3.48 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{16}} \, dx\)

Optimal. Leaf size=84 \[ \frac{b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}}-\frac{\left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}} \]

[Out]

-((a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(15*a*x^15) + (b*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6
])/(60*a^2*x^12)

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Rubi [A]  time = 0.0396448, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1355, 266, 45, 37} \[ \frac{b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}}-\frac{\left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^16,x]

[Out]

-((a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(15*a*x^15) + (b*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6
])/(60*a^2*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 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^{16}} \, 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^{16}} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^3}{x^6} \, dx,x,x^3\right )}{3 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}}{15 a x^{15}}-\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^3}{x^5} \, dx,x,x^3\right )}{15 a b \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}}{15 a x^{15}}+\frac{b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}}\\ \end{align*}

Mathematica [A]  time = 0.0140864, size = 61, normalized size = 0.73 \[ -\frac{\sqrt{\left (a+b x^3\right )^2} \left (15 a^2 b x^3+4 a^3+20 a b^2 x^6+10 b^3 x^9\right )}{60 x^{15} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^16,x]

[Out]

-(Sqrt[(a + b*x^3)^2]*(4*a^3 + 15*a^2*b*x^3 + 20*a*b^2*x^6 + 10*b^3*x^9))/(60*x^15*(a + b*x^3))

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Maple [A]  time = 0.007, size = 58, normalized size = 0.7 \begin{align*} -{\frac{10\,{b}^{3}{x}^{9}+20\,a{b}^{2}{x}^{6}+15\,{a}^{2}b{x}^{3}+4\,{a}^{3}}{60\,{x}^{15} \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^16,x)

[Out]

-1/60*(10*b^3*x^9+20*a*b^2*x^6+15*a^2*b*x^3+4*a^3)*((b*x^3+a)^2)^(3/2)/x^15/(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^16,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73807, size = 85, normalized size = 1.01 \begin{align*} -\frac{10 \, b^{3} x^{9} + 20 \, a b^{2} x^{6} + 15 \, a^{2} b x^{3} + 4 \, a^{3}}{60 \, x^{15}} \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^16,x, algorithm="fricas")

[Out]

-1/60*(10*b^3*x^9 + 20*a*b^2*x^6 + 15*a^2*b*x^3 + 4*a^3)/x^15

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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^{16}}\, 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**16,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**16, x)

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Giac [A]  time = 1.11682, size = 93, normalized size = 1.11 \begin{align*} -\frac{10 \, b^{3} x^{9} \mathrm{sgn}\left (b x^{3} + a\right ) + 20 \, a b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 15 \, a^{2} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + 4 \, a^{3} \mathrm{sgn}\left (b x^{3} + a\right )}{60 \, x^{15}} \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^16,x, algorithm="giac")

[Out]

-1/60*(10*b^3*x^9*sgn(b*x^3 + a) + 20*a*b^2*x^6*sgn(b*x^3 + a) + 15*a^2*b*x^3*sgn(b*x^3 + a) + 4*a^3*sgn(b*x^3
+ a))/x^15