Integrand size = 20, antiderivative size = 135 \[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=-\frac {3 b \left (x+x^3\right )^{2/3} \left (5-6 x^2+9 x^4\right )}{80 x^6}+\frac {1}{4} \left (a-i \sqrt {3} a\right ) \log \left (-i x+\sqrt {3} x-2 i \sqrt [3]{x+x^3}\right )+\frac {1}{4} \left (a+i \sqrt {3} a\right ) \log \left (i x+\sqrt {3} x+2 i \sqrt [3]{x+x^3}\right )-\frac {1}{2} a \log \left (-x+\sqrt [3]{x+x^3}\right ) \]
-3/80*b*(x^3+x)^(2/3)*(9*x^4-6*x^2+5)/x^6+1/4*(a-I*3^(1/2)*a)*ln(-I*x+x*3^ (1/2)-2*I*(x^3+x)^(1/3))+1/4*(a+I*3^(1/2)*a)*ln(I*x+x*3^(1/2)+2*I*(x^3+x)^ (1/3))-1/2*a*ln(-x+(x^3+x)^(1/3))
Time = 2.48 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.29 \[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=\frac {-15 b+3 b x^2-9 b x^4-27 b x^6+40 \sqrt {3} a x^{16/3} \sqrt [3]{1+x^2} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )-40 a x^{16/3} \sqrt [3]{1+x^2} \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )+20 a x^{16/3} \sqrt [3]{1+x^2} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )}{80 x^5 \sqrt [3]{x+x^3}} \]
(-15*b + 3*b*x^2 - 9*b*x^4 - 27*b*x^6 + 40*Sqrt[3]*a*x^(16/3)*(1 + x^2)^(1 /3)*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2*(1 + x^2)^(1/3))] - 40*a*x^(16/3 )*(1 + x^2)^(1/3)*Log[-x^(2/3) + (1 + x^2)^(1/3)] + 20*a*x^(16/3)*(1 + x^2 )^(1/3)*Log[x^(4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)])/(80*x^5* (x + x^3)^(1/3))
Time = 0.33 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2449, 2009}
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.
\(\displaystyle \int \frac {a x^6+b}{x^6 \sqrt [3]{x^3+x}} \, dx\) |
\(\Big \downarrow \) 2449 |
\(\displaystyle \int \left (\frac {a}{\sqrt [3]{x^3+x}}+\frac {b}{x^6 \sqrt [3]{x^3+x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {3} a \sqrt [3]{x} \sqrt [3]{x^2+1} \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{x^3+x}}-\frac {3 a \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 \sqrt [3]{x^3+x}}-\frac {3 b \left (x^3+x\right )^{2/3}}{16 x^6}+\frac {9 b \left (x^3+x\right )^{2/3}}{40 x^4}-\frac {27 b \left (x^3+x\right )^{2/3}}{80 x^2}\) |
(-3*b*(x + x^3)^(2/3))/(16*x^6) + (9*b*(x + x^3)^(2/3))/(40*x^4) - (27*b*( x + x^3)^(2/3))/(80*x^2) + (Sqrt[3]*a*x^(1/3)*(1 + x^2)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(2*(x + x^3)^(1/3)) - (3*a*x^(1/3)* (1 + x^2)^(1/3)*Log[x^(2/3) - (1 + x^2)^(1/3)])/(4*(x + x^3)^(1/3))
3.20.38.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_S ymbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ [{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) && !Integer Q[p] && NeQ[n, j]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 1.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.33
method | result | size |
meijerg | \(\frac {3 a \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}-\frac {3 b \left (\frac {9}{5} x^{4}-\frac {6}{5} x^{2}+1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}}{16 x^{\frac {16}{3}}}\) | \(44\) |
risch | \(-\frac {3 b \left (x^{2}+1\right ) \left (9 x^{4}-6 x^{2}+5\right )}{80 x^{5} {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}+\frac {3 a \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\) | \(51\) |
pseudoelliptic | \(\frac {20 a \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right ) x^{6}-40 a \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right ) x^{6}-40 a \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right ) x^{6}-27 b \left (x^{4}-\frac {2}{3} x^{2}+\frac {5}{9}\right ) {\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}}{80 x^{6}}\) | \(119\) |
trager | \(-\frac {3 b \left (x^{3}+x \right )^{\frac {2}{3}} \left (9 x^{4}-6 x^{2}+5\right )}{80 x^{6}}+\frac {a \left (6 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -174 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}+20 x^{2}-36 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+8\right )-6 \ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}-4 x^{2}+96 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-3\right ) \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+\ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}-4 x^{2}+96 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-3\right )\right )}{2}\) | \(447\) |
3/2*a*x^(2/3)*hypergeom([1/3,1/3],[4/3],-x^2)-3/16*b/x^(16/3)*(9/5*x^4-6/5 *x^2+1)*(x^2+1)^(2/3)
Time = 0.73 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=\frac {40 \, \sqrt {3} a x^{6} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) - 20 \, a x^{6} \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) - 3 \, {\left (9 \, b x^{4} - 6 \, b x^{2} + 5 \, b\right )} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{80 \, x^{6}} \]
1/80*(40*sqrt(3)*a*x^6*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(5 39*x^2 + 507) - 1274*sqrt(3)*(x^3 + x)^(2/3))/(2205*x^2 + 2197)) - 20*a*x^ 6*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1) - 3*(9*b*x^4 - 6*b*x^2 + 5*b)*(x^3 + x)^(2/3))/x^6
\[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=\int \frac {a x^{6} + b}{x^{6} \sqrt [3]{x \left (x^{2} + 1\right )}}\, dx \]
\[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=\int { \frac {a x^{6} + b}{{\left (x^{3} + x\right )}^{\frac {1}{3}} x^{6}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.65 \[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=-\frac {3}{16} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {8}{3}} + \frac {3}{5} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{3}} - \frac {1}{2} \, \sqrt {3} a \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{4} \, a \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{2} \, a \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) - \frac {3}{4} \, b {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} \]
-3/16*b*(1/x^2 + 1)^(8/3) + 3/5*b*(1/x^2 + 1)^(5/3) - 1/2*sqrt(3)*a*arctan (1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) + 1/4*a*log((1/x^2 + 1)^(2/3) + (1 /x^2 + 1)^(1/3) + 1) - 1/2*a*log(abs((1/x^2 + 1)^(1/3) - 1)) - 3/4*b*(1/x^ 2 + 1)^(2/3)
Time = 6.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.40 \[ \int \frac {b+a x^6}{x^6 \sqrt [3]{x+x^3}} \, dx=\frac {3\,a\,x\,{\left (x^2+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -x^2\right )}{2\,{\left (x^3+x\right )}^{1/3}}-\frac {3\,b\,{\left (x^3+x\right )}^{2/3}\,\left (9\,x^4-6\,x^2+5\right )}{80\,x^6} \]