Integrand size = 22, antiderivative size = 112 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {p (f x)^{2 n}}{4 f n}+\frac {d p x^{-n} (f x)^{2 n}}{2 e f n}-\frac {d^2 p x^{-2 n} (f x)^{2 n} \log \left (d+e x^n\right )}{2 e^2 f n}+\frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n} \] Output:
-1/4*p*(f*x)^(2*n)/f/n+1/2*d*p*(f*x)^(2*n)/e/f/n/(x^n)-1/2*d^2*p*(f*x)^(2* n)*ln(d+e*x^n)/e^2/f/n/(x^(2*n))+1/2*(f*x)^(2*n)*ln(c*(d+e*x^n)^p)/f/n
Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.66 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {x^{-2 n} (f x)^{2 n} \left (2 d^2 p \log \left (d+e x^n\right )+e x^n \left (-2 d p+e p x^n-2 e x^n \log \left (c \left (d+e x^n\right )^p\right )\right )\right )}{4 e^2 f n} \] Input:
Integrate[(f*x)^(-1 + 2*n)*Log[c*(d + e*x^n)^p],x]
Output:
-1/4*((f*x)^(2*n)*(2*d^2*p*Log[d + e*x^n] + e*x^n*(-2*d*p + e*p*x^n - 2*e* x^n*Log[c*(d + e*x^n)^p])))/(e^2*f*n*x^(2*n))
Time = 0.44 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81, 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, 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.
\(\displaystyle \int (f x)^{2 n-1} \log \left (c \left (d+e x^n\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {e p \int \frac {x^{n-1} (f x)^{2 n}}{e x^n+d}dx}{2 f}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {e p x^{-2 n} (f x)^{2 n} \int \frac {x^{3 n-1}}{e x^n+d}dx}{2 f}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {e p x^{-2 n} (f x)^{2 n} \int \frac {x^{2 n}}{e x^n+d}dx^n}{2 f n}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {e p x^{-2 n} (f x)^{2 n} \int \left (\frac {x^n}{e}+\frac {d^2}{e^2 \left (e x^n+d\right )}-\frac {d}{e^2}\right )dx^n}{2 f n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {e p x^{-2 n} (f x)^{2 n} \left (\frac {d^2 \log \left (d+e x^n\right )}{e^3}-\frac {d x^n}{e^2}+\frac {x^{2 n}}{2 e}\right )}{2 f n}\) |
Input:
Int[(f*x)^(-1 + 2*n)*Log[c*(d + e*x^n)^p],x]
Output:
-1/2*(e*p*(f*x)^(2*n)*(-((d*x^n)/e^2) + x^(2*n)/(2*e) + (d^2*Log[d + e*x^n ])/e^3))/(f*n*x^(2*n)) + ((f*x)^(2*n)*Log[c*(d + e*x^n)^p])/(2*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]
\[\int \left (f x \right )^{2 n -1} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )d x\]
Input:
int((f*x)^(2*n-1)*ln(c*(d+e*x^n)^p),x)
Output:
int((f*x)^(2*n-1)*ln(c*(d+e*x^n)^p),x)
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {2 \, d e f^{2 \, n - 1} p x^{n} - {\left (e^{2} p - 2 \, e^{2} \log \left (c\right )\right )} f^{2 \, n - 1} x^{2 \, n} + 2 \, {\left (e^{2} f^{2 \, n - 1} p x^{2 \, n} - d^{2} f^{2 \, n - 1} p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \] Input:
integrate((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p),x, algorithm="fricas")
Output:
1/4*(2*d*e*f^(2*n - 1)*p*x^n - (e^2*p - 2*e^2*log(c))*f^(2*n - 1)*x^(2*n) + 2*(e^2*f^(2*n - 1)*p*x^(2*n) - d^2*f^(2*n - 1)*p)*log(e*x^n + d))/(e^2*n )
\[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{2 n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}\, dx \] Input:
integrate((f*x)**(-1+2*n)*ln(c*(d+e*x**n)**p),x)
Output:
Integral((f*x)**(2*n - 1)*log(c*(d + e*x**n)**p), x)
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {2 \, d^{2} f^{2 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{3} n} + \frac {e f^{2 \, n} x^{2 \, n} - 2 \, d f^{2 \, n} x^{n}}{e^{2} n}\right )}}{4 \, f} + \frac {\left (f x\right )^{2 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, f n} \] Input:
integrate((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p),x, algorithm="maxima")
Output:
-1/4*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))/f + 1/2*(f*x)^(2*n)*log((e*x^n + d)^p*c)/(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. \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f\,x\right )}^{2\,n-1} \,d x \] 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 = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.67 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {f^{2 n} \left (2 x^{2 n} \mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) e^{2}-x^{2 n} e^{2} p +2 x^{n} d e p -2 \,\mathrm {log}\left (\left (x^{n} e +d \right )^{p} c \right ) d^{2}\right )}{4 e^{2} f n} \] Input:
int((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p),x)
Output:
(f**(2*n)*(2*x**(2*n)*log((x**n*e + d)**p*c)*e**2 - x**(2*n)*e**2*p + 2*x* *n*d*e*p - 2*log((x**n*e + d)**p*c)*d**2))/(4*e**2*f*n)