Integrand size = 22, antiderivative size = 120 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p x^n (f x)^{-2 n}}{2 d f n}-\frac {e^2 p x^{2 n} (f x)^{-2 n} \log (x)}{2 d^2 f}+\frac {e^2 p x^{2 n} (f x)^{-2 n} \log \left (d+e x^n\right )}{2 d^2 f n}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n} \] Output:
-1/2*e*p*x^n/d/f/n/((f*x)^(2*n))-1/2*e^2*p*x^(2*n)*ln(x)/d^2/f/((f*x)^(2*n ))+1/2*e^2*p*x^(2*n)*ln(d+e*x^n)/d^2/f/n/((f*x)^(2*n))-1/2*ln(c*(d+e*x^n)^ p)/f/n/((f*x)^(2*n))
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {(f x)^{-2 n} \left (e^2 n p x^{2 n} \log (x)-e^2 p x^{2 n} \log \left (d+e x^n\right )+d \left (e p x^n+d \log \left (c \left (d+e x^n\right )^p\right )\right )\right )}{2 d^2 f n} \] Input:
Integrate[(f*x)^(-1 - 2*n)*Log[c*(d + e*x^n)^p],x]
Output:
-1/2*(e^2*n*p*x^(2*n)*Log[x] - e^2*p*x^(2*n)*Log[d + e*x^n] + d*(e*p*x^n + d*Log[c*(d + e*x^n)^p]))/(d^2*f*n*(f*x)^(2*n))
Time = 0.44 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2905, 30, 798, 54, 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.
\(\displaystyle \int (f x)^{-2 n-1} \log \left (c \left (d+e x^n\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {e p \int \frac {x^{n-1} (f x)^{-2 n}}{e x^n+d}dx}{2 f}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {e p x^{2 n} (f x)^{-2 n} \int \frac {x^{-n-1}}{e x^n+d}dx}{2 f}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {e p x^{2 n} (f x)^{-2 n} \int \frac {x^{-2 n}}{e x^n+d}dx^n}{2 f n}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {e p x^{2 n} (f x)^{-2 n} \int \left (\frac {x^{-2 n}}{d}-\frac {e x^{-n}}{d^2}+\frac {e^2}{d^2 \left (e x^n+d\right )}\right )dx^n}{2 f n}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e p x^{2 n} (f x)^{-2 n} \left (-\frac {e \log \left (x^n\right )}{d^2}+\frac {e \log \left (d+e x^n\right )}{d^2}-\frac {x^{-n}}{d}\right )}{2 f n}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}\) |
Input:
Int[(f*x)^(-1 - 2*n)*Log[c*(d + e*x^n)^p],x]
Output:
(e*p*x^(2*n)*(-(1/(d*x^n)) - (e*Log[x^n])/d^2 + (e*Log[d + e*x^n])/d^2))/( 2*f*n*(f*x)^(2*n)) - Log[c*(d + e*x^n)^p]/(2*f*n*(f*x)^(2*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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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]
\[\int \left (f x \right )^{-1-2 n} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )d x\]
Input:
int((f*x)^(-1-2*n)*ln(c*(d+e*x^n)^p),x)
Output:
int((f*x)^(-1-2*n)*ln(c*(d+e*x^n)^p),x)
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e^{2} f^{-2 \, n - 1} n p x^{2 \, n} \log \left (x\right ) + d e f^{-2 \, n - 1} p x^{n} + d^{2} f^{-2 \, n - 1} \log \left (c\right ) - {\left (e^{2} f^{-2 \, n - 1} p x^{2 \, n} - d^{2} f^{-2 \, n - 1} p\right )} \log \left (e x^{n} + d\right )}{2 \, d^{2} n x^{2 \, n}} \] Input:
integrate((f*x)^(-1-2*n)*log(c*(d+e*x^n)^p),x, algorithm="fricas")
Output:
-1/2*(e^2*f^(-2*n - 1)*n*p*x^(2*n)*log(x) + d*e*f^(-2*n - 1)*p*x^n + d^2*f ^(-2*n - 1)*log(c) - (e^2*f^(-2*n - 1)*p*x^(2*n) - d^2*f^(-2*n - 1)*p)*log (e*x^n + d))/(d^2*n*x^(2*n))
Exception generated. \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\text {Exception raised: TypeError} \] Input:
integrate((f*x)**(-1-2*n)*ln(c*(d+e*x**n)**p),x)
Output:
Exception raised: TypeError >> Invalid comparison of non-real zoo
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {e \log \left (x\right )}{d^{2} f^{2 \, n}} - \frac {e \log \left (\frac {e x^{n} + d}{e}\right )}{d^{2} f^{2 \, n} n} + \frac {1}{d f^{2 \, n} n x^{n}}\right )}}{2 \, f} - \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, \left (f x\right )^{2 \, n} f n} \] Input:
integrate((f*x)^(-1-2*n)*log(c*(d+e*x^n)^p),x, algorithm="maxima")
Output:
-1/2*e*p*(e*log(x)/(d^2*f^(2*n)) - e*log((e*x^n + d)/e)/(d^2*f^(2*n)*n) + 1/(d*f^(2*n)*n*x^n))/f - 1/2*log((e*x^n + d)^p*c)/((f*x)^(2*n)*f*n)
\[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{-2 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \] Input:
integrate((f*x)^(-1-2*n)*log(c*(d+e*x^n)^p),x, algorithm="giac")
Output:
integrate((f*x)^(-2*n - 1)*log((e*x^n + d)^p*c), x)
Timed out. \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}{{\left (f\,x\right )}^{2\,n+1}} \,d x \] Input:
int(log(c*(d + e*x^n)^p)/(f*x)^(2*n + 1),x)
Output:
int(log(c*(d + e*x^n)^p)/(f*x)^(2*n + 1), x)
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{2 n} \mathrm {log}\left (x^{n} e +d \right ) e^{2} p -x^{2 n} \mathrm {log}\left (x \right ) e^{2} n p -x^{n} d e p -\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d^{2}}{2 x^{2 n} f^{2 n} d^{2} f n} \] Input:
int((f*x)^(-1-2*n)*log(c*(d+e*x^n)^p),x)
Output:
(x**(2*n)*log(x**n*e + d)*e**2*p - x**(2*n)*log(x)*e**2*n*p - x**n*d*e*p - log((x**n*e + d)**p*c)*d**2)/(2*x**(2*n)*f**(2*n)*d**2*f*n)