3.1042 \(\int x (a+b x^2)^p \, dx\)

Optimal. Leaf size=23 \[ \frac{\left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]

[Out]

(a + b*x^2)^(1 + p)/(2*b*(1 + p))

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Rubi [A]  time = 0.0047815, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {261} \[ \frac{\left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^p,x]

[Out]

(a + b*x^2)^(1 + p)/(2*b*(1 + p))

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0028317, size = 22, normalized size = 0.96 \[ \frac{\left (a+b x^2\right )^{p+1}}{2 b p+2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^2)^p,x]

[Out]

(a + b*x^2)^(1 + p)/(2*b + 2*b*p)

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Maple [A]  time = 0.001, size = 22, normalized size = 1. \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) ^{1+p}}{2\,b \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54206, size = 55, normalized size = 2.39 \begin{align*} \frac{{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{p}}{2 \,{\left (b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.709601, size = 97, normalized size = 4.22 \begin{align*} \begin{cases} \frac{x^{2}}{2 a} & \text{for}\: b = 0 \wedge p = -1 \\\frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b} + \frac{\log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b} & \text{for}\: p = -1 \\\frac{a \left (a + b x^{2}\right )^{p}}{2 b p + 2 b} + \frac{b x^{2} \left (a + b x^{2}\right )^{p}}{2 b p + 2 b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2/(2*a), Eq(b, 0) & Eq(p, -1)), (a**p*x**2/2, Eq(b, 0)), (log(-I*sqrt(a)*sqrt(1/b) + x)/(2*b) +
log(I*sqrt(a)*sqrt(1/b) + x)/(2*b), Eq(p, -1)), (a*(a + b*x**2)**p/(2*b*p + 2*b) + b*x**2*(a + b*x**2)**p/(2*b
*p + 2*b), True))

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Giac [A]  time = 1.77907, size = 28, normalized size = 1.22 \begin{align*} \frac{{\left (b x^{2} + a\right )}^{p + 1}}{2 \, b{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + a)^(p + 1)/(b*(p + 1))