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

Optimal. Leaf size=41 \[ \frac{\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0249834, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1107, 609} \[ \frac{\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1107

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

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

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

Mathematica [A]  time = 0.0042079, size = 29, normalized size = 0.71 \[ \frac{\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p}{4 b p+2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 40, normalized size = 1. \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p}}{2\,b \left ( 1+2\,p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.993721, size = 41, normalized size = 1. \begin{align*} \frac{{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{2 \, b{\left (2 \, p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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)/(b*(2*p + 1))

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Fricas [A]  time = 1.60884, size = 80, normalized size = 1.95 \begin{align*} \frac{{\left (b x^{2} + a\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{2 \,{\left (2 \, b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.18376, size = 78, normalized size = 1.9 \begin{align*} \frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b x^{2} +{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a}{2 \,{\left (2 \, b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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