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

Optimal. Leaf size=77 \[ \frac {b x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 77, 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 = {1355, 14} \begin {gather*} \frac {b x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \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^2,x]

[Out]

-((a*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x*(a + b*x^3))) + (b*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(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 \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^2} \, dx &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {a b+b^2 x^3}{x^2} \, dx}{a b+b^2 x^3}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a b}{x^2}+b^2 x\right ) \, dx}{a b+b^2 x^3}\\ &=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {b x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.49 \begin {gather*} \frac {\left (b x^3-2 a\right ) \sqrt {\left (a+b x^3\right )^2}}{2 x \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^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 16.23, size = 38, normalized size = 0.49 \begin {gather*} \frac {\left (b x^3-2 a\right ) \sqrt {\left (a+b x^3\right )^2}}{2 x \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.07, size = 14, normalized size = 0.18 \begin {gather*} \frac {b x^{3} - 2 \, a}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^3 - 2*a)/x

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giac [A]  time = 0.32, size = 29, normalized size = 0.38 \begin {gather*} \frac {1}{2} \, b x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {a \mathrm {sgn}\left (b x^{3} + a\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 36, normalized size = 0.47 \begin {gather*} -\frac {\left (-b \,x^{3}+2 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 \left (b \,x^{3}+a \right ) x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 14, normalized size = 0.18 \begin {gather*} \frac {b x^{3} - 2 \, a}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^3 - 2*a)/x

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

[Out]

int(((a + b*x^3)^2)^(1/2)/x^2, x)

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sympy [A]  time = 0.13, size = 8, normalized size = 0.10 \begin {gather*} - \frac {a}{x} + \frac {b x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/x + b*x**2/2

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