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

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

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Rubi [A]  time = 0.01, 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}{(2 p-3) 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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IntegrateAlgebraic [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x^2\right )^p}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x^{2}\right )^{p}}{x^{4}}\,{d x} \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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maple [A]  time = 0.00, size = 19, normalized size = 1.00 \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]

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

[Out]

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

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maxima [A]  time = 1.33, 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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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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {b^{p} \left (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**p*(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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