3.12.93 \(\int \frac {(2+x^8) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx\)

Optimal. Leaf size=87 \[ \frac {\sqrt {x^{16}-2 x^8+4} \left (x^8-2\right )}{8 x^8}-\frac {1}{8} \log \left (-x^8+\sqrt {x^{16}-2 x^8+4}+2\right )+\frac {1}{4} \tanh ^{-1}\left (\frac {2 x^8}{3}-\frac {2}{3} \sqrt {x^{16}-2 x^8+4}+\frac {1}{3}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 77, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1474, 812, 843, 619, 215, 724, 206} \begin {gather*} -\frac {1}{8} \sinh ^{-1}\left (\frac {1-x^8}{\sqrt {3}}\right )-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{8 x^8}-\frac {1}{8} \tanh ^{-1}\left (\frac {4-x^8}{2 \sqrt {x^{16}-2 x^8+4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + x^8)*Sqrt[4 - 2*x^8 + x^16])/x^9,x]

[Out]

-1/8*((2 - x^8)*Sqrt[4 - 2*x^8 + x^16])/x^8 - ArcSinh[(1 - x^8)/Sqrt[3]]/8 - ArcTanh[(4 - x^8)/(2*Sqrt[4 - 2*x
^8 + x^16])]/8

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx &=\frac {1}{8} \operatorname {Subst}\left (\int \frac {(2+x) \sqrt {4-2 x+x^2}}{x^2} \, dx,x,x^8\right )\\ &=-\frac {\left (2-x^8\right ) \sqrt {4-2 x^8+x^{16}}}{8 x^8}-\frac {1}{16} \operatorname {Subst}\left (\int \frac {-4-2 x}{x \sqrt {4-2 x+x^2}} \, dx,x,x^8\right )\\ &=-\frac {\left (2-x^8\right ) \sqrt {4-2 x^8+x^{16}}}{8 x^8}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt {4-2 x+x^2}} \, dx,x,x^8\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {4-2 x+x^2}} \, dx,x,x^8\right )\\ &=-\frac {\left (2-x^8\right ) \sqrt {4-2 x^8+x^{16}}}{8 x^8}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{16-x^2} \, dx,x,\frac {2 \left (4-x^8\right )}{\sqrt {4-2 x^8+x^{16}}}\right )+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{12}}} \, dx,x,2 \left (-1+x^8\right )\right )}{16 \sqrt {3}}\\ &=-\frac {\left (2-x^8\right ) \sqrt {4-2 x^8+x^{16}}}{8 x^8}-\frac {1}{8} \sinh ^{-1}\left (\frac {1-x^8}{\sqrt {3}}\right )-\frac {1}{8} \tanh ^{-1}\left (\frac {4-x^8}{2 \sqrt {4-2 x^8+x^{16}}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 68, normalized size = 0.78 \begin {gather*} \frac {1}{8} \left (\sinh ^{-1}\left (\frac {x^8-1}{\sqrt {3}}\right )+\frac {\sqrt {x^{16}-2 x^8+4} \left (x^8-2\right )}{x^8}-\tanh ^{-1}\left (\frac {4-x^8}{2 \sqrt {x^{16}-2 x^8+4}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + x^8)*Sqrt[4 - 2*x^8 + x^16])/x^9,x]

[Out]

(((-2 + x^8)*Sqrt[4 - 2*x^8 + x^16])/x^8 + ArcSinh[(-1 + x^8)/Sqrt[3]] - ArcTanh[(4 - x^8)/(2*Sqrt[4 - 2*x^8 +
x^16])])/8

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IntegrateAlgebraic [A]  time = 0.16, size = 87, normalized size = 1.00 \begin {gather*} \frac {\left (-2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{8 x^8}+\frac {1}{4} \tanh ^{-1}\left (\frac {1}{3}+\frac {2 x^8}{3}-\frac {2}{3} \sqrt {4-2 x^8+x^{16}}\right )-\frac {1}{8} \log \left (2-x^8+\sqrt {4-2 x^8+x^{16}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((2 + x^8)*Sqrt[4 - 2*x^8 + x^16])/x^9,x]

[Out]

((-2 + x^8)*Sqrt[4 - 2*x^8 + x^16])/(8*x^8) + ArcTanh[1/3 + (2*x^8)/3 - (2*Sqrt[4 - 2*x^8 + x^16])/3]/4 - Log[
2 - x^8 + Sqrt[4 - 2*x^8 + x^16]]/8

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fricas [A]  time = 0.45, size = 94, normalized size = 1.08 \begin {gather*} -\frac {2 \, x^{8} \log \left (2 \, x^{16} - 5 \, x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4} {\left (2 \, x^{8} - 3\right )} + 6\right ) - 2 \, x^{8} \log \left (-x^{8} + \sqrt {x^{16} - 2 \, x^{8} + 4} - 2\right ) + 5 \, x^{8} - 2 \, \sqrt {x^{16} - 2 \, x^{8} + 4} {\left (x^{8} - 2\right )}}{16 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+2)*(x^16-2*x^8+4)^(1/2)/x^9,x, algorithm="fricas")

[Out]

-1/16*(2*x^8*log(2*x^16 - 5*x^8 - sqrt(x^16 - 2*x^8 + 4)*(2*x^8 - 3) + 6) - 2*x^8*log(-x^8 + sqrt(x^16 - 2*x^8
+ 4) - 2) + 5*x^8 - 2*sqrt(x^16 - 2*x^8 + 4)*(x^8 - 2))/x^8

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giac [A]  time = 0.18, size = 126, normalized size = 1.45 \begin {gather*} \frac {1}{8} \, \sqrt {x^{16} - 2 \, x^{8} + 4} - \frac {x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4} - 4}{2 \, {\left ({\left (x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4}\right )}^{2} - 4\right )}} + \frac {1}{8} \, \log \left (x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4} + 2\right ) - \frac {1}{8} \, \log \left (-x^{8} + \sqrt {x^{16} - 2 \, x^{8} + 4} + 2\right ) - \frac {1}{8} \, \log \left (-x^{8} + \sqrt {x^{16} - 2 \, x^{8} + 4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+2)*(x^16-2*x^8+4)^(1/2)/x^9,x, algorithm="giac")

[Out]

1/8*sqrt(x^16 - 2*x^8 + 4) - 1/2*(x^8 - sqrt(x^16 - 2*x^8 + 4) - 4)/((x^8 - sqrt(x^16 - 2*x^8 + 4))^2 - 4) + 1
/8*log(x^8 - sqrt(x^16 - 2*x^8 + 4) + 2) - 1/8*log(-x^8 + sqrt(x^16 - 2*x^8 + 4) + 2) - 1/8*log(-x^8 + sqrt(x^
16 - 2*x^8 + 4) + 1)

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maple [A]  time = 1.07, size = 50, normalized size = 0.57

method	result	size
trager	\(\frac {\left (x^{8}-2\right ) \sqrt {x^{16}-2 x^{8}+4}}{8 x^{8}}-\frac {\ln \left (\frac {2-x^{8}+\sqrt {x^{16}-2 x^{8}+4}}{x^{4}}\right )}{4}\)	\(50\)
risch	\(\frac {x^{24}-4 x^{16}+8 x^{8}-8}{8 x^{8} \sqrt {x^{16}-2 x^{8}+4}}+\frac {\ln \left (\frac {x^{8}+\sqrt {x^{16}-2 x^{8}+4}-2}{x^{4}}\right )}{4}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8+2)*(x^16-2*x^8+4)^(1/2)/x^9,x,method=_RETURNVERBOSE)

[Out]

1/8*(x^8-2)*(x^16-2*x^8+4)^(1/2)/x^8-1/4*ln((2-x^8+(x^16-2*x^8+4)^(1/2))/x^4)

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maxima [A]  time = 0.45, size = 62, normalized size = 0.71 \begin {gather*} \frac {1}{8} \, \sqrt {x^{16} - 2 \, x^{8} + 4} - \frac {\sqrt {x^{16} - 2 \, x^{8} + 4}}{4 \, x^{8}} + \frac {1}{8} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (x^{8} - 1\right )}\right ) - \frac {1}{8} \, \operatorname {arsinh}\left (-\frac {1}{3} \, \sqrt {3} + \frac {4 \, \sqrt {3}}{3 \, x^{8}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+2)*(x^16-2*x^8+4)^(1/2)/x^9,x, algorithm="maxima")

[Out]

1/8*sqrt(x^16 - 2*x^8 + 4) - 1/4*sqrt(x^16 - 2*x^8 + 4)/x^8 + 1/8*arcsinh(1/3*sqrt(3)*(x^8 - 1)) - 1/8*arcsinh
(-1/3*sqrt(3) + 4/3*sqrt(3)/x^8)

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mupad [B]  time = 1.97, size = 80, normalized size = 0.92 \begin {gather*} \frac {\ln \left (\sqrt {x^{16}-2\,x^8+4}+x^8-1\right )}{8}-\frac {\ln \left (\frac {2\,\sqrt {x^{16}-2\,x^8+4}-x^8+4}{x^8}\right )}{8}-\frac {\sqrt {x^{16}-2\,x^8+4}}{4\,x^8}+\frac {\sqrt {x^{16}-2\,x^8+4}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^8 + 2)*(x^16 - 2*x^8 + 4)^(1/2))/x^9,x)

[Out]

log((x^16 - 2*x^8 + 4)^(1/2) + x^8 - 1)/8 - log((2*(x^16 - 2*x^8 + 4)^(1/2) - x^8 + 4)/x^8)/8 - (x^16 - 2*x^8
+ 4)^(1/2)/(4*x^8) + (x^16 - 2*x^8 + 4)^(1/2)/8

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{8} + 2\right ) \sqrt {x^{16} - 2 x^{8} + 4}}{x^{9}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**8+2)*(x**16-2*x**8+4)**(1/2)/x**9,x)

[Out]

Integral((x**8 + 2)*sqrt(x**16 - 2*x**8 + 4)/x**9, x)

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