 2 \, a b e m n + b^{2} e n^{2} + 2 \, {\left (2 \, a b e m^{2} - b^{2} e m n\right )} \log \left (c\right ) + 2 \, {\left (2 \, b^{2} e m^{2} n \log \left (c\right ) + 2 \, a b e m^{2} n - b^{2} e m n^{2}\right )} \log \left (x\right )\right )} f^{m - 1} x^{2 \, m} + 4 \, {\left (b^{2} d m^{2} n^{2} \log \left (x\right )^{2} + b^{2} d m^{2} \log \left (c\right )^{2} + a^{2} d m^{2} - 2 \, a b d m n + 2 \, b^{2} d n^{2} + 2 \, {\left (a b d m^{2} - b^{2} d m n\right )} \log \left (c\right ) + 2 \, {\left (b^{2} d m^{2} n \log \left (c\right ) + a b d m^{2} n - b^{2} d m n^{2}\right )} \log \left (x\right )\right )} f^{m - 1} x^{m}}{4 \, m^{3}} \]
1/4*((2*b^2*e*m^2*n^2*log(x)^2 + 2*b^2*e*m^2*log(c)^2 + 2*a^2*e*m^2 - 2*a* b*e*m*n + b^2*e*n^2 + 2*(2*a*b*e*m^2 - b^2*e*m*n)*log(c) + 2*(2*b^2*e*m^2* n*log(c) + 2*a*b*e*m^2*n - b^2*e*m*n^2)*log(x))*f^(m - 1)*x^(2*m) + 4*(b^2 *d*m^2*n^2*log(x)^2 + b^2*d*m^2*log(c)^2 + a^2*d*m^2 - 2*a*b*d*m*n + 2*b^2 *d*n^2 + 2*(a*b*d*m^2 - b^2*d*m*n)*log(c) + 2*(b^2*d*m^2*n*log(c) + a*b*d* m^2*n - b^2*d*m*n^2)*log(x))*f^(m - 1)*x^m)/m^3
Time = 9.28 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.56 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} d x \left (f x\right )^{m - 1}}{m} + \frac {a^{2} e x x^{m} \left (f x\right )^{m - 1}}{2 m} + \frac {2 a b d x \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m} - \frac {2 a b d n x \left (f x\right )^{m - 1}}{m^{2}} + \frac {a b e x x^{m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m} - \frac {a b e n x x^{m} \left (f x\right )^{m - 1}}{2 m^{2}} + \frac {b^{2} d x \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}^{2}}{m} - \frac {2 b^{2} d n x \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m^{2}} + \frac {2 b^{2} d n^{2} x \left (f x\right )^{m - 1}}{m^{3}} + \frac {b^{2} e x x^{m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}^{2}}{2 m} - \frac {b^{2} e n x x^{m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{2 m^{2}} + \frac {b^{2} e n^{2} x x^{m} \left (f x\right )^{m - 1}}{4 m^{3}} & \text {for}\: m \neq 0 \\\frac {\left (d + e\right ) \left (\begin {cases} \frac {a^{2} \log {\left (c x^{n} \right )} + a b \log {\left (c x^{n} \right )}^{2} + \frac {b^{2} \log {\left (c x^{n} \right )}^{3}}{3}}{n} & \text {for}\: n \neq 0 \\\left (a^{2} + 2 a b \log {\left (c \right )} + b^{2} \log {\left (c \right )}^{2}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases}\right )}{f} & \text {otherwise} \end {cases} \]
Piecewise((a**2*d*x*(f*x)**(m - 1)/m + a**2*e*x*x**m*(f*x)**(m - 1)/(2*m) + 2*a*b*d*x*(f*x)**(m - 1)*log(c*x**n)/m - 2*a*b*d*n*x*(f*x)**(m - 1)/m**2 + a*b*e*x*x**m*(f*x)**(m - 1)*log(c*x**n)/m - a*b*e*n*x*x**m*(f*x)**(m - 1)/(2*m**2) + b**2*d*x*(f*x)**(m - 1)*log(c*x**n)**2/m - 2*b**2*d*n*x*(f*x )**(m - 1)*log(c*x**n)/m**2 + 2*b**2*d*n**2*x*(f*x)**(m - 1)/m**3 + b**2*e *x*x**m*(f*x)**(m - 1)*log(c*x**n)**2/(2*m) - b**2*e*n*x*x**m*(f*x)**(m - 1)*log(c*x**n)/(2*m**2) + b**2*e*n**2*x*x**m*(f*x)**(m - 1)/(4*m**3), Ne(m , 0)), ((d + e)*Piecewise(((a**2*log(c*x**n) + a*b*log(c*x**n)**2 + b**2*l og(c*x**n)**3/3)/n, Ne(n, 0)), ((a**2 + 2*a*b*log(c) + b**2*log(c)**2)*log (x), True))/f, True))
Time = 0.21 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.14 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {b^{2} e f^{m - 1} x^{2 \, m} \log \left (c x^{n}\right )^{2}}{2 \, m} + \frac {a b e f^{m - 1} x^{2 \, m} \log \left (c x^{n}\right )}{m} - 2 \, {\left (\frac {f^{m - 1} n x^{m} \log \left (c x^{n}\right )}{m^{2}} - \frac {f^{m - 1} n^{2} x^{m}}{m^{3}}\right )} b^{2} d - \frac {1}{4} \, {\left (\frac {2 \, f^{m - 1} n x^{2 \, m} \log \left (c x^{n}\right )}{m^{2}} - \frac {f^{m - 1} n^{2} x^{2 \, m}}{m^{3}}\right )} b^{2} e + \frac {a^{2} e f^{m - 1} x^{2 \, m}}{2 \, m} - \frac {a b e f^{m - 1} n x^{2 \, m}}{2 \, m^{2}} - \frac {2 \, a b d f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b^{2} d \log \left (c x^{n}\right )^{2}}{f m} + \frac {2 \, \left (f x\right )^{m} a b d \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a^{2} d}{f m} \]
1/2*b^2*e*f^(m - 1)*x^(2*m)*log(c*x^n)^2/m + a*b*e*f^(m - 1)*x^(2*m)*log(c *x^n)/m - 2*(f^(m - 1)*n*x^m*log(c*x^n)/m^2 - f^(m - 1)*n^2*x^m/m^3)*b^2*d - 1/4*(2*f^(m - 1)*n*x^(2*m)*log(c*x^n)/m^2 - f^(m - 1)*n^2*x^(2*m)/m^3)* b^2*e + 1/2*a^2*e*f^(m - 1)*x^(2*m)/m - 1/2*a*b*e*f^(m - 1)*n*x^(2*m)/m^2 - 2*a*b*d*f^(m - 1)*n*x^m/m^2 + (f*x)^m*b^2*d*log(c*x^n)^2/(f*m) + 2*(f*x) ^m*a*b*d*log(c*x^n)/(f*m) + (f*x)^m*a^2*d/(f*m)
Time = 0.45 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.92 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {b^{2} e f^{m} n^{2} x^{2 \, m} \log \left (x\right )^{2}}{2 \, f m} + \frac {b^{2} d f^{m} n^{2} x^{m} \log \left (x\right )^{2}}{f m} + \frac {b^{2} e f^{m} n x^{2 \, m} \log \left (c\right ) \log \left (x\right )}{f m} + \frac {2 \, b^{2} d f^{m} n x^{m} \log \left (c\right ) \log \left (x\right )}{f m} + \frac {b^{2} e f^{m} x^{2 \, m} \log \left (c\right )^{2}}{2 \, f m} + \frac {b^{2} d f^{m} x^{m} \log \left (c\right )^{2}}{f m} + \frac {a b e f^{m} n x^{2 \, m} \log \left (x\right )}{f m} - \frac {b^{2} e f^{m} n^{2} x^{2 \, m} \log \left (x\right )}{2 \, f m^{2}} + \frac {2 \, a b d f^{m} n x^{m} \log \left (x\right )}{f m} - \frac {2 \, b^{2} d f^{m} n^{2} x^{m} \log \left (x\right )}{f m^{2}} + \frac {a b e f^{m} x^{2 \, m} \log \left (c\right )}{f m} - \frac {b^{2} e f^{m} n x^{2 \, m} \log \left (c\right )}{2 \, f m^{2}} + \frac {2 \, a b d f^{m} x^{m} \log \left (c\right )}{f m} - \frac {2 \, b^{2} d f^{m} n x^{m} \log \left (c\right )}{f m^{2}} + \frac {a^{2} e f^{m} x^{2 \, m}}{2 \, f m} - \frac {a b e f^{m} n x^{2 \, m}}{2 \, f m^{2}} + \frac {b^{2} e f^{m} n^{2} x^{2 \, m}}{4 \, f m^{3}} + \frac {a^{2} d f^{m} x^{m}}{f m} - \frac {2 \, a b d f^{m} n x^{m}}{f m^{2}} + \frac {2 \, b^{2} d f^{m} n^{2} x^{m}}{f m^{3}} \]
1/2*b^2*e*f^m*n^2*x^(2*m)*log(x)^2/(f*m) + b^2*d*f^m*n^2*x^m*log(x)^2/(f*m ) + b^2*e*f^m*n*x^(2*m)*log(c)*log(x)/(f*m) + 2*b^2*d*f^m*n*x^m*log(c)*log (x)/(f*m) + 1/2*b^2*e*f^m*x^(2*m)*log(c)^2/(f*m) + b^2*d*f^m*x^m*log(c)^2/ (f*m) + a*b*e*f^m*n*x^(2*m)*log(x)/(f*m) - 1/2*b^2*e*f^m*n^2*x^(2*m)*log(x )/(f*m^2) + 2*a*b*d*f^m*n*x^m*log(x)/(f*m) - 2*b^2*d*f^m*n^2*x^m*log(x)/(f *m^2) + a*b*e*f^m*x^(2*m)*log(c)/(f*m) - 1/2*b^2*e*f^m*n*x^(2*m)*log(c)/(f *m^2) + 2*a*b*d*f^m*x^m*log(c)/(f*m) - 2*b^2*d*f^m*n*x^m*log(c)/(f*m^2) + 1/2*a^2*e*f^m*x^(2*m)/(f*m) - 1/2*a*b*e*f^m*n*x^(2*m)/(f*m^2) + 1/4*b^2*e* f^m*n^2*x^(2*m)/(f*m^3) + a^2*d*f^m*x^m/(f*m) - 2*a*b*d*f^m*n*x^m/(f*m^2) + 2*b^2*d*f^m*n^2*x^m/(f*m^3)
Timed out. \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\int {\left (f\,x\right )}^{m-1}\,\left (d+e\,x^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2 \,d x \]