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

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

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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)]

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]

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*}

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Mathematica [A]  time = 0.00, size = 29, normalized size = 0.71 \begin {gather*} \frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p}{4 b p+2 b} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.11, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 37, normalized size = 0.90 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.26, size = 58, normalized size = 1.41 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.00, size = 40, normalized size = 0.98 \begin {gather*} \frac {\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{2 \left (2 p +1\right ) b} \end {gather*}

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 = 1.32, size = 30, normalized size = 0.73 \begin {gather*} \frac {{\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{2 \, b {\left (2 \, p + 1\right )}} \end {gather*}

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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mupad [B]  time = 4.67, size = 46, normalized size = 1.12 \begin {gather*} \left (\frac {x^2}{2\,\left (2\,p+1\right )}+\frac {a}{2\,b\,\left (2\,p+1\right )}\right )\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {x^{2}}{2 \sqrt {a^{2}}} & \text {for}\: b = 0 \wedge p = - \frac {1}{2} \\\frac {x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b = 0 \\\int \frac {x}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 2 b} + \frac {b x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 2 b} & \text {otherwise} \end {cases} \end {gather*}

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]

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

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