Integrand size = 20, antiderivative size = 82 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=-\frac {\arctan \left (\frac {1+x}{\sqrt {3} \sqrt {5+2 x+x^2}}\right )}{4 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {7+3 x}{\sqrt {13} \sqrt {5+2 x+x^2}}\right )}{12 \sqrt {13}}+\frac {1}{12} \text {arctanh}\left (\sqrt {5+2 x+x^2}\right ) \]
1/12*arctanh((x^2+2*x+5)^(1/2))-1/12*arctan(1/3*(1+x)*3^(1/2)/(x^2+2*x+5)^ (1/2))*3^(1/2)-1/156*arctanh(1/13*(7+3*x)*13^(1/2)/(x^2+2*x+5)^(1/2))*13^( 1/2)
Time = 0.37 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\frac {1}{156} \left (13 \sqrt {3} \arctan \left (\frac {4+2 x+x^2-(1+x) \sqrt {5+2 x+x^2}}{\sqrt {3}}\right )+13 \text {arctanh}\left (\sqrt {5+2 x+x^2}\right )-2 \sqrt {13} \text {arctanh}\left (\frac {2-x+\sqrt {5+2 x+x^2}}{\sqrt {13}}\right )\right ) \]
(13*Sqrt[3]*ArcTan[(4 + 2*x + x^2 - (1 + x)*Sqrt[5 + 2*x + x^2])/Sqrt[3]] + 13*ArcTanh[Sqrt[5 + 2*x + x^2]] - 2*Sqrt[13]*ArcTanh[(2 - x + Sqrt[5 + 2 *x + x^2])/Sqrt[13]])/156
Time = 0.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {2535, 1154, 219, 1358, 27, 1313, 217, 1357, 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.
\(\displaystyle \int \frac {1}{\sqrt {x^2+2 x+5} \left (x^3-8\right )} \, dx\) |
\(\Big \downarrow \) 2535 |
\(\displaystyle -\frac {1}{12} \int \frac {1}{(2-x) \sqrt {x^2+2 x+5}}dx-\frac {1}{12} \int \frac {x+4}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{6} \int \frac {1}{52-\frac {4 (3 x+7)^2}{x^2+2 x+5}}d\left (-\frac {2 (3 x+7)}{\sqrt {x^2+2 x+5}}\right )-\frac {1}{12} \int \frac {x+4}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {1}{12} \int \frac {x+4}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 1358 |
\(\displaystyle \frac {1}{12} \left (-3 \int \frac {1}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx-\frac {1}{2} \int \frac {2 (x+1)}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx\right )-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{12} \left (-3 \int \frac {1}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx-\int \frac {x+1}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx\right )-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 1313 |
\(\displaystyle \frac {1}{12} \left (12 \int \frac {1}{-\frac {8 (x+1)^2}{x^2+2 x+5}-24}d\frac {2 (x+1)}{\sqrt {x^2+2 x+5}}-\int \frac {x+1}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx\right )-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{12} \left (-\int \frac {x+1}{\left (x^2+2 x+4\right ) \sqrt {x^2+2 x+5}}dx-\sqrt {3} \arctan \left (\frac {x+1}{\sqrt {3} \sqrt {x^2+2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 1357 |
\(\displaystyle \frac {1}{12} \left (2 \int \frac {1}{2-2 \left (x^2+2 x+5\right )}d\sqrt {x^2+2 x+5}-\sqrt {3} \arctan \left (\frac {x+1}{\sqrt {3} \sqrt {x^2+2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{12} \left (\text {arctanh}\left (\sqrt {x^2+2 x+5}\right )-\sqrt {3} \arctan \left (\frac {x+1}{\sqrt {3} \sqrt {x^2+2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}\) |
-1/12*ArcTanh[(7 + 3*x)/(Sqrt[13]*Sqrt[5 + 2*x + x^2])]/Sqrt[13] + (-(Sqrt [3]*ArcTan[(1 + x)/(Sqrt[3]*Sqrt[5 + 2*x + x^2])]) + ArcTanh[Sqrt[5 + 2*x + x^2]])/12
3.3.44.3.1 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_)^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/(((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[1/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*( x_)^2]), x_Symbol] :> Simp[-2*e Subst[Int[1/(e*(b*e - 4*a*f) - (b*d - a*e )*x^2), x], x, (e + 2*f*x)/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[c*e - b*f, 0]
Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e _.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[-2*g Subst[Int[1/(b*d - a*e - b*x^2), x], x, Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[c*e - b*f, 0] & & EqQ[h*e - 2*g*f, 0]
Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + ( e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[-(h*e - 2*g*f)/(2*f) Int[1/ ((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[h/(2*f) Int[(e + 2*f*x)/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c , d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[c *e - b*f, 0] && NeQ[h*e - 2*g*f, 0]
Int[1/(Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]*((a_) + (b_.)*(x_)^3)), x_Sy mbol] :> With[{r = Numerator[Rt[-a/b, 3]], s = Denominator[Rt[-a/b, 3]]}, S imp[r/(3*a) Int[1/((r - s*x)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[r/(3*a ) Int[(2*r + s*x)/((r^2 + r*s*x + s^2*x^2)*Sqrt[d + e*x + f*x^2]), x], x] ] /; FreeQ[{a, b, d, e, f}, x] && NegQ[a/b]
Time = 0.78 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\sqrt {13}\, \operatorname {arctanh}\left (\frac {\left (14+6 x \right ) \sqrt {13}}{26 \sqrt {\left (-2+x \right )^{2}+1+6 x}}\right )}{156}+\frac {\operatorname {arctanh}\left (\sqrt {x^{2}+2 x +5}\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x +2\right )}{6 \sqrt {x^{2}+2 x +5}}\right )}{12}\) | \(69\) |
trager | \(\operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \ln \left (\frac {2880 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )^{2} x -126 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \sqrt {x^{2}+2 x +5}-306 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x -570 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+8 \sqrt {x^{2}+2 x +5}+7 x +19}{12 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x -x -2}\right )+\frac {\ln \left (\frac {5760 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )^{2} x +252 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \sqrt {x^{2}+2 x +5}-348 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1140 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )-5 \sqrt {x^{2}+2 x +5}+3 x -57}{6 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1}\right )}{12}-\ln \left (\frac {5760 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )^{2} x +252 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \sqrt {x^{2}+2 x +5}-348 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1140 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )-5 \sqrt {x^{2}+2 x +5}+3 x -57}{6 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1}\right ) \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) \ln \left (-\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) x +13 \sqrt {x^{2}+2 x +5}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right )}{-2+x}\right )}{156}\) | \(387\) |
-1/156*13^(1/2)*arctanh(1/26*(14+6*x)*13^(1/2)/((-2+x)^2+1+6*x)^(1/2))+1/1 2*arctanh((x^2+2*x+5)^(1/2))-1/12*3^(1/2)*arctan(1/6*3^(1/2)/(x^2+2*x+5)^( 1/2)*(2*x+2))
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (64) = 128\).
Time = 0.25 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x + 2\right )} + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} + 2 \, x + 5}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} x + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} + 2 \, x + 5}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (\frac {\sqrt {13} {\left (3 \, x + 7\right )} + \sqrt {x^{2} + 2 \, x + 5} {\left (3 \, \sqrt {13} - 13\right )} - 9 \, x - 21}{x - 2}\right ) - \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} + 2 \, x + 5} {\left (x + 2\right )} + 3 \, x + 6\right ) + \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} + 2 \, x + 5} x + x + 4\right ) \]
1/12*sqrt(3)*arctan(-1/3*sqrt(3)*(x + 2) + 1/3*sqrt(3)*sqrt(x^2 + 2*x + 5) ) - 1/12*sqrt(3)*arctan(-1/3*sqrt(3)*x + 1/3*sqrt(3)*sqrt(x^2 + 2*x + 5)) + 1/156*sqrt(13)*log((sqrt(13)*(3*x + 7) + sqrt(x^2 + 2*x + 5)*(3*sqrt(13) - 13) - 9*x - 21)/(x - 2)) - 1/24*log(x^2 - sqrt(x^2 + 2*x + 5)*(x + 2) + 3*x + 6) + 1/24*log(x^2 - sqrt(x^2 + 2*x + 5)*x + x + 4)
\[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\int \frac {1}{\left (x - 2\right ) \left (x^{2} + 2 x + 4\right ) \sqrt {x^{2} + 2 x + 5}}\, dx \]
\[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} - 8\right )} \sqrt {x^{2} + 2 \, x + 5}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (64) = 128\).
Time = 0.32 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} + 2 \, x + 5} + 2\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} + 2 \, x + 5}\right )}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} + 2 \, x + 5} + 4 \right |}}{{\left | -2 \, x + 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} + 2 \, x + 5} + 4 \right |}}\right ) - \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} + 2 \, x + 5}\right )}^{2} + 4 \, x - 4 \, \sqrt {x^{2} + 2 \, x + 5} + 7\right ) + \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} + 2 \, x + 5}\right )}^{2} + 3\right ) \]
1/12*sqrt(3)*arctan(-1/3*sqrt(3)*(x - sqrt(x^2 + 2*x + 5) + 2)) - 1/12*sqr t(3)*arctan(-1/3*sqrt(3)*(x - sqrt(x^2 + 2*x + 5))) + 1/156*sqrt(13)*log(a bs(-2*x - 2*sqrt(13) + 2*sqrt(x^2 + 2*x + 5) + 4)/abs(-2*x + 2*sqrt(13) + 2*sqrt(x^2 + 2*x + 5) + 4)) - 1/24*log((x - sqrt(x^2 + 2*x + 5))^2 + 4*x - 4*sqrt(x^2 + 2*x + 5) + 7) + 1/24*log((x - sqrt(x^2 + 2*x + 5))^2 + 3)
Timed out. \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\int \frac {1}{\left (x^3-8\right )\,\sqrt {x^2+2\,x+5}} \,d x \]