Integrand size = 24, antiderivative size = 63 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {5+4 x+4 x^2}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (1+2 x)}{\sqrt {5+4 x+4 x^2}}\right )}{\sqrt {165}} \]
1/11*arctan(1/11*(4*x^2+4*x+5)^(1/2)*11^(1/2))*11^(1/2)-1/165*arctanh(1/15 *(1+2*x)*165^(1/2)/(4*x^2+4*x+5)^(1/2))*165^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.62 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{2} \text {RootSum}\left [69-108 \text {$\#$1}+58 \text {$\#$1}^2-4 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {-5 \log \left (-2 x+\sqrt {5+4 x+4 x^2}-\text {$\#$1}\right )+\log \left (-2 x+\sqrt {5+4 x+4 x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-27+29 \text {$\#$1}-3 \text {$\#$1}^2+\text {$\#$1}^3}\&\right ] \]
RootSum[69 - 108*#1 + 58*#1^2 - 4*#1^3 + #1^4 & , (-5*Log[-2*x + Sqrt[5 + 4*x + 4*x^2] - #1] + Log[-2*x + Sqrt[5 + 4*x + 4*x^2] - #1]*#1^2)/(-27 + 2 9*#1 - 3*#1^2 + #1^3) & ]/2
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1358, 27, 1313, 220, 1357, 217}
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.
| \(\displaystyle \int \frac {x}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}} \, dx\) |
| \(\Big \downarrow \) 1358 |
| \(\displaystyle \frac {1}{8} \int \frac {4 (2 x+1)}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}}dx-\frac {1}{2} \int \frac {1}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {2 x+1}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}}dx-\frac {1}{2} \int \frac {1}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}}dx\) |
| \(\Big \downarrow \) 1313 |
| \(\displaystyle \frac {1}{2} \int \frac {2 x+1}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}}dx+4 \int \frac {1}{\frac {176 (2 x+1)^2}{4 x^2+4 x+5}-240}d\frac {4 (2 x+1)}{\sqrt {4 x^2+4 x+5}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{2} \int \frac {2 x+1}{\left (x^2+x+4\right ) \sqrt {4 x^2+4 x+5}}dx-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}}\) |
| \(\Big \downarrow \) 1357 |
| \(\displaystyle -\int \frac {1}{-4 x^2-4 x-16}d\sqrt {4 x^2+4 x+5}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\arctan \left (\frac {\sqrt {4 x^2+4 x+5}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}}\) |
ArcTan[Sqrt[5 + 4*x + 4*x^2]/Sqrt[11]]/Sqrt[11] - ArcTanh[(Sqrt[11/15]*(1 + 2*x))/Sqrt[5 + 4*x + 4*x^2]]/Sqrt[165]
3.3.45.3.1 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[(-(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/(((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 = 1.44 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84
| method | result | size |
| default | \(\frac {\arctan \left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {11}}{11}\right ) \sqrt {11}}{11}-\frac {\sqrt {165}\, \operatorname {arctanh}\left (\frac {\sqrt {165}\, \left (8 x +4\right )}{60 \sqrt {4 x^{2}+4 x +5}}\right )}{165}\) | \(53\) |
| trager | \(-\operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) \ln \left (\frac {3524400 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{5} x +111270 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3} x -41385 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}-3420 \sqrt {4 x^{2}+4 x +5}\, \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+754 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x -899 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )-52 \sqrt {4 x^{2}+4 x +5}}{165 x \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+3 x -4}\right )-\frac {165 \ln \left (\frac {8276400 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{5} x +385770 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3} x +97185 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}+9405 \sqrt {4 x^{2}+4 x +5}\, \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+3784 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +1364 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )+256 \sqrt {4 x^{2}+4 x +5}}{165 x \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+4 x +4}\right ) \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}}{4}-\frac {7 \ln \left (\frac {8276400 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{5} x +385770 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3} x +97185 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}+9405 \sqrt {4 x^{2}+4 x +5}\, \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+3784 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +1364 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )+256 \sqrt {4 x^{2}+4 x +5}}{165 x \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+4 x +4}\right ) \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )}{4}\) | \(514\) |
1/11*arctan(1/11*(4*x^2+4*x+5)^(1/2)*11^(1/2))*11^(1/2)-1/165*165^(1/2)*ar ctanh(1/60*165^(1/2)*(8*x+4)/(4*x^2+4*x+5)^(1/2))
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 237, normalized size of antiderivative = 3.76 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{330} \, \sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} {\left (i \, \sqrt {15} + 15\right )} - 480 \, x - 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) - \frac {1}{330} \, \sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} {\left (-i \, \sqrt {15} - 15\right )} - 480 \, x - 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) - \frac {1}{330} \, \sqrt {165} \sqrt {-2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} {\left (i \, \sqrt {15} - 15\right )} \sqrt {-2 i \, \sqrt {15} - 14} - 480 \, x + 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) + \frac {1}{330} \, \sqrt {165} \sqrt {-2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} {\left (-i \, \sqrt {15} + 15\right )} \sqrt {-2 i \, \sqrt {15} - 14} - 480 \, x + 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) \]
1/330*sqrt(165)*sqrt(2*I*sqrt(15) - 14)*log(sqrt(165)*sqrt(2*I*sqrt(15) - 14)*(I*sqrt(15) + 15) - 480*x - 240*I*sqrt(15) + 240*sqrt(4*x^2 + 4*x + 5) - 240) - 1/330*sqrt(165)*sqrt(2*I*sqrt(15) - 14)*log(sqrt(165)*sqrt(2*I*s qrt(15) - 14)*(-I*sqrt(15) - 15) - 480*x - 240*I*sqrt(15) + 240*sqrt(4*x^2 + 4*x + 5) - 240) - 1/330*sqrt(165)*sqrt(-2*I*sqrt(15) - 14)*log(sqrt(165 )*(I*sqrt(15) - 15)*sqrt(-2*I*sqrt(15) - 14) - 480*x + 240*I*sqrt(15) + 24 0*sqrt(4*x^2 + 4*x + 5) - 240) + 1/330*sqrt(165)*sqrt(-2*I*sqrt(15) - 14)* log(sqrt(165)*(-I*sqrt(15) + 15)*sqrt(-2*I*sqrt(15) - 14) - 480*x + 240*I* sqrt(15) + 240*sqrt(4*x^2 + 4*x + 5) - 240)
\[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int \frac {x}{\left (x^{2} + x + 4\right ) \sqrt {4 x^{2} + 4 x + 5}}\, dx \]
\[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int { \frac {x}{\sqrt {4 \, x^{2} + 4 \, x + 5} {\left (x^{2} + x + 4\right )}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (52) = 104\).
Time = 0.33 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.62 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} + \sqrt {11}}\right ) - \frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} - \sqrt {11}}\right ) - \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} + \sqrt {11}\right )}^{2}\right ) + \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} - \sqrt {11}\right )}^{2}\right ) \]
1/165*sqrt(165)*sqrt(15)*arctan(-(2*x - sqrt(4*x^2 + 4*x + 5) + 1)/(sqrt(1 5) + sqrt(11))) - 1/165*sqrt(165)*sqrt(15)*arctan(-(2*x - sqrt(4*x^2 + 4*x + 5) + 1)/(sqrt(15) - sqrt(11))) - 1/330*sqrt(165)*log(90000*(2*x - sqrt( 4*x^2 + 4*x + 5) + 1)^2 + 90000*(sqrt(15) + sqrt(11))^2) + 1/330*sqrt(165) *log(90000*(2*x - sqrt(4*x^2 + 4*x + 5) + 1)^2 + 90000*(sqrt(15) - sqrt(11 ))^2)
Timed out. \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int \frac {x}{\sqrt {4\,x^2+4\,x+5}\,\left (x^2+x+4\right )} \,d x \]
Time = 0.00 (sec) , antiderivative size = 227, normalized size of antiderivative = 3.60 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {\sqrt {11}\, \left (30 \mathit {atan} \left (\frac {\sqrt {4 x^{2}+4 x +5}+2 x +1}{\sqrt {11}+\sqrt {15}}\right )+\sqrt {15}\, \mathrm {log}\left (2 \sqrt {4 x^{2}+4 x +5}\, x +\sqrt {4 x^{2}+4 x +5}+\sqrt {165}+4 x^{2}+4 x +16\right )-\sqrt {15}\, \mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}-\frac {\sqrt {22}\, i}{2}+\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right )-\sqrt {15}\, \mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}+\frac {\sqrt {22}\, i}{2}-\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right )-15 \,\mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}-\frac {\sqrt {22}\, i}{2}+\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right ) i +15 \,\mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}+\frac {\sqrt {22}\, i}{2}-\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right ) i \right )}{330} \]
(sqrt(11)*(30*atan((sqrt(4*x**2 + 4*x + 5) + 2*x + 1)/(sqrt(11) + sqrt(15) )) + sqrt(15)*log(2*sqrt(4*x**2 + 4*x + 5)*x + sqrt(4*x**2 + 4*x + 5) + sq rt(165) + 4*x**2 + 4*x + 16) - sqrt(15)*log((sqrt(4*x**2 + 4*x + 5)*sqrt(2 ) - sqrt(22)*i + sqrt(30)*i + 2*sqrt(2)*x + sqrt(2))/2) - sqrt(15)*log((sq rt(4*x**2 + 4*x + 5)*sqrt(2) + sqrt(22)*i - sqrt(30)*i + 2*sqrt(2)*x + sqr t(2))/2) - 15*log((sqrt(4*x**2 + 4*x + 5)*sqrt(2) - sqrt(22)*i + sqrt(30)* i + 2*sqrt(2)*x + sqrt(2))/2)*i + 15*log((sqrt(4*x**2 + 4*x + 5)*sqrt(2) + sqrt(22)*i - sqrt(30)*i + 2*sqrt(2)*x + sqrt(2))/2)*i))/330