Integrand size = 20, antiderivative size = 84 \[ \int \frac {1}{\sqrt {5-2 x+x^2} \left (8+x^3\right )} \, dx=-\frac {\arctan \left (\frac {1-x}{\sqrt {3} \sqrt {5-2 x+x^2}}\right )}{4 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {5-2 x+x^2}}\right )}{12 \sqrt {13}}+\frac {1}{12} \text {arctanh}\left (\sqrt {5-2 x+x^2}\right ) \]
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 = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt {5-2 x+x^2} \left (8+x^3\right )} \, dx=\frac {1}{156} \left (-13 \sqrt {3} \arctan \left (\frac {4-2 x+x^2-(-1+x) \sqrt {5-2 x+x^2}}{\sqrt {3}}\right )+13 \text {arctanh}\left (\sqrt {5-2 x+x^2}\right )+2 \sqrt {13} \text {arctanh}\left (\frac {2+x-\sqrt {5-2 x+x^2}}{\sqrt {13}}\right )\right ) \]
(-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 = 0.35 (sec) , antiderivative size = 83, 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 = {2533, 1154, 219, 1358, 27, 1313, 216, 1357, 220}
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.
\(\displaystyle \int \frac {1}{\sqrt {x^2-2 x+5} \left (x^3+8\right )} \, dx\) |
\(\Big \downarrow \) 2533 |
\(\displaystyle \frac {1}{12} \int \frac {1}{(x+2) \sqrt {x^2-2 x+5}}dx+\frac {1}{12} \int \frac {4-x}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{12} \int \frac {4-x}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx-\frac {1}{6} \int \frac {1}{52-\frac {4 (7-3 x)^2}{x^2-2 x+5}}d\frac {2 (7-3 x)}{\sqrt {x^2-2 x+5}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{12} \int \frac {4-x}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 1358 |
\(\displaystyle \frac {1}{12} \left (3 \int \frac {1}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx-\frac {1}{2} \int -\frac {2 (1-x)}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx\right )-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{12} \left (3 \int \frac {1}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx+\int \frac {1-x}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx\right )-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 1313 |
\(\displaystyle \frac {1}{12} \left (\int \frac {1-x}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx+12 \int \frac {1}{\frac {8 (1-x)^2}{x^2-2 x+5}+24}d\left (-\frac {2 (1-x)}{\sqrt {x^2-2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{12} \left (\int \frac {1-x}{\left (x^2-2 x+4\right ) \sqrt {x^2-2 x+5}}dx-\sqrt {3} \arctan \left (\frac {1-x}{\sqrt {3} \sqrt {x^2-2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 1357 |
\(\displaystyle \frac {1}{12} \left (-2 \int \frac {1}{2 \left (x^2-2 x+5\right )-2}d\sqrt {x^2-2 x+5}-\sqrt {3} \arctan \left (\frac {1-x}{\sqrt {3} \sqrt {x^2-2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{12} \left (\text {arctanh}\left (\sqrt {x^2-2 x+5}\right )-\sqrt {3} \arctan \left (\frac {1-x}{\sqrt {3} \sqrt {x^2-2 x+5}}\right )\right )-\frac {\text {arctanh}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}\) |
-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
3.11.6.3.1 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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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]]}, Sim p[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] && PosQ[a/b]
Time = 1.49 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {\sqrt {13}\, \operatorname {arctanh}\left (\frac {\left (14-6 x \right ) \sqrt {13}}{26 \sqrt {\left (x +2\right )^{2}-6 x +1}}\right )}{156}+\frac {\operatorname {arctanh}\left (\sqrt {x^{2}-2 x +5}\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x -2\right )}{6 \sqrt {x^{2}-2 x +5}}\right )}{12}\) | \(69\) |
trager | \(\operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) \ln \left (\frac {2880 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )^{2} x +126 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) \sqrt {x^{2}-2 x +5}+306 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) x +8 \sqrt {x^{2}-2 x +5}-570 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )+7 x -19}{12 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) x +x -2}\right )-\frac {\ln \left (-\frac {-5760 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )^{2} x +252 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) \sqrt {x^{2}-2 x +5}-348 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) x +5 \sqrt {x^{2}-2 x +5}-1140 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )-3 x -57}{6 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) x +1}\right )}{12}-\ln \left (-\frac {-5760 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )^{2} x +252 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) \sqrt {x^{2}-2 x +5}-348 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) x +5 \sqrt {x^{2}-2 x +5}-1140 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )-3 x -57}{6 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right ) x +1}\right ) \operatorname {RootOf}\left (144 \textit {\_Z}^{2}+12 \textit {\_Z} +1\right )+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) x +13 \sqrt {x^{2}-2 x +5}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right )}{x +2}\right )}{156}\) | \(386\) |
-1/156*13^(1/2)*arctanh(1/26*(14-6*x)*13^(1/2)/((x+2)^2-6*x+1)^(1/2))+1/12 *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. 154 vs. \(2 (64) = 128\).
Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.83 \[ \int \frac {1}{\sqrt {5-2 x+x^2} \left (8+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - 2\right )} + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} - 2 \, x + 5}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} x + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} - 2 \, x + 5}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (-\frac {\sqrt {13} {\left (3 \, x - 7\right )} + \sqrt {x^{2} - 2 \, x + 5} {\left (3 \, \sqrt {13} + 13\right )} + 9 \, x - 21}{x + 2}\right ) + \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} - 2 \, x + 5} {\left (x - 2\right )} - 3 \, x + 6\right ) - \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} - 2 \, x + 5} x - x + 4\right ) \]
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 = 0.34 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.95 \[ \int \frac {1}{\sqrt {5-2 x+x^2} \left (8+x^3\right )} \, dx=-\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} - 2 \, x + 5}\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} - 2 \, x + 5} - 2\right )}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} - 2 \, x + 5} - 4 \right |}}{{\left | -2 \, x + 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} - 2 \, x + 5} - 4 \right |}}\right ) + \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} - 2 \, x + 5}\right )}^{2} - 4 \, x + 4 \, \sqrt {x^{2} - 2 \, x + 5} + 7\right ) - \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} - 2 \, x + 5}\right )}^{2} + 3\right ) \]
-1/12*sqrt(3)*arctan(-1/3*sqrt(3)*(x - sqrt(x^2 - 2*x + 5))) + 1/12*sqrt(3 )*arctan(-1/3*sqrt(3)*(x - sqrt(x^2 - 2*x + 5) - 2)) + 1/156*sqrt(13)*log( abs(-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. \[ \int \frac {1}{\sqrt {5-2 x+x^2} \left (8+x^3\right )} \, dx=\int \frac {1}{\left (x^3+8\right )\,\sqrt {x^2-2\,x+5}} \,d x \]