Integrand size = 22, antiderivative size = 141 \[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {p (f x)^{3 n}}{9 f n}-\frac {d^2 p x^{-2 n} (f x)^{3 n}}{3 e^2 f n}+\frac {d p x^{-n} (f x)^{3 n}}{6 e f n}+\frac {d^3 p x^{-3 n} (f x)^{3 n} \log \left (d+e x^n\right )}{3 e^3 f n}+\frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n} \]
-1/9*p*(f*x)^(3*n)/f/n-1/3*d^2*p*(f*x)^(3*n)/e^2/f/n/(x^(2*n))+1/6*d*p*(f* x)^(3*n)/e/f/n/(x^n)+1/3*d^3*p*(f*x)^(3*n)*ln(d+e*x^n)/e^3/f/n/(x^(3*n))+1 /3*(f*x)^(3*n)*ln(c*(d+e*x^n)^p)/f/n
Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.65 \[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{-3 n} (f x)^{3 n} \left (-e p x^n \left (6 d^2-3 d e x^n+2 e^2 x^{2 n}\right )+6 d^3 p \log \left (d+e x^n\right )+6 e^3 x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )\right )}{18 e^3 f n} \]
((f*x)^(3*n)*(-(e*p*x^n*(6*d^2 - 3*d*e*x^n + 2*e^2*x^(2*n))) + 6*d^3*p*Log [d + e*x^n] + 6*e^3*x^(3*n)*Log[c*(d + e*x^n)^p]))/(18*e^3*f*n*x^(3*n))
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.75, 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.
\(\displaystyle \int (f x)^{3 n-1} \log \left (c \left (d+e x^n\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {e p \int \frac {x^{n-1} (f x)^{3 n}}{e x^n+d}dx}{3 f}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {e p x^{-3 n} (f x)^{3 n} \int \frac {x^{4 n-1}}{e x^n+d}dx}{3 f}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {e p x^{-3 n} (f x)^{3 n} \int \frac {x^{3 n}}{e x^n+d}dx^n}{3 f n}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {e p x^{-3 n} (f x)^{3 n} \int \left (-\frac {d x^n}{e^2}+\frac {x^{2 n}}{e}-\frac {d^3}{e^3 \left (e x^n+d\right )}+\frac {d^2}{e^3}\right )dx^n}{3 f n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {e p x^{-3 n} (f x)^{3 n} \left (-\frac {d^3 \log \left (d+e x^n\right )}{e^4}+\frac {d^2 x^n}{e^3}-\frac {d x^{2 n}}{2 e^2}+\frac {x^{3 n}}{3 e}\right )}{3 f n}\) |
-1/3*(e*p*(f*x)^(3*n)*((d^2*x^n)/e^3 - (d*x^(2*n))/(2*e^2) + x^(3*n)/(3*e) - (d^3*Log[d + e*x^n])/e^4))/(f*n*x^(3*n)) + ((f*x)^(3*n)*Log[c*(d + e*x^ n)^p])/(3*f*n)
3.1.64.3.1 Defintions of rubi rules used
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]
\[\int \left (f x \right )^{-1+3 n} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )d x\]
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {3 \, d e^{2} f^{3 \, n - 1} p x^{2 \, n} - 6 \, d^{2} e f^{3 \, n - 1} p x^{n} - 2 \, {\left (e^{3} p - 3 \, e^{3} \log \left (c\right )\right )} f^{3 \, n - 1} x^{3 \, n} + 6 \, {\left (e^{3} f^{3 \, n - 1} p x^{3 \, n} + d^{3} f^{3 \, n - 1} p\right )} \log \left (e x^{n} + d\right )}{18 \, e^{3} n} \]
1/18*(3*d*e^2*f^(3*n - 1)*p*x^(2*n) - 6*d^2*e*f^(3*n - 1)*p*x^n - 2*(e^3*p - 3*e^3*log(c))*f^(3*n - 1)*x^(3*n) + 6*(e^3*f^(3*n - 1)*p*x^(3*n) + d^3* f^(3*n - 1)*p)*log(e*x^n + d))/(e^3*n)
\[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{3 n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}\, dx \]
Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {e p {\left (\frac {6 \, d^{3} f^{3 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{4} n} - \frac {2 \, e^{2} f^{3 \, n} x^{3 \, n} - 3 \, d e f^{3 \, n} x^{2 \, n} + 6 \, d^{2} f^{3 \, n} x^{n}}{e^{3} n}\right )}}{18 \, f} + \frac {\left (f x\right )^{3 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{3 \, f n} \]
1/18*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))/f + 1/3*(f*x)^(3*n )*log((e*x^n + d)^p*c)/(f*n)
\[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{3 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \]
Timed out. \[ \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f\,x\right )}^{3\,n-1} \,d x \]