3.34 \(\int \frac {x}{(3+x+x^2) \sqrt {5+x+x^2}} \, dx\)

Optimal. Leaf size=56 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {2}{11}} (2 x+1)}{\sqrt {x^2+x+5}}\right )}{\sqrt {22}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^2+x+5}}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

-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)

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Rubi [A]  time = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1025, 982, 204, 1024, 206} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {2}{11}} (2 x+1)}{\sqrt {x^2+x+5}}\right )}{\sqrt {22}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^2+x+5}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[x/((3 + x + x^2)*Sqrt[5 + x + x^2]),x]

[Out]

-(ArcTan[(Sqrt[2/11]*(1 + 2*x))/Sqrt[5 + x + x^2]]/Sqrt[22]) - ArcTanh[Sqrt[5 + x + x^2]/Sqrt[2]]/Sqrt[2]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx &=-\left (\frac {1}{2} \int \frac {1}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx\right )+\frac {1}{2} \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{-11-2 x^2} \, dx,x,\frac {1+2 x}{\sqrt {5+x+x^2}}\right )-\operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {5+x+x^2}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {2}{11}} (1+2 x)}{\sqrt {5+x+x^2}}\right )}{\sqrt {22}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {5+x+x^2}}{\sqrt {2}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 114, normalized size = 2.04 \[ \frac {-\left (\left (\sqrt {11}-i\right ) \tanh ^{-1}\left (\frac {-2 i \sqrt {11} x-i \sqrt {11}+19}{4 \sqrt {2} \sqrt {x^2+x+5}}\right )\right )-\left (\sqrt {11}+i\right ) \tanh ^{-1}\left (\frac {2 i \sqrt {11} x+i \sqrt {11}+19}{4 \sqrt {2} \sqrt {x^2+x+5}}\right )}{2 \sqrt {22}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((3 + x + x^2)*Sqrt[5 + x + x^2]),x]

[Out]

(-((-I + Sqrt[11])*ArcTanh[(19 - I*Sqrt[11] - (2*I)*Sqrt[11]*x)/(4*Sqrt[2]*Sqrt[5 + x + x^2])]) - (I + Sqrt[11
])*ArcTanh[(19 + I*Sqrt[11] + (2*I)*Sqrt[11]*x)/(4*Sqrt[2]*Sqrt[5 + x + x^2])])/(2*Sqrt[22])

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fricas [B]  time = 1.10, size = 307, normalized size = 5.48 \[ -\frac {1}{33} \, \sqrt {11} \sqrt {6} \sqrt {3} \arctan \left (\frac {2}{33} \, \sqrt {11} \sqrt {3} \sqrt {\sqrt {6} \sqrt {3} {\left (2 \, x + 1\right )} + 6 \, x^{2} - \sqrt {x^{2} + x + 5} {\left (2 \, \sqrt {6} \sqrt {3} + 6 \, x + 3\right )} + 6 \, x + 30} + \frac {1}{33} \, \sqrt {11} {\left (2 \, \sqrt {6} \sqrt {3} + 6 \, x + 3\right )} - \frac {2}{11} \, \sqrt {11} \sqrt {x^{2} + x + 5}\right ) + \frac {1}{33} \, \sqrt {11} \sqrt {6} \sqrt {3} \arctan \left (-\frac {1}{33} \, \sqrt {11} {\left (2 \, \sqrt {6} \sqrt {3} - 6 \, x - 3\right )} + \frac {1}{33} \, \sqrt {11} \sqrt {-12 \, \sqrt {6} \sqrt {3} {\left (2 \, x + 1\right )} + 72 \, x^{2} + 12 \, \sqrt {x^{2} + x + 5} {\left (2 \, \sqrt {6} \sqrt {3} - 6 \, x - 3\right )} + 72 \, x + 360} - \frac {2}{11} \, \sqrt {11} \sqrt {x^{2} + x + 5}\right ) + \frac {1}{12} \, \sqrt {6} \sqrt {3} \log \left (12 \, \sqrt {6} \sqrt {3} {\left (2 \, x + 1\right )} + 72 \, x^{2} - 12 \, \sqrt {x^{2} + x + 5} {\left (2 \, \sqrt {6} \sqrt {3} + 6 \, x + 3\right )} + 72 \, x + 360\right ) - \frac {1}{12} \, \sqrt {6} \sqrt {3} \log \left (-12 \, \sqrt {6} \sqrt {3} {\left (2 \, x + 1\right )} + 72 \, x^{2} + 12 \, \sqrt {x^{2} + x + 5} {\left (2 \, \sqrt {6} \sqrt {3} - 6 \, x - 3\right )} + 72 \, x + 360\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+x+3)/(x^2+x+5)^(1/2),x, algorithm="fricas")

[Out]

-1/33*sqrt(11)*sqrt(6)*sqrt(3)*arctan(2/33*sqrt(11)*sqrt(3)*sqrt(sqrt(6)*sqrt(3)*(2*x + 1) + 6*x^2 - sqrt(x^2
+ x + 5)*(2*sqrt(6)*sqrt(3) + 6*x + 3) + 6*x + 30) + 1/33*sqrt(11)*(2*sqrt(6)*sqrt(3) + 6*x + 3) - 2/11*sqrt(1
1)*sqrt(x^2 + x + 5)) + 1/33*sqrt(11)*sqrt(6)*sqrt(3)*arctan(-1/33*sqrt(11)*(2*sqrt(6)*sqrt(3) - 6*x - 3) + 1/
33*sqrt(11)*sqrt(-12*sqrt(6)*sqrt(3)*(2*x + 1) + 72*x^2 + 12*sqrt(x^2 + x + 5)*(2*sqrt(6)*sqrt(3) - 6*x - 3) +
72*x + 360) - 2/11*sqrt(11)*sqrt(x^2 + x + 5)) + 1/12*sqrt(6)*sqrt(3)*log(12*sqrt(6)*sqrt(3)*(2*x + 1) + 72*x
^2 - 12*sqrt(x^2 + x + 5)*(2*sqrt(6)*sqrt(3) + 6*x + 3) + 72*x + 360) - 1/12*sqrt(6)*sqrt(3)*log(-12*sqrt(6)*s
qrt(3)*(2*x + 1) + 72*x^2 + 12*sqrt(x^2 + x + 5)*(2*sqrt(6)*sqrt(3) - 6*x - 3) + 72*x + 360)

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giac [B]  time = 0.32, size = 133, normalized size = 2.38 \[ \frac {1}{22} \, \sqrt {11} \sqrt {2} \arctan \left (-\frac {1}{11} \, \sqrt {11} {\left (2 \, x + 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}\right ) - \frac {1}{22} \, \sqrt {11} \sqrt {2} \arctan \left (-\frac {1}{11} \, \sqrt {11} {\left (2 \, x - 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (324 \, {\left (2 \, x + 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}^{2} + 3564\right ) - \frac {1}{4} \, \sqrt {2} \log \left (324 \, {\left (2 \, x - 2 \, \sqrt {2} - 2 \, \sqrt {x^{2} + x + 5} + 1\right )}^{2} + 3564\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+x+3)/(x^2+x+5)^(1/2),x, algorithm="giac")

[Out]

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
)

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maple [A]  time = 0.01, size = 45, normalized size = 0.80 \[ -\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {x^{2}+x +5}\, \sqrt {2}}{2}\right )}{2}-\frac {\sqrt {22}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {22}}{11 \sqrt {x^{2}+x +5}}\right )}{22} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2+x+3)/(x^2+x+5)^(1/2),x)

[Out]

-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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {x^{2} + x + 5} {\left (x^{2} + x + 3\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+x+3)/(x^2+x+5)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/(sqrt(x^2 + x + 5)*(x^2 + x + 3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{\left (x^2+x+3\right )\,\sqrt {x^2+x+5}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((x + x^2 + 3)*(x + x^2 + 5)^(1/2)),x)

[Out]

int(x/((x + x^2 + 3)*(x + x^2 + 5)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (x^{2} + x + 3\right ) \sqrt {x^{2} + x + 5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**2+x+3)/(x**2+x+5)**(1/2),x)

[Out]

Integral(x/((x**2 + x + 3)*sqrt(x**2 + x + 5)), x)

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