\(\int (f x)^{-1-n} \log (c (d+e x^n)^p) \, dx\) [68]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int (f x)^{-1-n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {e p x^n (f x)^{-n} \log (x)}{d f}-\frac {e p x^n (f x)^{-n} \log \left (d+e x^n\right )}{d f n}-\frac {(f x)^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{f n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2505, 19, 272, 36, 29, 31} \[ \int (f x)^{-1-n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {(f x)^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{f n}+\frac {e p x^n \log (x) (f x)^{-n}}{d f}-\frac {e p x^n (f x)^{-n} \log \left (d+e x^n\right )}{d f n} \]

[In]

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

[Out]

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

Rule 19

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + n)*((b*v)^n/(a*v)^n), Int[u*v^(m + n),
x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[m + n]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rule 2505

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] - Dist[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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int (f x)^{-1-n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {(f x)^{-n} \left (-e n p x^n \log (x)+e p x^n \log \left (d+e x^n\right )+d \log \left (c \left (d+e x^n\right )^p\right )\right )}{d f n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int (f x)^{-1-n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {e f^{-n - 1} n p x^{n} \log \left (x\right ) - d f^{-n - 1} \log \left (c\right ) - {\left (e f^{-n - 1} p x^{n} + d f^{-n - 1} p\right )} \log \left (e x^{n} + d\right )}{d n x^{n}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int (f x)^{-1-n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int (f x)^{-1-n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {e p {\left (\frac {\log \left (x\right )}{d f^{n}} - \frac {\log \left (\frac {e x^{n} + d}{e}\right )}{d f^{n} n}\right )}}{f} - \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{\left (f x\right )^{n} f n} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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