3.45 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{13}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{\left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{12 a x^{12}} \]

[Out]

-((a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(12*a*x^12)

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Rubi [A]  time = 0.0182668, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {1355, 264} \[ -\frac{\left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{12 a x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^13,x]

[Out]

-((a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(12*a*x^12)

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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, 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^{13}} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac{\left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{12 a x^{12}}\\ \end{align*}

Mathematica [A]  time = 0.0129933, size = 59, normalized size = 1.44 \[ -\frac{\sqrt{\left (a+b x^3\right )^2} \left (4 a^2 b x^3+a^3+6 a b^2 x^6+4 b^3 x^9\right )}{12 x^{12} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^13,x]

[Out]

-(Sqrt[(a + b*x^3)^2]*(a^3 + 4*a^2*b*x^3 + 6*a*b^2*x^6 + 4*b^3*x^9))/(12*x^12*(a + b*x^3))

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Maple [A]  time = 0.008, size = 56, normalized size = 1.4 \begin{align*} -{\frac{4\,{b}^{3}{x}^{9}+6\,a{b}^{2}{x}^{6}+4\,{a}^{2}b{x}^{3}+{a}^{3}}{12\,{x}^{12} \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^13,x)

[Out]

-1/12*(4*b^3*x^9+6*a*b^2*x^6+4*a^2*b*x^3+a^3)*((b*x^3+a)^2)^(3/2)/x^12/(b*x^3+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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^13,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77512, size = 78, normalized size = 1.9 \begin{align*} -\frac{4 \, b^{3} x^{9} + 6 \, a b^{2} x^{6} + 4 \, a^{2} b x^{3} + a^{3}}{12 \, x^{12}} \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^13,x, algorithm="fricas")

[Out]

-1/12*(4*b^3*x^9 + 6*a*b^2*x^6 + 4*a^2*b*x^3 + a^3)/x^12

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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^{13}}\, 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**13,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**13, x)

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Giac [B]  time = 1.08694, size = 92, normalized size = 2.24 \begin{align*} -\frac{4 \, b^{3} x^{9} \mathrm{sgn}\left (b x^{3} + a\right ) + 6 \, a b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 4 \, a^{2} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + a^{3} \mathrm{sgn}\left (b x^{3} + a\right )}{12 \, x^{12}} \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^13,x, algorithm="giac")

[Out]

-1/12*(4*b^3*x^9*sgn(b*x^3 + a) + 6*a*b^2*x^6*sgn(b*x^3 + a) + 4*a^2*b*x^3*sgn(b*x^3 + a) + a^3*sgn(b*x^3 + a)
)/x^12