3.1.75 \(\int \frac {(b x^2)^p}{x^3} \, dx\) [75]

Optimal. Leaf size=21 \[ -\frac {\left (b x^2\right )^p}{2 (1-p) x^2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 21, 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}{2 (1-p) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2)^p/x^3,x]

[Out]

-1/2*(b*x^2)^p/((1 - p)*x^2)

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^3} \, dx &=\left (x^{-2 p} \left (b x^2\right )^p\right ) \int x^{-3+2 p} \, dx\\ &=-\frac {\left (b x^2\right )^p}{2 (1-p) x^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 18, normalized size = 0.86 \begin {gather*} \frac {\left (b x^2\right )^p}{(-2+2 p) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2)^p/x^3,x]

[Out]

(b*x^2)^p/((-2 + 2*p)*x^2)

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Maple [A]
time = 0.02, size = 18, normalized size = 0.86

method	result	size
gosper	\(\frac {\left (b \,x^{2}\right )^{p}}{2 x^{2} \left (-1+p \right )}\)	\(18\)
risch	\(\frac {\left (b \,x^{2}\right )^{p}}{2 x^{2} \left (-1+p \right )}\)	\(18\)
norman	\(\frac {{\mathrm e}^{p \ln \left (b \,x^{2}\right )}}{2 \left (-1+p \right ) x^{2}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2)^p/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.35, size = 18, normalized size = 0.86 \begin {gather*} \frac {b^{p} {\left (x^{2}\right )}^{p}}{2 \, {\left (p - 1\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2)^p/x^3,x, algorithm="maxima")

[Out]

1/2*b^p*(x^2)^p/((p - 1)*x^2)

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Fricas [A]
time = 0.36, size = 17, normalized size = 0.81 \begin {gather*} \frac {\left (b x^{2}\right )^{p}}{2 \, {\left (p - 1\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2)^p/x^3,x, algorithm="fricas")

[Out]

1/2*(b*x^2)^p/((p - 1)*x^2)

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Sympy [A]
time = 0.17, size = 22, normalized size = 1.05 \begin {gather*} \begin {cases} \frac {\left (b x^{2}\right )^{p}}{2 p x^{2} - 2 x^{2}} & \text {for}\: p \neq 1 \\b \log {\left (x \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2)**p/x**3,x)

[Out]

Piecewise(((b*x**2)**p/(2*p*x**2 - 2*x**2), Ne(p, 1)), (b*log(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^3,x, algorithm="giac")

[Out]

integrate((b*x^2)^p/x^3, x)

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Mupad [B]
time = 0.98, size = 18, normalized size = 0.86 \begin {gather*} \frac {{\left (b\,x^2\right )}^p}{x^2\,\left (2\,p-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2)^p/x^3,x)

[Out]

(b*x^2)^p/(x^2*(2*p - 2))

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