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]

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

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Rubi [A]  time = 0.00, antiderivative size = 23, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.00, 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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fricas [A]  time = 0.67, size = 25, normalized size = 1.09 \[ \frac {{\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b p + b\right )}} \]

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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giac [A]  time = 0.57, size = 21, normalized size = 0.91 \[ \frac {{\left (b x^{2} + a\right )}^{p + 1}}{2 \, b {\left (p + 1\right )}} \]

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))

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maple [A]  time = 0.00, size = 22, normalized size = 0.96 \[ \frac {\left (b \,x^{2}+a \right )^{p +1}}{2 \left (p +1\right ) b} \]

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)^(p+1)/b/(p+1)

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maxima [A]  time = 1.38, size = 21, normalized size = 0.91 \[ \frac {{\left (b x^{2} + a\right )}^{p + 1}}{2 \, b {\left (p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + a)^(p + 1)/(b*(p + 1))

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mupad [B]  time = 4.89, size = 21, normalized size = 0.91 \[ \frac {{\left (b\,x^2+a\right )}^{p+1}}{2\,b\,\left (p+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2)^p,x)

[Out]

(a + b*x^2)^(p + 1)/(2*b*(p + 1))

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sympy [A]  time = 0.73, size = 97, normalized size = 4.22 \[ \begin {cases} \frac {x^{2}}{2 a} & \text {for}\: b = 0 \wedge p = -1 \\\frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b} + \frac {\log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b} & \text {for}\: p = -1 \\\frac {a \left (a + b x^{2}\right )^{p}}{2 b p + 2 b} + \frac {b x^{2} \left (a + b x^{2}\right )^{p}}{2 b p + 2 b} & \text {otherwise} \end {cases} \]

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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