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

Optimal. Leaf size=63 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {4 x^2+4 x+5}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}} \]

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1025, 982, 207, 1024, 204} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {4 x^2+4 x+5}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {11}{15}} (2 x+1)}{\sqrt {4 x^2+4 x+5}}\right )}{\sqrt {165}} \]

Antiderivative was successfully verified.

[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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 982

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 1024

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 1025

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*} \int \frac {x}{\left (4+x+x^2\right ) \sqrt {5+4 x+4 x^2}} \, dx &=\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 \operatorname {Subst}\left (\int \frac {1}{-240+11 x^2} \, dx,x,\frac {4+8 x}{\sqrt {5+4 x+4 x^2}}\right )-\operatorname {Subst}\left (\int \frac {1}{-11-x^2} \, dx,x,\sqrt {5+4 x+4 x^2}\right )\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {5+4 x+4 x^2}}{\sqrt {11}}\right )}{\sqrt {11}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {11}{15}} (1+2 x)}{\sqrt {5+4 x+4 x^2}}\right )}{\sqrt {165}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 114, normalized size = 1.81 \[ \frac {\left (\sqrt {15}-i\right ) \tan ^{-1}\left (\frac {-2 i \sqrt {15} x-i \sqrt {15}+4}{\sqrt {11} \sqrt {4 x^2+4 x+5}}\right )+\left (\sqrt {15}+i\right ) \tan ^{-1}\left (\frac {2 i \sqrt {15} x+i \sqrt {15}+4}{\sqrt {11} \sqrt {4 x^2+4 x+5}}\right )}{2 \sqrt {165}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I + Sqrt[15])*ArcTan[(4 - I*Sqrt[15] - (2*I)*Sqrt[15]*x)/(Sqrt[11]*Sqrt[5 + 4*x + 4*x^2])] + (I + Sqrt[15])
*ArcTan[(4 + I*Sqrt[15] + (2*I)*Sqrt[15]*x)/(Sqrt[11]*Sqrt[5 + 4*x + 4*x^2])])/(2*Sqrt[165])

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IntegrateAlgebraic [C]  time = 0.24, size = 102, normalized size = 1.62 \[ \frac {1}{2} \text {RootSum}\left [\text {$\#$1}^4-4 \text {$\#$1}^3+58 \text {$\#$1}^2-108 \text {$\#$1}+69\& ,\frac {\text {$\#$1}^2 \log \left (-\text {$\#$1}+\sqrt {4 x^2+4 x+5}-2 x\right )-5 \log \left (-\text {$\#$1}+\sqrt {4 x^2+4 x+5}-2 x\right )}{\text {$\#$1}^3-3 \text {$\#$1}^2+29 \text {$\#$1}-27}\& \right ] \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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

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fricas [B]  time = 1.17, size = 307, normalized size = 4.87 \[ \frac {2}{165} \, \sqrt {165} \sqrt {15} \arctan \left (\frac {1}{60} \, \sqrt {2} \sqrt {4 \, x^{2} - \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (2 \, x + 1\right )} + 4 \, x - \sqrt {165} + 16} {\left (\sqrt {165} \sqrt {15} + 15 \, \sqrt {15}\right )} + \frac {1}{60} \, \sqrt {165} \sqrt {15} {\left (2 \, x + 1\right )} - \frac {1}{60} \, \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (\sqrt {165} \sqrt {15} + 15 \, \sqrt {15}\right )} + \frac {1}{4} \, \sqrt {15} {\left (2 \, x + 1\right )}\right ) + \frac {2}{165} \, \sqrt {165} \sqrt {15} \arctan \left (\frac {1}{60} \, \sqrt {2} \sqrt {4 \, x^{2} - \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (2 \, x + 1\right )} + 4 \, x + \sqrt {165} + 16} {\left (\sqrt {165} \sqrt {15} - 15 \, \sqrt {15}\right )} + \frac {1}{60} \, \sqrt {165} \sqrt {15} {\left (2 \, x + 1\right )} - \frac {1}{60} \, \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (\sqrt {165} \sqrt {15} - 15 \, \sqrt {15}\right )} - \frac {1}{4} \, \sqrt {15} {\left (2 \, x + 1\right )}\right ) - \frac {1}{330} \, \sqrt {165} \log \left (460800 \, x^{2} - 115200 \, \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (2 \, x + 1\right )} + 460800 \, x + 115200 \, \sqrt {165} + 1843200\right ) + \frac {1}{330} \, \sqrt {165} \log \left (460800 \, x^{2} - 115200 \, \sqrt {4 \, x^{2} + 4 \, x + 5} {\left (2 \, x + 1\right )} + 460800 \, x - 115200 \, \sqrt {165} + 1843200\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/165*sqrt(165)*sqrt(15)*arctan(1/60*sqrt(2)*sqrt(4*x^2 - sqrt(4*x^2 + 4*x + 5)*(2*x + 1) + 4*x - sqrt(165) +
16)*(sqrt(165)*sqrt(15) + 15*sqrt(15)) + 1/60*sqrt(165)*sqrt(15)*(2*x + 1) - 1/60*sqrt(4*x^2 + 4*x + 5)*(sqrt(
165)*sqrt(15) + 15*sqrt(15)) + 1/4*sqrt(15)*(2*x + 1)) + 2/165*sqrt(165)*sqrt(15)*arctan(1/60*sqrt(2)*sqrt(4*x
^2 - sqrt(4*x^2 + 4*x + 5)*(2*x + 1) + 4*x + sqrt(165) + 16)*(sqrt(165)*sqrt(15) - 15*sqrt(15)) + 1/60*sqrt(16
5)*sqrt(15)*(2*x + 1) - 1/60*sqrt(4*x^2 + 4*x + 5)*(sqrt(165)*sqrt(15) - 15*sqrt(15)) - 1/4*sqrt(15)*(2*x + 1)
) - 1/330*sqrt(165)*log(460800*x^2 - 115200*sqrt(4*x^2 + 4*x + 5)*(2*x + 1) + 460800*x + 115200*sqrt(165) + 18
43200) + 1/330*sqrt(165)*log(460800*x^2 - 115200*sqrt(4*x^2 + 4*x + 5)*(2*x + 1) + 460800*x - 115200*sqrt(165)
+ 1843200)

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giac [B]  time = 0.71, size = 165, normalized size = 2.62 \[ \frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} + \sqrt {11}}\right ) - \frac {1}{165} \, \sqrt {165} \sqrt {15} \arctan \left (-\frac {2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1}{\sqrt {15} - \sqrt {11}}\right ) - \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} + \sqrt {11}\right )}^{2}\right ) + \frac {1}{330} \, \sqrt {165} \log \left (90000 \, {\left (2 \, x - \sqrt {4 \, x^{2} + 4 \, x + 5} + 1\right )}^{2} + 90000 \, {\left (\sqrt {15} - \sqrt {11}\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.75, size = 53, normalized size = 0.84

method	result	size
default	\(\frac {\arctan \left (\frac {\sqrt {4 x^{2}+4 x +5}\, \sqrt {11}}{11}\right ) \sqrt {11}}{11}-\frac {\sqrt {165}\, \arctanh \left (\frac {\sqrt {165}\, \left (8 x +4\right )}{60 \sqrt {4 x^{2}+4 x +5}}\right )}{165}\)	\(53\)
trager	\(\RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) \ln \left (-\frac {3524400 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{5} x +111270 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3} x -41385 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}+3420 \sqrt {4 x^{2}+4 x +5}\, \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+754 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x -899 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )+52 \sqrt {4 x^{2}+4 x +5}}{165 x \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+3 x -4}\right )+\frac {165 \ln \left (-\frac {8276400 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{5} x +385770 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3} x +97185 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}-9405 \sqrt {4 x^{2}+4 x +5}\, \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+3784 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +1364 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )-256 \sqrt {4 x^{2}+4 x +5}}{165 x \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+4 x +4}\right ) \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}}{4}+\frac {7 \ln \left (-\frac {8276400 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{5} x +385770 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3} x +97185 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{3}-9405 \sqrt {4 x^{2}+4 x +5}\, \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+3784 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right ) x +1364 \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )-256 \sqrt {4 x^{2}+4 x +5}}{165 x \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )^{2}+4 x +4}\right ) \RootOf \left (27225 \textit {\_Z}^{4}+1155 \textit {\_Z}^{2}+16\right )}{4}\)	\(516\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {4 \, x^{2} + 4 \, x + 5} {\left (x^{2} + x + 4\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{\sqrt {4\,x^2+4\,x+5}\,\left (x^2+x+4\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (x^{2} + x + 4\right ) \sqrt {4 x^{2} + 4 x + 5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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