\(\int \frac {1}{\sqrt {1+x} (1+x^2)} \, dx\) [655]

Optimal result

Integrand size = 15, antiderivative size = 198 \[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=-\frac {1}{2} \sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )+\frac {1}{2} \sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )-\frac {\log \left (1+\sqrt {2}+x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\log \left (1+\sqrt {2}+x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )}{4 \sqrt {1+\sqrt {2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {722, 1108, 648, 632, 210, 642} \[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=-\frac {1}{2} \sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {1}{2} \sqrt {1+\sqrt {2}} \arctan \left (\frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )-\frac {\log \left (x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\log \left (x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}} \]

[In]

Int[1/(Sqrt[1 + x]*(1 + x^2)),x]

[Out]

-1/2*(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]) + (Sqrt[1 + Sq
rt[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 210

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

Rule 632

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 642

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]

Rule 648

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 722

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 1108

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]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{2-2 x^2+x^4} \, dx,x,\sqrt {1+x}\right ) \\ & = \frac {\text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=\sqrt {\frac {1}{2}+\frac {i}{2}} \arctan \left (\sqrt {-\frac {1}{2}-\frac {i}{2}} \sqrt {1+x}\right )+\sqrt {\frac {1}{2}-\frac {i}{2}} \arctan \left (\sqrt {-\frac {1}{2}+\frac {i}{2}} \sqrt {1+x}\right ) \]

[In]

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

[Out]

Sqrt[1/2 + I/2]*ArcTan[Sqrt[-1/2 - I/2]*Sqrt[1 + x]] + Sqrt[1/2 - I/2]*ArcTan[Sqrt[-1/2 + I/2]*Sqrt[1 + x]]

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{8}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{8}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}\)	\(269\)
default	\(\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{8}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{8}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}\)	\(269\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+1\right ) \ln \left (-\frac {64 \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+1\right ) \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{4} x +48 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+1\right ) x +40 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+1\right )-48 \sqrt {1+x}\, \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+1\right ) x +15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+1\right )+2 \sqrt {1+x}}{8 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2} x +x -1}\right )}{2}+\operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right ) \ln \left (-\frac {64 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{5} x -16 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{3} x -40 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{3}+24 \sqrt {1+x}\, \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2}+\operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right ) x +5 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )+7 \sqrt {1+x}}{8 \operatorname {RootOf}\left (32 \textit {\_Z}^{4}+8 \textit {\_Z}^{2}+1\right )^{2} x +x +1}\right )\)	\(409\)

[In]

int(1/(x^2+1)/(1+x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(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^(1/2)-1/2*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2)*a
rctan((2*(1+x)^(1/2)+(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))-1/8*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(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^(1/2)+1/2*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2
*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(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))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=\frac {1}{4} \, \sqrt {2} \sqrt {i - 1} \log \left (-\left (i - 1\right ) \, \sqrt {2} \sqrt {i - 1} + 2 \, \sqrt {x + 1}\right ) - \frac {1}{4} \, \sqrt {2} \sqrt {i - 1} \log \left (\left (i - 1\right ) \, \sqrt {2} \sqrt {i - 1} + 2 \, \sqrt {x + 1}\right ) + \frac {1}{4} \, \sqrt {2} \sqrt {-i - 1} \log \left (\left (i + 1\right ) \, \sqrt {2} \sqrt {-i - 1} + 2 \, \sqrt {x + 1}\right ) - \frac {1}{4} \, \sqrt {2} \sqrt {-i - 1} \log \left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {-i - 1} + 2 \, \sqrt {x + 1}\right ) \]

[In]

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

[Out]

1/4*sqrt(2)*sqrt(I - 1)*log(-(I - 1)*sqrt(2)*sqrt(I - 1) + 2*sqrt(x + 1)) - 1/4*sqrt(2)*sqrt(I - 1)*log((I - 1
)*sqrt(2)*sqrt(I - 1) + 2*sqrt(x + 1)) + 1/4*sqrt(2)*sqrt(-I - 1)*log((I + 1)*sqrt(2)*sqrt(-I - 1) + 2*sqrt(x
+ 1)) - 1/4*sqrt(2)*sqrt(-I - 1)*log(-(I + 1)*sqrt(2)*sqrt(-I - 1) + 2*sqrt(x + 1))

Sympy [F]

\[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=\int \frac {1}{\sqrt {x + 1} \left (x^{2} + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )} \sqrt {x + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=\frac {1}{2} \, \sqrt {\sqrt {2} + 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{2} \, \sqrt {\sqrt {2} + 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{4} \, \sqrt {\sqrt {2} - 1} \log \left (2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) - \frac {1}{4} \, \sqrt {\sqrt {2} - 1} \log \left (-2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.66 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx=\mathrm {atanh}\left (\frac {16\,\sqrt {2}\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {x+1}}{128\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}-8}-\frac {16\,\sqrt {2}\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {x+1}}{128\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}-8}\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}+2\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\right )-\mathrm {atanh}\left (\frac {16\,\sqrt {2}\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {x+1}}{128\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}+8}+\frac {16\,\sqrt {2}\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {x+1}}{128\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}+8}\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{16}-\frac {1}{16}}-2\,\sqrt {\frac {\sqrt {2}}{16}-\frac {1}{16}}\right ) \]

[In]

int(1/((x^2 + 1)*(x + 1)^(1/2)),x)

[Out]

atanh((16*2^(1/2)*(- 2^(1/2)/16 - 1/16)^(1/2)*(x + 1)^(1/2))/(128*(2^(1/2)/16 - 1/16)^(1/2)*(- 2^(1/2)/16 - 1/
16)^(1/2) - 8) - (16*2^(1/2)*(2^(1/2)/16 - 1/16)^(1/2)*(x + 1)^(1/2))/(128*(2^(1/2)/16 - 1/16)^(1/2)*(- 2^(1/2
)/16 - 1/16)^(1/2) - 8))*(2*(- 2^(1/2)/16 - 1/16)^(1/2) + 2*(2^(1/2)/16 - 1/16)^(1/2)) - atanh((16*2^(1/2)*(-
2^(1/2)/16 - 1/16)^(1/2)*(x + 1)^(1/2))/(128*(2^(1/2)/16 - 1/16)^(1/2)*(- 2^(1/2)/16 - 1/16)^(1/2) + 8) + (16*
2^(1/2)*(2^(1/2)/16 - 1/16)^(1/2)*(x + 1)^(1/2))/(128*(2^(1/2)/16 - 1/16)^(1/2)*(- 2^(1/2)/16 - 1/16)^(1/2) +
8))*(2*(- 2^(1/2)/16 - 1/16)^(1/2) - 2*(2^(1/2)/16 - 1/16)^(1/2))