3.1.71 \(\int x (b x^2)^p \, dx\) [71]

Optimal. Leaf size=19 \[ \frac {x^2 \left (b x^2\right )^p}{2 (1+p)} \]

[Out]

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

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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 = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \begin {gather*} \frac {x^2 \left (b x^2\right )^p}{2 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(b*x^2)^p)/(2*(1 + p))

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(b*x^2)^p)/(2 + 2*p)

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Maple [A]
time = 0.01, size = 20, normalized size = 1.05

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (b x^{2}\right )^{p + 1}}{2 \, b {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2)^(p + 1)/(b*(p + 1))

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Fricas [A]
time = 0.38, size = 17, normalized size = 0.89 \begin {gather*} \frac {\left (b x^{2}\right )^{p} x^{2}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2)^p*x^2/(p + 1)

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Sympy [A]
time = 0.08, size = 20, normalized size = 1.05 \begin {gather*} \begin {cases} \frac {x^{2} \left (b x^{2}\right )^{p}}{2 p + 2} & \text {for}\: p \neq -1 \\\frac {\log {\left (x \right )}}{b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*(b*x**2)**p/(2*p + 2), Ne(p, -1)), (log(x)/b, True))

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Giac [A]
time = 1.03, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (b x^{2}\right )^{p + 1}}{2 \, b {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2)^(p + 1)/(b*(p + 1))

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Mupad [B]
time = 0.99, size = 18, normalized size = 0.95 \begin {gather*} \frac {x^2\,{\left (b\,x^2\right )}^p}{2\,\left (p+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*(b*x^2)^p)/(2*(p + 1))

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