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

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

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Rubi [A]  time = 0.13, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {708, 1094, 634, 618, 204, 628} \begin {gather*} -\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 ) \end {gather*}

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 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 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.02, size = 55, normalized size = 0.28 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [C]  time = 0.11, size = 65, normalized size = 0.33 \begin {gather*} \sqrt {\frac {1}{2}+\frac {i}{2}} \tan ^{-1}\left (\sqrt {-\frac {1}{2}-\frac {i}{2}} \sqrt {x+1}\right )+\sqrt {\frac {1}{2}-\frac {i}{2}} \tan ^{-1}\left (\sqrt {-\frac {1}{2}+\frac {i}{2}} \sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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]]

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fricas [A]  time = 0.43, size = 269, normalized size = 1.36 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.54, size = 152, normalized size = 0.77 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.13, size = 420, normalized size = 2.12 \begin {gather*} \frac {\sqrt {2}\, \left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{4 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2}\, \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2}\, \left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{4 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2}\, \arctan \left (\frac {2 \sqrt {x +1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {2}-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{4}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x +1+\sqrt {2}-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{8}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {2}+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{4}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x +1+\sqrt {2}+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2+2*2^(1/2))^(1/2)*ln(x+1+2^(1/2)+(x+1)^(1/2)*(2+2*2^(1/2))^(1/2))-1/8*(2+2*2^(1/2))^(1/2)*2^(1/2)*ln(x+1
+2^(1/2)+(x+1)^(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*(x+1)^(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*(x+1)^(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*(x+1)^(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(x+1+2^(1/2)-(x+1)^(1/2)*(2+2*2^(1/2))^(1/2))+1/8*(2+2*2^(1/
2))^(1/2)*2^(1/2)*ln(x+1+2^(1/2)-(x+1)^(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*(x+1)^(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*(x+1)^(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*(x+1)^(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} + 1\right )} \sqrt {x + 1}}\,{d x} \end {gather*}

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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mupad [B]  time = 0.46, size = 226, normalized size = 1.14 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x + 1} \left (x^{2} + 1\right )}\, dx \end {gather*}

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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