\(\int (f x)^{-1-2 n} \log (c (d+e x^n)^p) \, dx\) [69]

Optimal result

Integrand size = 22, antiderivative size = 120 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p x^n (f x)^{-2 n}}{2 d f n}-\frac {e^2 p x^{2 n} (f x)^{-2 n} \log (x)}{2 d^2 f}+\frac {e^2 p x^{2 n} (f x)^{-2 n} \log \left (d+e x^n\right )}{2 d^2 f n}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2505, 20, 272, 46} \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {e^2 p x^{2 n} \log (x) (f x)^{-2 n}}{2 d^2 f}+\frac {e^2 p x^{2 n} (f x)^{-2 n} \log \left (d+e x^n\right )}{2 d^2 f n}-\frac {e p x^n (f x)^{-2 n}}{2 d f n} \]

[In]

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

[Out]

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

Rule 20

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

Rule 46

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}+\frac {(e p) \int \frac {x^{-1+n} (f x)^{-2 n}}{d+e x^n} \, dx}{2 f} \\ & = -\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}+\frac {\left (e p x^{2 n} (f x)^{-2 n}\right ) \int \frac {x^{-1-n}}{d+e x^n} \, dx}{2 f} \\ & = -\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}+\frac {\left (e p x^{2 n} (f x)^{-2 n}\right ) \text {Subst}\left (\int \frac {1}{x^2 (d+e x)} \, dx,x,x^n\right )}{2 f n} \\ & = -\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}+\frac {\left (e p x^{2 n} (f x)^{-2 n}\right ) \text {Subst}\left (\int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx,x,x^n\right )}{2 f n} \\ & = -\frac {e p x^n (f x)^{-2 n}}{2 d f n}-\frac {e^2 p x^{2 n} (f x)^{-2 n} \log (x)}{2 d^2 f}+\frac {e^2 p x^{2 n} (f x)^{-2 n} \log \left (d+e x^n\right )}{2 d^2 f n}-\frac {(f x)^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {(f x)^{-2 n} \left (e^2 n p x^{2 n} \log (x)-e^2 p x^{2 n} \log \left (d+e x^n\right )+d \left (e p x^n+d \log \left (c \left (d+e x^n\right )^p\right )\right )\right )}{2 d^2 f n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e^{2} f^{-2 \, n - 1} n p x^{2 \, n} \log \left (x\right ) + d e f^{-2 \, n - 1} p x^{n} + d^{2} f^{-2 \, n - 1} \log \left (c\right ) - {\left (e^{2} f^{-2 \, n - 1} p x^{2 \, n} - d^{2} f^{-2 \, n - 1} p\right )} \log \left (e x^{n} + d\right )}{2 \, d^{2} n x^{2 \, n}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((f*x)**(-1-2*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 = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1-2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {e \log \left (x\right )}{d^{2} f^{2 \, n}} - \frac {e \log \left (\frac {e x^{n} + d}{e}\right )}{d^{2} f^{2 \, n} n} + \frac {1}{d f^{2 \, n} n x^{n}}\right )}}{2 \, f} - \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, \left (f x\right )^{2 \, n} f n} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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