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

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

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Rubi [A]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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 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 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^{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*}

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Mathematica [A]  time = 0.01, size = 61, normalized size = 0.73 \begin {gather*} -\frac {\sqrt {\left (a+b x^3\right )^2} \left (4 a^3+15 a^2 b x^3+20 a b^2 x^6+10 b^3 x^9\right )}{60 x^{15} \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^16,x]

[Out]

-1/60*(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))/(x^15*(a + b*x^3))

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IntegrateAlgebraic [B]  time = 1.02, size = 356, normalized size = 4.24 \begin {gather*} \frac {4 b^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \left (-4 a^7 b-31 a^6 b^2 x^3-104 a^5 b^3 x^6-196 a^4 b^4 x^9-224 a^3 b^5 x^{12}-155 a^2 b^6 x^{15}-60 a b^7 x^{18}-10 b^8 x^{21}\right )+4 \sqrt {b^2} b^4 \left (4 a^8+35 a^7 b x^3+135 a^6 b^2 x^6+300 a^5 b^3 x^9+420 a^4 b^4 x^{12}+379 a^3 b^5 x^{15}+215 a^2 b^6 x^{18}+70 a b^7 x^{21}+10 b^8 x^{24}\right )}{15 \sqrt {b^2} x^{15} \sqrt {a^2+2 a b x^3+b^2 x^6} \left (-16 a^4 b^4-64 a^3 b^5 x^3-96 a^2 b^6 x^6-64 a b^7 x^9-16 b^8 x^{12}\right )+15 x^{15} \left (16 a^5 b^5+80 a^4 b^6 x^3+160 a^3 b^7 x^6+160 a^2 b^8 x^9+80 a b^9 x^{12}+16 b^{10} x^{15}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-4*a^7*b - 31*a^6*b^2*x^3 - 104*a^5*b^3*x^6 - 196*a^4*b^4*x^9 - 224*a^
3*b^5*x^12 - 155*a^2*b^6*x^15 - 60*a*b^7*x^18 - 10*b^8*x^21) + 4*b^4*Sqrt[b^2]*(4*a^8 + 35*a^7*b*x^3 + 135*a^6
*b^2*x^6 + 300*a^5*b^3*x^9 + 420*a^4*b^4*x^12 + 379*a^3*b^5*x^15 + 215*a^2*b^6*x^18 + 70*a*b^7*x^21 + 10*b^8*x
^24))/(15*Sqrt[b^2]*x^15*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-16*a^4*b^4 - 64*a^3*b^5*x^3 - 96*a^2*b^6*x^6 - 64*a
*b^7*x^9 - 16*b^8*x^12) + 15*x^15*(16*a^5*b^5 + 80*a^4*b^6*x^3 + 160*a^3*b^7*x^6 + 160*a^2*b^8*x^9 + 80*a*b^9*
x^12 + 16*b^10*x^15))

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fricas [A]  time = 0.85, size = 37, normalized size = 0.44 \begin {gather*} -\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 {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^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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giac [A]  time = 0.41, size = 69, normalized size = 0.82 \begin {gather*} -\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 {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^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

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maple [A]  time = 0.01, size = 58, normalized size = 0.69 \begin {gather*} -\frac {\left (10 b^{3} x^{9}+20 a \,b^{2} x^{6}+15 a^{2} b \,x^{3}+4 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b \,x^{3}+a \right )^{3} x^{15}} \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^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 [B]  time = 0.52, size = 179, normalized size = 2.13 \begin {gather*} -\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5}}{12 \, a^{5}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4}}{12 \, a^{4} x^{3}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{12 \, a^{5} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{12 \, a^{4} x^{9}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{12 \, a^{3} x^{12}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{15 \, a^{2} x^{15}} \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^16,x, algorithm="maxima")

[Out]

-1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^5/a^5 - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^4/(a^4*x^3) + 1/12*
(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^3/(a^5*x^6) - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/(a^4*x^9) + 1/12*
(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a^3*x^12) - 1/15*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^15)

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mupad [B]  time = 1.21, size = 151, normalized size = 1.80 \begin {gather*} -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{15\,x^{15}\,\left (b\,x^3+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^6\,\left (b\,x^3+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^9\,\left (b\,x^3+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{4\,x^{12}\,\left (b\,x^3+a\right )} \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^16,x)

[Out]

- (a^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(15*x^15*(a + b*x^3)) - (b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x
^6*(a + b*x^3)) - (a*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(3*x^9*(a + b*x^3)) - (a^2*b*(a^2 + b^2*x^6 + 2*a*
b*x^3)^(1/2))/(4*x^12*(a + b*x^3))

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

[Out]

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

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