3.291 \(\int \frac{1+x}{-\sqrt{1+x+x^2}+\sqrt{4+2 x+x^2}} \, dx\)

Optimal. Leaf size=158 \[ \frac{1}{2} \sqrt{x^2+2 x+4} (x+1)+\frac{1}{4} (2 x+1) \sqrt{x^2+x+1}-2 \sqrt{x^2+x+1}-2 \sqrt{x^2+2 x+4}-2 \sqrt{7} \tanh ^{-1}\left (\frac{5 x+1}{2 \sqrt{7} \sqrt{x^2+x+1}}\right )+2 \sqrt{7} \tanh ^{-1}\left (\frac{1-2 x}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )+\frac{11}{2} \sinh ^{-1}\left (\frac{x+1}{\sqrt{3}}\right )+\frac{43}{8} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

-2*Sqrt[1 + x + x^2] + ((1 + 2*x)*Sqrt[1 + x + x^2])/4 - 2*Sqrt[4 + 2*x + x^2] + ((1 + x)*Sqrt[4 + 2*x + x^2])
/2 + (11*ArcSinh[(1 + x)/Sqrt[3]])/2 + (43*ArcSinh[(1 + 2*x)/Sqrt[3]])/8 - 2*Sqrt[7]*ArcTanh[(1 + 5*x)/(2*Sqrt
[7]*Sqrt[1 + x + x^2])] + 2*Sqrt[7]*ArcTanh[(1 - 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])]

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Rubi [A]  time = 0.525733, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 36, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {6742, 734, 843, 619, 215, 724, 206, 612} \[ \frac{1}{2} \sqrt{x^2+2 x+4} (x+1)+\frac{1}{4} (2 x+1) \sqrt{x^2+x+1}-2 \sqrt{x^2+x+1}-2 \sqrt{x^2+2 x+4}-2 \sqrt{7} \tanh ^{-1}\left (\frac{5 x+1}{2 \sqrt{7} \sqrt{x^2+x+1}}\right )+2 \sqrt{7} \tanh ^{-1}\left (\frac{1-2 x}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )+\frac{11}{2} \sinh ^{-1}\left (\frac{x+1}{\sqrt{3}}\right )+\frac{43}{8} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)/(-Sqrt[1 + x + x^2] + Sqrt[4 + 2*x + x^2]),x]

[Out]

-2*Sqrt[1 + x + x^2] + ((1 + 2*x)*Sqrt[1 + x + x^2])/4 - 2*Sqrt[4 + 2*x + x^2] + ((1 + x)*Sqrt[4 + 2*x + x^2])
/2 + (11*ArcSinh[(1 + x)/Sqrt[3]])/2 + (43*ArcSinh[(1 + 2*x)/Sqrt[3]])/8 - 2*Sqrt[7]*ArcTanh[(1 + 5*x)/(2*Sqrt
[7]*Sqrt[1 + x + x^2])] + 2*Sqrt[7]*ArcTanh[(1 - 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 206

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

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rubi steps

\begin{align*} \int \frac{1+x}{-\sqrt{1+x+x^2}+\sqrt{4+2 x+x^2}} \, dx &=\int \left (-\frac{1}{\sqrt{1+x+x^2}-\sqrt{4+2 x+x^2}}-\frac{x}{\sqrt{1+x+x^2}-\sqrt{4+2 x+x^2}}\right ) \, dx\\ &=-\int \frac{1}{\sqrt{1+x+x^2}-\sqrt{4+2 x+x^2}} \, dx-\int \frac{x}{\sqrt{1+x+x^2}-\sqrt{4+2 x+x^2}} \, dx\\ &=-\int \left (-\frac{\sqrt{1+x+x^2}}{3+x}-\frac{\sqrt{4+2 x+x^2}}{3+x}\right ) \, dx-\int \left (-\sqrt{1+x+x^2}+\frac{3 \sqrt{1+x+x^2}}{3+x}-\sqrt{4+2 x+x^2}+\frac{3 \sqrt{4+2 x+x^2}}{3+x}\right ) \, dx\\ &=-\left (3 \int \frac{\sqrt{1+x+x^2}}{3+x} \, dx\right )-3 \int \frac{\sqrt{4+2 x+x^2}}{3+x} \, dx+\int \sqrt{1+x+x^2} \, dx+\int \frac{\sqrt{1+x+x^2}}{3+x} \, dx+\int \sqrt{4+2 x+x^2} \, dx+\int \frac{\sqrt{4+2 x+x^2}}{3+x} \, dx\\ &=-2 \sqrt{1+x+x^2}+\frac{1}{4} (1+2 x) \sqrt{1+x+x^2}-2 \sqrt{4+2 x+x^2}+\frac{1}{2} (1+x) \sqrt{4+2 x+x^2}+\frac{3}{8} \int \frac{1}{\sqrt{1+x+x^2}} \, dx-\frac{1}{2} \int \frac{1+5 x}{(3+x) \sqrt{1+x+x^2}} \, dx-\frac{1}{2} \int \frac{-2+4 x}{(3+x) \sqrt{4+2 x+x^2}} \, dx+\frac{3}{2} \int \frac{1+5 x}{(3+x) \sqrt{1+x+x^2}} \, dx+\frac{3}{2} \int \frac{1}{\sqrt{4+2 x+x^2}} \, dx+\frac{3}{2} \int \frac{-2+4 x}{(3+x) \sqrt{4+2 x+x^2}} \, dx\\ &=-2 \sqrt{1+x+x^2}+\frac{1}{4} (1+2 x) \sqrt{1+x+x^2}-2 \sqrt{4+2 x+x^2}+\frac{1}{2} (1+x) \sqrt{4+2 x+x^2}-2 \int \frac{1}{\sqrt{4+2 x+x^2}} \, dx-\frac{5}{2} \int \frac{1}{\sqrt{1+x+x^2}} \, dx+6 \int \frac{1}{\sqrt{4+2 x+x^2}} \, dx+7 \int \frac{1}{(3+x) \sqrt{1+x+x^2}} \, dx+7 \int \frac{1}{(3+x) \sqrt{4+2 x+x^2}} \, dx+\frac{15}{2} \int \frac{1}{\sqrt{1+x+x^2}} \, dx-21 \int \frac{1}{(3+x) \sqrt{1+x+x^2}} \, dx-21 \int \frac{1}{(3+x) \sqrt{4+2 x+x^2}} \, dx+\frac{1}{8} \sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )+\frac{1}{4} \sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{12}}} \, dx,x,2+2 x\right )\\ &=-2 \sqrt{1+x+x^2}+\frac{1}{4} (1+2 x) \sqrt{1+x+x^2}-2 \sqrt{4+2 x+x^2}+\frac{1}{2} (1+x) \sqrt{4+2 x+x^2}+\frac{3}{2} \sinh ^{-1}\left (\frac{1+x}{\sqrt{3}}\right )+\frac{3}{8} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )-14 \operatorname{Subst}\left (\int \frac{1}{28-x^2} \, dx,x,\frac{-1-5 x}{\sqrt{1+x+x^2}}\right )-14 \operatorname{Subst}\left (\int \frac{1}{28-x^2} \, dx,x,\frac{2-4 x}{\sqrt{4+2 x+x^2}}\right )+42 \operatorname{Subst}\left (\int \frac{1}{28-x^2} \, dx,x,\frac{-1-5 x}{\sqrt{1+x+x^2}}\right )+42 \operatorname{Subst}\left (\int \frac{1}{28-x^2} \, dx,x,\frac{2-4 x}{\sqrt{4+2 x+x^2}}\right )-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{12}}} \, dx,x,2+2 x\right )}{\sqrt{3}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )}{2 \sqrt{3}}+\sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{12}}} \, dx,x,2+2 x\right )+\frac{1}{2} \left (5 \sqrt{3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )\\ &=-2 \sqrt{1+x+x^2}+\frac{1}{4} (1+2 x) \sqrt{1+x+x^2}-2 \sqrt{4+2 x+x^2}+\frac{1}{2} (1+x) \sqrt{4+2 x+x^2}+\frac{11}{2} \sinh ^{-1}\left (\frac{1+x}{\sqrt{3}}\right )+\frac{43}{8} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )-2 \sqrt{7} \tanh ^{-1}\left (\frac{1+5 x}{2 \sqrt{7} \sqrt{1+x+x^2}}\right )+2 \sqrt{7} \tanh ^{-1}\left (\frac{1-2 x}{\sqrt{7} \sqrt{4+2 x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.277175, size = 151, normalized size = 0.96 \[ \frac{1}{8} \left (2 \left (2 \sqrt{x^2+x+1} x+2 \sqrt{x^2+2 x+4} x-7 \sqrt{x^2+x+1}-6 \sqrt{x^2+2 x+4}-8 \sqrt{7} \tanh ^{-1}\left (\frac{5 x+1}{2 \sqrt{7} \sqrt{x^2+x+1}}\right )+8 \sqrt{7} \tanh ^{-1}\left (\frac{1-2 x}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )\right )+44 \sinh ^{-1}\left (\frac{x+1}{\sqrt{3}}\right )+43 \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/(-Sqrt[1 + x + x^2] + Sqrt[4 + 2*x + x^2]),x]

[Out]

(44*ArcSinh[(1 + x)/Sqrt[3]] + 43*ArcSinh[(1 + 2*x)/Sqrt[3]] + 2*(-7*Sqrt[1 + x + x^2] + 2*x*Sqrt[1 + x + x^2]
- 6*Sqrt[4 + 2*x + x^2] + 2*x*Sqrt[4 + 2*x + x^2] - 8*Sqrt[7]*ArcTanh[(1 + 5*x)/(2*Sqrt[7]*Sqrt[1 + x + x^2])
] + 8*Sqrt[7]*ArcTanh[(1 - 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])]))/8

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Maple [A]  time = 0.012, size = 140, normalized size = 0.9 \begin{align*} -2\,\sqrt{ \left ( 3+x \right ) ^{2}-5\,x-8}+{\frac{43}{8}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ( x+{\frac{1}{2}} \right ) } \right ) }+2\,\sqrt{7}{\it Artanh} \left ( 1/14\,{\frac{ \left ( -1-5\,x \right ) \sqrt{7}}{\sqrt{ \left ( 3+x \right ) ^{2}-5\,x-8}}} \right ) -2\,\sqrt{ \left ( 3+x \right ) ^{2}-4\,x-5}+{\frac{11}{2}{\it Arcsinh} \left ({\frac{ \left ( 1+x \right ) \sqrt{3}}{3}} \right ) }+2\,\sqrt{7}{\it Artanh} \left ( 1/14\,{\frac{ \left ( 2-4\,x \right ) \sqrt{7}}{\sqrt{ \left ( 3+x \right ) ^{2}-4\,x-5}}} \right ) +{\frac{1+2\,x}{4}\sqrt{{x}^{2}+x+1}}+{\frac{2\,x+2}{4}\sqrt{{x}^{2}+2\,x+4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)/(-(x^2+x+1)^(1/2)+(x^2+2*x+4)^(1/2)),x)

[Out]

-2*((3+x)^2-5*x-8)^(1/2)+43/8*arcsinh(2/3*3^(1/2)*(x+1/2))+2*7^(1/2)*arctanh(1/14*(-1-5*x)*7^(1/2)/((3+x)^2-5*
x-8)^(1/2))-2*((3+x)^2-4*x-5)^(1/2)+11/2*arcsinh(1/3*(1+x)*3^(1/2))+2*7^(1/2)*arctanh(1/14*(2-4*x)*7^(1/2)/((3
+x)^2-4*x-5)^(1/2))+1/4*(1+2*x)*(x^2+x+1)^(1/2)+1/4*(2*x+2)*(x^2+2*x+4)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x + 1}{\sqrt{x^{2} + 2 \, x + 4} - \sqrt{x^{2} + x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-(x^2+x+1)^(1/2)+(x^2+2*x+4)^(1/2)),x, algorithm="maxima")

[Out]

integrate((x + 1)/(sqrt(x^2 + 2*x + 4) - sqrt(x^2 + x + 1)), x)

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Fricas [A]  time = 1.88251, size = 460, normalized size = 2.91 \begin{align*} \frac{1}{4} \, \sqrt{x^{2} + x + 1}{\left (2 \, x - 7\right )} + \frac{1}{2} \, \sqrt{x^{2} + 2 \, x + 4}{\left (x - 3\right )} + 2 \, \sqrt{7} \log \left (\frac{2 \, \sqrt{7}{\left (5 \, x + 1\right )} + 2 \, \sqrt{x^{2} + x + 1}{\left (5 \, \sqrt{7} - 14\right )} - 25 \, x - 5}{x + 3}\right ) + 2 \, \sqrt{7} \log \left (\frac{\sqrt{7}{\left (2 \, x - 1\right )} + \sqrt{x^{2} + 2 \, x + 4}{\left (2 \, \sqrt{7} - 7\right )} - 4 \, x + 2}{x + 3}\right ) - \frac{11}{2} \, \log \left (-x + \sqrt{x^{2} + 2 \, x + 4} - 1\right ) - \frac{43}{8} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-(x^2+x+1)^(1/2)+(x^2+2*x+4)^(1/2)),x, algorithm="fricas")

[Out]

1/4*sqrt(x^2 + x + 1)*(2*x - 7) + 1/2*sqrt(x^2 + 2*x + 4)*(x - 3) + 2*sqrt(7)*log((2*sqrt(7)*(5*x + 1) + 2*sqr
t(x^2 + x + 1)*(5*sqrt(7) - 14) - 25*x - 5)/(x + 3)) + 2*sqrt(7)*log((sqrt(7)*(2*x - 1) + sqrt(x^2 + 2*x + 4)*
(2*sqrt(7) - 7) - 4*x + 2)/(x + 3)) - 11/2*log(-x + sqrt(x^2 + 2*x + 4) - 1) - 43/8*log(-2*x + 2*sqrt(x^2 + x
+ 1) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x + 1}{- \sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 2 x + 4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-(x**2+x+1)**(1/2)+(x**2+2*x+4)**(1/2)),x)

[Out]

Integral((x + 1)/(-sqrt(x**2 + x + 1) + sqrt(x**2 + 2*x + 4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x + 1}{\sqrt{x^{2} + 2 \, x + 4} - \sqrt{x^{2} + x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(-(x^2+x+1)^(1/2)+(x^2+2*x+4)^(1/2)),x, algorithm="giac")

[Out]

integrate((x + 1)/(sqrt(x^2 + 2*x + 4) - sqrt(x^2 + x + 1)), x)