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

Optimal. Leaf size=165 \[ \frac {3 a b^2 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {3 a^2 b x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b^3 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 165, 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, 270} \begin {gather*} \frac {b^3 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {3 a b^2 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {3 a^2 b x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}-\frac {a^3 \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[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^2,x]

[Out]

-((a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x*(a + b*x^3))) + (3*a^2*b*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a
+ b*x^3)) + (3*a*b^2*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(a + b*x^3)) + (b^3*x^8*Sqrt[a^2 + 2*a*b*x^3 + b
^2*x^6])/(8*(a + b*x^3))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^2} \, dx &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (a b+b^2 x^3\right )^3}{x^2} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^3 b^3}{x^2}+3 a^2 b^4 x+3 a b^5 x^4+b^6 x^7\right ) \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {3 a^2 b x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {3 a b^2 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {b^3 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-40 a^3+60 a^2 b x^3+24 a b^2 x^6+5 b^3 x^9\right )}{40 x \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(-40*a^3 + 60*a^2*b*x^3 + 24*a*b^2*x^6 + 5*b^3*x^9))/(40*x*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 17.20, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-40 a^3+60 a^2 b x^3+24 a b^2 x^6+5 b^3 x^9\right )}{40 x \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(-40*a^3 + 60*a^2*b*x^3 + 24*a*b^2*x^6 + 5*b^3*x^9))/(40*x*(a + b*x^3))

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fricas [A]  time = 1.38, size = 37, normalized size = 0.22 \begin {gather*} \frac {5 \, b^{3} x^{9} + 24 \, a b^{2} x^{6} + 60 \, a^{2} b x^{3} - 40 \, a^{3}}{40 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^2,x, algorithm="fricas")

[Out]

1/40*(5*b^3*x^9 + 24*a*b^2*x^6 + 60*a^2*b*x^3 - 40*a^3)/x

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giac [A]  time = 0.34, size = 67, normalized size = 0.41 \begin {gather*} \frac {1}{8} \, b^{3} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{5} \, a b^{2} x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{2} \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^2,x, algorithm="giac")

[Out]

1/8*b^3*x^8*sgn(b*x^3 + a) + 3/5*a*b^2*x^5*sgn(b*x^3 + a) + 3/2*a^2*b*x^2*sgn(b*x^3 + a) - a^3*sgn(b*x^3 + a)/
x

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \begin {gather*} -\frac {\left (-5 b^{3} x^{9}-24 a \,b^{2} x^{6}-60 a^{2} b \,x^{3}+40 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{40 \left (b \,x^{3}+a \right )^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^2,x)

[Out]

-1/40*(-5*b^3*x^9-24*a*b^2*x^6-60*a^2*b*x^3+40*a^3)*((b*x^3+a)^2)^(3/2)/x/(b*x^3+a)^3

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maxima [A]  time = 0.50, size = 37, normalized size = 0.22 \begin {gather*} \frac {5 \, b^{3} x^{9} + 24 \, a b^{2} x^{6} + 60 \, a^{2} b x^{3} - 40 \, a^{3}}{40 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^2,x, algorithm="maxima")

[Out]

1/40*(5*b^3*x^9 + 24*a*b^2*x^6 + 60*a^2*b*x^3 - 40*a^3)/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^2,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**2,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**2, x)

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