\(\int \frac {\log ^2(a x^n)^p}{x} \, dx\) [149]

Optimal result

Integrand size = 14, antiderivative size = 27 \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\frac {\log \left (a x^n\right ) \log ^2\left (a x^n\right )^p}{n (1+2 p)} \]

[Out]

ln(a*x^n)*(ln(a*x^n)^2)^p/n/(1+2*p)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {15, 30} \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\frac {\log \left (a x^n\right ) \log ^2\left (a x^n\right )^p}{n (2 p+1)} \]

[In]

Int[(Log[a*x^n]^2)^p/x,x]

[Out]

(Log[a*x^n]*(Log[a*x^n]^2)^p)/(n*(1 + 2*p))

Rule 15

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\frac {\log \left (a x^n\right ) \log ^2\left (a x^n\right )^p}{n (1+2 p)} \]

[In]

Integrate[(Log[a*x^n]^2)^p/x,x]

[Out]

(Log[a*x^n]*(Log[a*x^n]^2)^p)/(n*(1 + 2*p))

Maple [A] (verified)

Time = 5.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\ln \left (a \,x^{n}\right ) {\mathrm e}^{p \ln \left (\ln \left (a \,x^{n}\right )^{2}\right )}}{n \left (1+2 p \right )}\)	\(30\)
default	\(\frac {\ln \left (a \,x^{n}\right ) {\mathrm e}^{p \ln \left (\ln \left (a \,x^{n}\right )^{2}\right )}}{n \left (1+2 p \right )}\)	\(30\)

[In]

int((ln(a*x^n)^2)^p/x,x,method=_RETURNVERBOSE)

[Out]

1/n/(1+2*p)*ln(a*x^n)*exp(p*ln(ln(a*x^n)^2))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\frac {{\left (n \log \left (x\right ) + \log \left (a\right )\right )} {\left (n^{2} \log \left (x\right )^{2} + 2 \, n \log \left (a\right ) \log \left (x\right ) + \log \left (a\right )^{2}\right )}^{p}}{2 \, n p + n} \]

[In]

integrate((log(a*x^n)^2)^p/x,x, algorithm="fricas")

[Out]

(n*log(x) + log(a))*(n^2*log(x)^2 + 2*n*log(a)*log(x) + log(a)^2)^p/(2*n*p + n)

Sympy [F]

\[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\int \frac {\left (\log {\left (a x^{n} \right )}^{2}\right )^{p}}{x}\, dx \]

[In]

integrate((ln(a*x**n)**2)**p/x,x)

[Out]

Integral((log(a*x**n)**2)**p/x, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((log(a*x^n)^2)^p/x,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 [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\frac {{\left (n \log \left (x\right ) \mathrm {sgn}\left (\log \left (a x^{n}\right )\right ) + \log \left (a\right ) \mathrm {sgn}\left (\log \left (a x^{n}\right )\right )\right )}^{2 \, p + 1}}{n {\left (2 \, p + 1\right )} \mathrm {sgn}\left (\log \left (a x^{n}\right )\right )} \]

[In]

integrate((log(a*x^n)^2)^p/x,x, algorithm="giac")

[Out]

(n*log(x)*sgn(log(a*x^n)) + log(a)*sgn(log(a*x^n)))^(2*p + 1)/(n*(2*p + 1)*sgn(log(a*x^n)))

Mupad [B] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\log ^2\left (a x^n\right )^p}{x} \, dx=\frac {\ln \left (a\,x^n\right )\,{\left ({\ln \left (a\,x^n\right )}^2\right )}^p}{n\,\left (2\,p+1\right )} \]

[In]

int((log(a*x^n)^2)^p/x,x)

[Out]

(log(a*x^n)*(log(a*x^n)^2)^p)/(n*(2*p + 1))