\(\int \frac {x}{(4+x+x^2) \sqrt {5+4 x+4 x^2}} \, dx\) [245]

Optimal result

Integrand size = 24, antiderivative size = 63 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {5+4 x+4 x^2}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (1+2 x)}{\sqrt {5+4 x+4 x^2}}\right )}{\sqrt {165}} \] Output:

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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.62 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{2} \text {RootSum}\left [69-108 \text {$\#$1}+58 \text {$\#$1}^2-4 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {-5 \log \left (-2 x+\sqrt {5+4 x+4 x^2}-\text {$\#$1}\right )+\log \left (-2 x+\sqrt {5+4 x+4 x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-27+29 \text {$\#$1}-3 \text {$\#$1}^2+\text {$\#$1}^3}\&\right ] \] Input:

Integrate[x/((4 + x + x^2)*Sqrt[5 + 4*x + 4*x^2]),x]
 

Output:

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
 

Rubi [A] (verified)

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

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {4 x^2+4 x+5}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}}\)

Input:

Int[x/((4 + x + x^2)*Sqrt[5 + 4*x + 4*x^2]),x]
 

Output:

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]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 220

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])
 

rule 1313

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]
 

rule 1357

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]
 

rule 1358

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]
 

Maple [A] (verified)

Time = 0.85 (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 +3420 \sqrt {4 x^{2}+4 x +5}\, \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}-41385 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}+754 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +52 \sqrt {4 x^{2}+4 x +5}-899 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )}{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 +9405 \sqrt {4 x^{2}+4 x +5}\, \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}-97185 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}-3784 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +256 \sqrt {4 x^{2}+4 x +5}-1364 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )}{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 +9405 \sqrt {4 x^{2}+4 x +5}\, \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}-97185 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}-3784 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +256 \sqrt {4 x^{2}+4 x +5}-1364 \operatorname {RootOf}\left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )}{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}\)	\(514\)

Input:

int(x/(x^2+x+4)/(4*x^2+4*x+5)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (52) = 104\).

Time = 0.07 (sec) , antiderivative size = 193, normalized size of antiderivative = 3.06 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=-\frac {1}{11} \, \sqrt {11} \arctan \left (-\frac {1}{44} \, \sqrt {165} \sqrt {11} {\left (2 \, x + 1\right )} + \frac {1}{44} \, \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (\sqrt {165} \sqrt {11} + 11 \, \sqrt {11}\right )} - \frac {1}{4} \, \sqrt {11} {\left (2 \, x + 1\right )}\right ) + \frac {1}{11} \, \sqrt {11} \arctan \left (-\frac {1}{44} \, \sqrt {165} \sqrt {11} {\left (2 \, x + 1\right )} + \frac {1}{44} \, \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (\sqrt {165} \sqrt {11} - 11 \, \sqrt {11}\right )} + \frac {1}{4} \, \sqrt {11} {\left (2 \, x + 1\right )}\right ) - \frac {1}{330} \, \sqrt {165} \log \left (4 \, x^{2} - \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (2 \, x + 1\right )} + 4 \, x + \sqrt {165} + 16\right ) + \frac {1}{330} \, \sqrt {165} \log \left (4 \, x^{2} - \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (2 \, x + 1\right )} + 4 \, x - \sqrt {165} + 16\right ) \] Input:

integrate(x/(x^2+x+4)/(4*x^2+4*x+5)^(1/2),x, algorithm="fricas")
 

Output:

-1/11*sqrt(11)*arctan(-1/44*sqrt(165)*sqrt(11)*(2*x + 1) + 1/44*sqrt(4*x^2 
 + 4*x + 5)*(sqrt(165)*sqrt(11) + 11*sqrt(11)) - 1/4*sqrt(11)*(2*x + 1)) + 
 1/11*sqrt(11)*arctan(-1/44*sqrt(165)*sqrt(11)*(2*x + 1) + 1/44*sqrt(4*x^2 
 + 4*x + 5)*(sqrt(165)*sqrt(11) - 11*sqrt(11)) + 1/4*sqrt(11)*(2*x + 1)) - 
 1/330*sqrt(165)*log(4*x^2 - sqrt(4*x^2 + 4*x + 5)*(2*x + 1) + 4*x + sqrt( 
165) + 16) + 1/330*sqrt(165)*log(4*x^2 - sqrt(4*x^2 + 4*x + 5)*(2*x + 1) + 
 4*x - sqrt(165) + 16)
 

Sympy [F]

\[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int \frac {x}{\left (x^{2} + x + 4\right ) \sqrt {4 x^{2} + 4 x + 5}}\, dx \] Input:

integrate(x/(x**2+x+4)/(4*x**2+4*x+5)**(1/2),x)
 

Output:

Integral(x/((x**2 + x + 4)*sqrt(4*x**2 + 4*x + 5)), x)
 

Maxima [F]

\[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int { \frac {x}{\sqrt {4 \, x^{2} + 4 \, x + 5} {\left (x^{2} + x + 4\right )}} \,d x } \] Input:

integrate(x/(x^2+x+4)/(4*x^2+4*x+5)^(1/2),x, algorithm="maxima")
 

Output:

integrate(x/(sqrt(4*x^2 + 4*x + 5)*(x^2 + x + 4)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (52) = 104\).

Time = 0.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.62 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} + \sqrt {11}}\right ) - \frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} - \sqrt {11}}\right ) - \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} + \sqrt {11}\right )}^{2}\right ) + \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} - \sqrt {11}\right )}^{2}\right ) \] Input:

integrate(x/(x^2+x+4)/(4*x^2+4*x+5)^(1/2),x, algorithm="giac")
 

Output:

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)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int \frac {x}{\sqrt {4\,x^2+4\,x+5}\,\left (x^2+x+4\right )} \,d x \] Input:

int(x/((4*x + 4*x^2 + 5)^(1/2)*(x + x^2 + 4)),x)
 

Output:

int(x/((4*x + 4*x^2 + 5)^(1/2)*(x + x^2 + 4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 227, normalized size of antiderivative = 3.60 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {\sqrt {11}\, \left (30 \mathit {atan} \left (\frac {\sqrt {4 x^{2}+4 x +5}+2 x +1}{\sqrt {11}+\sqrt {15}}\right )+\sqrt {15}\, \mathrm {log}\left (2 \sqrt {4 x^{2}+4 x +5}\, x +\sqrt {4 x^{2}+4 x +5}+\sqrt {165}+4 x^{2}+4 x +16\right )-\sqrt {15}\, \mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}-\frac {\sqrt {22}\, i}{2}+\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right )-\sqrt {15}\, \mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}+\frac {\sqrt {22}\, i}{2}-\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right )-15 \,\mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}-\frac {\sqrt {22}\, i}{2}+\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right ) i +15 \,\mathrm {log}\left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {2}}{2}+\frac {\sqrt {22}\, i}{2}-\frac {\sqrt {30}\, i}{2}+\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right ) i \right )}{330} \] Input:

int(x/(x^2+x+4)/(4*x^2+4*x+5)^(1/2),x)
 

Output:

(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