3.41 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=162 \[ -\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac{b^3 x \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

[Out]

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

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Rubi [A]  time = 0.0409398, antiderivative size = 162, 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, 270} \[ -\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac{b^3 x \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^9,x]

[Out]

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

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^9} \, 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^9} \, 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 (b^6+\frac{a^3 b^3}{x^9}+\frac{3 a^2 b^4}{x^6}+\frac{3 a b^5}{x^3}\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}}{8 x^8 \left (a+b x^3\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac{b^3 x \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3}\\ \end{align*}

Mathematica [A]  time = 0.01548, size = 61, normalized size = 0.38 \[ -\frac{\sqrt{\left (a+b x^3\right )^2} \left (24 a^2 b x^3+5 a^3+60 a b^2 x^6-40 b^3 x^9\right )}{40 x^8 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^9,x]

[Out]

-(Sqrt[(a + b*x^3)^2]*(5*a^3 + 24*a^2*b*x^3 + 60*a*b^2*x^6 - 40*b^3*x^9))/(40*x^8*(a + b*x^3))

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Maple [A]  time = 0.004, size = 58, normalized size = 0.4 \begin{align*} -{\frac{-40\,{b}^{3}{x}^{9}+60\,a{b}^{2}{x}^{6}+24\,{a}^{2}b{x}^{3}+5\,{a}^{3}}{40\,{x}^{8} \left ( b{x}^{3}+a \right ) ^{3}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

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^9,x)

[Out]

-1/40*(-40*b^3*x^9+60*a*b^2*x^6+24*a^2*b*x^3+5*a^3)*((b*x^3+a)^2)^(3/2)/x^8/(b*x^3+a)^3

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Maxima [A]  time = 1.02377, size = 50, normalized size = 0.31 \begin{align*} \frac{40 \, b^{3} x^{9} - 60 \, a b^{2} x^{6} - 24 \, a^{2} b x^{3} - 5 \, a^{3}}{40 \, x^{8}} \end{align*}

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^9,x, algorithm="maxima")

[Out]

1/40*(40*b^3*x^9 - 60*a*b^2*x^6 - 24*a^2*b*x^3 - 5*a^3)/x^8

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Fricas [A]  time = 1.79179, size = 82, normalized size = 0.51 \begin{align*} \frac{40 \, b^{3} x^{9} - 60 \, a b^{2} x^{6} - 24 \, a^{2} b x^{3} - 5 \, a^{3}}{40 \, x^{8}} \end{align*}

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^9,x, algorithm="fricas")

[Out]

1/40*(40*b^3*x^9 - 60*a*b^2*x^6 - 24*a^2*b*x^3 - 5*a^3)/x^8

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac{3}{2}}}{x^{9}}\, dx \end{align*}

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**9,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**9, x)

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Giac [A]  time = 1.1176, size = 90, normalized size = 0.56 \begin{align*} b^{3} x \mathrm{sgn}\left (b x^{3} + a\right ) - \frac{60 \, a b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 24 \, a^{2} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + 5 \, a^{3} \mathrm{sgn}\left (b x^{3} + a\right )}{40 \, x^{8}} \end{align*}

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^9,x, algorithm="giac")

[Out]

b^3*x*sgn(b*x^3 + a) - 1/40*(60*a*b^2*x^6*sgn(b*x^3 + a) + 24*a^2*b*x^3*sgn(b*x^3 + a) + 5*a^3*sgn(b*x^3 + a))
/x^8