3.4.70 \(\int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=67 \[ \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2}-\frac {a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 640, 609} \begin {gather*} \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2}-\frac {a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

-(a*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*b^2) + (a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(6*b^2)

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 640

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

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps

\begin {align*} \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x \sqrt {a^2+2 a b x+b^2 x^2} \, dx,x,x^2\right )\\ &=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2}-\frac {a \operatorname {Subst}\left (\int \sqrt {a^2+2 a b x+b^2 x^2} \, dx,x,x^2\right )}{2 b}\\ &=-\frac {a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.58 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a x^4+2 b x^6\right )}{12 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(3*a*x^4 + 2*b*x^6))/(12*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 6.19, size = 39, normalized size = 0.58 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a x^4+2 b x^6\right )}{12 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(3*a*x^4 + 2*b*x^6))/(12*(a + b*x^2))

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fricas [A]  time = 1.07, size = 13, normalized size = 0.19 \begin {gather*} \frac {1}{6} \, b x^{6} + \frac {1}{4} \, a x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*b*x^6 + 1/4*a*x^4

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giac [A]  time = 0.18, size = 23, normalized size = 0.34 \begin {gather*} \frac {1}{12} \, {\left (2 \, b x^{6} + 3 \, a x^{4}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/12*(2*b*x^6 + 3*a*x^4)*sgn(b*x^2 + a)

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maple [A]  time = 0.00, size = 36, normalized size = 0.54 \begin {gather*} \frac {\left (2 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{4}}{12 b \,x^{2}+12 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*((b*x^2+a)^2)^(1/2),x)

[Out]

1/12*x^4*(2*b*x^2+3*a)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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maxima [A]  time = 1.28, size = 13, normalized size = 0.19 \begin {gather*} \frac {1}{6} \, b x^{6} + \frac {1}{4} \, a x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*b*x^6 + 1/4*a*x^4

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mupad [B]  time = 4.31, size = 59, normalized size = 0.88 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (8\,b^2\,\left (a^2+b^2\,x^4\right )-12\,a^2\,b^2+4\,a\,b^3\,x^2\right )}{48\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*((a + b*x^2)^2)^(1/2),x)

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*(8*b^2*(a^2 + b^2*x^4) - 12*a^2*b^2 + 4*a*b^3*x^2))/(48*b^4)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.18 \begin {gather*} \frac {a x^{4}}{4} + \frac {b x^{6}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*((b*x**2+a)**2)**(1/2),x)

[Out]

a*x**4/4 + b*x**6/6

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