Integrand size = 13, antiderivative size = 203 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=-\frac {1}{2 a x^2}+\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{b} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} a^{5/4}} \]
-1/2/a/x^2-1/8*b^(1/4)*arctan(-1+b^(1/4)*x^2*2^(1/2)/a^(1/4))/a^(5/4)*2^(1 /2)-1/8*b^(1/4)*arctan(1+b^(1/4)*x^2*2^(1/2)/a^(1/4))/a^(5/4)*2^(1/2)-1/16 *b^(1/4)*ln(-a^(1/4)*b^(1/4)*x^2*2^(1/2)+a^(1/2)+x^4*b^(1/2))/a^(5/4)*2^(1 /2)+1/16*b^(1/4)*ln(a^(1/4)*b^(1/4)*x^2*2^(1/2)+a^(1/2)+x^4*b^(1/2))/a^(5/ 4)*2^(1/2)
Time = 0.09 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.90 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=\frac {-8 \sqrt [4]{a}+2 \sqrt {2} \sqrt [4]{b} x^2 \arctan \left (\cot \left (\frac {\pi }{8}\right )-\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )+2 \sqrt {2} \sqrt [4]{b} x^2 \arctan \left (\cot \left (\frac {\pi }{8}\right )+\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )-2 \sqrt {2} \sqrt [4]{b} x^2 \arctan \left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}-\tan \left (\frac {\pi }{8}\right )\right )+2 \sqrt {2} \sqrt [4]{b} x^2 \arctan \left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \sqrt [4]{b} x^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2-2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \sqrt [4]{b} x^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2+2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \sqrt [4]{b} x^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2-2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \sqrt [4]{b} x^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2+2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )\right )}{16 a^{5/4} x^2} \]
(-8*a^(1/4) + 2*Sqrt[2]*b^(1/4)*x^2*ArcTan[Cot[Pi/8] - (b^(1/8)*x*Csc[Pi/8 ])/a^(1/8)] + 2*Sqrt[2]*b^(1/4)*x^2*ArcTan[Cot[Pi/8] + (b^(1/8)*x*Csc[Pi/8 ])/a^(1/8)] - 2*Sqrt[2]*b^(1/4)*x^2*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) - Tan[Pi/8]] + 2*Sqrt[2]*b^(1/4)*x^2*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) + Tan[Pi/8]] - Sqrt[2]*b^(1/4)*x^2*Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^ (1/8)*x*Cos[Pi/8]] - Sqrt[2]*b^(1/4)*x^2*Log[a^(1/4) + b^(1/4)*x^2 + 2*a^( 1/8)*b^(1/8)*x*Cos[Pi/8]] + Sqrt[2]*b^(1/4)*x^2*Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]] + Sqrt[2]*b^(1/4)*x^2*Log[a^(1/4) + b^(1/ 4)*x^2 + 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]])/(16*a^(5/4)*x^2)
Time = 0.42 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {807, 847, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^8+a\right )}dx^2\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \int \frac {x^4}{b x^8+a}dx^2}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\int \frac {\sqrt {b} x^4+\sqrt {a}}{b x^8+a}dx^2}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\int \frac {1}{x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt {b}}+\frac {\int \frac {1}{x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt {b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\int \frac {1}{-x^4-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int \frac {1}{-x^4-1}d\left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x^2}{\sqrt [4]{b} \left (x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} x^2+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x^2}{\sqrt [4]{b} \left (x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} x^2+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x^2}{x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt {b}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{b} x^2+\sqrt [4]{a}}{x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt [4]{a} \sqrt {b}}}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (-\frac {b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x^2}\right )\) |
(-(1/(a*x^2)) - (b*((-(ArcTan[1 - (Sqrt[2]*b^(1/4)*x^2)/a^(1/4)]/(Sqrt[2]* a^(1/4)*b^(1/4))) + ArcTan[1 + (Sqrt[2]*b^(1/4)*x^2)/a^(1/4)]/(Sqrt[2]*a^( 1/4)*b^(1/4)))/(2*Sqrt[b]) - (-1/2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x ^2 + Sqrt[b]*x^4]/(Sqrt[2]*a^(1/4)*b^(1/4)) + Log[Sqrt[a] + Sqrt[2]*a^(1/4 )*b^(1/4)*x^2 + Sqrt[b]*x^4]/(2*Sqrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[b])))/a) /2
3.15.57.3.1 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[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.26
| method | result | size |
| risch | \(-\frac {1}{2 a \,x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} \textit {\_Z}^{4}+b \right )}{\sum }\textit {\_R} \ln \left (\left (-9 \textit {\_R}^{4} a^{5}-8 b \right ) x^{2}-a^{4} \textit {\_R}^{3}\right )\right )}{8}\) | \(53\) |
| default | \(-\frac {1}{2 a \,x^{2}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{4}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x^{2} \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{4}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x^{2} \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 a \left (\frac {a}{b}\right )^{\frac {1}{4}}}\) | \(119\) |
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=-\frac {a x^{2} \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (a^{4} \left (-\frac {b}{a^{5}}\right )^{\frac {3}{4}} + b x^{2}\right ) - i \, a x^{2} \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (i \, a^{4} \left (-\frac {b}{a^{5}}\right )^{\frac {3}{4}} + b x^{2}\right ) + i \, a x^{2} \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a^{4} \left (-\frac {b}{a^{5}}\right )^{\frac {3}{4}} + b x^{2}\right ) - a x^{2} \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a^{4} \left (-\frac {b}{a^{5}}\right )^{\frac {3}{4}} + b x^{2}\right ) + 4}{8 \, a x^{2}} \]
-1/8*(a*x^2*(-b/a^5)^(1/4)*log(a^4*(-b/a^5)^(3/4) + b*x^2) - I*a*x^2*(-b/a ^5)^(1/4)*log(I*a^4*(-b/a^5)^(3/4) + b*x^2) + I*a*x^2*(-b/a^5)^(1/4)*log(- I*a^4*(-b/a^5)^(3/4) + b*x^2) - a*x^2*(-b/a^5)^(1/4)*log(-a^4*(-b/a^5)^(3/ 4) + b*x^2) + 4)/(a*x^2)
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.17 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=\operatorname {RootSum} {\left (4096 t^{4} a^{5} + b, \left ( t \mapsto t \log {\left (- \frac {512 t^{3} a^{4}}{b} + x^{2} \right )} \right )\right )} - \frac {1}{2 a x^{2}} \]
Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=-\frac {b {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{4} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{4} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, a} - \frac {1}{2 \, a x^{2}} \]
-1/16*b*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^( 1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*a rctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*s qrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(b)*x^4 + sqrt( 2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)* x^4 - sqrt(2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/(a^(1/4)*b^(3/4)))/a - 1/2/(a *x^2)
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=-\frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \log \left (x^{4} + \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, a^{2}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \log \left (x^{4} - \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, a^{2}} - \frac {1}{2 \, a x^{2}} \]
-1/8*sqrt(2)*(a*b^3)^(1/4)*x^4*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2)*(a/b)^( 1/4))/(a/b)^(1/4))/a^2 - 1/8*sqrt(2)*(a*b^3)^(1/4)*x^4*arctan(1/2*sqrt(2)* (2*x^2 - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/a^2 - 1/16*sqrt(2)*(a*b^3)^(1/4 )*x^4*log(x^4 + sqrt(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/a^2 + 1/16*sqrt(2)*(a *b^3)^(1/4)*x^4*log(x^4 - sqrt(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/a^2 - 1/2/( a*x^2)
Time = 5.81 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.27 \[ \int \frac {1}{x^3 \left (a+b x^8\right )} \, dx=\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,x^2}{a^{1/4}}\right )}{4\,a^{5/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,x^2}{a^{1/4}}\right )}{4\,a^{5/4}}-\frac {1}{2\,a\,x^2} \]