3.92 \(\int x (a+b x^2) (a^2+2 a b x^2+b^2 x^4)^p \, dx\)

Optimal. Leaf size=34 \[ \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{4 b (p+1)} \]

[Out]

(a^2 + 2*a*b*x^2 + b^2*x^4)^(1 + p)/(4*b*(1 + p))

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Rubi [A]  time = 0.0289665, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {1247, 629} \[ \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{4 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

(a^2 + 2*a*b*x^2 + b^2*x^4)^(1 + p)/(4*b*(1 + p))

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int x \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,x^2\right )\\ &=\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{1+p}}{4 b (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0088728, size = 25, normalized size = 0.74 \[ \frac{\left (\left (a+b x^2\right )^2\right )^{p+1}}{4 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

((a + b*x^2)^2)^(1 + p)/(4*b*(1 + p))

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Maple [A]  time = 0.003, size = 40, normalized size = 1.2 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) ^{2} \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p}}{4\,b \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

1/4*(b*x^2+a)^2/b/(1+p)*(b^2*x^4+2*a*b*x^2+a^2)^p

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Maxima [B]  time = 0.994282, size = 116, normalized size = 3.41 \begin{align*} \frac{{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{2 \, p} a}{2 \, b{\left (2 \, p + 1\right )}} + \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{4 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

1/2*(b*x^2 + a)*(b*x^2 + a)^(2*p)*a/(b*(2*p + 1)) + 1/4*(b^2*(2*p + 1)*x^4 + 2*a*b*p*x^2 - a^2)*(b*x^2 + a)^(2
*p)/((2*p^2 + 3*p + 1)*b)

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Fricas [A]  time = 1.60822, size = 99, normalized size = 2.91 \begin{align*} \frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \,{\left (b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

1/4*(b^2*x^4 + 2*a*b*x^2 + a^2)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(b*p + b)

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Sympy [A]  time = 11.9969, size = 165, normalized size = 4.85 \begin{align*} \begin{cases} \frac{x^{2}}{2 a} & \text{for}\: b = 0 \wedge p = -1 \\\frac{a x^{2} \left (a^{2}\right )^{p}}{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^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 4 b} + \frac{2 a b x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 4 b} + \frac{b^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 4 b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Piecewise((x**2/(2*a), Eq(b, 0) & Eq(p, -1)), (a*x**2*(a**2)**p/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**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(4*b*p + 4*b)
+ 2*a*b*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(4*b*p + 4*b) + b**2*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(
4*b*p + 4*b), True))

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Giac [B]  time = 1.12262, size = 119, normalized size = 3.5 \begin{align*} \frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} x^{4} + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b x^{2} +{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2}}{4 \,{\left (b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

1/4*((b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^2*x^4 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b*x^2 + (b^2*x^4 + 2*a*b*x^2 +
a^2)^p*a^2)/(b*p + b)