### 3.655 $$\int \frac{1}{\sqrt{1+x} (1+x^2)} \, dx$$

Optimal. Leaf size=198 $-\frac{\log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}}}+\frac{\log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}}}-\frac{1}{2} \sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\frac{1}{2} \sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )$

[Out]

-(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]])/2 + (Sqrt[1 + Sqrt
[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]])/2 - Log[1 + Sqrt[2] + x - Sqrt[2*
(1 + Sqrt[2])]*Sqrt[1 + x]]/(4*Sqrt[1 + Sqrt[2]]) + Log[1 + Sqrt[2] + x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(
4*Sqrt[1 + Sqrt[2]])

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Rubi [A]  time = 0.132041, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {708, 1094, 634, 618, 204, 628} $-\frac{\log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}}}+\frac{\log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}}}-\frac{1}{2} \sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\frac{1}{2} \sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(Sqrt[1 + x]*(1 + x^2)),x]

[Out]

-(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]])/2 + (Sqrt[1 + Sqrt
[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]])/2 - Log[1 + Sqrt[2] + x - Sqrt[2*
(1 + Sqrt[2])]*Sqrt[1 + x]]/(4*Sqrt[1 + Sqrt[2]]) + Log[1 + Sqrt[2] + x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(
4*Sqrt[1 + Sqrt[2]])

Rule 708

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(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 1094

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

Rule 634

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

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

Mathematica [C]  time = 0.0230761, size = 55, normalized size = 0.28 $\frac{1}{2} (1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1-i}}\right )+\frac{1}{2} (1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1+i}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - I)^(3/2)*ArcTanh[Sqrt[1 + x]/Sqrt[1 - I]])/2 + ((1 + I)^(3/2)*ArcTanh[Sqrt[1 + x]/Sqrt[1 + I]])/2

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Maple [B]  time = 0.058, size = 420, normalized size = 2.1 \begin{align*}{\frac{\sqrt{2+2\,\sqrt{2}}}{4}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }-{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}}{8}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{ \left ( 2+2\,\sqrt{2} \right ) \sqrt{2}}{4\,\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{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}}{\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{\sqrt{2+2\,\sqrt{2}}\sqrt{2}}{8}\ln \left ( 1+x+\sqrt{2}-\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{ \left ( 2+2\,\sqrt{2} \right ) \sqrt{2}}{4\,\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{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}}{\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^2+1)/(1+x)^(1/2),x)

[Out]

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))-1/8*(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))+1/4*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))-1/2*(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/(-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))*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))+1/8*(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))+1/4*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))-1/2*(2+2*2^(1/2))/(-2+2*2^(1/2))^(1/2)*arc
tan((2*(1+x)^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))+1/(-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))*2^(1/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.29097, size = 868, normalized size = 4.38 \begin{align*} \frac{1}{8} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4}{\left (\sqrt{2} - 1\right )} \log \left (2 \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + 4 \, x + 4 \, \sqrt{2} + 4\right ) - \frac{1}{8} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4}{\left (\sqrt{2} - 1\right )} \log \left (-2 \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + 4 \, x + 4 \, \sqrt{2} + 4\right ) - \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4} \arctan \left (\frac{1}{4} \cdot 2^{\frac{3}{4}} \sqrt{2 \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + 4 \, x + 4 \, \sqrt{2} + 4} \sqrt{2 \, \sqrt{2} + 4} - \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} - \sqrt{2} - 1\right ) - \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4} \arctan \left (\frac{1}{4} \cdot 2^{\frac{3}{4}} \sqrt{-2 \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \sqrt{2 \, \sqrt{2} + 4} + 4 \, x + 4 \, \sqrt{2} + 4} \sqrt{2 \, \sqrt{2} + 4} - \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1} \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^2+1)/(1+x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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