\(\int \frac {(a (b x^n)^p)^q}{x^2} \, dx\) [194]

Optimal result

Integrand size = 15, antiderivative size = 25 \[ \int \frac {\left (a \left (b x^n\right )^p\right )^q}{x^2} \, dx=-\frac {\left (a \left (b x^n\right )^p\right )^q}{(1-n p q) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1971, 30} \[ \int \frac {\left (a \left (b x^n\right )^p\right )^q}{x^2} \, dx=-\frac {\left (a \left (b x^n\right )^p\right )^q}{x (1-n p q)} \]

[In]

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

[Out]

-((a*(b*x^n)^p)^q/((1 - n*p*q)*x))

Rule 30

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

Rule 1971

Int[(u_.)*((c_.)*((d_.)*((a_.) + (b_.)*(x_))^(n_))^(q_))^(p_), x_Symbol] :> Dist[(c*(d*(a + b*x)^n)^q)^p/(a +
b*x)^(n*p*q), Int[u*(a + b*x)^(n*p*q), x], x] /; FreeQ[{a, b, c, d, n, q, p}, x] &&  !IntegerQ[q] &&  !Integer
Q[p]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a \left (b x^n\right )^p\right )^q}{x^2} \, dx=\frac {\left (a \left (b x^n\right )^p\right )^q}{(-1+n p q) x} \]

[In]

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

[Out]

(a*(b*x^n)^p)^q/((-1 + n*p*q)*x)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
gosper	\(\frac {{\left (a \left (b \,x^{n}\right )^{p}\right )}^{q}}{x \left (n p q -1\right )}\)	\(24\)
parallelrisch	\(\frac {{\left (a \left (b \,x^{n}\right )^{p}\right )}^{q}}{x \left (n p q -1\right )}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a \left (b x^n\right )^p\right )^q}{x^2} \, dx=\frac {e^{\left (n p q \log \left (x\right ) + p q \log \left (b\right ) + q \log \left (a\right )\right )}}{{\left (n p q - 1\right )} x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a \left (b x^n\right )^p\right )^q}{x^2} \, dx=\begin {cases} \frac {\left (a \left (b x^{n}\right )^{p}\right )^{q}}{x \left (n p q - 1\right )} & \text {for}\: n p q \neq 1 \\\frac {\left (a \left (b x^{n}\right )^{p}\right )^{q} \log {\left (x \right )}}{x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a \left (b x^n\right )^p\right )^q}{x^2} \, dx=\frac {a^{q} b^{p q} {\left ({\left (x^{n}\right )}^{p}\right )}^{q}}{{\left (n p q - 1\right )} x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(((b*x^n)^p*a)^q/x^2, x)

Mupad [F(-1)]

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

[In]

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

[Out]

int((a*(b*x^n)^p)^q/x^2, x)