\(\int \frac {\log ^q(c (d+e x^n)^p)}{x (f+g x^{-2 n})} \, dx\) [388]

Optimal result

Integrand size = 29, antiderivative size = 29 \[ \int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\text {Int}\left (\frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )},x\right ) \]

[Out]

Unintegrable(ln(c*(d+e*x^n)^p)^q/x/(f+g/(x^(2*n))),x)

Rubi [N/A]

Not integrable

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx \]

[In]

Int[Log[c*(d + e*x^n)^p]^q/(x*(f + g/x^(2*n))),x]

[Out]

Defer[Int][Log[c*(d + e*x^n)^p]^q/(x*(f + g/x^(2*n))), x]

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

Mathematica [N/A]

Not integrable

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx \]

[In]

Integrate[Log[c*(d + e*x^n)^p]^q/(x*(f + g/x^(2*n))),x]

[Out]

Integrate[Log[c*(d + e*x^n)^p]^q/(x*(f + g/x^(2*n))), x]

Maple [N/A]

Not integrable

Time = 0.64 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07

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

[In]

int(ln(c*(d+e*x^n)^p)^q/x/(f+g/(x^(2*n))),x)

[Out]

int(ln(c*(d+e*x^n)^p)^q/x/(f+g/(x^(2*n))),x)

Fricas [N/A]

Not integrable

Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{q}}{{\left (f + \frac {g}{x^{2 \, n}}\right )} x} \,d x } \]

[In]

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

[Out]

integral(x^(2*n)*log((e*x^n + d)^p*c)^q/(f*x*x^(2*n) + g*x), x)

Sympy [F(-1)]

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

[In]

integrate(ln(c*(d+e*x**n)**p)**q/x/(f+g/(x**(2*n))),x)

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [N/A]

Not integrable

Time = 3.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{q}}{{\left (f + \frac {g}{x^{2 \, n}}\right )} x} \,d x } \]

[In]

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

[Out]

integrate(log((e*x^n + d)^p*c)^q/((f + g/x^(2*n))*x), x)

Mupad [N/A]

Not integrable

Time = 1.63 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\log ^q\left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\int \frac {{\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}^q}{x\,\left (f+\frac {g}{x^{2\,n}}\right )} \,d x \]

[In]

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

[Out]

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