3.1.76 \(\int \frac {(b x^2)^p}{x^4} \, dx\) [76]

Optimal. Leaf size=19 \[ -\frac {\left (b x^2\right )^p}{(3-2 p) x^3} \]

[Out]

-(b*x^2)^p/(3-2*p)/x^3

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 30} \begin {gather*} -\frac {\left (b x^2\right )^p}{(3-2 p) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2)^p/x^4,x]

[Out]

-((b*x^2)^p/((3 - 2*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*} \int \frac {\left (b x^2\right )^p}{x^4} \, dx &=\left (x^{-2 p} \left (b x^2\right )^p\right ) \int x^{-4+2 p} \, dx\\ &=-\frac {\left (b x^2\right )^p}{(3-2 p) x^3}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 18, normalized size = 0.95 \begin {gather*} \frac {\left (b x^2\right )^p}{(-3+2 p) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2)^p/x^4,x]

[Out]

(b*x^2)^p/((-3 + 2*p)*x^3)

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Maple [A]
time = 0.02, size = 19, normalized size = 1.00

method	result	size
gosper	\(\frac {\left (b \,x^{2}\right )^{p}}{x^{3} \left (-3+2 p \right )}\)	\(19\)
risch	\(\frac {\left (b \,x^{2}\right )^{p}}{x^{3} \left (-3+2 p \right )}\)	\(19\)
norman	\(\frac {{\mathrm e}^{p \ln \left (b \,x^{2}\right )}}{\left (-3+2 p \right ) x^{3}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2)^p/x^4,x,method=_RETURNVERBOSE)

[Out]

1/x^3/(-3+2*p)*(b*x^2)^p

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Maxima [A]
time = 0.30, size = 19, normalized size = 1.00 \begin {gather*} \frac {b^{p} x^{2 \, p}}{{\left (2 \, p - 3\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2)^p/x^4,x, algorithm="maxima")

[Out]

b^p*x^(2*p)/((2*p - 3)*x^3)

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Fricas [A]
time = 0.36, size = 18, normalized size = 0.95 \begin {gather*} \frac {\left (b x^{2}\right )^{p}}{{\left (2 \, p - 3\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2)^p/x^4,x, algorithm="fricas")

[Out]

(b*x^2)^p/((2*p - 3)*x^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {\left (b x^{2}\right )^{p}}{2 p x^{3} - 3 x^{3}} & \text {for}\: p \neq \frac {3}{2} \\\int \frac {\left (b x^{2}\right )^{\frac {3}{2}}}{x^{4}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2)**p/x**4,x)

[Out]

Piecewise(((b*x**2)**p/(2*p*x**3 - 3*x**3), Ne(p, 3/2)), (Integral((b*x**2)**(3/2)/x**4, x), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2)^p/x^4,x, algorithm="giac")

[Out]

integrate((b*x^2)^p/x^4, x)

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Mupad [B]
time = 0.97, size = 18, normalized size = 0.95 \begin {gather*} \frac {{\left (b\,x^2\right )}^p}{x^3\,\left (2\,p-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2)^p/x^4,x)

[Out]

(b*x^2)^p/(x^3*(2*p - 3))

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