\(\int x^{2-n p q} (a (b x^n)^p)^q \, dx\) [201]

Optimal result

Integrand size = 21, antiderivative size = 24 \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {1}{3} x^{3-n p q} \left (a \left (b x^n\right )^p\right )^q \]

[Out]

1/3*x^(-n*p*q+3)*(a*(b*x^n)^p)^q

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1971, 30} \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {1}{3} x^{3-n p q} \left (a \left (b x^n\right )^p\right )^q \]

[In]

Int[x^(2 - n*p*q)*(a*(b*x^n)^p)^q,x]

[Out]

(x^(3 - n*p*q)*(a*(b*x^n)^p)^q)/3

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 \, dx \\ & = \frac {1}{3} x^{3-n p q} \left (a \left (b x^n\right )^p\right )^q \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {1}{3} x^{3-n p q} \left (a \left (b x^n\right )^p\right )^q \]

[In]

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

[Out]

(x^(3 - n*p*q)*(a*(b*x^n)^p)^q)/3

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
gosper	\(\frac {x^{-n p q +3} {\left (a \left (b \,x^{n}\right )^{p}\right )}^{q}}{3}\)	\(23\)
parallelrisch	\(\frac {x \,x^{-n p q +2} {\left (a \left (b \,x^{n}\right )^{p}\right )}^{q}}{3}\)	\(24\)

[In]

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

[Out]

1/3*x^(-n*p*q+3)*(a*(b*x^n)^p)^q

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {1}{3} \, x^{3} e^{\left (p q \log \left (b\right ) + q \log \left (a\right )\right )} \]

[In]

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

[Out]

1/3*x^3*e^(p*q*log(b) + q*log(a))

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {x x^{- n p q + 2} \left (a \left (b x^{n}\right )^{p}\right )^{q}}{3} \]

[In]

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

[Out]

x*x**(-n*p*q + 2)*(a*(b*x**n)**p)**q/3

Maxima [F]

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

[In]

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

[Out]

integrate(((b*x^n)^p*a)^q*x^(-n*p*q + 2), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {1}{3} \, x e^{\left (p q \log \left (b\right ) + q \log \left (a\right ) + 2 \, \log \left (x\right )\right )} \]

[In]

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

[Out]

1/3*x*e^(p*q*log(b) + q*log(a) + 2*log(x))

Mupad [B] (verification not implemented)

Time = 5.66 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int x^{2-n p q} \left (a \left (b x^n\right )^p\right )^q \, dx=\frac {x^{3-n\,p\,q}\,{\left (a\,{\left (b\,x^n\right )}^p\right )}^q}{3} \]

[In]

int(x^(2 - n*p*q)*(a*(b*x^n)^p)^q,x)

[Out]

(x^(3 - n*p*q)*(a*(b*x^n)^p)^q)/3