3.24 \(\int \frac{1-x+3 x^2}{\sqrt{1-x+x^2} (1+x+x^2)^2} \, dx\)

Optimal. Leaf size=86 \[ \frac{\sqrt{x^2-x+1} (x+1)}{x^2+x+1}+\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} (x+1)}{\sqrt{x^2-x+1}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{2}{3}} (1-x)}{\sqrt{x^2-x+1}}\right )}{\sqrt{6}} \]

[Out]

((1 + x)*Sqrt[1 - x + x^2])/(1 + x + x^2) + Sqrt[2]*ArcTan[(Sqrt[2]*(1 + x))/Sqrt[1 - x + x^2]] - ArcTanh[(Sqr
t[2/3]*(1 - x))/Sqrt[1 - x + x^2]]/Sqrt[6]

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Rubi [A]  time = 0.0839424, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {1060, 1035, 1029, 206, 204} \[ \frac{\sqrt{x^2-x+1} (x+1)}{x^2+x+1}+\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} (x+1)}{\sqrt{x^2-x+1}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{2}{3}} (1-x)}{\sqrt{x^2-x+1}}\right )}{\sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x + 3*x^2)/(Sqrt[1 - x + x^2]*(1 + x + x^2)^2),x]

[Out]

((1 + x)*Sqrt[1 - x + x^2])/(1 + x + x^2) + Sqrt[2]*ArcTan[(Sqrt[2]*(1 + x))/Sqrt[1 - x + x^2]] - ArcTanh[(Sqr
t[2/3]*(1 - x))/Sqrt[1 - x + x^2]]/Sqrt[6]

Rule 1060

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_
)^2)^(q_), x_Symbol] :> Simp[((a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*((A*c - a*C)*(2*a*c*e - b*(c
*d + a*f)) + (A*b - a*B)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) + c*(A*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) - B*(b
*c*d - 2*a*c*e + a*b*f) + C*(b^2*d - a*b*e - 2*a*(c*d - a*f)))*x))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)
*(c*e - b*f))*(p + 1)), x] + Dist[1/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), Int[(a
+ b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(b*B - 2*A*c - 2*a*C)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f)
)*(p + 1) + (b^2*(C*d + A*f) - b*(B*c*d + A*c*e + a*C*e + a*B*f) + 2*(A*c*(c*d - a*f) - a*(c*C*d - B*c*e - a*C
*f)))*(a*f*(p + 1) - c*d*(p + 2)) - e*((A*c - a*C)*(2*a*c*e - b*(c*d + a*f)) + (A*b - a*B)*(2*c^2*d + b^2*f -
c*(b*e + 2*a*f)))*(p + q + 2) - (2*f*((A*c - a*C)*(2*a*c*e - b*(c*d + a*f)) + (A*b - a*B)*(2*c^2*d + b^2*f - c
*(b*e + 2*a*f)))*(p + q + 2) - (b^2*(C*d + A*f) - b*(B*c*d + A*c*e + a*C*e + a*B*f) + 2*(A*c*(c*d - a*f) - a*(
c*C*d - B*c*e - a*C*f)))*(b*f*(p + 1) - c*e*(2*p + q + 4)))*x - c*f*(b^2*(C*d + A*f) - b*(B*c*d + A*c*e + a*C*
e + a*B*f) + 2*(A*c*(c*d - a*f) - a*(c*C*d - B*c*e - a*C*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b,
c, d, e, f, A, B, C, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && NeQ[(c*d - a*f)^2 -
(b*d - a*e)*(c*e - b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

Rule 1035

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> With[{q = Rt[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[Simp[h*(b*d - a*e) - g*(c*
d - a*f - q) - (g*(c*e - b*f) - h*(c*d - a*f + q))*x, x]/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - D
ist[1/(2*q), Int[Simp[h*(b*d - a*e) - g*(c*d - a*f + q) - (g*(c*e - b*f) - h*(c*d - a*f - q))*x, 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] && NeQ[b*d - a*e, 0] && NegQ[b^2 - 4*a*c]

Rule 1029

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> Dist[-2*g*(g*b - 2*a*h), Subst[Int[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - (b*d - a*e)*x^2, x], x], x, S
imp[g*b - 2*a*h - (b*h - 2*g*c)*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] && NeQ[b*d - a*e, 0] && EqQ[h^2*(b*d - a*e) - 2*g*h*(c*d - a*f) + g^2*(
c*e - b*f), 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 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])

Rubi steps

\begin{align*} \int \frac{1-x+3 x^2}{\sqrt{1-x+x^2} \left (1+x+x^2\right )^2} \, dx &=\frac{(1+x) \sqrt{1-x+x^2}}{1+x+x^2}+\frac{1}{12} \int \frac{18-6 x}{\sqrt{1-x+x^2} \left (1+x+x^2\right )} \, dx\\ &=\frac{(1+x) \sqrt{1-x+x^2}}{1+x+x^2}+\frac{1}{48} \int \frac{24+24 x}{\sqrt{1-x+x^2} \left (1+x+x^2\right )} \, dx-\frac{1}{48} \int \frac{-48+48 x}{\sqrt{1-x+x^2} \left (1+x+x^2\right )} \, dx\\ &=\frac{(1+x) \sqrt{1-x+x^2}}{1+x+x^2}+24 \operatorname{Subst}\left (\int \frac{1}{1728-2 x^2} \, dx,x,\frac{-24+24 x}{\sqrt{1-x+x^2}}\right )+288 \operatorname{Subst}\left (\int \frac{1}{-20736-2 x^2} \, dx,x,\frac{-144-144 x}{\sqrt{1-x+x^2}}\right )\\ &=\frac{(1+x) \sqrt{1-x+x^2}}{1+x+x^2}+\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} (1+x)}{\sqrt{1-x+x^2}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{2}{3}} (1-x)}{\sqrt{1-x+x^2}}\right )}{\sqrt{6}}\\ \end{align*}

Mathematica [C]  time = 2.38024, size = 961, normalized size = 11.17 \[ \frac{\sqrt{x^2-x+1} (x+1)}{x^2+x+1}+\frac{\left (7-i \sqrt{3}\right ) \tan ^{-1}\left (\frac{3 \left (\left (-21-4 i \sqrt{3}\right ) x^4+14 \left (7-2 i \sqrt{3}\right ) x^3+\left (-103-36 i \sqrt{3}\right ) x^2+\left (94+32 i \sqrt{3}\right ) x-64 i \sqrt{3}-17\right )}{\left (84 i-113 \sqrt{3}\right ) x^4+2 \left (52 \sqrt{3-3 i \sqrt{3}} \sqrt{x^2-x+1}+21 \sqrt{3}+138 i\right ) x^3+\left (52 \sqrt{3-3 i \sqrt{3}} \sqrt{x^2-x+1}-59 \sqrt{3}-180 i\right ) x^2+2 \left (26 \sqrt{3-3 i \sqrt{3}} \sqrt{x^2-x+1}-69 \sqrt{3}+132 i\right ) x-52 \sqrt{3-3 i \sqrt{3}} \sqrt{x^2-x+1}+67 \sqrt{3}+96 i}\right )}{4 \sqrt{3-3 i \sqrt{3}}}-\frac{i \left (-7 i+\sqrt{3}\right ) \tan ^{-1}\left (\frac{3 \left (\left (-21+4 i \sqrt{3}\right ) x^4+14 \left (7+2 i \sqrt{3}\right ) x^3+\left (-103+36 i \sqrt{3}\right ) x^2+\left (94-32 i \sqrt{3}\right ) x+64 i \sqrt{3}-17\right )}{\left (84 i+113 \sqrt{3}\right ) x^4-2 \left (52 \sqrt{3+3 i \sqrt{3}} \sqrt{x^2-x+1}+21 \sqrt{3}-138 i\right ) x^3+\left (-52 \sqrt{3+3 i \sqrt{3}} \sqrt{x^2-x+1}+59 \sqrt{3}-180 i\right ) x^2+\left (-52 \sqrt{3+3 i \sqrt{3}} \sqrt{x^2-x+1}+138 \sqrt{3}+264 i\right ) x+52 \sqrt{3+3 i \sqrt{3}} \sqrt{x^2-x+1}-67 \sqrt{3}+96 i}\right )}{4 \sqrt{3+3 i \sqrt{3}}}-\frac{\left (7 i+\sqrt{3}\right ) \log \left (16 \left (x^2+x+1\right )^2\right )}{8 \sqrt{3-3 i \sqrt{3}}}-\frac{\left (-7 i+\sqrt{3}\right ) \log \left (16 \left (x^2+x+1\right )^2\right )}{8 \sqrt{3+3 i \sqrt{3}}}+\frac{\left (7 i+\sqrt{3}\right ) \log \left (\left (x^2+x+1\right ) \left (\left (11 i+4 \sqrt{3}\right ) x^2-\left (8 i \sqrt{1-i \sqrt{3}} \sqrt{x^2-x+1}+4 \sqrt{3}+17 i\right ) x+10 i \sqrt{1-i \sqrt{3}} \sqrt{x^2-x+1}+4 \sqrt{3}+11 i\right )\right )}{8 \sqrt{3-3 i \sqrt{3}}}+\frac{\left (-7 i+\sqrt{3}\right ) \log \left (\left (x^2+x+1\right ) \left (\left (-11 i+4 \sqrt{3}\right ) x^2+\left (8 i \sqrt{1+i \sqrt{3}} \sqrt{x^2-x+1}-4 \sqrt{3}+17 i\right ) x-10 i \sqrt{1+i \sqrt{3}} \sqrt{x^2-x+1}+4 \sqrt{3}-11 i\right )\right )}{8 \sqrt{3+3 i \sqrt{3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x + 3*x^2)/(Sqrt[1 - x + x^2]*(1 + x + x^2)^2),x]

[Out]

((1 + x)*Sqrt[1 - x + x^2])/(1 + x + x^2) + ((7 - I*Sqrt[3])*ArcTan[(3*(-17 - (64*I)*Sqrt[3] + (94 + (32*I)*Sq
rt[3])*x + (-103 - (36*I)*Sqrt[3])*x^2 + 14*(7 - (2*I)*Sqrt[3])*x^3 + (-21 - (4*I)*Sqrt[3])*x^4))/(96*I + 67*S
qrt[3] + (84*I - 113*Sqrt[3])*x^4 - 52*Sqrt[3 - (3*I)*Sqrt[3]]*Sqrt[1 - x + x^2] + 2*x*(132*I - 69*Sqrt[3] + 2
6*Sqrt[3 - (3*I)*Sqrt[3]]*Sqrt[1 - x + x^2]) + x^2*(-180*I - 59*Sqrt[3] + 52*Sqrt[3 - (3*I)*Sqrt[3]]*Sqrt[1 -
x + x^2]) + 2*x^3*(138*I + 21*Sqrt[3] + 52*Sqrt[3 - (3*I)*Sqrt[3]]*Sqrt[1 - x + x^2]))])/(4*Sqrt[3 - (3*I)*Sqr
t[3]]) - ((I/4)*(-7*I + Sqrt[3])*ArcTan[(3*(-17 + (64*I)*Sqrt[3] + (94 - (32*I)*Sqrt[3])*x + (-103 + (36*I)*Sq
rt[3])*x^2 + 14*(7 + (2*I)*Sqrt[3])*x^3 + (-21 + (4*I)*Sqrt[3])*x^4))/(96*I - 67*Sqrt[3] + (84*I + 113*Sqrt[3]
)*x^4 + 52*Sqrt[3 + (3*I)*Sqrt[3]]*Sqrt[1 - x + x^2] + x^2*(-180*I + 59*Sqrt[3] - 52*Sqrt[3 + (3*I)*Sqrt[3]]*S
qrt[1 - x + x^2]) + x*(264*I + 138*Sqrt[3] - 52*Sqrt[3 + (3*I)*Sqrt[3]]*Sqrt[1 - x + x^2]) - 2*x^3*(-138*I + 2
1*Sqrt[3] + 52*Sqrt[3 + (3*I)*Sqrt[3]]*Sqrt[1 - x + x^2]))])/Sqrt[3 + (3*I)*Sqrt[3]] - ((-7*I + Sqrt[3])*Log[1
6*(1 + x + x^2)^2])/(8*Sqrt[3 + (3*I)*Sqrt[3]]) - ((7*I + Sqrt[3])*Log[16*(1 + x + x^2)^2])/(8*Sqrt[3 - (3*I)*
Sqrt[3]]) + ((7*I + Sqrt[3])*Log[(1 + x + x^2)*(11*I + 4*Sqrt[3] + (11*I + 4*Sqrt[3])*x^2 + (10*I)*Sqrt[1 - I*
Sqrt[3]]*Sqrt[1 - x + x^2] - x*(17*I + 4*Sqrt[3] + (8*I)*Sqrt[1 - I*Sqrt[3]]*Sqrt[1 - x + x^2]))])/(8*Sqrt[3 -
(3*I)*Sqrt[3]]) + ((-7*I + Sqrt[3])*Log[(1 + x + x^2)*(-11*I + 4*Sqrt[3] + (-11*I + 4*Sqrt[3])*x^2 - (10*I)*S
qrt[1 + I*Sqrt[3]]*Sqrt[1 - x + x^2] + x*(17*I - 4*Sqrt[3] + (8*I)*Sqrt[1 + I*Sqrt[3]]*Sqrt[1 - x + x^2]))])/(
8*Sqrt[3 + (3*I)*Sqrt[3]])

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Maple [B]  time = 0.034, size = 454, normalized size = 5.3 \begin{align*}{\frac{1}{6} \left ( 6\,{\frac{\sqrt{6} \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}{\it Artanh} \left ( 1/4\,\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}\sqrt{6} \right ) \sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}}-9\,{\frac{\sqrt{2} \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}\arctan \left ( 2\,{\frac{ \left ( 1+x \right ) \sqrt{2}}{1-x}{\frac{1}{\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}}}} \right ) \sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}}+2\,\sqrt{6}{\it Artanh} \left ( 1/4\,\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}\sqrt{6} \right ) \sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}-3\,\sqrt{2}\arctan \left ( 2\,{\frac{ \left ( 1+x \right ) \sqrt{2}}{1-x}{\frac{1}{\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}}}} \right ) \sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}+12\,{\frac{ \left ( 1+x \right ) ^{3}}{ \left ( 1-x \right ) ^{3}}}+36\,{\frac{1+x}{1-x}} \right ){\frac{1}{\sqrt{{ \left ({\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3 \right ) \left ( 1+{\frac{1+x}{1-x}} \right ) ^{-2}}}}} \left ( 1+{\frac{1+x}{1-x}} \right ) ^{-1} \left ( 3\,{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+1 \right ) ^{-1}}-{\frac{1}{2}\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3} \left ( \sqrt{6}{\it Artanh} \left ({\frac{\sqrt{6}}{4}\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}} \right ) -3\,\sqrt{2}\arctan \left ( 2\,{\frac{ \left ( 1+x \right ) \sqrt{2}}{1-x}{\frac{1}{\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3}}}} \right ) \right ){\frac{1}{\sqrt{{ \left ({\frac{ \left ( 1+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+3 \right ) \left ( 1+{\frac{1+x}{1-x}} \right ) ^{-2}}}}} \left ( 1+{\frac{1+x}{1-x}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*6^(1/2)*arctanh(1/4*((1+x)^2/(1-x)^2+3)^(1/2)*6^(1/2))*((1+x)^2/(1-x)^2+3)^(1/2)*(1+x)^2/(1-x)^2-9*arct
an(2*(1+x)/(1-x)/((1+x)^2/(1-x)^2+3)^(1/2)*2^(1/2))*2^(1/2)*((1+x)^2/(1-x)^2+3)^(1/2)*(1+x)^2/(1-x)^2+2*6^(1/2
)*arctanh(1/4*((1+x)^2/(1-x)^2+3)^(1/2)*6^(1/2))*((1+x)^2/(1-x)^2+3)^(1/2)-3*2^(1/2)*arctan(2*(1+x)/(1-x)/((1+
x)^2/(1-x)^2+3)^(1/2)*2^(1/2))*((1+x)^2/(1-x)^2+3)^(1/2)+12*(1+x)^3/(1-x)^3+36*(1+x)/(1-x))/(((1+x)^2/(1-x)^2+
3)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(1-x))/(3*(1+x)^2/(1-x)^2+1)-1/2*((1+x)^2/(1-x)^2+3)^(1/2)*(6^(1/2)*arcta
nh(1/4*((1+x)^2/(1-x)^2+3)^(1/2)*6^(1/2))-3*2^(1/2)*arctan(2*(1+x)/(1-x)/((1+x)^2/(1-x)^2+3)^(1/2)*2^(1/2)))/(
1+(1+x)/(1-x))/(((1+x)^2/(1-x)^2+3)/(1+(1+x)/(1-x))^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 \, x^{2} - x + 1}{{\left (x^{2} + x + 1\right )}^{2} \sqrt{x^{2} - x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((3*x^2 - x + 1)/((x^2 + x + 1)^2*sqrt(x^2 - x + 1)), x)

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Fricas [B]  time = 2.26406, size = 1095, normalized size = 12.73 \begin{align*} -\frac{8 \, \sqrt{6} \sqrt{3}{\left (x^{2} + x + 1\right )} \arctan \left (\frac{2}{3} \, \sqrt{6} \sqrt{3}{\left (x - 1\right )} + \frac{2}{3} \, \sqrt{2 \, x^{2} - \sqrt{x^{2} - x + 1}{\left (2 \, x - \sqrt{6} + 1\right )} - \sqrt{6}{\left (x + 1\right )} + 4}{\left (\sqrt{6} \sqrt{3} + 3 \, \sqrt{3}\right )} - \frac{2}{3} \, \sqrt{x^{2} - x + 1}{\left (\sqrt{6} \sqrt{3} + 3 \, \sqrt{3}\right )} + \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 8 \, \sqrt{6} \sqrt{3}{\left (x^{2} + x + 1\right )} \arctan \left (\frac{2}{3} \, \sqrt{6} \sqrt{3}{\left (x - 1\right )} + \frac{2}{3} \, \sqrt{2 \, x^{2} - \sqrt{x^{2} - x + 1}{\left (2 \, x + \sqrt{6} + 1\right )} + \sqrt{6}{\left (x + 1\right )} + 4}{\left (\sqrt{6} \sqrt{3} - 3 \, \sqrt{3}\right )} - \frac{2}{3} \, \sqrt{x^{2} - x + 1}{\left (\sqrt{6} \sqrt{3} - 3 \, \sqrt{3}\right )} - \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \sqrt{6}{\left (x^{2} + x + 1\right )} \log \left (12168 \, x^{2} - 6084 \, \sqrt{x^{2} - x + 1}{\left (2 \, x + \sqrt{6} + 1\right )} + 6084 \, \sqrt{6}{\left (x + 1\right )} + 24336\right ) + \sqrt{6}{\left (x^{2} + x + 1\right )} \log \left (12168 \, x^{2} - 6084 \, \sqrt{x^{2} - x + 1}{\left (2 \, x - \sqrt{6} + 1\right )} - 6084 \, \sqrt{6}{\left (x + 1\right )} + 24336\right ) - 12 \, x^{2} - 12 \, \sqrt{x^{2} - x + 1}{\left (x + 1\right )} - 12 \, x - 12}{12 \,{\left (x^{2} + x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(8*sqrt(6)*sqrt(3)*(x^2 + x + 1)*arctan(2/3*sqrt(6)*sqrt(3)*(x - 1) + 2/3*sqrt(2*x^2 - sqrt(x^2 - x + 1)
*(2*x - sqrt(6) + 1) - sqrt(6)*(x + 1) + 4)*(sqrt(6)*sqrt(3) + 3*sqrt(3)) - 2/3*sqrt(x^2 - x + 1)*(sqrt(6)*sqr
t(3) + 3*sqrt(3)) + sqrt(3)*(2*x - 1)) + 8*sqrt(6)*sqrt(3)*(x^2 + x + 1)*arctan(2/3*sqrt(6)*sqrt(3)*(x - 1) +
2/3*sqrt(2*x^2 - sqrt(x^2 - x + 1)*(2*x + sqrt(6) + 1) + sqrt(6)*(x + 1) + 4)*(sqrt(6)*sqrt(3) - 3*sqrt(3)) -
2/3*sqrt(x^2 - x + 1)*(sqrt(6)*sqrt(3) - 3*sqrt(3)) - sqrt(3)*(2*x - 1)) - sqrt(6)*(x^2 + x + 1)*log(12168*x^2
- 6084*sqrt(x^2 - x + 1)*(2*x + sqrt(6) + 1) + 6084*sqrt(6)*(x + 1) + 24336) + sqrt(6)*(x^2 + x + 1)*log(1216
8*x^2 - 6084*sqrt(x^2 - x + 1)*(2*x - sqrt(6) + 1) - 6084*sqrt(6)*(x + 1) + 24336) - 12*x^2 - 12*sqrt(x^2 - x
+ 1)*(x + 1) - 12*x - 12)/(x^2 + x + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 x^{2} - x + 1}{\sqrt{x^{2} - x + 1} \left (x^{2} + x + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((3*x**2 - x + 1)/(sqrt(x**2 - x + 1)*(x**2 + x + 1)**2), x)

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Giac [C]  time = 1.13719, size = 424, normalized size = 4.93 \begin{align*} -\frac{1}{12} \, \sqrt{6}{\left (2 i \, \sqrt{3} + 1\right )} \log \left (2 i \, \sqrt{6} \sqrt{3} - 12 \, x + 6 \, \sqrt{6} - 6 i \, \sqrt{3} + 12 \, \sqrt{x^{2} - x + 1} - 6\right ) + \frac{1}{12} \, \sqrt{6}{\left (-2 i \, \sqrt{3} + 1\right )} \log \left (2 i \, \sqrt{6} \sqrt{3} - 12 \, x - 6 \, \sqrt{6} + 6 i \, \sqrt{3} + 12 \, \sqrt{x^{2} - x + 1} - 6\right ) - \frac{1}{12} \, \sqrt{6}{\left (-2 i \, \sqrt{3} + 1\right )} \log \left (-2 i \, \sqrt{6} \sqrt{3} - 12 \, x + 6 \, \sqrt{6} + 6 i \, \sqrt{3} + 12 \, \sqrt{x^{2} - x + 1} - 6\right ) + \frac{1}{12} \, \sqrt{6}{\left (2 i \, \sqrt{3} + 1\right )} \log \left (-2 i \, \sqrt{6} \sqrt{3} - 12 \, x - 6 \, \sqrt{6} - 6 i \, \sqrt{3} + 12 \, \sqrt{x^{2} - x + 1} - 6\right ) + \frac{{\left (x - \sqrt{x^{2} - x + 1}\right )}^{3} + 4 \,{\left (x - \sqrt{x^{2} - x + 1}\right )}^{2} - 10 \, x + 10 \, \sqrt{x^{2} - x + 1} + 5}{{\left (x - \sqrt{x^{2} - x + 1}\right )}^{4} + 2 \,{\left (x - \sqrt{x^{2} - x + 1}\right )}^{3} +{\left (x - \sqrt{x^{2} - x + 1}\right )}^{2} - 6 \, x + 6 \, \sqrt{x^{2} - x + 1} + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(6)*(2*I*sqrt(3) + 1)*log(2*I*sqrt(6)*sqrt(3) - 12*x + 6*sqrt(6) - 6*I*sqrt(3) + 12*sqrt(x^2 - x + 1
) - 6) + 1/12*sqrt(6)*(-2*I*sqrt(3) + 1)*log(2*I*sqrt(6)*sqrt(3) - 12*x - 6*sqrt(6) + 6*I*sqrt(3) + 12*sqrt(x^
2 - x + 1) - 6) - 1/12*sqrt(6)*(-2*I*sqrt(3) + 1)*log(-2*I*sqrt(6)*sqrt(3) - 12*x + 6*sqrt(6) + 6*I*sqrt(3) +
12*sqrt(x^2 - x + 1) - 6) + 1/12*sqrt(6)*(2*I*sqrt(3) + 1)*log(-2*I*sqrt(6)*sqrt(3) - 12*x - 6*sqrt(6) - 6*I*s
qrt(3) + 12*sqrt(x^2 - x + 1) - 6) + ((x - sqrt(x^2 - x + 1))^3 + 4*(x - sqrt(x^2 - x + 1))^2 - 10*x + 10*sqrt
(x^2 - x + 1) + 5)/((x - sqrt(x^2 - x + 1))^4 + 2*(x - sqrt(x^2 - x + 1))^3 + (x - sqrt(x^2 - x + 1))^2 - 6*x
+ 6*sqrt(x^2 - x + 1) + 3)