\(\int \frac {(b x^n)^p}{x^2} \, dx\) [2717]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=-\frac {\left (b x^n\right )^p}{(1-n p) x} \]

[Out]

-(b*x^n)^p/(-n*p+1)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 30} \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=-\frac {\left (b x^n\right )^p}{x (1-n p)} \]

[In]

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

[Out]

-((b*x^n)^p/((1 - n*p)*x))

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}& = \left (x^{-n p} \left (b x^n\right )^p\right ) \int x^{-2+n p} \, dx \\ & = -\frac {\left (b x^n\right )^p}{(1-n p) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=\frac {\left (b x^n\right )^p}{(-1+n p) x} \]

[In]

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

[Out]

(b*x^n)^p/((-1 + n*p)*x)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {\left (b \,x^{n}\right )^{p}}{x \left (n p -1\right )}\)	\(19\)
parallelrisch	\(\frac {\left (b \,x^{n}\right )^{p}}{x \left (n p -1\right )}\)	\(19\)

[In]

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

[Out]

1/x/(n*p-1)*(b*x^n)^p

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=\frac {e^{\left (n p \log \left (x\right ) + p \log \left (b\right )\right )}}{{\left (n p - 1\right )} x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=\begin {cases} \frac {\left (b x^{n}\right )^{p}}{n p x - x} & \text {for}\: n \neq \frac {1}{p} \\\frac {\left (b x^{\frac {1}{p}}\right )^{p} \log {\left (x \right )}}{x} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x**n)**p/x**2,x)

[Out]

Piecewise(((b*x**n)**p/(n*p*x - x), Ne(n, 1/p)), ((b*x**(1/p))**p*log(x)/x, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=\frac {b^{p} {\left (x^{n}\right )}^{p}}{{\left (n p - 1\right )} x} \]

[In]

integrate((b*x^n)^p/x^2,x, algorithm="maxima")

[Out]

b^p*(x^n)^p/((n*p - 1)*x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x^n)^p/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^n\right )^p}{x^2} \, dx=\int \frac {{\left (b\,x^n\right )}^p}{x^2} \,d x \]

[In]

int((b*x^n)^p/x^2,x)

[Out]

int((b*x^n)^p/x^2, x)