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

Optimal. Leaf size=205 \[ \frac {\log \left (x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\log \left (x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {700, 1127, 1161, 618, 204, 1164, 628} \begin {gather*} \frac {\log \left (x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\log \left (x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \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[Sqrt[1 + x]/(1 + x^2),x]

[Out]

-(Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]) + Sqrt[(1 + Sq
rt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]] + Log[1 + Sqrt[2] + x - Sqrt[
2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(2*Sqrt[2*(1 + Sqrt[2])]) - Log[1 + Sqrt[2] + x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 +
x]]/(2*Sqrt[2*(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 700

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

Rule 1127

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

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1164

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

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 55, normalized size = 0.27 \begin {gather*} i \sqrt {1+i} \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {1+i}}\right )-i \sqrt {1-i} \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {1-i}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x]/(1 + x^2),x]

[Out]

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

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IntegrateAlgebraic [C]  time = 0.11, size = 57, normalized size = 0.28 \begin {gather*} \sqrt {1-i} \tan ^{-1}\left (\sqrt {-\frac {1}{2}-\frac {i}{2}} \sqrt {x+1}\right )+\sqrt {1+i} \tan ^{-1}\left (\sqrt {-\frac {1}{2}+\frac {i}{2}} \sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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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fricas [A]  time = 0.43, size = 260, normalized size = 1.27 \begin {gather*} \frac {1}{8} \cdot 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + x + \sqrt {2} + 1\right ) - \frac {1}{8} \cdot 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (-\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {x + 1} \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}{2} \cdot 2^{\frac {3}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + \frac {1}{2} \cdot 2^{\frac {1}{4}} \sqrt {2^{\frac {3}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 2 \, x + 2 \, \sqrt {2} + 2} \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}{2} \cdot 2^{\frac {3}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + \frac {1}{2} \cdot 2^{\frac {1}{4}} \sqrt {-2^{\frac {3}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 2 \, x + 2 \, \sqrt {2} + 2} \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)^(1/2)/(x^2+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.94, size = 160, 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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maple [B]  time = 0.70, size = 336, normalized size = 1.64 \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 )}{2 \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}\, \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 {\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+2 \sqrt {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}}\, \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 )}{4}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {2}+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x + 1}}{x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + 1)/(x^2 + 1), x)

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mupad [B]  time = 0.12, size = 109, normalized size = 0.53 \begin {gather*} \mathrm {atanh}\left (4\,{\left (\sqrt {-\frac {\sqrt {2}}{8}-\frac {1}{8}}+\sqrt {\frac {\sqrt {2}}{8}-\frac {1}{8}}\right )}^3\,\sqrt {x+1}\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{8}-\frac {1}{8}}+2\,\sqrt {\frac {\sqrt {2}}{8}-\frac {1}{8}}\right )+\mathrm {atanh}\left (4\,{\left (\sqrt {-\frac {\sqrt {2}}{8}-\frac {1}{8}}-\sqrt {\frac {\sqrt {2}}{8}-\frac {1}{8}}\right )}^3\,\sqrt {x+1}\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{8}-\frac {1}{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 + 1)^(1/2)/(x^2 + 1),x)

[Out]

atanh(4*((- 2^(1/2)/8 - 1/8)^(1/2) + (2^(1/2)/8 - 1/8)^(1/2))^3*(x + 1)^(1/2))*(2*(- 2^(1/2)/8 - 1/8)^(1/2) +
2*(2^(1/2)/8 - 1/8)^(1/2)) + atanh(4*((- 2^(1/2)/8 - 1/8)^(1/2) - (2^(1/2)/8 - 1/8)^(1/2))^3*(x + 1)^(1/2))*(2
*(- 2^(1/2)/8 - 1/8)^(1/2) - 2*(2^(1/2)/8 - 1/8)^(1/2))

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sympy [A]  time = 3.92, size = 31, normalized size = 0.15 \begin {gather*} 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((1+x)**(1/2)/(x**2+1),x)

[Out]

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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