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

Optimal. Leaf size=74 \[ \frac {a x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {b x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1088} \[ \frac {b x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1088

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.49 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a x+b x^3\right )}{3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 10, normalized size = 0.14 \[ \frac {1}{3} \, b x^{3} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 + a*x

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giac [A]  time = 0.18, size = 20, normalized size = 0.27 \[ \frac {1}{3} \, {\left (b x^{3} + 3 \, a x\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^3 + 3*a*x)*sgn(b*x^2 + a)

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maple [A]  time = 0.00, size = 33, normalized size = 0.45 \[ \frac {\left (b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x}{3 b \,x^{2}+3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 10, normalized size = 0.14 \[ \frac {1}{3} \, b x^{3} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 + a*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2)^(1/2),x)

[Out]

int(((a + b*x^2)^2)^(1/2), x)

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sympy [A]  time = 0.10, size = 8, normalized size = 0.11 \[ a x + \frac {b x^{3}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*x**3/3

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