\(\int \frac {1+2 x}{(3+x+x^2) \sqrt {5+x+x^2}} \, dx\) [33]

Optimal result

Integrand size = 24, antiderivative size = 24 \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {5+x+x^2}}{\sqrt {2}}\right ) \]

[Out]

-arctanh(1/2*(x^2+x+5)^(1/2)*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1038, 212} \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {x^2+x+5}}{\sqrt {2}}\right ) \]

[In]

Int[(1 + 2*x)/((3 + x + x^2)*Sqrt[5 + x + x^2]),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[5 + x + x^2]/Sqrt[2]])

Rule 212

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

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {5+x+x^2}}{\sqrt {2}}\right ) \]

[In]

Integrate[(1 + 2*x)/((3 + x + x^2)*Sqrt[5 + x + x^2]),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[5 + x + x^2]/Sqrt[2]])

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83

method	result	size
default	\(-\operatorname {arctanh}\left (\frac {\sqrt {x^{2}+x +5}\, \sqrt {2}}{2}\right ) \sqrt {2}\)	\(20\)
pseudoelliptic	\(-\operatorname {arctanh}\left (\frac {\sqrt {x^{2}+x +5}\, \sqrt {2}}{2}\right ) \sqrt {2}\)	\(20\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {x^{2}+x +5}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{x^{2}+x +3}\right )}{2}\)	\(56\)

[In]

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

[Out]

-arctanh(1/2*(x^2+x+5)^(1/2)*2^(1/2))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\frac {1}{2} \, \sqrt {2} \log \left (\frac {x^{2} - 2 \, \sqrt {2} \sqrt {x^{2} + x + 5} + x + 7}{x^{2} + x + 3}\right ) \]

[In]

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

[Out]

1/2*sqrt(2)*log((x^2 - 2*sqrt(2)*sqrt(x^2 + x + 5) + x + 7)/(x^2 + x + 3))

Sympy [A] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\frac {\sqrt {2} \left (\log {\left (\sqrt {x^{2} + x + 5} - \sqrt {2} \right )} - \log {\left (\sqrt {x^{2} + x + 5} + \sqrt {2} \right )}\right )}{2} \]

[In]

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

[Out]

sqrt(2)*(log(sqrt(x**2 + x + 5) - sqrt(2)) - log(sqrt(x**2 + x + 5) + sqrt(2)))/2

Maxima [F]

\[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=\int { \frac {2 \, x + 1}{\sqrt {x^{2} + x + 5} {\left (x^{2} + x + 3\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\frac {1}{2} \, \sqrt {2} \log \left (\sqrt {2} + \sqrt {x^{2} + x + 5}\right ) + \frac {1}{2} \, \sqrt {2} \log \left (-\sqrt {2} + \sqrt {x^{2} + x + 5}\right ) \]

[In]

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

[Out]

-1/2*sqrt(2)*log(sqrt(2) + sqrt(x^2 + x + 5)) + 1/2*sqrt(2)*log(-sqrt(2) + sqrt(x^2 + x + 5))

Mupad [B] (verification not implemented)

Time = 12.55 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {1+2 x}{\left (3+x+x^2\right ) \sqrt {5+x+x^2}} \, dx=-\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {x^2+x+5}}{2}\right ) \]

[In]

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

[Out]

-2^(1/2)*atanh((2^(1/2)*(x + x^2 + 5)^(1/2))/2)