3.4.71 \(\int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=36 \[ \frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b} \]

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1107, 609} \begin {gather*} \frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

((a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*b)

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

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.06 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (2 a x^2+b x^4\right )}{4 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(2*a*x^2 + b*x^4))/(4*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 5.68, size = 38, normalized size = 1.06 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (2 a x^2+b x^4\right )}{4 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(2*a*x^2 + b*x^4))/(4*(a + b*x^2))

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fricas [A]  time = 0.67, size = 13, normalized size = 0.36 \begin {gather*} \frac {1}{4} \, b x^{4} + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/4*b*x^4 + 1/2*a*x^2

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giac [A]  time = 0.16, size = 22, normalized size = 0.61 \begin {gather*} \frac {1}{4} \, {\left (b x^{4} + 2 \, a x^{2}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/4*(b*x^4 + 2*a*x^2)*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 35, normalized size = 0.97 \begin {gather*} \frac {\left (b \,x^{2}+2 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2}}{4 b \,x^{2}+4 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((b*x^2+a)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.32, size = 14, normalized size = 0.39 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{2}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/4*(b*x^2 + a)^2/b

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mupad [B]  time = 4.35, size = 33, normalized size = 0.92 \begin {gather*} \left (\frac {a}{4\,b}+\frac {x^2}{4}\right )\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((a + b*x^2)^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.10, size = 12, normalized size = 0.33 \begin {gather*} \frac {a x^{2}}{2} + \frac {b x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x**2+a)**2)**(1/2),x)

[Out]

a*x**2/2 + b*x**4/4

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