3.1.10 \(\int x \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx\)

Optimal. Leaf size=79 \[ \frac {b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {a x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )} \]

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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 = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1355, 14} \begin {gather*} \frac {b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {a x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a + b*x^3)) + (b*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(a + b*x^
3))

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 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x \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 x \left (a b+b^2 x^3\right ) \, dx}{a b+b^2 x^3}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (a b x+b^2 x^4\right ) \, dx}{a b+b^2 x^3}\\ &=\frac {a x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.49 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (5 a x^2+2 b x^5\right )}{10 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(5*a*x^2 + 2*b*x^5))/(10*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 8.16, size = 39, normalized size = 0.49 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (5 a x^2+2 b x^5\right )}{10 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(5*a*x^2 + 2*b*x^5))/(10*(a + b*x^3))

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fricas [A]  time = 1.01, size = 13, normalized size = 0.16 \begin {gather*} \frac {1}{5} \, b x^{5} + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*x^5 + 1/2*a*x^2

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giac [A]  time = 0.33, size = 29, normalized size = 0.37 \begin {gather*} \frac {1}{5} \, b x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 36, normalized size = 0.46 \begin {gather*} \frac {\left (2 b \,x^{3}+5 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}\, x^{2}}{10 b \,x^{3}+10 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*x^2*(2*b*x^3+5*a)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

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maxima [A]  time = 0.45, size = 13, normalized size = 0.16 \begin {gather*} \frac {1}{5} \, b x^{5} + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*x^5 + 1/2*a*x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((a + b*x^3)^2)^(1/2),x)

[Out]

int(x*((a + b*x^3)^2)^(1/2), x)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.15 \begin {gather*} \frac {a x^{2}}{2} + \frac {b x^{5}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**2/2 + b*x**5/5

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