\(\int (f x)^{-1+2 n} \log ^2(c (d+e x^n)^p) \, dx\) [164]

Optimal result

Integrand size = 24, antiderivative size = 255 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {2 d p^2 x^{1-n} (f x)^{-1+2 n}}{e n}+\frac {p^2 x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2}{4 e^2 n}+\frac {2 d p x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2 n}-\frac {p x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}-\frac {d x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2 n}+\frac {x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2 n} \] Output:

-2*d*p^2*x^(1-n)*(f*x)^(-1+2*n)/e/n+1/4*p^2*x^(1-2*n)*(f*x)^(-1+2*n)*(d+e* 
x^n)^2/e^2/n+2*d*p*x^(1-2*n)*(f*x)^(-1+2*n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)/e^ 
2/n-1/2*p*x^(1-2*n)*(f*x)^(-1+2*n)*(d+e*x^n)^2*ln(c*(d+e*x^n)^p)/e^2/n-d*x 
^(1-2*n)*(f*x)^(-1+2*n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)^2/e^2/n+1/2*x^(1-2*n)* 
(f*x)^(-1+2*n)*(d+e*x^n)^2*ln(c*(d+e*x^n)^p)^2/e^2/n
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.55 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{-2 n} (f x)^{2 n} \left (2 d^2 p^2 \log ^2\left (d+e x^n\right )+2 d^2 p \log \left (d+e x^n\right ) \left (3 p-2 \log \left (c \left (d+e x^n\right )^p\right )\right )+e x^n \left (p^2 \left (-6 d+e x^n\right )+2 p \left (2 d-e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+2 e x^n \log ^2\left (c \left (d+e x^n\right )^p\right )\right )\right )}{4 e^2 f n} \] Input:

Integrate[(f*x)^(-1 + 2*n)*Log[c*(d + e*x^n)^p]^2,x]
 

Output:

((f*x)^(2*n)*(2*d^2*p^2*Log[d + e*x^n]^2 + 2*d^2*p*Log[d + e*x^n]*(3*p - 2 
*Log[c*(d + e*x^n)^p]) + e*x^n*(p^2*(-6*d + e*x^n) + 2*p*(2*d - e*x^n)*Log 
[c*(d + e*x^n)^p] + 2*e*x^n*Log[c*(d + e*x^n)^p]^2)))/(4*e^2*f*n*x^(2*n))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.64, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2906, 2904, 2848, 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.

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

\(\Big \downarrow \) 2906

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

\(\Big \downarrow \) 2904

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

\(\Big \downarrow \) 2848

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^{1-2 n} (f x)^{2 n-1} \left (\frac {\left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2}-\frac {d \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2}-\frac {p \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2}+\frac {2 d p \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2}+\frac {p^2 \left (d+e x^n\right )^2}{4 e^2}-\frac {2 d p^2 x^n}{e}\right )}{n}\)

Input:

Int[(f*x)^(-1 + 2*n)*Log[c*(d + e*x^n)^p]^2,x]
 

Output:

(x^(1 - 2*n)*(f*x)^(-1 + 2*n)*((-2*d*p^2*x^n)/e + (p^2*(d + e*x^n)^2)/(4*e 
^2) + (2*d*p*(d + e*x^n)*Log[c*(d + e*x^n)^p])/e^2 - (p*(d + e*x^n)^2*Log[ 
c*(d + e*x^n)^p])/(2*e^2) - (d*(d + e*x^n)*Log[c*(d + e*x^n)^p]^2)/e^2 + ( 
(d + e*x^n)^2*Log[c*(d + e*x^n)^p]^2)/(2*e^2)))/n
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2848

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d 
 + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - 
 d*g, 0] && IGtQ[q, 0]
 

rule 2904

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L 
og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, 
 x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & 
&  !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
 

rule 2906

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_)*( 
x_))^(m_), x_Symbol] :> Simp[(f*x)^m/x^m   Int[x^m*(a + b*Log[c*(d + e*x^n) 
^p])^q, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q}, x] && IntegerQ[Simp 
lify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])
 

Maple [F]

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

Input:

int((f*x)^(2*n-1)*ln(c*(d+e*x^n)^p)^2,x)
 

Output:

int((f*x)^(2*n-1)*ln(c*(d+e*x^n)^p)^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.80 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {{\left (e^{2} p^{2} - 2 \, e^{2} p \log \left (c\right ) + 2 \, e^{2} \log \left (c\right )^{2}\right )} f^{2 \, n - 1} x^{2 \, n} - 2 \, {\left (3 \, d e p^{2} - 2 \, d e p \log \left (c\right )\right )} f^{2 \, n - 1} x^{n} + 2 \, {\left (e^{2} f^{2 \, n - 1} p^{2} x^{2 \, n} - d^{2} f^{2 \, n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} + 2 \, {\left (2 \, d e f^{2 \, n - 1} p^{2} x^{n} - {\left (e^{2} p^{2} - 2 \, e^{2} p \log \left (c\right )\right )} f^{2 \, n - 1} x^{2 \, n} + {\left (3 \, d^{2} p^{2} - 2 \, d^{2} p \log \left (c\right )\right )} f^{2 \, n - 1}\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \] Input:

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

Output:

1/4*((e^2*p^2 - 2*e^2*p*log(c) + 2*e^2*log(c)^2)*f^(2*n - 1)*x^(2*n) - 2*( 
3*d*e*p^2 - 2*d*e*p*log(c))*f^(2*n - 1)*x^n + 2*(e^2*f^(2*n - 1)*p^2*x^(2* 
n) - d^2*f^(2*n - 1)*p^2)*log(e*x^n + d)^2 + 2*(2*d*e*f^(2*n - 1)*p^2*x^n 
- (e^2*p^2 - 2*e^2*p*log(c))*f^(2*n - 1)*x^(2*n) + (3*d^2*p^2 - 2*d^2*p*lo 
g(c))*f^(2*n - 1))*log(e*x^n + d))/(e^2*n)
 

Sympy [F]

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

integrate((f*x)**(-1+2*n)*ln(c*(d+e*x**n)**p)**2,x)
 

Output:

Integral((f*x)**(2*n - 1)*log(c*(d + e*x**n)**p)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.78 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {2 \, d^{2} f^{2 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{3} n} + \frac {e f^{2 \, n} x^{2 \, n} - 2 \, d f^{2 \, n} x^{n}}{e^{2} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, f} + \frac {\left (f x\right )^{2 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{2 \, f n} + \frac {{\left (2 \, d^{2} f^{2 \, n} \log \left (e x^{n} + d\right )^{2} + e^{2} f^{2 \, n} x^{2 \, n} - 6 \, d e f^{2 \, n} x^{n} - 2 \, {\left (2 \, f^{2 \, n} \log \left (e\right ) - 3 \, f^{2 \, n}\right )} d^{2} \log \left (e x^{n} + d\right )\right )} p^{2}}{4 \, e^{2} f n} \] Input:

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

Output:

-1/2*e*p*(2*d^2*f^(2*n)*log((e*x^n + d)/e)/(e^3*n) + (e*f^(2*n)*x^(2*n) - 
2*d*f^(2*n)*x^n)/(e^2*n))*log((e*x^n + d)^p*c)/f + 1/2*(f*x)^(2*n)*log((e* 
x^n + d)^p*c)^2/(f*n) + 1/4*(2*d^2*f^(2*n)*log(e*x^n + d)^2 + e^2*f^(2*n)* 
x^(2*n) - 6*d*e*f^(2*n)*x^n - 2*(2*f^(2*n)*log(e) - 3*f^(2*n))*d^2*log(e*x 
^n + d))*p^2/(e^2*f*n)
 

Giac [F]

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

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

Output:

integrate((f*x)^(2*n - 1)*log((e*x^n + d)^p*c)^2, x)
 

Mupad [F(-1)]

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

int(log(c*(d + e*x^n)^p)^2*(f*x)^(2*n - 1),x)
 

Output:

int(log(c*(d + e*x^n)^p)^2*(f*x)^(2*n - 1), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.56 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {f^{2 n} \left (2 x^{2 n} {\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right )}^{2} e^{2}-2 x^{2 n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) e^{2} p +x^{2 n} e^{2} p^{2}+4 x^{n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d e p -6 x^{n} d e \,p^{2}-2 {\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right )}^{2} d^{2}+6 \,\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d^{2} p \right )}{4 e^{2} f n} \] Input:

int((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p)^2,x)
 

Output:

(f**(2*n)*(2*x**(2*n)*log((x**n*e + d)**p*c)**2*e**2 - 2*x**(2*n)*log((x** 
n*e + d)**p*c)*e**2*p + x**(2*n)*e**2*p**2 + 4*x**n*log((x**n*e + d)**p*c) 
*d*e*p - 6*x**n*d*e*p**2 - 2*log((x**n*e + d)**p*c)**2*d**2 + 6*log((x**n* 
e + d)**p*c)*d**2*p))/(4*e**2*f*n)