\(\int \frac {1+2 x}{(1+x^2) \sqrt {-1+x+x^2}} \, dx\) [11]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=-\sqrt {\frac {1}{2} \left (2+\sqrt {5}\right )} \arctan \left (\frac {5+2 \sqrt {5}-\sqrt {5} x}{\sqrt {10 \left (2+\sqrt {5}\right )} \sqrt {-1+x+x^2}}\right )+\sqrt {\frac {1}{2} \left (-2+\sqrt {5}\right )} \text {arctanh}\left (\frac {5-2 \sqrt {5}+\sqrt {5} x}{\sqrt {10 \left (-2+\sqrt {5}\right )} \sqrt {-1+x+x^2}}\right ) \]

[Out]

1/2*arctanh((5-2*5^(1/2)+x*5^(1/2))/(x^2+x-1)^(1/2)/(-20+10*5^(1/2))^(1/2))*(-4+2*5^(1/2))^(1/2)-1/2*arctan((5
+2*5^(1/2)-x*5^(1/2))/(x^2+x-1)^(1/2)/(20+10*5^(1/2))^(1/2))*(4+2*5^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1050, 1044, 213, 209} \[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\sqrt {\frac {1}{2} \left (\sqrt {5}-2\right )} \text {arctanh}\left (\frac {\sqrt {5} x-2 \sqrt {5}+5}{\sqrt {10 \left (\sqrt {5}-2\right )} \sqrt {x^2+x-1}}\right )-\sqrt {\frac {1}{2} \left (2+\sqrt {5}\right )} \arctan \left (\frac {-\sqrt {5} x+2 \sqrt {5}+5}{\sqrt {10 \left (2+\sqrt {5}\right )} \sqrt {x^2+x-1}}\right ) \]

[In]

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

[Out]

-(Sqrt[(2 + Sqrt[5])/2]*ArcTan[(5 + 2*Sqrt[5] - Sqrt[5]*x)/(Sqrt[10*(2 + Sqrt[5])]*Sqrt[-1 + x + x^2])]) + Sqr
t[(-2 + Sqrt[5])/2]*ArcTanh[(5 - 2*Sqrt[5] + Sqrt[5]*x)/(Sqrt[10*(-2 + Sqrt[5])]*Sqrt[-1 + x + x^2])]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 1044

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
a*g*h, Subst[Int[1/Simp[2*a^2*g*h*c + a*e*x^2, x], x], x, Simp[a*h - g*c*x, x]/Sqrt[d + e*x + f*x^2]], x] /; F
reeQ[{a, c, d, e, f, g, h}, x] && EqQ[a*h^2*e + 2*g*h*(c*d - a*f) - g^2*c*e, 0]

Rule 1050

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \frac {-\sqrt {5}+\left (-5-2 \sqrt {5}\right ) x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx}{2 \sqrt {5}}+\frac {\int \frac {\sqrt {5}+\left (-5+2 \sqrt {5}\right ) x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx}{2 \sqrt {5}} \\ & = -\left (\left (-5+2 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{10 \left (2-\sqrt {5}\right )+x^2} \, dx,x,\frac {-5+2 \sqrt {5}-\sqrt {5} x}{\sqrt {-1+x+x^2}}\right )\right )+\left (5+2 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{10 \left (2+\sqrt {5}\right )+x^2} \, dx,x,\frac {-5-2 \sqrt {5}+\sqrt {5} x}{\sqrt {-1+x+x^2}}\right ) \\ & = -\sqrt {\frac {1}{2} \left (2+\sqrt {5}\right )} \tan ^{-1}\left (\frac {5+2 \sqrt {5}-\sqrt {5} x}{\sqrt {10 \left (2+\sqrt {5}\right )} \sqrt {-1+x+x^2}}\right )+\sqrt {\frac {1}{2} \left (-2+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {5-2 \sqrt {5}+\sqrt {5} x}{\sqrt {10 \left (-2+\sqrt {5}\right )} \sqrt {-1+x+x^2}}\right ) \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\frac {1}{2} \text {RootSum}\left [2-4 \text {$\#$1}+6 \text {$\#$1}^2+\text {$\#$1}^4\&,\frac {3 \log \left (-x+\sqrt {-1+x+x^2}-\text {$\#$1}\right )-2 \log \left (-x+\sqrt {-1+x+x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 \log \left (-x+\sqrt {-1+x+x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-1+3 \text {$\#$1}+\text {$\#$1}^3}\&\right ] \]

[In]

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

[Out]

RootSum[2 - 4*#1 + 6*#1^2 + #1^4 & , (3*Log[-x + Sqrt[-1 + x + x^2] - #1] - 2*Log[-x + Sqrt[-1 + x + x^2] - #1
]*#1 + 2*Log[-x + Sqrt[-1 + x + x^2] - #1]*#1^2)/(-1 + 3*#1 + #1^3) & ]/2

Maple [C] (warning: unable to verify)

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

Time = 2.08 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.10

method	result	size
trager	\(\operatorname {RootOf}\left (\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2}+\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2}+\textit {\_Z}^{2}+1\right ) \operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2} x +2 \operatorname {RootOf}\left (\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2}+\textit {\_Z}^{2}+1\right ) x +\sqrt {x^{2}+x -1}+\operatorname {RootOf}\left (\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2}+\textit {\_Z}^{2}+1\right )}{4 x \operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2}+2 x -1}\right )-\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{3} x +2 \operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right ) x +\sqrt {x^{2}+x -1}-\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )}{4 x \operatorname {RootOf}\left (16 \textit {\_Z}^{4}+16 \textit {\_Z}^{2}+5\right )^{2}+2 x +1}\right )\)	\(246\)
default	\(\frac {\sqrt {\frac {10 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}-\frac {5 \sqrt {5}\, \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+10+5 \sqrt {5}}\, \sqrt {5}\, \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {\left (-2+\sqrt {5}\right ) \left (-\frac {\left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+4 \sqrt {5}+9\right )}\, \sqrt {20+10 \sqrt {5}}\, \left (\frac {\sqrt {5}\, \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+\frac {2 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}-\sqrt {5}+2\right ) \left (-\sqrt {5}-2+x \right ) \left (-2+\sqrt {5}\right )}{5 \left (-\sqrt {5}+2-x \right ) \left (\frac {\left (-\sqrt {5}-2+x \right )^{4}}{\left (-\sqrt {5}+2-x \right )^{4}}-\frac {18 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+1\right )}\right ) \sqrt {5}+\operatorname {arctanh}\left (\frac {\sqrt {\frac {10 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}-\frac {5 \sqrt {5}\, \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+10+5 \sqrt {5}}}{\sqrt {20+10 \sqrt {5}}}\right )+2 \arctan \left (\frac {\sqrt {5}\, \sqrt {\left (-2+\sqrt {5}\right ) \left (-\frac {\left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+4 \sqrt {5}+9\right )}\, \sqrt {20+10 \sqrt {5}}\, \left (\frac {\sqrt {5}\, \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+\frac {2 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}-\sqrt {5}+2\right ) \left (-\sqrt {5}-2+x \right ) \left (-2+\sqrt {5}\right )}{5 \left (-\sqrt {5}+2-x \right ) \left (\frac {\left (-\sqrt {5}-2+x \right )^{4}}{\left (-\sqrt {5}+2-x \right )^{4}}-\frac {18 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}+1\right )}\right )\right )}{\sqrt {-\frac {5 \left (\frac {\sqrt {5}\, \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}-\frac {2 \left (-\sqrt {5}-2+x \right )^{2}}{\left (-\sqrt {5}+2-x \right )^{2}}-\sqrt {5}-2\right )}{\left (1+\frac {-\sqrt {5}-2+x}{-\sqrt {5}+2-x}\right )^{2}}}\, \left (1+\frac {-\sqrt {5}-2+x}{-\sqrt {5}+2-x}\right ) \sqrt {20+10 \sqrt {5}}}\)	\(637\)

[In]

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

[Out]

RootOf(RootOf(16*_Z^4+16*_Z^2+5)^2+_Z^2+1)*ln(-(4*RootOf(RootOf(16*_Z^4+16*_Z^2+5)^2+_Z^2+1)*RootOf(16*_Z^4+16
*_Z^2+5)^2*x+2*RootOf(RootOf(16*_Z^4+16*_Z^2+5)^2+_Z^2+1)*x+(x^2+x-1)^(1/2)+RootOf(RootOf(16*_Z^4+16*_Z^2+5)^2
+_Z^2+1))/(4*x*RootOf(16*_Z^4+16*_Z^2+5)^2+2*x-1))-RootOf(16*_Z^4+16*_Z^2+5)*ln(-(4*RootOf(16*_Z^4+16*_Z^2+5)^
3*x+2*RootOf(16*_Z^4+16*_Z^2+5)*x+(x^2+x-1)^(1/2)-RootOf(16*_Z^4+16*_Z^2+5))/(4*x*RootOf(16*_Z^4+16*_Z^2+5)^2+
2*x+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\frac {1}{2} \, \sqrt {i - 2} \log \left (-x + \sqrt {i - 2} + \sqrt {x^{2} + x - 1} + i\right ) - \frac {1}{2} \, \sqrt {i - 2} \log \left (-x - \sqrt {i - 2} + \sqrt {x^{2} + x - 1} + i\right ) + \frac {1}{2} \, \sqrt {-i - 2} \log \left (-x + \sqrt {-i - 2} + \sqrt {x^{2} + x - 1} - i\right ) - \frac {1}{2} \, \sqrt {-i - 2} \log \left (-x - \sqrt {-i - 2} + \sqrt {x^{2} + x - 1} - i\right ) \]

[In]

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

[Out]

1/2*sqrt(I - 2)*log(-x + sqrt(I - 2) + sqrt(x^2 + x - 1) + I) - 1/2*sqrt(I - 2)*log(-x - sqrt(I - 2) + sqrt(x^
2 + x - 1) + I) + 1/2*sqrt(-I - 2)*log(-x + sqrt(-I - 2) + sqrt(x^2 + x - 1) - I) - 1/2*sqrt(-I - 2)*log(-x -
sqrt(-I - 2) + sqrt(x^2 + x - 1) - I)

Sympy [F]

\[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\int \frac {2 x + 1}{\left (x^{2} + 1\right ) \sqrt {x^{2} + x - 1}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\int { \frac {2 \, x + 1}{\sqrt {x^{2} + x - 1} {\left (x^{2} + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (86) = 172\).

Time = 0.34 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.91 \[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 4} \log \left (16 \, {\left (15 \, \sqrt {5} {\left (x - \sqrt {x^{2} + x - 1}\right )} + 33 \, x + 5 \, \sqrt {5} - 33 \, \sqrt {x^{2} + x - 1} + 2 \, \sqrt {5 \, \sqrt {5} + 11} + 11\right )}^{2} + 16 \, {\left (5 \, \sqrt {5} {\left (x - \sqrt {x^{2} + x - 1}\right )} + 11 \, x - 5 \, \sqrt {5} \sqrt {5 \, \sqrt {5} + 11} - 15 \, \sqrt {5} - 11 \, \sqrt {x^{2} + x - 1} - 11 \, \sqrt {5 \, \sqrt {5} + 11} - 33\right )}^{2}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 4} \log \left (16 \, {\left (15 \, \sqrt {5} {\left (x - \sqrt {x^{2} + x - 1}\right )} + 33 \, x + 5 \, \sqrt {5} - 33 \, \sqrt {x^{2} + x - 1} - 2 \, \sqrt {5 \, \sqrt {5} + 11} + 11\right )}^{2} + 16 \, {\left (5 \, \sqrt {5} {\left (x - \sqrt {x^{2} + x - 1}\right )} + 11 \, x + 5 \, \sqrt {5} \sqrt {5 \, \sqrt {5} + 11} - 15 \, \sqrt {5} - 11 \, \sqrt {x^{2} + x - 1} + 11 \, \sqrt {5 \, \sqrt {5} + 11} - 33\right )}^{2}\right ) + \frac {\sqrt {2 \, \sqrt {5} - 4} {\left (\arctan \left (3\right ) + \arctan \left (\frac {1}{10} \, {\left (x - \sqrt {x^{2} + x - 1}\right )} {\left (\sqrt {5} \sqrt {5 \, \sqrt {5} + 11} + 4 \, \sqrt {5} - 5 \, \sqrt {5 \, \sqrt {5} + 11}\right )} - \frac {7}{10} \, \sqrt {5} \sqrt {5 \, \sqrt {5} + 11} + \frac {1}{5} \, \sqrt {5} + \frac {3}{2} \, \sqrt {5 \, \sqrt {5} + 11}\right )\right )}}{2 \, {\left (\sqrt {5} - 2\right )}} - \frac {\sqrt {2 \, \sqrt {5} - 4} {\left (\arctan \left (3\right ) + \arctan \left (-\frac {1}{10} \, {\left (x - \sqrt {x^{2} + x - 1}\right )} {\left (\sqrt {5} \sqrt {5 \, \sqrt {5} + 11} - 4 \, \sqrt {5} - 5 \, \sqrt {5 \, \sqrt {5} + 11}\right )} + \frac {7}{10} \, \sqrt {5} \sqrt {5 \, \sqrt {5} + 11} + \frac {1}{5} \, \sqrt {5} - \frac {3}{2} \, \sqrt {5 \, \sqrt {5} + 11}\right )\right )}}{2 \, {\left (\sqrt {5} - 2\right )}} \]

[In]

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

[Out]

1/4*sqrt(2*sqrt(5) - 4)*log(16*(15*sqrt(5)*(x - sqrt(x^2 + x - 1)) + 33*x + 5*sqrt(5) - 33*sqrt(x^2 + x - 1) +
2*sqrt(5*sqrt(5) + 11) + 11)^2 + 16*(5*sqrt(5)*(x - sqrt(x^2 + x - 1)) + 11*x - 5*sqrt(5)*sqrt(5*sqrt(5) + 11
) - 15*sqrt(5) - 11*sqrt(x^2 + x - 1) - 11*sqrt(5*sqrt(5) + 11) - 33)^2) - 1/4*sqrt(2*sqrt(5) - 4)*log(16*(15*
sqrt(5)*(x - sqrt(x^2 + x - 1)) + 33*x + 5*sqrt(5) - 33*sqrt(x^2 + x - 1) - 2*sqrt(5*sqrt(5) + 11) + 11)^2 + 1
6*(5*sqrt(5)*(x - sqrt(x^2 + x - 1)) + 11*x + 5*sqrt(5)*sqrt(5*sqrt(5) + 11) - 15*sqrt(5) - 11*sqrt(x^2 + x -
1) + 11*sqrt(5*sqrt(5) + 11) - 33)^2) + 1/2*sqrt(2*sqrt(5) - 4)*(arctan(3) + arctan(1/10*(x - sqrt(x^2 + x - 1
))*(sqrt(5)*sqrt(5*sqrt(5) + 11) + 4*sqrt(5) - 5*sqrt(5*sqrt(5) + 11)) - 7/10*sqrt(5)*sqrt(5*sqrt(5) + 11) + 1
/5*sqrt(5) + 3/2*sqrt(5*sqrt(5) + 11)))/(sqrt(5) - 2) - 1/2*sqrt(2*sqrt(5) - 4)*(arctan(3) + arctan(-1/10*(x -
sqrt(x^2 + x - 1))*(sqrt(5)*sqrt(5*sqrt(5) + 11) - 4*sqrt(5) - 5*sqrt(5*sqrt(5) + 11)) + 7/10*sqrt(5)*sqrt(5*
sqrt(5) + 11) + 1/5*sqrt(5) - 3/2*sqrt(5*sqrt(5) + 11)))/(sqrt(5) - 2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {-1+x+x^2}} \, dx=\int \frac {2\,x+1}{\left (x^2+1\right )\,\sqrt {x^2+x-1}} \,d x \]

[In]

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

[Out]

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