Integrand size = 22, antiderivative size = 101 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {2 p^2 x (f x)^{-1+n}}{n}-\frac {2 p x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n} \] Output:
2*p^2*x*(f*x)^(-1+n)/n-2*p*x^(1-n)*(f*x)^(-1+n)*(d+e*x^n)*ln(c*(d+e*x^n)^p )/e/n+x^(1-n)*(f*x)^(-1+n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)^2/e/n
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{-n} (f x)^n \left (2 e p^2 x^n-2 p \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+\left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )\right )}{e f n} \] Input:
Integrate[(f*x)^(-1 + n)*Log[c*(d + e*x^n)^p]^2,x]
Output:
((f*x)^n*(2*e*p^2*x^n - 2*p*(d + e*x^n)*Log[c*(d + e*x^n)^p] + (d + e*x^n) *Log[c*(d + e*x^n)^p]^2))/(e*f*n*x^n)
Time = 0.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2906, 2904, 2836, 2733, 2732}
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.
\(\displaystyle \int (f x)^{n-1} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2906 |
\(\displaystyle x^{1-n} (f x)^{n-1} \int x^{n-1} \log ^2\left (c \left (e x^n+d\right )^p\right )dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {x^{1-n} (f x)^{n-1} \int \log ^2\left (c \left (e x^n+d\right )^p\right )dx^n}{n}\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {x^{1-n} (f x)^{n-1} \int \log ^2\left (c \left (e x^n+d\right )^p\right )d\left (e x^n+d\right )}{e n}\) |
\(\Big \downarrow \) 2733 |
\(\displaystyle \frac {x^{1-n} (f x)^{n-1} \left (\left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )-2 p \int \log \left (c \left (e x^n+d\right )^p\right )d\left (e x^n+d\right )\right )}{e n}\) |
\(\Big \downarrow \) 2732 |
\(\displaystyle \frac {x^{1-n} (f x)^{n-1} \left (\left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )-2 p \left (\left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )-p \left (d+e x^n\right )\right )\right )}{e n}\) |
Input:
Int[(f*x)^(-1 + n)*Log[c*(d + e*x^n)^p]^2,x]
Output:
(x^(1 - n)*(f*x)^(-1 + n)*((d + e*x^n)*Log[c*(d + e*x^n)^p]^2 - 2*p*(-(p*( d + e*x^n)) + (d + e*x^n)*Log[c*(d + e*x^n)^p])))/(e*n)
Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x ] /; FreeQ[{c, n}, x]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b *Log[c*x^n])^p, x] - Simp[b*n*p Int[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[1/e Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ a, b, c, d, e, n, p}, x]
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])
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])
\[\int \left (f x \right )^{-1+n} {\ln \left (c \left (d +e \,x^{n}\right )^{p}\right )}^{2}d x\]
Input:
int((f*x)^(-1+n)*ln(c*(d+e*x^n)^p)^2,x)
Output:
int((f*x)^(-1+n)*ln(c*(d+e*x^n)^p)^2,x)
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.20 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {{\left (2 \, e p^{2} - 2 \, e p \log \left (c\right ) + e \log \left (c\right )^{2}\right )} f^{n - 1} x^{n} + {\left (e f^{n - 1} p^{2} x^{n} + d f^{n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} - 2 \, {\left ({\left (e p^{2} - e p \log \left (c\right )\right )} f^{n - 1} x^{n} + {\left (d p^{2} - d p \log \left (c\right )\right )} f^{n - 1}\right )} \log \left (e x^{n} + d\right )}{e n} \] Input:
integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x, algorithm="fricas")
Output:
((2*e*p^2 - 2*e*p*log(c) + e*log(c)^2)*f^(n - 1)*x^n + (e*f^(n - 1)*p^2*x^ n + d*f^(n - 1)*p^2)*log(e*x^n + d)^2 - 2*((e*p^2 - e*p*log(c))*f^(n - 1)* x^n + (d*p^2 - d*p*log(c))*f^(n - 1))*log(e*x^n + d))/(e*n)
\[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}^{2}\, dx \] Input:
integrate((f*x)**(-1+n)*ln(c*(d+e*x**n)**p)**2,x)
Output:
Integral((f*x)**(n - 1)*log(c*(d + e*x**n)**p)**2, x)
Time = 0.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.45 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {2 \, e p {\left (\frac {f^{n} x^{n}}{e n} - \frac {d f^{n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{2} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{f} + \frac {\left (f x\right )^{n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{f n} - \frac {{\left (d f^{n} \log \left (e x^{n} + d\right )^{2} - 2 \, e f^{n} x^{n} - 2 \, {\left (f^{n} \log \left (e\right ) - f^{n}\right )} d \log \left (e x^{n} + d\right )\right )} p^{2}}{e f n} \] Input:
integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x, algorithm="maxima")
Output:
-2*e*p*(f^n*x^n/(e*n) - d*f^n*log((e*x^n + d)/e)/(e^2*n))*log((e*x^n + d)^ p*c)/f + (f*x)^n*log((e*x^n + d)^p*c)^2/(f*n) - (d*f^n*log(e*x^n + d)^2 - 2*e*f^n*x^n - 2*(f^n*log(e) - f^n)*d*log(e*x^n + d))*p^2/(e*f*n)
\[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2} \,d x } \] Input:
integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x, algorithm="giac")
Output:
integrate((f*x)^(n - 1)*log((e*x^n + d)^p*c)^2, x)
Timed out. \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}^2\,{\left (f\,x\right )}^{n-1} \,d x \] Input:
int(log(c*(d + e*x^n)^p)^2*(f*x)^(n - 1),x)
Output:
int(log(c*(d + e*x^n)^p)^2*(f*x)^(n - 1), x)
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {f^{n} \left (x^{n} {\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right )}^{2} e -2 x^{n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) e p +2 x^{n} e \,p^{2}+{\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right )}^{2} d -2 \,\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d p \right )}{e f n} \] Input:
int((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x)
Output:
(f**n*(x**n*log((x**n*e + d)**p*c)**2*e - 2*x**n*log((x**n*e + d)**p*c)*e* p + 2*x**n*e*p**2 + log((x**n*e + d)**p*c)**2*d - 2*log((x**n*e + d)**p*c) *d*p))/(e*f*n)