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]
-(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])]
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Rubi [A] time = 0.156796, antiderivative size = 126, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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 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)]
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 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 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])
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 = 0.0361806, 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])])
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Maple [B] time = 0.075, size = 753, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1+2*x)/(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)*(3*5^(1/2)*(-10+10*5^(1/2))^(1/2)*(-22+10*5^(1/2))^(1/2)*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)*(5^(1/2)-5)*(-1/
2*5^(1/2)+1/2+x)/(-1/2*5^(1/2)-1/2-x)/((-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))+5*(-10+10*5^(1/2))^(1/2)*(-22+10*5^(1/2))^(1/2)*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)*(5^(1/2)-5)*(-1/2*5
^(1/2)+1/2+x)/(-1/2*5^(1/2)-1/2-x)/((-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))+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)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 \, x + 1}{\sqrt{x^{2} + 2 \, x + 2}{\left (x^{2} + 1\right )}}\,{d x} \end{align*}
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)
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Fricas [B] time = 2.02381, size = 2399, normalized size = 19.04 \begin{align*} \text{result too large to display} \end{align*}
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)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 x + 1}{\left (x^{2} + 1\right ) \sqrt{x^{2} + 2 x + 2}}\, dx \end{align*}
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)
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Giac [B] time = 1.34969, size = 369, normalized size = 2.93 \begin{align*} -\frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (\frac{2 \, i}{\sqrt{5} - 1} + 1\right )} \log \left (-16 \,{\left (i + 1\right )}{\left (x - \sqrt{x^{2} + 2 \, x + 2}\right )} + 16 \, \sqrt{\sqrt{5} + 2}{\left (\frac{i}{\sqrt{5} + 2} + 1\right )} - 16 \, i + 16\right ) + \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (\frac{2 \, i}{\sqrt{5} - 1} + 1\right )} \log \left (-16 \,{\left (i + 1\right )}{\left (x - \sqrt{x^{2} + 2 \, x + 2}\right )} - 16 \, \sqrt{\sqrt{5} + 2}{\left (\frac{i}{\sqrt{5} + 2} + 1\right )} - 16 \, i + 16\right ) + \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (\frac{2 \, i}{\sqrt{5} - 1} - 1\right )} \log \left (-16 \,{\left (i + 1\right )}{\left (x - \sqrt{x^{2} + 2 \, x + 2}\right )} + 16 \, \sqrt{\sqrt{5} - 2}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} + 16 \, i - 16\right ) - \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 2}{\left (\frac{2 \, i}{\sqrt{5} - 1} - 1\right )} \log \left (-16 \,{\left (i + 1\right )}{\left (x - \sqrt{x^{2} + 2 \, x + 2}\right )} - 16 \, \sqrt{\sqrt{5} - 2}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} + 16 \, i - 16\right ) \end{align*}
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)*(2*i/(sqrt(5) - 1) + 1)*log(-16*(i + 1)*(x - sqrt(x^2 + 2*x + 2)) + 16*sqrt(sqrt(5) +
2)*(i/(sqrt(5) + 2) + 1) - 16*i + 16) + 1/4*sqrt(2*sqrt(5) - 2)*(2*i/(sqrt(5) - 1) + 1)*log(-16*(i + 1)*(x -
sqrt(x^2 + 2*x + 2)) - 16*sqrt(sqrt(5) + 2)*(i/(sqrt(5) + 2) + 1) - 16*i + 16) + 1/4*sqrt(2*sqrt(5) - 2)*(2*i/
(sqrt(5) - 1) - 1)*log(-16*(i + 1)*(x - sqrt(x^2 + 2*x + 2)) + 16*sqrt(sqrt(5) - 2)*(i/(sqrt(5) - 2) + 1) + 16
*i - 16) - 1/4*sqrt(2*sqrt(5) - 2)*(2*i/(sqrt(5) - 1) - 1)*log(-16*(i + 1)*(x - sqrt(x^2 + 2*x + 2)) - 16*sqrt
(sqrt(5) - 2)*(i/(sqrt(5) - 2) + 1) + 16*i - 16)