\(\int (f x)^{-1+3 n} \log ^2(c (d+e x^n)^p) \, dx\) [163]

Optimal result

Integrand size = 24, antiderivative size = 372 \[ \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {2 d^2 p^2 x^{1-2 n} (f x)^{-1+3 n}}{e^2 n}-\frac {d p^2 x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^2}{2 e^3 n}+\frac {2 p^2 x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^3}{27 e^3 n}-\frac {d^3 p^2 x^{1-3 n} (f x)^{-1+3 n} \log ^2\left (d+e x^n\right )}{3 e^3 n}-\frac {2 d^2 p x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^3 n}+\frac {d p x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{e^3 n}-\frac {2 p x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^3 \log \left (c \left (d+e x^n\right )^p\right )}{9 e^3 n}+\frac {2 d^3 p x^{1-3 n} (f x)^{-1+3 n} \log \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{3 e^3 n}+\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n} \] Output:

2*d^2*p^2*x^(1-2*n)*(f*x)^(-1+3*n)/e^2/n-1/2*d*p^2*x^(1-3*n)*(f*x)^(-1+3*n 
)*(d+e*x^n)^2/e^3/n+2/27*p^2*x^(1-3*n)*(f*x)^(-1+3*n)*(d+e*x^n)^3/e^3/n-1/ 
3*d^3*p^2*x^(1-3*n)*(f*x)^(-1+3*n)*ln(d+e*x^n)^2/e^3/n-2*d^2*p*x^(1-3*n)*( 
f*x)^(-1+3*n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)/e^3/n+d*p*x^(1-3*n)*(f*x)^(-1+3* 
n)*(d+e*x^n)^2*ln(c*(d+e*x^n)^p)/e^3/n-2/9*p*x^(1-3*n)*(f*x)^(-1+3*n)*(d+e 
*x^n)^3*ln(c*(d+e*x^n)^p)/e^3/n+2/3*d^3*p*x^(1-3*n)*(f*x)^(-1+3*n)*ln(d+e* 
x^n)*ln(c*(d+e*x^n)^p)/e^3/n+1/3*x*(f*x)^(-1+3*n)*ln(c*(d+e*x^n)^p)^2/n
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.46 \[ \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{-3 n} (f x)^{3 n} \left (-18 d^3 p^2 \log ^2\left (d+e x^n\right )+6 d^3 p \log \left (d+e x^n\right ) \left (-11 p+6 \log \left (c \left (d+e x^n\right )^p\right )\right )+e x^n \left (p^2 \left (66 d^2-15 d e x^n+4 e^2 x^{2 n}\right )-6 p \left (6 d^2-3 d e x^n+2 e^2 x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )+18 e^2 x^{2 n} \log ^2\left (c \left (d+e x^n\right )^p\right )\right )\right )}{54 e^3 f n} \] Input:

Integrate[(f*x)^(-1 + 3*n)*Log[c*(d + e*x^n)^p]^2,x]
 

Output:

((f*x)^(3*n)*(-18*d^3*p^2*Log[d + e*x^n]^2 + 6*d^3*p*Log[d + e*x^n]*(-11*p 
 + 6*Log[c*(d + e*x^n)^p]) + e*x^n*(p^2*(66*d^2 - 15*d*e*x^n + 4*e^2*x^(2* 
n)) - 6*p*(6*d^2 - 3*d*e*x^n + 2*e^2*x^(2*n))*Log[c*(d + e*x^n)^p] + 18*e^ 
2*x^(2*n)*Log[c*(d + e*x^n)^p]^2)))/(54*e^3*f*n*x^(3*n))
 

Rubi [A] (warning: unable to verify)

Time = 0.88 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.53, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2906, 2904, 2845, 2858, 25, 27, 2772, 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.

\(\displaystyle \int (f x)^{3 n-1} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2906

\(\displaystyle x^{1-3 n} (f x)^{3 n-1} \int x^{3 n-1} \log ^2\left (c \left (e x^n+d\right )^p\right )dx\)

\(\Big \downarrow \) 2904

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \int x^{2 n} \log ^2\left (c \left (e x^n+d\right )^p\right )dx^n}{n}\)

\(\Big \downarrow \) 2845

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \left (\frac {1}{3} x^{3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )-\frac {2}{3} e p \int \frac {x^{3 n} \log \left (c \left (e x^n+d\right )^p\right )}{e x^n+d}dx^n\right )}{n}\)

\(\Big \downarrow \) 2858

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \left (\frac {1}{3} x^{3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )-\frac {2}{3} p \int x^{2 n} \log \left (c \left (e x^n+d\right )^p\right )d\left (e x^n+d\right )\right )}{n}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \left (\frac {2}{3} p \int -x^{2 n} \log \left (c \left (e x^n+d\right )^p\right )d\left (e x^n+d\right )+\frac {1}{3} x^{3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )\right )}{n}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \left (\frac {2 p \int -e^3 x^{2 n} \log \left (c \left (e x^n+d\right )^p\right )d\left (e x^n+d\right )}{3 e^3}+\frac {1}{3} x^{3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )\right )}{n}\)

\(\Big \downarrow \) 2772

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \left (\frac {2 p \left (-p \int \left (d^3 \log \left (e x^n+d\right ) x^{-n}-\frac {x^{2 n}}{3}-3 d^2+\frac {3}{2} d \left (e x^n+d\right )\right )d\left (e x^n+d\right )+d^3 \log \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )-3 d^2 \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+\frac {3}{2} d x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {1}{3} x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )\right )}{3 e^3}+\frac {1}{3} x^{3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )\right )}{n}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^{1-3 n} (f x)^{3 n-1} \left (\frac {2 p \left (d^3 \log \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )-3 d^2 \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+\frac {3}{2} d x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {1}{3} x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )-p \left (\frac {1}{2} d^3 \log ^2\left (d+e x^n\right )-3 d^2 \left (d+e x^n\right )+\frac {3}{4} d x^{2 n}-\frac {x^{3 n}}{9}\right )\right )}{3 e^3}+\frac {1}{3} x^{3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )\right )}{n}\)

Input:

Int[(f*x)^(-1 + 3*n)*Log[c*(d + e*x^n)^p]^2,x]
 

Output:

(x^(1 - 3*n)*(f*x)^(-1 + 3*n)*((x^(3*n)*Log[c*(d + e*x^n)^p]^2)/3 + (2*p*( 
-(p*((3*d*x^(2*n))/4 - x^(3*n)/9 - 3*d^2*(d + e*x^n) + (d^3*Log[d + e*x^n] 
^2)/2)) + (3*d*x^(2*n)*Log[c*(d + e*x^n)^p])/2 - (x^(3*n)*Log[c*(d + e*x^n 
)^p])/3 - 3*d^2*(d + e*x^n)*Log[c*(d + e*x^n)^p] + d^3*Log[d + e*x^n]*Log[ 
c*(d + e*x^n)^p]))/(3*e^3)))/n
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2772

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])
 

rule 2845

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^ 
n])^p/(g*(q + 1))), x] - Simp[b*e*n*(p/(g*(q + 1)))   Int[(f + g*x)^(q + 1) 
*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, 
d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && In 
tegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))
 

rule 2858

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ 
.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e   Subst[In 
t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + 
e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - 
d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
 

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]

\[\int \left (f x \right )^{-1+3 n} {\ln \left (c \left (d +e \,x^{n}\right )^{p}\right )}^{2}d x\]

Input:

int((f*x)^(-1+3*n)*ln(c*(d+e*x^n)^p)^2,x)
 

Output:

int((f*x)^(-1+3*n)*ln(c*(d+e*x^n)^p)^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.72 \[ \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {2 \, {\left (2 \, e^{3} p^{2} - 6 \, e^{3} p \log \left (c\right ) + 9 \, e^{3} \log \left (c\right )^{2}\right )} f^{3 \, n - 1} x^{3 \, n} - 3 \, {\left (5 \, d e^{2} p^{2} - 6 \, d e^{2} p \log \left (c\right )\right )} f^{3 \, n - 1} x^{2 \, n} + 6 \, {\left (11 \, d^{2} e p^{2} - 6 \, d^{2} e p \log \left (c\right )\right )} f^{3 \, n - 1} x^{n} + 18 \, {\left (e^{3} f^{3 \, n - 1} p^{2} x^{3 \, n} + d^{3} f^{3 \, n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} + 6 \, {\left (3 \, d e^{2} f^{3 \, n - 1} p^{2} x^{2 \, n} - 6 \, d^{2} e f^{3 \, n - 1} p^{2} x^{n} - 2 \, {\left (e^{3} p^{2} - 3 \, e^{3} p \log \left (c\right )\right )} f^{3 \, n - 1} x^{3 \, n} - {\left (11 \, d^{3} p^{2} - 6 \, d^{3} p \log \left (c\right )\right )} f^{3 \, n - 1}\right )} \log \left (e x^{n} + d\right )}{54 \, e^{3} n} \] Input:

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/54*(2*(2*e^3*p^2 - 6*e^3*p*log(c) + 9*e^3*log(c)^2)*f^(3*n - 1)*x^(3*n) 
- 3*(5*d*e^2*p^2 - 6*d*e^2*p*log(c))*f^(3*n - 1)*x^(2*n) + 6*(11*d^2*e*p^2 
 - 6*d^2*e*p*log(c))*f^(3*n - 1)*x^n + 18*(e^3*f^(3*n - 1)*p^2*x^(3*n) + d 
^3*f^(3*n - 1)*p^2)*log(e*x^n + d)^2 + 6*(3*d*e^2*f^(3*n - 1)*p^2*x^(2*n) 
- 6*d^2*e*f^(3*n - 1)*p^2*x^n - 2*(e^3*p^2 - 3*e^3*p*log(c))*f^(3*n - 1)*x 
^(3*n) - (11*d^3*p^2 - 6*d^3*p*log(c))*f^(3*n - 1))*log(e*x^n + d))/(e^3*n 
)
 

Sympy [F]

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

integrate((f*x)**(-1+3*n)*ln(c*(d+e*x**n)**p)**2,x)
 

Output:

Integral((f*x)**(3*n - 1)*log(c*(d + e*x**n)**p)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.64 \[ \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {e p {\left (\frac {6 \, d^{3} f^{3 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{4} n} - \frac {2 \, e^{2} f^{3 \, n} x^{3 \, n} - 3 \, d e f^{3 \, n} x^{2 \, n} + 6 \, d^{2} f^{3 \, n} x^{n}}{e^{3} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{9 \, f} + \frac {\left (f x\right )^{3 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{3 \, f n} - \frac {{\left (18 \, d^{3} f^{3 \, n} \log \left (e x^{n} + d\right )^{2} - 4 \, e^{3} f^{3 \, n} x^{3 \, n} + 15 \, d e^{2} f^{3 \, n} x^{2 \, n} - 66 \, d^{2} e f^{3 \, n} x^{n} - 6 \, {\left (6 \, f^{3 \, n} \log \left (e\right ) - 11 \, f^{3 \, n}\right )} d^{3} \log \left (e x^{n} + d\right )\right )} p^{2}}{54 \, e^{3} f n} \] Input:

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x, algorithm="maxima")
 

Output:

1/9*e*p*(6*d^3*f^(3*n)*log((e*x^n + d)/e)/(e^4*n) - (2*e^2*f^(3*n)*x^(3*n) 
 - 3*d*e*f^(3*n)*x^(2*n) + 6*d^2*f^(3*n)*x^n)/(e^3*n))*log((e*x^n + d)^p*c 
)/f + 1/3*(f*x)^(3*n)*log((e*x^n + d)^p*c)^2/(f*n) - 1/54*(18*d^3*f^(3*n)* 
log(e*x^n + d)^2 - 4*e^3*f^(3*n)*x^(3*n) + 15*d*e^2*f^(3*n)*x^(2*n) - 66*d 
^2*e*f^(3*n)*x^n - 6*(6*f^(3*n)*log(e) - 11*f^(3*n))*d^3*log(e*x^n + d))*p 
^2/(e^3*f*n)
 

Giac [F]

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

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x, algorithm="giac")
 

Output:

integrate((f*x)^(3*n - 1)*log((e*x^n + d)^p*c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}^2\,{\left (f\,x\right )}^{3\,n-1} \,d x \] Input:

int(log(c*(d + e*x^n)^p)^2*(f*x)^(3*n - 1),x)
 

Output:

int(log(c*(d + e*x^n)^p)^2*(f*x)^(3*n - 1), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.50 \[ \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {f^{3 n} \left (18 x^{3 n} {\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right )}^{2} e^{3}-12 x^{3 n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) e^{3} p +4 x^{3 n} e^{3} p^{2}+18 x^{2 n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d \,e^{2} p -15 x^{2 n} d \,e^{2} p^{2}-36 x^{n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d^{2} e p +66 x^{n} d^{2} e \,p^{2}+18 {\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right )}^{2} d^{3}-66 \,\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d^{3} p \right )}{54 e^{3} f n} \] Input:

int((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x)
 

Output:

(f**(3*n)*(18*x**(3*n)*log((x**n*e + d)**p*c)**2*e**3 - 12*x**(3*n)*log((x 
**n*e + d)**p*c)*e**3*p + 4*x**(3*n)*e**3*p**2 + 18*x**(2*n)*log((x**n*e + 
 d)**p*c)*d*e**2*p - 15*x**(2*n)*d*e**2*p**2 - 36*x**n*log((x**n*e + d)**p 
*c)*d**2*e*p + 66*x**n*d**2*e*p**2 + 18*log((x**n*e + d)**p*c)**2*d**3 - 6 
6*log((x**n*e + d)**p*c)*d**3*p))/(54*e**3*f*n)