\(\int \frac {x \sqrt {1+x}}{1+x^2} \, dx\) [619]

Optimal result

Integrand size = 16, antiderivative size = 168 \[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=2 \sqrt {1+x}+\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )-\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}}{1+\sqrt {2}+x}\right ) \] Output:

2*(1+x)^(1/2)+1/2*(-2+2*2^(1/2))^(1/2)*arctan(((2+2*2^(1/2))^(1/2)-2*(1+x) 
^(1/2))/(-2+2*2^(1/2))^(1/2))-1/2*(-2+2*2^(1/2))^(1/2)*arctan(((2+2*2^(1/2 
))^(1/2)+2*(1+x)^(1/2))/(-2+2*2^(1/2))^(1/2))-1/2*(2+2*2^(1/2))^(1/2)*arct 
anh((2+2*2^(1/2))^(1/2)*(1+x)^(1/2)/(1+2^(1/2)+x))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.40 \[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=2 \sqrt {1+x}-\sqrt {-1+i} \arctan \left (\sqrt {-\frac {1}{2}-\frac {i}{2}} \sqrt {1+x}\right )-\sqrt {-1-i} \arctan \left (\sqrt {-\frac {1}{2}+\frac {i}{2}} \sqrt {1+x}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.28, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {561, 25, 1602, 25, 1483, 1142, 25, 1083, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x \sqrt {x+1}}{x^2+1} \, dx\)

\(\Big \downarrow \) 561

\(\displaystyle 2 \int \frac {x (x+1)}{(x+1)^2-2 (x+1)+2}d\sqrt {x+1}\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int -\frac {x (x+1)}{(x+1)^2-2 (x+1)+2}d\sqrt {x+1}\)

\(\Big \downarrow \) 1602

\(\displaystyle 2 \left (\int -\frac {1-x}{(x+1)^2-2 (x+1)+2}d\sqrt {x+1}+\sqrt {x+1}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\sqrt {x+1}-\int \frac {1-x}{(x+1)^2-2 (x+1)+2}d\sqrt {x+1}\right )\)

\(\Big \downarrow \) 1483

\(\displaystyle 2 \left (-\frac {\int \frac {2 \sqrt {2 \left (1+\sqrt {2}\right )}-\left (2+\sqrt {2}\right ) \sqrt {x+1}}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}}{4 \sqrt {1+\sqrt {2}}}-\frac {\int \frac {\left (2+\sqrt {2}\right ) \sqrt {x+1}+2 \sqrt {2 \left (1+\sqrt {2}\right )}}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}}{4 \sqrt {1+\sqrt {2}}}+\sqrt {x+1}\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle 2 \left (-\frac {\sqrt {\sqrt {2}-1} \int \frac {1}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}-\frac {1}{2} \left (2+\sqrt {2}\right ) \int -\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}}{4 \sqrt {1+\sqrt {2}}}-\frac {\sqrt {\sqrt {2}-1} \int \frac {1}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}+\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}}{4 \sqrt {1+\sqrt {2}}}+\sqrt {x+1}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (-\frac {\sqrt {\sqrt {2}-1} \int \frac {1}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}+\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}}{4 \sqrt {1+\sqrt {2}}}-\frac {\sqrt {\sqrt {2}-1} \int \frac {1}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}+\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}}{4 \sqrt {1+\sqrt {2}}}+\sqrt {x+1}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle 2 \left (-\frac {\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}-2 \sqrt {\sqrt {2}-1} \int \frac {1}{-x+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {x+1}-\sqrt {2 \left (1+\sqrt {2}\right )}\right )}{4 \sqrt {1+\sqrt {2}}}-\frac {\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}-2 \sqrt {\sqrt {2}-1} \int \frac {1}{-x+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}\right )}{4 \sqrt {1+\sqrt {2}}}+\sqrt {x+1}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle 2 \left (-\frac {\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}+\sqrt {2} \arctan \left (\frac {2 \sqrt {x+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{4 \sqrt {1+\sqrt {2}}}-\frac {\frac {1}{2} \left (2+\sqrt {2}\right ) \int \frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1}d\sqrt {x+1}+\sqrt {2} \arctan \left (\frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{4 \sqrt {1+\sqrt {2}}}+\sqrt {x+1}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle 2 \left (-\frac {\sqrt {2} \arctan \left (\frac {2 \sqrt {x+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )-\frac {1}{2} \left (2+\sqrt {2}\right ) \log \left (x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}-\frac {\sqrt {2} \arctan \left (\frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {1}{2} \left (2+\sqrt {2}\right ) \log \left (x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}+\sqrt {x+1}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217

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 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1083

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

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1483

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r)   In 
t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[1/(2*c*q*r)   Int[(d*r 
 + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && N 
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
 

rule 1602

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 
1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3))   Int[(f*x)^(m - 2)* 
(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p 
+ 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c 
, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | 
| IntegerQ[m])
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(2 \sqrt {x +1}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {x +1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(168\)
default	\(2 \sqrt {x +1}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {x +1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(168\)
risch	\(2 \sqrt {x +1}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {x +1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(168\)
trager	\(2 \sqrt {x +1}-2 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-256 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{5} x +16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{3} x +48 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \sqrt {x +1}+80 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{3}+6 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right ) x +\sqrt {x +1}+10 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )}{16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} x -x +1}\right )-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) \ln \left (\frac {64 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{4} x \operatorname {RootOf}\left (\textit {\_Z}^{2}+16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right )-12 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) x +20 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right )+48 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \sqrt {x +1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right )-7 \sqrt {x +1}}{16 \operatorname {RootOf}\left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} x -x -1}\right )}{2}\)	\(419\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.15 \[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=-\sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}} \arctan \left (2 \, \sqrt {x + 1} {\left (\sqrt {2} + 1\right )} \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}} + 2 \, {\left (\sqrt {2} + 1\right )} \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}}\right ) + \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}} \arctan \left (-2 \, \sqrt {x + 1} {\left (\sqrt {2} + 1\right )} \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}} + 2 \, {\left (\sqrt {2} + 1\right )} \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} \log \left (x + 2 \, \sqrt {x + 1} \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} + \sqrt {2} + 1\right ) + \frac {1}{2} \, \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} \log \left (x - 2 \, \sqrt {x + 1} \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} + \sqrt {2} + 1\right ) + 2 \, \sqrt {x + 1} \] Input:

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

Output:

-sqrt(1/2*sqrt(2) - 1/2)*arctan(2*sqrt(x + 1)*(sqrt(2) + 1)*sqrt(1/2*sqrt( 
2) - 1/2) + 2*(sqrt(2) + 1)*sqrt(1/2*sqrt(2) + 1/2)*sqrt(1/2*sqrt(2) - 1/2 
)) + sqrt(1/2*sqrt(2) - 1/2)*arctan(-2*sqrt(x + 1)*(sqrt(2) + 1)*sqrt(1/2* 
sqrt(2) - 1/2) + 2*(sqrt(2) + 1)*sqrt(1/2*sqrt(2) + 1/2)*sqrt(1/2*sqrt(2) 
- 1/2)) - 1/2*sqrt(1/2*sqrt(2) + 1/2)*log(x + 2*sqrt(x + 1)*sqrt(1/2*sqrt( 
2) + 1/2) + sqrt(2) + 1) + 1/2*sqrt(1/2*sqrt(2) + 1/2)*log(x - 2*sqrt(x + 
1)*sqrt(1/2*sqrt(2) + 1/2) + sqrt(2) + 1) + 2*sqrt(x + 1)
 

Sympy [F]

\[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=\int \frac {x \sqrt {x + 1}}{x^{2} + 1}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=\int { \frac {\sqrt {x + 1} x}{x^{2} + 1} \,d x } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.99 \[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=-\frac {1}{2} \, \sqrt {2 \, \sqrt {2} - 2} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {2} - 2} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {2} + 2} \log \left (2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {2} + 2} \log \left (-2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) + 2 \, \sqrt {x + 1} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.20 \[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=2\,\sqrt {x+1}+\mathrm {atanh}\left (\frac {\sqrt {x+1}}{4\,\sqrt {\frac {\sqrt {2}}{8}+\frac {1}{8}}}-\frac {\sqrt {x+1}}{4\,\sqrt {\frac {1}{8}-\frac {\sqrt {2}}{8}}}+\frac {\sqrt {2}\,\sqrt {x+1}}{8\,\sqrt {\frac {1}{8}-\frac {\sqrt {2}}{8}}}+\frac {\sqrt {2}\,\sqrt {x+1}}{8\,\sqrt {\frac {\sqrt {2}}{8}+\frac {1}{8}}}\right )\,\left (2\,\sqrt {\frac {1}{8}-\frac {\sqrt {2}}{8}}-2\,\sqrt {\frac {\sqrt {2}}{8}+\frac {1}{8}}\right )-\mathrm {atanh}\left (\frac {\sqrt {x+1}}{4\,\sqrt {\frac {1}{8}-\frac {\sqrt {2}}{8}}}+\frac {\sqrt {x+1}}{4\,\sqrt {\frac {\sqrt {2}}{8}+\frac {1}{8}}}-\frac {\sqrt {2}\,\sqrt {x+1}}{8\,\sqrt {\frac {1}{8}-\frac {\sqrt {2}}{8}}}+\frac {\sqrt {2}\,\sqrt {x+1}}{8\,\sqrt {\frac {\sqrt {2}}{8}+\frac {1}{8}}}\right )\,\left (2\,\sqrt {\frac {1}{8}-\frac {\sqrt {2}}{8}}+2\,\sqrt {\frac {\sqrt {2}}{8}+\frac {1}{8}}\right ) \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.81 \[ \int \frac {x \sqrt {1+x}}{1+x^2} \, dx=\frac {\sqrt {\sqrt {2}-1}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+1}\, \sqrt {2}-2 \sqrt {x +1}}{\sqrt {\sqrt {2}-1}\, \sqrt {2}}\right )}{2}-\frac {\sqrt {\sqrt {2}-1}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+1}\, \sqrt {2}+2 \sqrt {x +1}}{\sqrt {\sqrt {2}-1}\, \sqrt {2}}\right )}{2}+\frac {\sqrt {\sqrt {2}+1}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {x +1}\, \sqrt {\sqrt {2}+1}\, \sqrt {2}+\sqrt {2}+x +1\right )}{4}-\frac {\sqrt {\sqrt {2}+1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {x +1}\, \sqrt {\sqrt {2}+1}\, \sqrt {2}+\sqrt {2}+x +1\right )}{4}+2 \sqrt {x +1} \] Input:

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

Output:

(2*sqrt(sqrt(2) - 1)*sqrt(2)*atan((sqrt(sqrt(2) + 1)*sqrt(2) - 2*sqrt(x + 
1))/(sqrt(sqrt(2) - 1)*sqrt(2))) - 2*sqrt(sqrt(2) - 1)*sqrt(2)*atan((sqrt( 
sqrt(2) + 1)*sqrt(2) + 2*sqrt(x + 1))/(sqrt(sqrt(2) - 1)*sqrt(2))) + sqrt( 
sqrt(2) + 1)*sqrt(2)*log( - sqrt(x + 1)*sqrt(sqrt(2) + 1)*sqrt(2) + sqrt(2 
) + x + 1) - sqrt(sqrt(2) + 1)*sqrt(2)*log(sqrt(x + 1)*sqrt(sqrt(2) + 1)*s 
qrt(2) + sqrt(2) + x + 1) + 8*sqrt(x + 1))/4