Integrand size = 24, antiderivative size = 255 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {2 d p^2 x^{1-n} (f x)^{-1+2 n}}{e n}+\frac {p^2 x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2}{4 e^2 n}+\frac {2 d p x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2 n}-\frac {p x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}-\frac {d x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2 n}+\frac {x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2 n} \]
-2*d*p^2*x^(1-n)*(f*x)^(-1+2*n)/e/n+1/4*p^2*x^(1-2*n)*(f*x)^(-1+2*n)*(d+e* x^n)^2/e^2/n+2*d*p*x^(1-2*n)*(f*x)^(-1+2*n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)/e^ 2/n-1/2*p*x^(1-2*n)*(f*x)^(-1+2*n)*(d+e*x^n)^2*ln(c*(d+e*x^n)^p)/e^2/n-d*x ^(1-2*n)*(f*x)^(-1+2*n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)^2/e^2/n+1/2*x^(1-2*n)* (f*x)^(-1+2*n)*(d+e*x^n)^2*ln(c*(d+e*x^n)^p)^2/e^2/n
Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.55 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{-2 n} (f x)^{2 n} \left (2 d^2 p^2 \log ^2\left (d+e x^n\right )+2 d^2 p \log \left (d+e x^n\right ) \left (3 p-2 \log \left (c \left (d+e x^n\right )^p\right )\right )+e x^n \left (p^2 \left (-6 d+e x^n\right )+2 p \left (2 d-e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+2 e x^n \log ^2\left (c \left (d+e x^n\right )^p\right )\right )\right )}{4 e^2 f n} \]
((f*x)^(2*n)*(2*d^2*p^2*Log[d + e*x^n]^2 + 2*d^2*p*Log[d + e*x^n]*(3*p - 2 *Log[c*(d + e*x^n)^p]) + e*x^n*(p^2*(-6*d + e*x^n) + 2*p*(2*d - e*x^n)*Log [c*(d + e*x^n)^p] + 2*e*x^n*Log[c*(d + e*x^n)^p]^2)))/(4*e^2*f*n*x^(2*n))
Time = 0.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.64, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2906, 2904, 2848, 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.
\(\displaystyle \int (f x)^{2 n-1} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2906 |
\(\displaystyle x^{1-2 n} (f x)^{2 n-1} \int x^{2 n-1} \log ^2\left (c \left (e x^n+d\right )^p\right )dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {x^{1-2 n} (f x)^{2 n-1} \int x^n \log ^2\left (c \left (e x^n+d\right )^p\right )dx^n}{n}\) |
\(\Big \downarrow \) 2848 |
\(\displaystyle \frac {x^{1-2 n} (f x)^{2 n-1} \int \left (\frac {\left (e x^n+d\right ) \log ^2\left (c \left (e x^n+d\right )^p\right )}{e}-\frac {d \log ^2\left (c \left (e x^n+d\right )^p\right )}{e}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^{1-2 n} (f x)^{2 n-1} \left (\frac {\left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2}-\frac {d \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2}-\frac {p \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2}+\frac {2 d p \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2}+\frac {p^2 \left (d+e x^n\right )^2}{4 e^2}-\frac {2 d p^2 x^n}{e}\right )}{n}\) |
(x^(1 - 2*n)*(f*x)^(-1 + 2*n)*((-2*d*p^2*x^n)/e + (p^2*(d + e*x^n)^2)/(4*e ^2) + (2*d*p*(d + e*x^n)*Log[c*(d + e*x^n)^p])/e^2 - (p*(d + e*x^n)^2*Log[ c*(d + e*x^n)^p])/(2*e^2) - (d*(d + e*x^n)*Log[c*(d + e*x^n)^p]^2)/e^2 + ( (d + e*x^n)^2*Log[c*(d + e*x^n)^p]^2)/(2*e^2)))/n
3.2.64.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - d*g, 0] && IGtQ[q, 0]
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])
\[\int \left (f x \right )^{-1+2 n} {\ln \left (c \left (d +e \,x^{n}\right )^{p}\right )}^{2}d x\]
Time = 0.33 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.80 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {{\left (e^{2} p^{2} - 2 \, e^{2} p \log \left (c\right ) + 2 \, e^{2} \log \left (c\right )^{2}\right )} f^{2 \, n - 1} x^{2 \, n} - 2 \, {\left (3 \, d e p^{2} - 2 \, d e p \log \left (c\right )\right )} f^{2 \, n - 1} x^{n} + 2 \, {\left (e^{2} f^{2 \, n - 1} p^{2} x^{2 \, n} - d^{2} f^{2 \, n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} + 2 \, {\left (2 \, d e f^{2 \, n - 1} p^{2} x^{n} - {\left (e^{2} p^{2} - 2 \, e^{2} p \log \left (c\right )\right )} f^{2 \, n - 1} x^{2 \, n} + {\left (3 \, d^{2} p^{2} - 2 \, d^{2} p \log \left (c\right )\right )} f^{2 \, n - 1}\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \]
1/4*((e^2*p^2 - 2*e^2*p*log(c) + 2*e^2*log(c)^2)*f^(2*n - 1)*x^(2*n) - 2*( 3*d*e*p^2 - 2*d*e*p*log(c))*f^(2*n - 1)*x^n + 2*(e^2*f^(2*n - 1)*p^2*x^(2* n) - d^2*f^(2*n - 1)*p^2)*log(e*x^n + d)^2 + 2*(2*d*e*f^(2*n - 1)*p^2*x^n - (e^2*p^2 - 2*e^2*p*log(c))*f^(2*n - 1)*x^(2*n) + (3*d^2*p^2 - 2*d^2*p*lo g(c))*f^(2*n - 1))*log(e*x^n + d))/(e^2*n)
\[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{2 n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}^{2}\, dx \]
Time = 0.22 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.78 \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {2 \, d^{2} f^{2 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{3} n} + \frac {e f^{2 \, n} x^{2 \, n} - 2 \, d f^{2 \, n} x^{n}}{e^{2} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, f} + \frac {\left (f x\right )^{2 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{2 \, f n} + \frac {{\left (2 \, d^{2} f^{2 \, n} \log \left (e x^{n} + d\right )^{2} + e^{2} f^{2 \, n} x^{2 \, n} - 6 \, d e f^{2 \, n} x^{n} - 2 \, {\left (2 \, f^{2 \, n} \log \left (e\right ) - 3 \, f^{2 \, n}\right )} d^{2} \log \left (e x^{n} + d\right )\right )} p^{2}}{4 \, e^{2} f n} \]
-1/2*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))*log((e*x^n + d)^p*c)/f + 1/2*(f*x)^(2*n)*log((e* x^n + d)^p*c)^2/(f*n) + 1/4*(2*d^2*f^(2*n)*log(e*x^n + d)^2 + e^2*f^(2*n)* x^(2*n) - 6*d*e*f^(2*n)*x^n - 2*(2*f^(2*n)*log(e) - 3*f^(2*n))*d^2*log(e*x ^n + d))*p^2/(e^2*f*n)
\[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{2 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2} \,d x } \]
Timed out. \[ \int (f x)^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}^2\,{\left (f\,x\right )}^{2\,n-1} \,d x \]