3.11 \(\int \frac{1+2 x}{(1+x^2) \sqrt{-1+x+x^2}} \, dx\)
Optimal. Leaf size=117 \[ \sqrt{\frac{1}{2} \left (\sqrt{5}-2\right )} \tanh ^{-1}\left (\frac{\sqrt{5} x-2 \sqrt{5}+5}{\sqrt{10 \left (\sqrt{5}-2\right )} \sqrt{x^2+x-1}}\right )-\sqrt{\frac{1}{2} \left (2+\sqrt{5}\right )} \tan ^{-1}\left (\frac{-\sqrt{5} x+2 \sqrt{5}+5}{\sqrt{10 \left (2+\sqrt{5}\right )} \sqrt{x^2+x-1}}\right ) \]
[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])]
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Rubi [A] time = 0.168886, antiderivative size = 117, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{1036, 1030, 207, 203} \[ \sqrt{\frac{1}{2} \left (\sqrt{5}-2\right )} \tanh ^{-1}\left (\frac{\sqrt{5} x-2 \sqrt{5}+5}{\sqrt{10 \left (\sqrt{5}-2\right )} \sqrt{x^2+x-1}}\right )-\sqrt{\frac{1}{2} \left (2+\sqrt{5}\right )} \tan ^{-1}\left (\frac{-\sqrt{5} x+2 \sqrt{5}+5}{\sqrt{10 \left (2+\sqrt{5}\right )} \sqrt{x^2+x-1}}\right ) \]
Antiderivative was successfully verified.
[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 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{-1+x+x^2}} \, dx &=-\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 ) \operatorname{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 ) \operatorname{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] time = 0.0338414, size = 78, normalized size = 0.67 \[ -\frac{1}{2} i \left (\sqrt{2+i} \tanh ^{-1}\left (\frac{\sqrt{2+i} (x-i)}{2 \sqrt{x^2+x-1}}\right )-\sqrt{2-i} \tanh ^{-1}\left (\frac{\sqrt{2-i} (x+i)}{2 \sqrt{x^2+x-1}}\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(1 + 2*x)/((1 + x^2)*Sqrt[-1 + x + x^2]),x]
[Out]
(-I/2)*(Sqrt[2 + I]*ArcTanh[(Sqrt[2 + I]*(-I + x))/(2*Sqrt[-1 + x + x^2])] - Sqrt[2 - I]*ArcTanh[(Sqrt[2 - I]*
(I + x))/(2*Sqrt[-1 + x + x^2])])
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Maple [B] time = 0.161, size = 637, normalized size = 5.4 \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+x-1)^(1/2),x)
[Out]
(10*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2-5*5^(1/2)*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+10+5*5^(1/2))^(1/2)*5^(1/2)*
(arctan(1/5*5^(1/2)*((-2+5^(1/2))*(-(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+4*5^(1/2)+9))^(1/2)*(20+10*5^(1/2))^(1/2
)*(5^(1/2)*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+2*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2-5^(1/2)+2)*(-2+5^(1/2))*(-5^(
1/2)-2+x)/(-5^(1/2)+2-x)/((-5^(1/2)-2+x)^4/(-5^(1/2)+2-x)^4-18*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+1))*5^(1/2)+a
rctanh((10*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2-5*5^(1/2)*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+10+5*5^(1/2))^(1/2)/(
20+10*5^(1/2))^(1/2))+2*arctan(1/5*5^(1/2)*((-2+5^(1/2))*(-(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+4*5^(1/2)+9))^(1/
2)*(20+10*5^(1/2))^(1/2)*(5^(1/2)*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2+2*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2-5^(1/2
)+2)*(-2+5^(1/2))*(-5^(1/2)-2+x)/(-5^(1/2)+2-x)/((-5^(1/2)-2+x)^4/(-5^(1/2)+2-x)^4-18*(-5^(1/2)-2+x)^2/(-5^(1/
2)+2-x)^2+1)))/(-5*(5^(1/2)*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2-2*(-5^(1/2)-2+x)^2/(-5^(1/2)+2-x)^2-5^(1/2)-2)/(
1+(-5^(1/2)-2+x)/(-5^(1/2)+2-x))^2)^(1/2)/(1+(-5^(1/2)-2+x)/(-5^(1/2)+2-x))/(20+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} + x - 1}{\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+x-1)^(1/2),x, algorithm="maxima")
[Out]
integrate((2*x + 1)/(sqrt(x^2 + x - 1)*(x^2 + 1)), x)
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Fricas [B] time = 1.20553, size = 2379, normalized size = 20.33 \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+x-1)^(1/2),x, algorithm="fricas")
[Out]
1/20*5^(1/4)*sqrt(4*sqrt(5) + 10)*(2*sqrt(5) - 5)*log(2*x^2 - 2*sqrt(x^2 + x - 1)*x + 1/5*(5^(1/4)*sqrt(x^2 +
x - 1)*(2*sqrt(5) - 5) - 5^(1/4)*(sqrt(5)*(2*x + 1) - 5*x))*sqrt(4*sqrt(5) + 10) + x + sqrt(5)) - 1/20*5^(1/4)
*sqrt(4*sqrt(5) + 10)*(2*sqrt(5) - 5)*log(2*x^2 - 2*sqrt(x^2 + x - 1)*x - 1/5*(5^(1/4)*sqrt(x^2 + x - 1)*(2*sq
rt(5) - 5) - 5^(1/4)*(sqrt(5)*(2*x + 1) - 5*x))*sqrt(4*sqrt(5) + 10) + x + sqrt(5)) - 1/5*5^(3/4)*sqrt(4*sqrt(
5) + 10)*arctan(2/55*sqrt(5)*(sqrt(5)*(2*x - 1) + 3*x + 4) + 1/275*sqrt(10*x^2 - 10*sqrt(x^2 + x - 1)*x + (5^(
1/4)*sqrt(x^2 + x - 1)*(2*sqrt(5) - 5) - 5^(1/4)*(sqrt(5)*(2*x + 1) - 5*x))*sqrt(4*sqrt(5) + 10) + 5*x + 5*sqr
t(5))*((5^(3/4)*(2*sqrt(5) + 3) + 2*5^(1/4)*(4*sqrt(5) - 5))*sqrt(4*sqrt(5) + 10) + 2*sqrt(5)*(3*sqrt(5) + 10)
- 20*sqrt(5) + 80) - 2/55*sqrt(x^2 + x - 1)*(sqrt(5)*(2*sqrt(5) + 3) + 8*sqrt(5) - 10) + 1/55*sqrt(5)*(16*x +
3) + 1/275*(5^(3/4)*(sqrt(5)*(3*x + 4) + 10*x - 5) - sqrt(x^2 + x - 1)*(5^(3/4)*(3*sqrt(5) + 10) - 10*5^(1/4)
*(sqrt(5) - 4)) - 10*5^(1/4)*(sqrt(5)*(x - 6) - 4*x + 13))*sqrt(4*sqrt(5) + 10) - 4/11*x + 2/11) - 1/5*5^(3/4)
*sqrt(4*sqrt(5) + 10)*arctan(-2/55*sqrt(5)*(sqrt(5)*(2*x - 1) + 3*x + 4) + 1/275*sqrt(10*x^2 - 10*sqrt(x^2 + x
- 1)*x - (5^(1/4)*sqrt(x^2 + x - 1)*(2*sqrt(5) - 5) - 5^(1/4)*(sqrt(5)*(2*x + 1) - 5*x))*sqrt(4*sqrt(5) + 10)
+ 5*x + 5*sqrt(5))*((5^(3/4)*(2*sqrt(5) + 3) + 2*5^(1/4)*(4*sqrt(5) - 5))*sqrt(4*sqrt(5) + 10) - 2*sqrt(5)*(3
*sqrt(5) + 10) + 20*sqrt(5) - 80) + 2/55*sqrt(x^2 + x - 1)*(sqrt(5)*(2*sqrt(5) + 3) + 8*sqrt(5) - 10) - 1/55*s
qrt(5)*(16*x + 3) + 1/275*(5^(3/4)*(sqrt(5)*(3*x + 4) + 10*x - 5) - sqrt(x^2 + x - 1)*(5^(3/4)*(3*sqrt(5) + 10
) - 10*5^(1/4)*(sqrt(5) - 4)) - 10*5^(1/4)*(sqrt(5)*(x - 6) - 4*x + 13))*sqrt(4*sqrt(5) + 10) + 4/11*x - 2/11)
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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} + x - 1}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Giac [C] time = 1.29809, size = 336, normalized size = 2.87 \begin{align*} \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 4}{\left (-\frac{i}{\sqrt{5} - 2} + 1\right )} \log \left (-4 \, \sqrt{5 \, \sqrt{5} + 11}{\left (\frac{2 i}{5 \, \sqrt{5} + 11} - 1\right )} - \left (12 i + 4\right ) \, x + \left (12 i + 4\right ) \, \sqrt{x^{2} + x - 1} - 4 i + 12\right ) - \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 4}{\left (-\frac{i}{\sqrt{5} - 2} + 1\right )} \log \left (-4 \, \sqrt{5 \, \sqrt{5} + 11}{\left (-\frac{2 i}{5 \, \sqrt{5} + 11} + 1\right )} - \left (12 i + 4\right ) \, x + \left (12 i + 4\right ) \, \sqrt{x^{2} + x - 1} - 4 i + 12\right ) - \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 4}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} \log \left (-4 \, \sqrt{5 \, \sqrt{5} - 11}{\left (\frac{2 i}{5 \, \sqrt{5} - 11} - 1\right )} - \left (4 i + 12\right ) \, x + \left (4 i + 12\right ) \, \sqrt{x^{2} + x - 1} + 12 i - 4\right ) + \frac{1}{4} \, \sqrt{2 \, \sqrt{5} - 4}{\left (\frac{i}{\sqrt{5} - 2} + 1\right )} \log \left (-4 \, \sqrt{5 \, \sqrt{5} - 11}{\left (-\frac{2 i}{5 \, \sqrt{5} - 11} + 1\right )} - \left (4 i + 12\right ) \, x + \left (4 i + 12\right ) \, \sqrt{x^{2} + x - 1} + 12 i - 4\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)*(-I/(sqrt(5) - 2) + 1)*log(-4*sqrt(5*sqrt(5) + 11)*(2*I/(5*sqrt(5) + 11) - 1) - (12*I
+ 4)*x + (12*I + 4)*sqrt(x^2 + x - 1) - 4*I + 12) - 1/4*sqrt(2*sqrt(5) - 4)*(-I/(sqrt(5) - 2) + 1)*log(-4*sqrt
(5*sqrt(5) + 11)*(-2*I/(5*sqrt(5) + 11) + 1) - (12*I + 4)*x + (12*I + 4)*sqrt(x^2 + x - 1) - 4*I + 12) - 1/4*s
qrt(2*sqrt(5) - 4)*(I/(sqrt(5) - 2) + 1)*log(-4*sqrt(5*sqrt(5) - 11)*(2*I/(5*sqrt(5) - 11) - 1) - (4*I + 12)*x
+ (4*I + 12)*sqrt(x^2 + x - 1) + 12*I - 4) + 1/4*sqrt(2*sqrt(5) - 4)*(I/(sqrt(5) - 2) + 1)*log(-4*sqrt(5*sqrt
(5) - 11)*(-2*I/(5*sqrt(5) - 11) + 1) - (4*I + 12)*x + (4*I + 12)*sqrt(x^2 + x - 1) + 12*I - 4)