\(\int (f x)^{-1+n} \log ^2(c (d+e x^n)^p) \, dx\) [165]

Optimal result

Integrand size = 22, antiderivative size = 101 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {2 p^2 x (f x)^{-1+n}}{n}-\frac {2 p x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n} \]

[Out]

2*p^2*x*(f*x)^(-1+n)/n-2*p*x^(1-n)*(f*x)^(-1+n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)/e/n+x^(1-n)*(f*x)^(-1+n)*(d+e*x^n)
*ln(c*(d+e*x^n)^p)^2/e/n

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2506, 2504, 2436, 2333, 2332} \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{1-n} (f x)^{n-1} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n}-\frac {2 p x^{1-n} (f x)^{n-1} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {2 p^2 x (f x)^{n-1}}{n} \]

[In]

Int[(f*x)^(-1 + n)*Log[c*(d + e*x^n)^p]^2,x]

[Out]

(2*p^2*x*(f*x)^(-1 + n))/n - (2*p*x^(1 - n)*(f*x)^(-1 + n)*(d + e*x^n)*Log[c*(d + e*x^n)^p])/(e*n) + (x^(1 - n
)*(f*x)^(-1 + n)*(d + e*x^n)*Log[c*(d + e*x^n)^p]^2)/(e*n)

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rule 2506

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_)*(x_))^(m_), x_Symbol] :> Dist[(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[
Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps \begin{align*} \text {integral}& = \left (x^{1-n} (f x)^{-1+n}\right ) \int x^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx \\ & = \frac {\left (x^{1-n} (f x)^{-1+n}\right ) \text {Subst}\left (\int \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n} \\ & = \frac {\left (x^{1-n} (f x)^{-1+n}\right ) \text {Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,d+e x^n\right )}{e n} \\ & = \frac {x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n}-\frac {\left (2 p x^{1-n} (f x)^{-1+n}\right ) \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^n\right )}{e n} \\ & = \frac {2 p^2 x (f x)^{-1+n}}{n}-\frac {2 p x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {x^{1-n} (f x)^{-1+n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {x^{-n} (f x)^n \left (2 e p^2 x^n-2 p \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+\left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )\right )}{e f n} \]

[In]

Integrate[(f*x)^(-1 + n)*Log[c*(d + e*x^n)^p]^2,x]

[Out]

((f*x)^n*(2*e*p^2*x^n - 2*p*(d + e*x^n)*Log[c*(d + e*x^n)^p] + (d + e*x^n)*Log[c*(d + e*x^n)^p]^2))/(e*f*n*x^n
)

Maple [F]

\[\int \left (f x \right )^{n -1} {\ln \left (c \left (d +e \,x^{n}\right )^{p}\right )}^{2}d x\]

[In]

int((f*x)^(n-1)*ln(c*(d+e*x^n)^p)^2,x)

[Out]

int((f*x)^(n-1)*ln(c*(d+e*x^n)^p)^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.20 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {{\left (2 \, e p^{2} - 2 \, e p \log \left (c\right ) + e \log \left (c\right )^{2}\right )} f^{n - 1} x^{n} + {\left (e f^{n - 1} p^{2} x^{n} + d f^{n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} - 2 \, {\left ({\left (e p^{2} - e p \log \left (c\right )\right )} f^{n - 1} x^{n} + {\left (d p^{2} - d p \log \left (c\right )\right )} f^{n - 1}\right )} \log \left (e x^{n} + d\right )}{e n} \]

[In]

integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x, algorithm="fricas")

[Out]

((2*e*p^2 - 2*e*p*log(c) + e*log(c)^2)*f^(n - 1)*x^n + (e*f^(n - 1)*p^2*x^n + d*f^(n - 1)*p^2)*log(e*x^n + d)^
2 - 2*((e*p^2 - e*p*log(c))*f^(n - 1)*x^n + (d*p^2 - d*p*log(c))*f^(n - 1))*log(e*x^n + d))/(e*n)

Sympy [F]

\[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}^{2}\, dx \]

[In]

integrate((f*x)**(-1+n)*ln(c*(d+e*x**n)**p)**2,x)

[Out]

Integral((f*x)**(n - 1)*log(c*(d + e*x**n)**p)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.45 \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {2 \, e p {\left (\frac {f^{n} x^{n}}{e n} - \frac {d f^{n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{2} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{f} + \frac {\left (f x\right )^{n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{f n} - \frac {{\left (d f^{n} \log \left (e x^{n} + d\right )^{2} - 2 \, e f^{n} x^{n} - 2 \, {\left (f^{n} \log \left (e\right ) - f^{n}\right )} d \log \left (e x^{n} + d\right )\right )} p^{2}}{e f n} \]

[In]

integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x, algorithm="maxima")

[Out]

-2*e*p*(f^n*x^n/(e*n) - d*f^n*log((e*x^n + d)/e)/(e^2*n))*log((e*x^n + d)^p*c)/f + (f*x)^n*log((e*x^n + d)^p*c
)^2/(f*n) - (d*f^n*log(e*x^n + d)^2 - 2*e*f^n*x^n - 2*(f^n*log(e) - f^n)*d*log(e*x^n + d))*p^2/(e*f*n)

Giac [F]

\[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2} \,d x } \]

[In]

integrate((f*x)^(-1+n)*log(c*(d+e*x^n)^p)^2,x, algorithm="giac")

[Out]

integrate((f*x)^(n - 1)*log((e*x^n + d)^p*c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^{-1+n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx=\int {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}^2\,{\left (f\,x\right )}^{n-1} \,d x \]

[In]

int(log(c*(d + e*x^n)^p)^2*(f*x)^(n - 1),x)

[Out]

int(log(c*(d + e*x^n)^p)^2*(f*x)^(n - 1), x)