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

Optimal. Leaf size=79 \[ \frac {b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )} \]

[Out]

1/3*a*x^3*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/5*b*x^5*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]  time = 0.02, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1112, 14} \[ \frac {b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (b*x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*(a + b*x^
2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^2 \left (a b+b^2 x^2\right ) \, dx}{a b+b^2 x^2}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a b x^2+b^2 x^4\right ) \, dx}{a b+b^2 x^2}\\ &=\frac {a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.49 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \left (5 a x^3+3 b x^5\right )}{15 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(5*a*x^3 + 3*b*x^5))/(15*(a + b*x^2))

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fricas [A]  time = 0.88, size = 13, normalized size = 0.16 \[ \frac {1}{5} \, b x^{5} + \frac {1}{3} \, a x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*x^5 + 1/3*a*x^3

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giac [A]  time = 0.15, size = 29, normalized size = 0.37 \[ \frac {1}{5} \, b x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{3} \, a x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*x^5*sgn(b*x^2 + a) + 1/3*a*x^3*sgn(b*x^2 + a)

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maple [A]  time = 0.00, size = 36, normalized size = 0.46 \[ \frac {\left (3 b \,x^{2}+5 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{3}}{15 b \,x^{2}+15 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*x^3*(3*b*x^2+5*a)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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maxima [A]  time = 1.31, size = 13, normalized size = 0.16 \[ \frac {1}{5} \, b x^{5} + \frac {1}{3} \, a x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*x^5 + 1/3*a*x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\sqrt {{\left (b\,x^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*((a + b*x^2)^2)^(1/2),x)

[Out]

int(x^2*((a + b*x^2)^2)^(1/2), x)

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sympy [A]  time = 0.15, size = 12, normalized size = 0.15 \[ \frac {a x^{3}}{3} + \frac {b x^{5}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 + b*x**5/5

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