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

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1039, 996, 213, 1038, 210} \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {4 x^2+4 x+5}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}} \]

[In]

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

[Out]

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]

Rule 210

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 213

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 996

Int[1/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*e, Su
bst[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 1038

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol
] :> Dist[-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 1039

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbo
l] :> Dist[-(h*e - 2*g*f)/(2*f), Int[1/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {4+8 x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx-\frac {1}{2} \int \frac {1}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx \\ & = 4 \text {Subst}\left (\int \frac {1}{-240+11 x^2} \, dx,x,\frac {4+8 x}{\sqrt {5+4 x+4 x^2}}\right )-\text {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,\sqrt {5+4 x+4 x^2}\right ) \\ & = \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}} \\ \end{align*}

Mathematica [C] (verified)

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 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{2} \text {RootSum}\left [69-108 \text {$\#$1}+58 \text {$\#$1}^2-4 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {-5 \log \left (-2 x+\sqrt {5+4 x+4 x^2}-\text {$\#$1}\right )+\log \left (-2 x+\sqrt {5+4 x+4 x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-27+29 \text {$\#$1}-3 \text {$\#$1}^2+\text {$\#$1}^3}\&\right ] \]

[In]

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

[Out]

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 + 29*#1 - 3*#1^2 + #1^3) & ]/2

Maple [A] (verified)

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}\)	\(514\)

[In]

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

[Out]

1/11*arctan(1/11*(4*x^2+4*x+5)^(1/2)*11^(1/2))*11^(1/2)-1/165*165^(1/2)*arctanh(1/60*165^(1/2)*(8*x+4)/(4*x^2+
4*x+5)^(1/2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 237, normalized size of antiderivative = 3.76 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{330} \, \sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} {\left (i \, \sqrt {15} + 15\right )} - 480 \, x - 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) - \frac {1}{330} \, \sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} \sqrt {2 i \, \sqrt {15} - 14} {\left (-i \, \sqrt {15} - 15\right )} - 480 \, x - 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) - \frac {1}{330} \, \sqrt {165} \sqrt {-2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} {\left (i \, \sqrt {15} - 15\right )} \sqrt {-2 i \, \sqrt {15} - 14} - 480 \, x + 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) + \frac {1}{330} \, \sqrt {165} \sqrt {-2 i \, \sqrt {15} - 14} \log \left (\sqrt {165} {\left (-i \, \sqrt {15} + 15\right )} \sqrt {-2 i \, \sqrt {15} - 14} - 480 \, x + 240 i \, \sqrt {15} + 240 \, \sqrt {4 \, x^{2} + 4 \, x + 5} - 240\right ) \]

[In]

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

[Out]

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
*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(16
5)*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) + 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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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 = 0.33 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.62 \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} + \sqrt {11}}\right ) - \frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} - \sqrt {11}}\right ) - \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} + \sqrt {11}\right )}^{2}\right ) + \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} - \sqrt {11}\right )}^{2}\right ) \]

[In]

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

[Out]

1/165*sqrt(165)*sqrt(15)*arctan(-(2*x - sqrt(4*x^2 + 4*x + 5) + 1)/(sqrt(15) + sqrt(11))) - 1/165*sqrt(165)*sq
rt(15)*arctan(-(2*x - sqrt(4*x^2 + 4*x + 5) + 1)/(sqrt(15) - sqrt(11))) - 1/330*sqrt(165)*log(90000*(2*x - sqr
t(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. \[ \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx=\int \frac {x}{\sqrt {4\,x^2+4\,x+5}\,\left (x^2+x+4\right )} \,d x \]

[In]

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

[Out]

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