3.4.15 \(\int \frac {x \sqrt {1+x}}{1+x^2} \, dx\) [315]

Optimal. Leaf size=214 \[ 2 \sqrt {1+x}+\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {2 \left (1+\sqrt {2}\right )}}+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (1+\sqrt {2}+x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (1+\sqrt {2}+x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {839, 841, 1183, 648, 632, 210, 642} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\text {ArcTan}\left (\frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (1+\sqrt {2}\right )}}+2 \sqrt {x+1}+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[1 + x])/(1 + x^2),x]

[Out]

2*Sqrt[1 + x] + ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(1 + Sqrt[2])] -
ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(1 + Sqrt[2])] + (Sqrt[(1 + Sqr
t[2])/2]*Log[1 + Sqrt[2] + x - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]])/2 - (Sqrt[(1 + Sqrt[2])/2]*Log[1 + Sqrt[2]
+ x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]])/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 839

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(
c*m)), x] + Dist[1/c, Int[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x]/(a + c*x^2)), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 841

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

Rule 1183

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]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {x \sqrt {1+x}}{1+x^2} \, dx &=2 \sqrt {1+x}+\int \frac {-1+x}{\sqrt {1+x} \left (1+x^2\right )} \, dx\\ &=2 \sqrt {1+x}+2 \text {Subst}\left (\int \frac {-2+x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+x}\right )\\ &=2 \sqrt {1+x}+\frac {\text {Subst}\left (\int \frac {-2 \sqrt {2 \left (1+\sqrt {2}\right )}-\left (-2-\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 {-2 \sqrt {2 \left (1+\sqrt {2}\right )}+\left (-2-\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}}}\\ &=2 \sqrt {1+x}-\frac {1}{2} \sqrt {3-2 \sqrt {2}} \text {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} \sqrt {3-2 \sqrt {2}} \text {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} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \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 )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \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 )\\ &=2 \sqrt {1+x}+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (1+\sqrt {2}+x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (1+\sqrt {2}+x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )+\sqrt {3-2 \sqrt {2}} \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 {3-2 \sqrt {2}} \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 )\\ &=2 \sqrt {1+x}+\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \tan ^{-1}\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 )} \tan ^{-1}\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 {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (1+\sqrt {2}+x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \log \left (1+\sqrt {2}+x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.08, size = 68, normalized size = 0.32 \begin {gather*} 2 \sqrt {1+x}-\sqrt {-1+i} \tan ^{-1}\left (\sqrt {-\frac {1}{2}-\frac {i}{2}} \sqrt {1+x}\right )-\sqrt {-1-i} \tan ^{-1}\left (\sqrt {-\frac {1}{2}+\frac {i}{2}} \sqrt {1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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]]

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Maple [A]
time = 0.51, size = 167, normalized size = 0.78

method	result	size
derivativedivides	\(2 \sqrt {1+x}+\frac {\ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(167\)
default	\(2 \sqrt {1+x}+\frac {\ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(167\)
risch	\(2 \sqrt {1+x}+\frac {\ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{4}-\frac {\arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{\sqrt {-2+2 \sqrt {2}}}+\frac {\arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{4}+\frac {\arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{\sqrt {-2+2 \sqrt {2}}}\)	\(240\)
trager	\(2 \sqrt {1+x}+\frac {\RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) \ln \left (-\frac {64 \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{4} x -12 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) x -48 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \sqrt {1+x}+20 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right )-\RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right ) x +7 \sqrt {1+x}-5 \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2}-2\right )}{16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} x -x -1}\right )}{2}+2 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right ) \ln \left (-\frac {256 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{5} x -16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{3} x -80 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{3}+48 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} \sqrt {1+x}-6 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right ) x -10 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )+\sqrt {1+x}}{16 \RootOf \left (128 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}+1\right )^{2} x -x +1}\right )\)	\(420\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + 1)*x/(x^2 + 1), x)

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Fricas [A]
time = 2.72, size = 307, normalized size = 1.43 \begin {gather*} -\frac {1}{8} \cdot 2^{\frac {1}{4}} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} \log \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} \sqrt {x + 1} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} + x + \sqrt {2} + 1\right ) + \frac {1}{8} \cdot 2^{\frac {1}{4}} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} \log \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} \sqrt {x + 1} {\left (\sqrt {2} + 2\right )} \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}{4} \cdot 2^{\frac {3}{4}} \sqrt {2^{\frac {1}{4}} \sqrt {x + 1} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} + 2 \, x + 2 \, \sqrt {2} + 2} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} - \frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {x + 1} {\left (\sqrt {2} + 1\right )} \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}{4} \cdot 2^{\frac {3}{4}} \sqrt {-2^{\frac {1}{4}} \sqrt {x + 1} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} + 2 \, x + 2 \, \sqrt {2} + 2} {\left (\sqrt {2} + 2\right )} \sqrt {-2 \, \sqrt {2} + 4} - \frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {x + 1} {\left (\sqrt {2} + 1\right )} \sqrt {-2 \, \sqrt {2} + 4} + \sqrt {2} + 1\right ) + 2 \, \sqrt {x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 4.07, size = 68, normalized size = 0.32 \begin {gather*} 2 \sqrt {x + 1} - 4 \operatorname {RootSum} {\left (512 t^{4} + 32 t^{2} + 1, \left ( t \mapsto t \log {\left (- 128 t^{3} + \sqrt {x + 1} \right )} \right )\right )} + 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x + 1) - 4*RootSum(512*_t**4 + 32*_t**2 + 1, Lambda(_t, _t*log(-128*_t**3 + sqrt(x + 1)))) + 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 [A]
time = 1.19, size = 167, normalized size = 0.78 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Mupad [B]
time = 0.11, size = 201, normalized size = 0.94 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x + 1)^(1/2))/(x^2 + 1),x)

[Out]

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))

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