\(\int \frac {(b \log ^m(a x^n))^p}{x} \, dx\) [152]

Optimal result

Integrand size = 16, antiderivative size = 29 \[ \int \frac {\left (b \log ^m\left (a x^n\right )\right )^p}{x} \, dx=\frac {\log \left (a x^n\right ) \left (b \log ^m\left (a x^n\right )\right )^p}{n (1+m p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {15, 30} \[ \int \frac {\left (b \log ^m\left (a x^n\right )\right )^p}{x} \, dx=\frac {\log \left (a x^n\right ) \left (b \log ^m\left (a x^n\right )\right )^p}{n (m p+1)} \]

[In]

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

[Out]

(Log[a*x^n]*(b*Log[a*x^n]^m)^p)/(n*(1 + m*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 (b x^m\right )^p \, dx,x,\log \left (a x^n\right )\right )}{n} \\ & = \frac {\left (\log ^{-m p}\left (a x^n\right ) \left (b \log ^m\left (a x^n\right )\right )^p\right ) \text {Subst}\left (\int x^{m p} \, dx,x,\log \left (a x^n\right )\right )}{n} \\ & = \frac {\log \left (a x^n\right ) \left (b \log ^m\left (a x^n\right )\right )^p}{n (1+m p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b \log ^m\left (a x^n\right )\right )^p}{x} \, dx=\frac {\log \left (a x^n\right ) \left (b \log ^m\left (a x^n\right )\right )^p}{n (1+m p)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\ln \left (a \,x^{n}\right ) {\mathrm e}^{p \ln \left (b \,{\mathrm e}^{m \ln \left (\ln \left (a \,x^{n}\right )\right )}\right )}}{n \left (m p +1\right )}\)	\(34\)
default	\(\frac {\ln \left (a \,x^{n}\right ) {\mathrm e}^{p \ln \left (b \,{\mathrm e}^{m \ln \left (\ln \left (a \,x^{n}\right )\right )}\right )}}{n \left (m p +1\right )}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (b \log ^m\left (a x^n\right )\right )^p}{x} \, dx=\frac {{\left (n \log \left (x\right ) + \log \left (a\right )\right )} e^{\left (m p \log \left (n \log \left (x\right ) + \log \left (a\right )\right ) + p \log \left (b\right )\right )}}{m n p + n} \]

[In]

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

[Out]

(n*log(x) + log(a))*e^(m*p*log(n*log(x) + log(a)) + p*log(b))/(m*n*p + n)

Sympy [F]

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

[In]

integrate((b*ln(a*x**n)**m)**p/x,x)

[Out]

Integral((b*log(a*x**n)**m)**p/x, x)

Maxima [F(-2)]

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

[In]

integrate((b*log(a*x^n)^m)^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 = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\left (b \log ^m\left (a x^n\right )\right )^p}{x} \, dx=\frac {{\left (n \log \left (x\right ) + \log \left (a\right )\right )} e^{\left (m p \log \left (n \log \left (x\right ) + \log \left (a\right )\right ) + p \log \left (b\right )\right )}}{{\left (m p + 1\right )} n} \]

[In]

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

[Out]

(n*log(x) + log(a))*e^(m*p*log(n*log(x) + log(a)) + p*log(b))/((m*p + 1)*n)

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b \log ^m\left (a x^n\right )\right )^p}{x} \, dx=\frac {\ln \left (a\,x^n\right )\,{\left (b\,{\ln \left (a\,x^n\right )}^m\right )}^p}{n\,\left (m\,p+1\right )} \]

[In]

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

[Out]

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