3.1.11 \(\int \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx\) [11]

Optimal. Leaf size=74 \[ \frac {a x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )} \]

[Out]

a*x*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/4*b*x^4*((b*x^3+a)^2)^(1/2)/(b*x^3+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 = {1357} \begin {gather*} \frac {a x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

Rule 1357

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 36, normalized size = 0.49 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (4 a x+b x^4\right )}{4 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

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Maple [A]
time = 0.02, size = 33, normalized size = 0.45

method	result	size
gosper	\(\frac {x \left (b \,x^{3}+4 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 b \,x^{3}+4 a}\)	\(33\)
default	\(\frac {x \left (b \,x^{3}+4 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 b \,x^{3}+4 a}\)	\(33\)
risch	\(\frac {a x \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}+\frac {b \,x^{4} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 b \,x^{3}+4 a}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 10, normalized size = 0.14 \begin {gather*} \frac {1}{4} \, b x^{4} + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + a*x

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Fricas [A]
time = 0.34, size = 10, normalized size = 0.14 \begin {gather*} \frac {1}{4} \, b x^{4} + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + a*x

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Sympy [A]
time = 0.01, size = 8, normalized size = 0.11 \begin {gather*} a x + \frac {b x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*x**4/4

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Giac [A]
time = 6.19, size = 20, normalized size = 0.27 \begin {gather*} \frac {1}{4} \, {\left (b x^{4} + 4 \, a x\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^4 + 4*a*x)*sgn(b*x^3 + a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {{\left (b\,x^3+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^3)^2)^(1/2),x)

[Out]

int(((a + b*x^3)^2)^(1/2), x)

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