### 3.654 $$\int \frac{\sqrt{1+x}}{1+x^2} \, dx$$

Optimal. Leaf size=205 $\frac{\log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )$

[Out]

-(Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]) + Sqrt[(1 + Sq
rt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]] + Log[1 + Sqrt[2] + x - Sqrt[
2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(2*Sqrt[2*(1 + Sqrt[2])]) - Log[1 + Sqrt[2] + x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 +
x]]/(2*Sqrt[2*(1 + Sqrt[2])])

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Rubi [A]  time = 0.223013, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.467, Rules used = {700, 1127, 1161, 618, 204, 1164, 628} $\frac{\log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[1 + x]/(1 + x^2),x]

[Out]

-(Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]) + Sqrt[(1 + Sq
rt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]] + Log[1 + Sqrt[2] + x - Sqrt[
2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(2*Sqrt[2*(1 + Sqrt[2])]) - Log[1 + Sqrt[2] + x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 +
x]]/(2*Sqrt[2*(1 + Sqrt[2])])

Rule 700

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

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

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

Rule 204

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

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{1+x}}{1+x^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^2}{2-2 x^2+x^4} \, dx,x,\sqrt{1+x}\right )\\ &=-\operatorname{Subst}\left (\int \frac{\sqrt{2}-x^2}{2-2 x^2+x^4} \, dx,x,\sqrt{1+x}\right )+\operatorname{Subst}\left (\int \frac{\sqrt{2}+x^2}{2-2 x^2+x^4} \, dx,x,\sqrt{1+x}\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}+\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (1+\sqrt{2}\right )}+2 x}{-\sqrt{2}-\sqrt{2 \left (1+\sqrt{2}\right )} x-x^2} \, dx,x,\sqrt{1+x}\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 x}{-\sqrt{2}+\sqrt{2 \left (1+\sqrt{2}\right )} x-x^2} \, dx,x,\sqrt{1+x}\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}\\ &=\frac{\log \left (1+\sqrt{2}+x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\log \left (1+\sqrt{2}+x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\operatorname{Subst}\left (\int \frac{1}{2 \left (1-\sqrt{2}\right )-x^2} \, dx,x,-\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}\right )-\operatorname{Subst}\left (\int \frac{1}{2 \left (1-\sqrt{2}\right )-x^2} \, dx,x,\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}\right )\\ &=\frac{\tan ^{-1}\left (\frac{-\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}}{\sqrt{2 \left (-1+\sqrt{2}\right )}}\right )}{\sqrt{2 \left (-1+\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}}{\sqrt{2 \left (-1+\sqrt{2}\right )}}\right )}{\sqrt{2 \left (-1+\sqrt{2}\right )}}+\frac{\log \left (1+\sqrt{2}+x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\log \left (1+\sqrt{2}+x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )}{2 \sqrt{2 \left (1+\sqrt{2}\right )}}\\ \end{align*}

Mathematica [C]  time = 0.025796, size = 55, normalized size = 0.27 $i \sqrt{1+i} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1+i}}\right )-i \sqrt{1-i} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1-i}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[1 + x]/(1 + x^2),x]

[Out]

(-I)*Sqrt[1 - I]*ArcTanh[Sqrt[1 + x]/Sqrt[1 - I]] + I*Sqrt[1 + I]*ArcTanh[Sqrt[1 + x]/Sqrt[1 + I]]

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Maple [B]  time = 0.107, size = 336, normalized size = 1.6 \begin{align*} -{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}}{4}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{\sqrt{2} \left ( 2+2\,\sqrt{2} \right ) }{2\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}}{4}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }-{\frac{2+2\,\sqrt{2}}{2\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}}{4}\ln \left ( 1+x+\sqrt{2}-\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{\sqrt{2} \left ( 2+2\,\sqrt{2} \right ) }{2\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }-{\frac{\sqrt{2+2\,\sqrt{2}}}{4}\ln \left ( 1+x+\sqrt{2}-\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }-{\frac{2+2\,\sqrt{2}}{2\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}} \right ) } \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2+2*2^(1/2))^(1/2)*2^(1/2)*ln(1+x+2^(1/2)+(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))+1/2*2^(1/2)*(2+2*2^(1/2))/(-2
+2*2^(1/2))^(1/2)*arctan((2*(1+x)^(1/2)+(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))+1/4*(2+2*2^(1/2))^(1/2)*ln(
1+x+2^(1/2)+(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))-1/2*(2+2*2^(1/2))/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+x)^(1/2)+(2+2
*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))+1/4*(2+2*2^(1/2))^(1/2)*2^(1/2)*ln(1+x+2^(1/2)-(1+x)^(1/2)*(2+2*2^(1/2)
)^(1/2))+1/2*2^(1/2)*(2+2*2^(1/2))/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+x)^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/
2))^(1/2))-1/4*(2+2*2^(1/2))^(1/2)*ln(1+x+2^(1/2)-(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))-1/2*(2+2*2^(1/2))/(-2+2*2^(
1/2))^(1/2)*arctan((2*(1+x)^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + 1)/(x^2 + 1), x)

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Fricas [A]  time = 2.31457, size = 860, normalized size = 4.2 \begin{align*} \frac{1}{8} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4}{\left (\sqrt{2} - 2\right )} \log \left (\frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + x + \sqrt{2} + 1\right ) - \frac{1}{8} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4}{\left (\sqrt{2} - 2\right )} \log \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + x + \sqrt{2} + 1\right ) - \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{2 \, \sqrt{2} + 4} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + 2 \, x + 2 \, \sqrt{2} + 2} \sqrt{2 \, \sqrt{2} + 4} - \sqrt{2} - 1\right ) - \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{2 \, \sqrt{2} + 4} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{-2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + 2 \, x + 2 \, \sqrt{2} + 2} \sqrt{2 \, \sqrt{2} + 4} + \sqrt{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*2^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*log(1/2*2^(3/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + x + sqrt(2) +
1) - 1/8*2^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*log(-1/2*2^(3/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + x + sqrt
(2) + 1) - 1/2*2^(3/4)*sqrt(2*sqrt(2) + 4)*arctan(-1/2*2^(3/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + 1/2*2^(1/4)*s
qrt(2^(3/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + 2*x + 2*sqrt(2) + 2)*sqrt(2*sqrt(2) + 4) - sqrt(2) - 1) - 1/2*2^
(3/4)*sqrt(2*sqrt(2) + 4)*arctan(-1/2*2^(3/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + 1/2*2^(1/4)*sqrt(-2^(3/4)*sqrt
(x + 1)*sqrt(2*sqrt(2) + 4) + 2*x + 2*sqrt(2) + 2)*sqrt(2*sqrt(2) + 4) + sqrt(2) + 1)

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Sympy [A]  time = 2.7738, size = 31, normalized size = 0.15 \begin{align*} 2 \operatorname{RootSum}{\left (128 t^{4} + 16 t^{2} + 1, \left ( t \mapsto t \log{\left (64 t^{3} + 4 t + \sqrt{x + 1} \right )} \right )\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*RootSum(128*_t**4 + 16*_t**2 + 1, Lambda(_t, _t*log(64*_t**3 + 4*_t + sqrt(x + 1))))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + 1)/(x^2 + 1), x)