Integrand size = 20, antiderivative size = 56 \[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {\frac {2}{11}} (1+2 x)}{\sqrt {5+x+x^2}}\right )}{\sqrt {22}}-\frac {\text {arctanh}\left (\frac {\sqrt {5+x+x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \]
-1/2*arctanh(1/2*(x^2+x+5)^(1/2)*2^(1/2))*2^(1/2)-1/22*arctan(1/11*(1+2*x) *22^(1/2)/(x^2+x+5)^(1/2))*22^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.64 \[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\text {RootSum}\left [23-2 \text {$\#$1}+3 \text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {-5 \log \left (-x+\sqrt {5+x+x^2}-\text {$\#$1}\right )+\log \left (-x+\sqrt {5+x+x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-1+3 \text {$\#$1}-3 \text {$\#$1}^2+2 \text {$\#$1}^3}\&\right ] \]
RootSum[23 - 2*#1 + 3*#1^2 - 2*#1^3 + #1^4 & , (-5*Log[-x + Sqrt[5 + x + x ^2] - #1] + Log[-x + Sqrt[5 + x + x^2] - #1]*#1^2)/(-1 + 3*#1 - 3*#1^2 + 2 *#1^3) & ]
Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1358, 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.
\(\displaystyle \int \frac {x}{\left (x^2+x+3\right ) \sqrt {x^2+x+5}} \, dx\) |
\(\Big \downarrow \) 1358 |
\(\displaystyle \frac {1}{2} \int \frac {2 x+1}{\left (x^2+x+3\right ) \sqrt {x^2+x+5}}dx-\frac {1}{2} \int \frac {1}{\left (x^2+x+3\right ) \sqrt {x^2+x+5}}dx\) |
\(\Big \downarrow \) 1313 |
\(\displaystyle \frac {1}{2} \int \frac {2 x+1}{\left (x^2+x+3\right ) \sqrt {x^2+x+5}}dx+\int \frac {1}{-\frac {2 (2 x+1)^2}{x^2+x+5}-11}d\frac {2 x+1}{\sqrt {x^2+x+5}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \int \frac {2 x+1}{\left (x^2+x+3\right ) \sqrt {x^2+x+5}}dx-\frac {\arctan \left (\frac {\sqrt {\frac {2}{11}} (2 x+1)}{\sqrt {x^2+x+5}}\right )}{\sqrt {22}}\) |
\(\Big \downarrow \) 1357 |
\(\displaystyle -\int \frac {1}{-x^2-x-3}d\sqrt {x^2+x+5}-\frac {\arctan \left (\frac {\sqrt {\frac {2}{11}} (2 x+1)}{\sqrt {x^2+x+5}}\right )}{\sqrt {22}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {\frac {2}{11}} (2 x+1)}{\sqrt {x^2+x+5}}\right )}{\sqrt {22}}-\frac {\text {arctanh}\left (\frac {\sqrt {x^2+x+5}}{\sqrt {2}}\right )}{\sqrt {2}}\) |
-(ArcTan[(Sqrt[2/11]*(1 + 2*x))/Sqrt[5 + x + x^2]]/Sqrt[22]) - ArcTanh[Sqr t[5 + x + x^2]/Sqrt[2]]/Sqrt[2]
3.1.34.3.1 Defintions of rubi rules used
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/(((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]
Time = 0.99 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {x^{2}+x +5}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {22}}{11 \sqrt {x^{2}+x +5}}\right ) \sqrt {22}}{22}\) | \(45\) |
trager | \(-\operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right ) \ln \left (\frac {133342 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{5} x -34298 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3} x +2970 \sqrt {x^{2}+x +5}\, \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{2}+29667 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3}+2100 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right ) x -504 \sqrt {x^{2}+x +5}-4650 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )}{22 x \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{2}-2 x +3}\right )+\frac {22 \ln \left (-\frac {-22990 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{5} x +2079 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3} x +990 \sqrt {x^{2}+x +5}\, \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{2}+5115 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3}+106 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right ) x -57 \sqrt {x^{2}+x +5}+186 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )}{22 x \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{2}-3 x -3}\right ) \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3}}{3}-\frac {5 \ln \left (-\frac {-22990 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{5} x +2079 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3} x +990 \sqrt {x^{2}+x +5}\, \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{2}+5115 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{3}+106 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right ) x -57 \sqrt {x^{2}+x +5}+186 \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )}{22 x \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )^{2}-3 x -3}\right ) \operatorname {RootOf}\left (484 \textit {\_Z}^{4}-110 \textit {\_Z}^{2}+9\right )}{3}\) | \(492\) |
-1/2*arctanh(1/2*(x^2+x+5)^(1/2)*2^(1/2))*2^(1/2)-1/22*arctan(1/11*(1+2*x) *22^(1/2)/(x^2+x+5)^(1/2))*22^(1/2)
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 3.62 \[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\frac {1}{22} \, \sqrt {11} \sqrt {i \, \sqrt {11} + 5} \log \left ({\left (\sqrt {11} - i\right )} \sqrt {i \, \sqrt {11} + 5} - 6 \, x + 3 i \, \sqrt {11} + 6 \, \sqrt {x^{2} + x + 5} - 3\right ) + \frac {1}{22} \, \sqrt {11} \sqrt {i \, \sqrt {11} + 5} \log \left (-{\left (\sqrt {11} - i\right )} \sqrt {i \, \sqrt {11} + 5} - 6 \, x + 3 i \, \sqrt {11} + 6 \, \sqrt {x^{2} + x + 5} - 3\right ) - \frac {1}{22} \, \sqrt {11} \sqrt {-i \, \sqrt {11} + 5} \log \left ({\left (\sqrt {11} + i\right )} \sqrt {-i \, \sqrt {11} + 5} - 6 \, x - 3 i \, \sqrt {11} + 6 \, \sqrt {x^{2} + x + 5} - 3\right ) + \frac {1}{22} \, \sqrt {11} \sqrt {-i \, \sqrt {11} + 5} \log \left (-{\left (\sqrt {11} + i\right )} \sqrt {-i \, \sqrt {11} + 5} - 6 \, x - 3 i \, \sqrt {11} + 6 \, \sqrt {x^{2} + x + 5} - 3\right ) \]
-1/22*sqrt(11)*sqrt(I*sqrt(11) + 5)*log((sqrt(11) - I)*sqrt(I*sqrt(11) + 5 ) - 6*x + 3*I*sqrt(11) + 6*sqrt(x^2 + x + 5) - 3) + 1/22*sqrt(11)*sqrt(I*s qrt(11) + 5)*log(-(sqrt(11) - I)*sqrt(I*sqrt(11) + 5) - 6*x + 3*I*sqrt(11) + 6*sqrt(x^2 + x + 5) - 3) - 1/22*sqrt(11)*sqrt(-I*sqrt(11) + 5)*log((sqr t(11) + I)*sqrt(-I*sqrt(11) + 5) - 6*x - 3*I*sqrt(11) + 6*sqrt(x^2 + x + 5 ) - 3) + 1/22*sqrt(11)*sqrt(-I*sqrt(11) + 5)*log(-(sqrt(11) + I)*sqrt(-I*s qrt(11) + 5) - 6*x - 3*I*sqrt(11) + 6*sqrt(x^2 + x + 5) - 3)
\[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\int \frac {x}{\left (x^{2} + x + 3\right ) \sqrt {x^{2} + x + 5}}\, dx \]
\[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\int { \frac {x}{\sqrt {x^{2} + x + 5} {\left (x^{2} + x + 3\right )}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (44) = 88\).
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.38 \[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\frac {1}{22} \, \sqrt {11} \sqrt {2} \arctan \left (-\frac {1}{11} \, \sqrt {11} {\left (2 \, x + 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}\right ) - \frac {1}{22} \, \sqrt {11} \sqrt {2} \arctan \left (-\frac {1}{11} \, \sqrt {11} {\left (2 \, x - 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (324 \, {\left (2 \, x + 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}^{2} + 3564\right ) - \frac {1}{4} \, \sqrt {2} \log \left (324 \, {\left (2 \, x - 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}^{2} + 3564\right ) \]
1/22*sqrt(11)*sqrt(2)*arctan(-1/11*sqrt(11)*(2*x + 2*sqrt(2) - 2*sqrt(x^2 + x + 5) + 1)) - 1/22*sqrt(11)*sqrt(2)*arctan(-1/11*sqrt(11)*(2*x - 2*sqrt (2) - 2*sqrt(x^2 + x + 5) + 1)) + 1/4*sqrt(2)*log(324*(2*x + 2*sqrt(2) - 2 *sqrt(x^2 + x + 5) + 1)^2 + 3564) - 1/4*sqrt(2)*log(324*(2*x - 2*sqrt(2) - 2*sqrt(x^2 + x + 5) + 1)^2 + 3564)
Timed out. \[ \int \frac {x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\int \frac {x}{\left (x^2+x+3\right )\,\sqrt {x^2+x+5}} \,d x \]