Integrand size = 20, antiderivative size = 69 \[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {p (f x)^n}{f n}+\frac {d p x^{-n} (f x)^n \log \left (d+e x^n\right )}{e f n}+\frac {(f x)^n \log \left (c \left (d+e x^n\right )^p\right )}{f n} \] Output:
-p*(f*x)^n/f/n+d*p*(f*x)^n*ln(d+e*x^n)/e/f/n/(x^n)+(f*x)^n*ln(c*(d+e*x^n)^ p)/f/n
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{1-n} (f x)^{-1+n} \left (-p x^n+\frac {\left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e}\right )}{n} \] Input:
Integrate[(f*x)^(-1 + n)*Log[c*(d + e*x^n)^p],x]
Output:
(x^(1 - n)*(f*x)^(-1 + n)*(-(p*x^n) + ((d + e*x^n)*Log[c*(d + e*x^n)^p])/e ))/n
Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2905, 30, 798, 49, 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.
\(\displaystyle \int (f x)^{n-1} \log \left (c \left (d+e x^n\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {(f x)^n \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {e p \int \frac {x^{n-1} (f x)^n}{e x^n+d}dx}{f}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {(f x)^n \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {e p x^{-n} (f x)^n \int \frac {x^{2 n-1}}{e x^n+d}dx}{f}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {(f x)^n \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {e p x^{-n} (f x)^n \int \frac {x^n}{e x^n+d}dx^n}{f n}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(f x)^n \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {e p x^{-n} (f x)^n \int \left (\frac {1}{e}-\frac {d}{e \left (e x^n+d\right )}\right )dx^n}{f n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(f x)^n \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {e p x^{-n} (f x)^n \left (\frac {x^n}{e}-\frac {d \log \left (d+e x^n\right )}{e^2}\right )}{f n}\) |
Input:
Int[(f*x)^(-1 + n)*Log[c*(d + e*x^n)^p],x]
Output:
-((e*p*(f*x)^n*(x^n/e - (d*Log[d + e*x^n])/e^2))/(f*n*x^n)) + ((f*x)^n*Log [c*(d + e*x^n)^p])/(f*n)
Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & & !IntegerQ[p]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ (m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Simp[b*e*n*(p/(f*(m + 1))) Int[x^(n - 1)*((f*x)^(m + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
\[\int \left (f x \right )^{-1+n} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )d x\]
Input:
int((f*x)^(-1+n)*ln(c*(d+e*x^n)^p),x)
Output:
int((f*x)^(-1+n)*ln(c*(d+e*x^n)^p),x)
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {{\left (e p - e \log \left (c\right )\right )} f^{n - 1} x^{n} - {\left (e f^{n - 1} p x^{n} + d f^{n - 1} p\right )} \log \left (e x^{n} + d\right )}{e n} \] Input:
integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p),x, algorithm="fricas")
Output:
-((e*p - e*log(c))*f^(n - 1)*x^n - (e*f^(n - 1)*p*x^n + d*f^(n - 1)*p)*log (e*x^n + d))/(e*n)
\[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}\, dx \] Input:
integrate((f*x)**(-1+n)*ln(c*(d+e*x**n)**p),x)
Output:
Integral((f*x)**(n - 1)*log(c*(d + e*x**n)**p), x)
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01 \[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {f^{n} x^{n}}{e n} - \frac {d f^{n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{2} n}\right )}}{f} + \frac {\left (f x\right )^{n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{f n} \] Input:
integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p),x, algorithm="maxima")
Output:
-e*p*(f^n*x^n/(e*n) - d*f^n*log((e*x^n + d)/e)/(e^2*n))/f + (f*x)^n*log((e *x^n + d)^p*c)/(f*n)
\[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \] Input:
integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p),x, algorithm="giac")
Output:
integrate((f*x)^(n - 1)*log((e*x^n + d)^p*c), x)
Timed out. \[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f\,x\right )}^{n-1} \,d x \] Input:
int(log(c*(d + e*x^n)^p)*(f*x)^(n - 1),x)
Output:
int(log(c*(d + e*x^n)^p)*(f*x)^(n - 1), x)
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75 \[ \int (f x)^{-1+n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {f^{n} \left (x^{n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) e -x^{n} e p +\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d \right )}{e f n} \] Input:
int((f*x)^(-1+n)*log(c*(d+e*x^n)^p),x)
Output:
(f**n*(x**n*log((x**n*e + d)**p*c)*e - x**n*e*p + log((x**n*e + d)**p*c)*d ))/(e*f*n)