3.998 \(\int \frac {1+2 x}{(1+x^2) \sqrt {2+2 x+x^2}} \, dx\)

Optimal. Leaf size=126 \[ -\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) x}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {x^2+2 x+2}}\right )-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\left (5-\sqrt {5}\right ) x+2 \sqrt {5}}{\sqrt {10 \left (\sqrt {5}-1\right )} \sqrt {x^2+2 x+2}}\right ) \]

[Out]

-1/2*arctanh((x*(5-5^(1/2))+2*5^(1/2))/(x^2+2*x+2)^(1/2)/(-10+10*5^(1/2))^(1/2))*(-2+2*5^(1/2))^(1/2)-1/2*arct
an((2*5^(1/2)-x*(5+5^(1/2)))/(x^2+2*x+2)^(1/2)/(10+10*5^(1/2))^(1/2))*(2+2*5^(1/2))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1036, 1030, 207, 203} \[ -\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) x}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {x^2+2 x+2}}\right )-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\left (5-\sqrt {5}\right ) x+2 \sqrt {5}}{\sqrt {10 \left (\sqrt {5}-1\right )} \sqrt {x^2+2 x+2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 1030

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 1036

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*} \int \frac {1+2 x}{\left (1+x^2\right ) \sqrt {2+2 x+x^2}} \, dx &=-\frac {\int \frac {-5-\sqrt {5}-2 \sqrt {5} x}{\left (1+x^2\right ) \sqrt {2+2 x+x^2}} \, dx}{2 \sqrt {5}}+\frac {\int \frac {-5+\sqrt {5}+2 \sqrt {5} x}{\left (1+x^2\right ) \sqrt {2+2 x+x^2}} \, dx}{2 \sqrt {5}}\\ &=\left (2 \left (5-\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{20 \left (1-\sqrt {5}\right )+2 x^2} \, dx,x,\frac {2 \sqrt {5}+\left (5-\sqrt {5}\right ) x}{\sqrt {2+2 x+x^2}}\right )+\left (2 \left (5+\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{20 \left (1+\sqrt {5}\right )+2 x^2} \, dx,x,\frac {-2 \sqrt {5}+\left (5+\sqrt {5}\right ) x}{\sqrt {2+2 x+x^2}}\right )\\ &=-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {2 \sqrt {5}-\left (5+\sqrt {5}\right ) x}{\sqrt {10 \left (1+\sqrt {5}\right )} \sqrt {2+2 x+x^2}}\right )-\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {2 \sqrt {5}+\left (5-\sqrt {5}\right ) x}{\sqrt {10 \left (-1+\sqrt {5}\right )} \sqrt {2+2 x+x^2}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 87, normalized size = 0.69 \[ \frac {1}{2} i \left (\sqrt {1+2 i} \tanh ^{-1}\left (\frac {(1+i) x+(2+i)}{\sqrt {1+2 i} \sqrt {x^2+2 x+2}}\right )-\sqrt {1-2 i} \tanh ^{-1}\left (\frac {(2-2 i) x+(4-2 i)}{2 \sqrt {1-2 i} \sqrt {x^2+2 x+2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/2)*(Sqrt[1 + 2*I]*ArcTanh[((2 + I) + (1 + I)*x)/(Sqrt[1 + 2*I]*Sqrt[2 + 2*x + x^2])] - Sqrt[1 - 2*I]*ArcTan
h[((4 - 2*I) + (2 - 2*I)*x)/(2*Sqrt[1 - 2*I]*Sqrt[2 + 2*x + x^2])])

________________________________________________________________________________________

fricas [B]  time = 0.84, size = 770, normalized size = 6.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*5^(3/4)*sqrt(2)*sqrt(sqrt(5) + 5)*arctan(1/200*sqrt(20*x^2 - 20*sqrt(x^2 + 2*x + 2)*x - (2*5^(3/4)*sqrt(2)
*sqrt(x^2 + 2*x + 2) - 5^(1/4)*(sqrt(5)*sqrt(2)*(2*x + 1) - 5*sqrt(2)))*sqrt(sqrt(5) + 5) + 20*x + 10*sqrt(5)
+ 30)*(sqrt(10)*(5^(3/4)*(sqrt(5)*sqrt(2) - sqrt(2)) + 2*5^(3/4)*sqrt(2))*sqrt(sqrt(5) + 5) + 10*sqrt(10)*(sqr
t(5) + 3)) + 1/10*sqrt(5)*(sqrt(5)*(2*x + 1) + 5) + 1/2*sqrt(5)*x + 1/20*(5^(3/4)*(sqrt(5)*sqrt(2)*x - sqrt(2)
*(x - 2)) - sqrt(x^2 + 2*x + 2)*(5^(3/4)*(sqrt(5)*sqrt(2) - sqrt(2)) + 2*5^(3/4)*sqrt(2)) + 5^(1/4)*(sqrt(5)*s
qrt(2)*(2*x + 1) + 5*sqrt(2)))*sqrt(sqrt(5) + 5) - 1/2*sqrt(x^2 + 2*x + 2)*(sqrt(5) + 3) + 1/2*x + 1) + 1/5*5^
(3/4)*sqrt(2)*sqrt(sqrt(5) + 5)*arctan(1/200*sqrt(20*x^2 - 20*sqrt(x^2 + 2*x + 2)*x + (2*5^(3/4)*sqrt(2)*sqrt(
x^2 + 2*x + 2) - 5^(1/4)*(sqrt(5)*sqrt(2)*(2*x + 1) - 5*sqrt(2)))*sqrt(sqrt(5) + 5) + 20*x + 10*sqrt(5) + 30)*
(sqrt(10)*(5^(3/4)*(sqrt(5)*sqrt(2) - sqrt(2)) + 2*5^(3/4)*sqrt(2))*sqrt(sqrt(5) + 5) - 10*sqrt(10)*(sqrt(5) +
3)) - 1/10*sqrt(5)*(sqrt(5)*(2*x + 1) + 5) - 1/2*sqrt(5)*x + 1/20*(5^(3/4)*(sqrt(5)*sqrt(2)*x - sqrt(2)*(x -
2)) - sqrt(x^2 + 2*x + 2)*(5^(3/4)*(sqrt(5)*sqrt(2) - sqrt(2)) + 2*5^(3/4)*sqrt(2)) + 5^(1/4)*(sqrt(5)*sqrt(2)
*(2*x + 1) + 5*sqrt(2)))*sqrt(sqrt(5) + 5) + 1/2*sqrt(x^2 + 2*x + 2)*(sqrt(5) + 3) - 1/2*x - 1) + 1/40*5^(1/4)
*(sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(sqrt(5) + 5)*log(2*x^2 - 2*sqrt(x^2 + 2*x + 2)*x + 1/10*(2*5^(3/4)*sqrt(2)
*sqrt(x^2 + 2*x + 2) - 5^(1/4)*(sqrt(5)*sqrt(2)*(2*x + 1) - 5*sqrt(2)))*sqrt(sqrt(5) + 5) + 2*x + sqrt(5) + 3)
- 1/40*5^(1/4)*(sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(sqrt(5) + 5)*log(2*x^2 - 2*sqrt(x^2 + 2*x + 2)*x - 1/10*(2*
5^(3/4)*sqrt(2)*sqrt(x^2 + 2*x + 2) - 5^(1/4)*(sqrt(5)*sqrt(2)*(2*x + 1) - 5*sqrt(2)))*sqrt(sqrt(5) + 5) + 2*x
+ sqrt(5) + 3)

________________________________________________________________________________________

giac [B]  time = 0.72, size = 444, normalized size = 3.52 \[ \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (256 \, {\left (\sqrt {5} {\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )} - 2 \, x + \sqrt {5} \sqrt {\sqrt {5} - 2} + \sqrt {5} + 2 \, \sqrt {x^{2} + 2 \, x + 2} - 2 \, \sqrt {\sqrt {5} - 2} - 2\right )}^{2} + 256 \, {\left (\sqrt {5} {\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )} - 2 \, x - \sqrt {5} + 2 \, \sqrt {x^{2} + 2 \, x + 2} + \sqrt {\sqrt {5} - 2} + 2\right )}^{2}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (256 \, {\left (\sqrt {5} {\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )} - 2 \, x - \sqrt {5} \sqrt {\sqrt {5} - 2} + \sqrt {5} + 2 \, \sqrt {x^{2} + 2 \, x + 2} + 2 \, \sqrt {\sqrt {5} - 2} - 2\right )}^{2} + 256 \, {\left (\sqrt {5} {\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )} - 2 \, x - \sqrt {5} + 2 \, \sqrt {x^{2} + 2 \, x + 2} - \sqrt {\sqrt {5} - 2} + 2\right )}^{2}\right ) + \frac {{\left (\pi + 4 \, \arctan \left (\frac {1}{2} \, {\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )} {\left (2 \, \sqrt {5} \sqrt {\sqrt {5} - 2} + \sqrt {5} + 4 \, \sqrt {\sqrt {5} - 2} + 3\right )} + \frac {3}{2} \, \sqrt {5} \sqrt {\sqrt {5} - 2} + \frac {1}{2} \, \sqrt {5} + \frac {7}{2} \, \sqrt {\sqrt {5} - 2} + \frac {3}{2}\right )\right )} \sqrt {2 \, \sqrt {5} - 2}}{4 \, {\left (\sqrt {5} - 1\right )}} - \frac {{\left (\pi + 4 \, \arctan \left (-\frac {1}{2} \, {\left (x - \sqrt {x^{2} + 2 \, x + 2}\right )} {\left (2 \, \sqrt {5} \sqrt {\sqrt {5} - 2} - \sqrt {5} + 4 \, \sqrt {\sqrt {5} - 2} - 3\right )} - \frac {3}{2} \, \sqrt {5} \sqrt {\sqrt {5} - 2} + \frac {1}{2} \, \sqrt {5} - \frac {7}{2} \, \sqrt {\sqrt {5} - 2} + \frac {3}{2}\right )\right )} \sqrt {2 \, \sqrt {5} - 2}}{4 \, {\left (\sqrt {5} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2*sqrt(5) - 2)*log(256*(sqrt(5)*(x - sqrt(x^2 + 2*x + 2)) - 2*x + sqrt(5)*sqrt(sqrt(5) - 2) + sqrt(5)
+ 2*sqrt(x^2 + 2*x + 2) - 2*sqrt(sqrt(5) - 2) - 2)^2 + 256*(sqrt(5)*(x - sqrt(x^2 + 2*x + 2)) - 2*x - sqrt(5)
+ 2*sqrt(x^2 + 2*x + 2) + sqrt(sqrt(5) - 2) + 2)^2) - 1/4*sqrt(2*sqrt(5) - 2)*log(256*(sqrt(5)*(x - sqrt(x^2
+ 2*x + 2)) - 2*x - sqrt(5)*sqrt(sqrt(5) - 2) + sqrt(5) + 2*sqrt(x^2 + 2*x + 2) + 2*sqrt(sqrt(5) - 2) - 2)^2 +
256*(sqrt(5)*(x - sqrt(x^2 + 2*x + 2)) - 2*x - sqrt(5) + 2*sqrt(x^2 + 2*x + 2) - sqrt(sqrt(5) - 2) + 2)^2) +
1/4*(pi + 4*arctan(1/2*(x - sqrt(x^2 + 2*x + 2))*(2*sqrt(5)*sqrt(sqrt(5) - 2) + sqrt(5) + 4*sqrt(sqrt(5) - 2)
+ 3) + 3/2*sqrt(5)*sqrt(sqrt(5) - 2) + 1/2*sqrt(5) + 7/2*sqrt(sqrt(5) - 2) + 3/2))*sqrt(2*sqrt(5) - 2)/(sqrt(5
) - 1) - 1/4*(pi + 4*arctan(-1/2*(x - sqrt(x^2 + 2*x + 2))*(2*sqrt(5)*sqrt(sqrt(5) - 2) - sqrt(5) + 4*sqrt(sqr
t(5) - 2) - 3) - 3/2*sqrt(5)*sqrt(sqrt(5) - 2) + 1/2*sqrt(5) - 7/2*sqrt(sqrt(5) - 2) + 3/2))*sqrt(2*sqrt(5) -
2)/(sqrt(5) - 1)

________________________________________________________________________________________

maple [B]  time = 0.11, size = 753, normalized size = 5.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(10*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2-2*5^(1/2)*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2
+10+2*5^(1/2))^(1/2)*(5*arctan(1/80*(-22+10*5^(1/2))^(1/2)*((5-5^(1/2))*(2*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2
)-1/2-x)^2+5^(1/2)+3))^(1/2)*(11*5^(1/2)*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2+25*(-1/2*5^(1/2)+1/2+x)
^2/(-1/2*5^(1/2)-1/2-x)^2+4*5^(1/2)+10)*(-1/2*5^(1/2)+1/2+x)/(-1/2*5^(1/2)-1/2-x)*(5^(1/2)-5)/((-1/2*5^(1/2)+1
/2+x)^4/(-1/2*5^(1/2)-1/2-x)^4+3*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2+1))*(-10+10*5^(1/2))^(1/2)*(-22
+10*5^(1/2))^(1/2)+3*arctan(1/80*(-22+10*5^(1/2))^(1/2)*((5-5^(1/2))*(2*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1
/2-x)^2+5^(1/2)+3))^(1/2)*(11*5^(1/2)*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2+25*(-1/2*5^(1/2)+1/2+x)^2/
(-1/2*5^(1/2)-1/2-x)^2+4*5^(1/2)+10)*(-1/2*5^(1/2)+1/2+x)/(-1/2*5^(1/2)-1/2-x)*(5^(1/2)-5)/((-1/2*5^(1/2)+1/2+
x)^4/(-1/2*5^(1/2)-1/2-x)^4+3*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2+1))*(-10+10*5^(1/2))^(1/2)*5^(1/2)
*(-22+10*5^(1/2))^(1/2)+20*arctanh((10*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2-2*5^(1/2)*(-1/2*5^(1/2)+1
/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2+10+2*5^(1/2))^(1/2)/(-10+10*5^(1/2))^(1/2))*5^(1/2)-60*arctanh((10*(-1/2*5^(1/2
)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2-2*5^(1/2)*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2+10+2*5^(1/2))^(1/2)/
(-10+10*5^(1/2))^(1/2)))/(-2*(5^(1/2)*(-1/2*5^(1/2)+1/2+x)^2/(-1/2*5^(1/2)-1/2-x)^2-5*(-1/2*5^(1/2)+1/2+x)^2/(
-1/2*5^(1/2)-1/2-x)^2-5^(1/2)-5)/((-1/2*5^(1/2)+1/2+x)/(-1/2*5^(1/2)-1/2-x)+1)^2)^(1/2)/((-1/2*5^(1/2)+1/2+x)/
(-1/2*5^(1/2)-1/2-x)+1)/(5^(1/2)-5)/(-10+10*5^(1/2))^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2 \, x + 1}{\sqrt {x^{2} + 2 \, x + 2} {\left (x^{2} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {2\,x+1}{\left (x^2+1\right )\,\sqrt {x^2+2\,x+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2 x + 1}{\left (x^{2} + 1\right ) \sqrt {x^{2} + 2 x + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________