Integrand size = 23, antiderivative size = 87 \[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=\frac {\left (-2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{8 x^8}+\frac {1}{4} \text {arctanh}\left (\frac {1}{3}+\frac {2 x^8}{3}-\frac {2}{3} \sqrt {4-2 x^8+x^{16}}\right )-\frac {1}{8} \log \left (2-x^8+\sqrt {4-2 x^8+x^{16}}\right ) \]
1/8*(x^8-2)*(x^16-2*x^8+4)^(1/2)/x^8+1/4*arctanh(1/3+2/3*x^8-2/3*(x^16-2*x ^8+4)^(1/2))-1/8*ln(2-x^8+(x^16-2*x^8+4)^(1/2))
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=\frac {1}{8} \left (\frac {\left (-2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^8}+2 \text {arctanh}\left (\frac {1}{3} \left (1+2 x^8-2 \sqrt {4-2 x^8+x^{16}}\right )\right )-\log \left (2-x^8+\sqrt {4-2 x^8+x^{16}}\right )\right ) \]
(((-2 + x^8)*Sqrt[4 - 2*x^8 + x^16])/x^8 + 2*ArcTanh[(1 + 2*x^8 - 2*Sqrt[4 - 2*x^8 + x^16])/3] - Log[2 - x^8 + Sqrt[4 - 2*x^8 + x^16]])/8
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {1802, 1230, 27, 1269, 1090, 222, 1154, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^8+2\right ) \sqrt {x^{16}-2 x^8+4}}{x^9} \, dx\) |
\(\Big \downarrow \) 1802 |
\(\displaystyle \frac {1}{8} \int \frac {\left (x^8+2\right ) \sqrt {x^{16}-2 x^8+4}}{x^{16}}dx^8\) |
\(\Big \downarrow \) 1230 |
\(\displaystyle \frac {1}{8} \left (-\frac {1}{2} \int -\frac {2 \left (x^8+2\right )}{x^8 \sqrt {x^{16}-2 x^8+4}}dx^8-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{x^8}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (\int \frac {x^8+2}{x^8 \sqrt {x^{16}-2 x^8+4}}dx^8-\frac {\left (2-x^8\right ) \sqrt {x^{16}-2 x^8+4}}{x^8}\right )\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{8} \left (\int \frac {1}{\sqrt {x^{16}-2 x^8+4}}dx^8+2 \int \frac {1}{x^8 \sqrt {x^{16}-2 x^8+4}}dx^8-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{x^8}\right )\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{8} \left (\frac {\int \frac {1}{\sqrt {\frac {x^{16}}{12}+1}}d\left (2 x^8-2\right )}{2 \sqrt {3}}+2 \int \frac {1}{x^8 \sqrt {x^{16}-2 x^8+4}}dx^8-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{x^8}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{8} \left (2 \int \frac {1}{x^8 \sqrt {x^{16}-2 x^8+4}}dx^8+\text {arcsinh}\left (\frac {2 x^8-2}{2 \sqrt {3}}\right )-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{x^8}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{8} \left (-4 \int \frac {1}{16-x^{16}}d\frac {2 \left (4-x^8\right )}{\sqrt {x^{16}-2 x^8+4}}+\text {arcsinh}\left (\frac {2 x^8-2}{2 \sqrt {3}}\right )-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{x^8}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{8} \left (\text {arcsinh}\left (\frac {2 x^8-2}{2 \sqrt {3}}\right )-\text {arctanh}\left (\frac {4-x^8}{2 \sqrt {x^{16}-2 x^8+4}}\right )-\frac {\sqrt {x^{16}-2 x^8+4} \left (2-x^8\right )}{x^8}\right )\) |
(-(((2 - x^8)*Sqrt[4 - 2*x^8 + x^16])/x^8) + ArcSinh[(-2 + 2*x^8)/(2*Sqrt[ 3])] - ArcTanh[(4 - x^8)/(2*Sqrt[4 - 2*x^8 + x^16])])/8
3.12.93.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[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]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(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] + Simp[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] && GtQ[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])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(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] && !IGtQ[m, 0]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[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]]
Time = 1.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.55
method | result | size |
trager | \(\frac {\left (x^{8}-2\right ) \sqrt {x^{16}-2 x^{8}+4}}{8 x^{8}}+\frac {\ln \left (\frac {x^{8}+\sqrt {x^{16}-2 x^{8}+4}-2}{x^{4}}\right )}{4}\) | \(48\) |
risch | \(\frac {x^{24}-4 x^{16}+8 x^{8}-8}{8 x^{8} \sqrt {x^{16}-2 x^{8}+4}}-\frac {\ln \left (\frac {2-x^{8}+\sqrt {x^{16}-2 x^{8}+4}}{x^{4}}\right )}{4}\) | \(60\) |
pseudoelliptic | \(\frac {\operatorname {arcsinh}\left (\frac {\sqrt {3}\, \left (x^{8}-1\right )}{3}\right ) x^{8}+\operatorname {arctanh}\left (\frac {x^{8}-4}{2 \sqrt {x^{16}-2 x^{8}+4}}\right ) x^{8}+x^{8} \sqrt {x^{16}-2 x^{8}+4}-2 \sqrt {x^{16}-2 x^{8}+4}}{8 x^{8}}\) | \(76\) |
Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08 \[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=-\frac {2 \, x^{8} \log \left (2 \, x^{16} - 5 \, x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4} {\left (2 \, x^{8} - 3\right )} + 6\right ) - 2 \, x^{8} \log \left (-x^{8} + \sqrt {x^{16} - 2 \, x^{8} + 4} - 2\right ) + 5 \, x^{8} - 2 \, \sqrt {x^{16} - 2 \, x^{8} + 4} {\left (x^{8} - 2\right )}}{16 \, x^{8}} \]
-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
\[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=\int \frac {\left (x^{8} + 2\right ) \sqrt {x^{16} - 2 x^{8} + 4}}{x^{9}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71 \[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=\frac {1}{8} \, \sqrt {x^{16} - 2 \, x^{8} + 4} - \frac {\sqrt {x^{16} - 2 \, x^{8} + 4}}{4 \, x^{8}} + \frac {1}{8} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (x^{8} - 1\right )}\right ) - \frac {1}{8} \, \operatorname {arsinh}\left (-\frac {1}{3} \, \sqrt {3} + \frac {4 \, \sqrt {3}}{3 \, x^{8}}\right ) \]
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)
Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=\frac {1}{8} \, \sqrt {x^{16} - 2 \, x^{8} + 4} - \frac {x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4} - 4}{2 \, {\left ({\left (x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4}\right )}^{2} - 4\right )}} + \frac {1}{8} \, \log \left (x^{8} - \sqrt {x^{16} - 2 \, x^{8} + 4} + 2\right ) - \frac {1}{8} \, \log \left (-x^{8} + \sqrt {x^{16} - 2 \, x^{8} + 4} + 2\right ) - \frac {1}{8} \, \log \left (-x^{8} + \sqrt {x^{16} - 2 \, x^{8} + 4} + 1\right ) \]
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)
Time = 7.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {\left (2+x^8\right ) \sqrt {4-2 x^8+x^{16}}}{x^9} \, dx=\frac {\ln \left (\sqrt {x^{16}-2\,x^8+4}+x^8-1\right )}{8}-\frac {\ln \left (\frac {2\,\sqrt {x^{16}-2\,x^8+4}-x^8+4}{x^8}\right )}{8}-\frac {\sqrt {x^{16}-2\,x^8+4}}{4\,x^8}+\frac {\sqrt {x^{16}-2\,x^8+4}}{8} \]