\(\int \frac {(b x^n)^p}{x^4} \, dx\) [2719]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=-\frac {\left (b x^n\right )^p}{(3-n p) x^3} \]

[Out]

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

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} \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=-\frac {\left (b x^n\right )^p}{x^3 (3-n p)} \]

[In]

Int[(b*x^n)^p/x^4,x]

[Out]

-((b*x^n)^p/((3 - n*p)*x^3))

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^{-4+n p} \, dx \\ & = -\frac {\left (b x^n\right )^p}{(3-n p) x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=\frac {\left (b x^n\right )^p}{(-3+n p) x^3} \]

[In]

Integrate[(b*x^n)^p/x^4,x]

[Out]

(b*x^n)^p/((-3 + n*p)*x^3)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {\left (b \,x^{n}\right )^{p}}{x^{3} \left (n p -3\right )}\)	\(19\)
parallelrisch	\(\frac {\left (b \,x^{n}\right )^{p}}{x^{3} \left (n p -3\right )}\)	\(19\)

[In]

int((b*x^n)^p/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=\frac {e^{\left (n p \log \left (x\right ) + p \log \left (b\right )\right )}}{{\left (n p - 3\right )} x^{3}} \]

[In]

integrate((b*x^n)^p/x^4,x, algorithm="fricas")

[Out]

e^(n*p*log(x) + p*log(b))/((n*p - 3)*x^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).

Time = 0.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=\begin {cases} \frac {\left (b x^{n}\right )^{p}}{n p x^{3} - 3 x^{3}} & \text {for}\: n \neq \frac {3}{p} \\\frac {\left (b x^{\frac {3}{p}}\right )^{p} \log {\left (x \right )}}{x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x**n)**p/x**4,x)

[Out]

Piecewise(((b*x**n)**p/(n*p*x**3 - 3*x**3), Ne(n, 3/p)), ((b*x**(3/p))**p*log(x)/x**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=\frac {b^{p} {\left (x^{n}\right )}^{p}}{{\left (n p - 3\right )} x^{3}} \]

[In]

integrate((b*x^n)^p/x^4,x, algorithm="maxima")

[Out]

b^p*(x^n)^p/((n*p - 3)*x^3)

Giac [F]

\[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=\int { \frac {\left (b x^{n}\right )^{p}}{x^{4}} \,d x } \]

[In]

integrate((b*x^n)^p/x^4,x, algorithm="giac")

[Out]

integrate((b*x^n)^p/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^n\right )^p}{x^4} \, dx=\int \frac {{\left (b\,x^n\right )}^p}{x^4} \,d x \]

[In]

int((b*x^n)^p/x^4,x)

[Out]

int((b*x^n)^p/x^4, x)