Integrand size = 13, antiderivative size = 203 \[ \int \frac {x^9}{a+b x^8} \, dx=\frac {x^2}{2 b}+\frac {\sqrt [4]{a} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}} \]
1/2*x^2/b-1/8*a^(1/4)*arctan(-1+b^(1/4)*x^2*2^(1/2)/a^(1/4))/b^(5/4)*2^(1/ 2)-1/8*a^(1/4)*arctan(1+b^(1/4)*x^2*2^(1/2)/a^(1/4))/b^(5/4)*2^(1/2)+1/16* a^(1/4)*ln(-a^(1/4)*b^(1/4)*x^2*2^(1/2)+a^(1/2)+x^4*b^(1/2))/b^(5/4)*2^(1/ 2)-1/16*a^(1/4)*ln(a^(1/4)*b^(1/4)*x^2*2^(1/2)+a^(1/2)+x^4*b^(1/2))/b^(5/4 )*2^(1/2)
Time = 0.44 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.78 \[ \int \frac {x^9}{a+b x^8} \, dx=\frac {8 \sqrt [4]{b} x^2+2 \sqrt {2} \sqrt [4]{a} \arctan \left (\cot \left (\frac {\pi }{8}\right )-\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )+2 \sqrt {2} \sqrt [4]{a} \arctan \left (\cot \left (\frac {\pi }{8}\right )+\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )-2 \sqrt {2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}-\tan \left (\frac {\pi }{8}\right )\right )+2 \sqrt {2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2-2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2+2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2-2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2+2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )\right )}{16 b^{5/4}} \]
(8*b^(1/4)*x^2 + 2*Sqrt[2]*a^(1/4)*ArcTan[Cot[Pi/8] - (b^(1/8)*x*Csc[Pi/8] )/a^(1/8)] + 2*Sqrt[2]*a^(1/4)*ArcTan[Cot[Pi/8] + (b^(1/8)*x*Csc[Pi/8])/a^ (1/8)] - 2*Sqrt[2]*a^(1/4)*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) - Tan[Pi/8 ]] + 2*Sqrt[2]*a^(1/4)*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) + Tan[Pi/8]] + Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Cos[Pi/8] ] + Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x*Cos[Pi /8]] - Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Sin [Pi/8]] - Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x* Sin[Pi/8]])/(16*b^(5/4))
Time = 0.44 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {807, 843, 755, 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.
| \(\displaystyle \int \frac {x^9}{a+b x^8} \, dx\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {1}{2} \int \frac {x^8}{b x^8+a}dx^2\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \int \frac {1}{b x^8+a}dx^2}{b}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {a}}+\frac {\int \frac {\sqrt {b} x^4+\sqrt {a}}{b x^8+a}dx^2}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\frac {\int \frac {1}{x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt {b}}+\frac {\int \frac {1}{x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt {b}}}{2 \sqrt {a}}+\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\frac {\int \frac {1}{-x^4-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int \frac {1}{-x^4-1}d\left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\int \frac {\sqrt {a}-\sqrt {b} x^4}{b x^8+a}dx^2}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x^2}{\sqrt [4]{b} \left (x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} x^2+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x^2}{\sqrt [4]{b} \left (x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} x^2+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x^2}{x^4-\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt {2} \sqrt [4]{a} \sqrt {b}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{b} x^2+\sqrt [4]{a}}{x^4+\frac {\sqrt {2} \sqrt [4]{a} x^2}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx^2}{2 \sqrt [4]{a} \sqrt {b}}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2}{b}-\frac {a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )\) |
(x^2/b - (a*((-(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[a]) + (-1/2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x^2 + Sq rt[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[a])))/b)/2
3.15.51.3.1 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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) 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[(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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && 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 = 3.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.18
| method | result | size |
| risch | \(\frac {x^{2}}{2 b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\textit {\_R} \ln \left (x^{2}-\textit {\_R} \right )}{8 b}\) | \(36\) |
| default | \(\frac {x^{2}}{2 b}-\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{4}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x^{2} \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{4}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x^{2} \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}\) | \(119\) |
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.58 \[ \int \frac {x^9}{a+b x^8} \, dx=-\frac {b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \log \left (x^{2} + b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}}\right ) + i \, b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \log \left (x^{2} + i \, b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}}\right ) - i \, b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \log \left (x^{2} - i \, b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}}\right ) - b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \log \left (x^{2} - b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}}\right ) - 4 \, x^{2}}{8 \, b} \]
-1/8*(b*(-a/b^5)^(1/4)*log(x^2 + b*(-a/b^5)^(1/4)) + I*b*(-a/b^5)^(1/4)*lo g(x^2 + I*b*(-a/b^5)^(1/4)) - I*b*(-a/b^5)^(1/4)*log(x^2 - I*b*(-a/b^5)^(1 /4)) - b*(-a/b^5)^(1/4)*log(x^2 - b*(-a/b^5)^(1/4)) - 4*x^2)/b
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.13 \[ \int \frac {x^9}{a+b x^8} \, dx=\operatorname {RootSum} {\left (4096 t^{4} b^{5} + a, \left ( t \mapsto t \log {\left (- 8 t b + x^{2} \right )} \right )\right )} + \frac {x^{2}}{2 b} \]
Time = 0.30 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.94 \[ \int \frac {x^9}{a+b x^8} \, dx=\frac {x^{2}}{2 \, b} - \frac {\frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {b} x^{4} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {b} x^{4} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{b^{\frac {1}{4}}}}{16 \, b} \]
1/2*x^2/b - 1/16*(2*sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 + sq rt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/sqrt(sqrt(a)*sqrt(b)) + 2*sq rt(2)*sqrt(a)*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(2)*a^(1/4)*log(sqrt(b )*x^4 + sqrt(2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/b^(1/4) - sqrt(2)*a^(1/4)*l og(sqrt(b)*x^4 - sqrt(2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/b^(1/4))/b
Time = 0.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.90 \[ \int \frac {x^9}{a+b x^8} \, dx=\frac {x^{2}}{2 \, b} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (x^{4} + \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, b^{2}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (x^{4} - \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, b^{2}} \]
1/2*x^2/b - 1/8*sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2)* (a/b)^(1/4))/(a/b)^(1/4))/b^2 - 1/8*sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt( 2)*(2*x^2 - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/b^2 - 1/16*sqrt(2)*(a*b^3)^( 1/4)*log(x^4 + sqrt(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/b^2 + 1/16*sqrt(2)*(a* b^3)^(1/4)*log(x^4 - sqrt(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/b^2
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.27 \[ \int \frac {x^9}{a+b x^8} \, dx=\frac {x^2}{2\,b}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,x^2}{{\left (-a\right )}^{1/4}}\right )}{4\,b^{5/4}}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atanh}\left (\frac {b^{1/4}\,x^2}{{\left (-a\right )}^{1/4}}\right )}{4\,b^{5/4}} \]