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

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

[Out]

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

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Rubi [A]  time = 0.020172, antiderivative size = 74, normalized size of antiderivative = 1., 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} \[ \frac{b x \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac{a \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]]

Rubi steps

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

Mathematica [A]  time = 0.0075755, size = 37, normalized size = 0.5 \[ -\frac{\left (a-2 b x^3\right ) \sqrt{\left (a+b x^3\right )^2}}{2 x^2 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 34, normalized size = 0.5 \begin{align*} -{\frac{-2\,b{x}^{3}+a}{2\,{x}^{2} \left ( b{x}^{3}+a \right ) }\sqrt{ \left ( b{x}^{3}+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01841, size = 20, normalized size = 0.27 \begin{align*} \frac{2 \, b x^{3} - a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x^3 - a)/x^2

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Fricas [A]  time = 1.73779, size = 31, normalized size = 0.42 \begin{align*} \frac{2 \, b x^{3} - a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x^3 - a)/x^2

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Sympy [A]  time = 0.273665, size = 8, normalized size = 0.11 \begin{align*} - \frac{a}{2 x^{2}} + b x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(2*x**2) + b*x

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Giac [A]  time = 1.10113, size = 35, normalized size = 0.47 \begin{align*} b x \mathrm{sgn}\left (b x^{3} + a\right ) - \frac{a \mathrm{sgn}\left (b x^{3} + a\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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