3.12 \(\int \frac{\sqrt{x^2+\sqrt{1+x^4}}}{(1+x)^2 \sqrt{1+x^4}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{\sqrt{1-i x^2}}{2 (x+1)}-\frac{\sqrt{1+i x^2}}{2 (x+1)}-\frac{1}{4} (1-i)^{3/2} \tanh ^{-1}\left (\frac{1+i x}{\sqrt{1-i} \sqrt{1-i x^2}}\right )-\frac{1}{4} (1+i)^{3/2} \tanh ^{-1}\left (\frac{1-i x}{\sqrt{1+i} \sqrt{1+i x^2}}\right ) \]

[Out]

-Sqrt[1 - I*x^2]/(2*(1 + x)) - Sqrt[1 + I*x^2]/(2*(1 + x)) - ((1 - I)^(3/2)*ArcTanh[(1 + I*x)/(Sqrt[1 - I]*Sqr
t[1 - I*x^2])])/4 - ((1 + I)^(3/2)*ArcTanh[(1 - I*x)/(Sqrt[1 + I]*Sqrt[1 + I*x^2])])/4

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Rubi [A]  time = 0.185294, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2133, 731, 725, 206} \[ -\frac{\sqrt{1-i x^2}}{2 (x+1)}-\frac{\sqrt{1+i x^2}}{2 (x+1)}-\frac{1}{4} (1-i)^{3/2} \tanh ^{-1}\left (\frac{1+i x}{\sqrt{1-i} \sqrt{1-i x^2}}\right )-\frac{1}{4} (1+i)^{3/2} \tanh ^{-1}\left (\frac{1-i x}{\sqrt{1+i} \sqrt{1+i x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^2 + Sqrt[1 + x^4]]/((1 + x)^2*Sqrt[1 + x^4]),x]

[Out]

-Sqrt[1 - I*x^2]/(2*(1 + x)) - Sqrt[1 + I*x^2]/(2*(1 + x)) - ((1 - I)^(3/2)*ArcTanh[(1 + I*x)/(Sqrt[1 - I]*Sqr
t[1 - I*x^2])])/4 - ((1 + I)^(3/2)*ArcTanh[(1 - I*x)/(Sqrt[1 + I]*Sqrt[1 + I*x^2])])/4

Rule 2133

Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_S
ymbol] :> Dist[(1 - I)/2, Int[(c + d*x)^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Dist[(1 + I)/2, Int[(c + d*x)^m/Sq
rt[Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && GtQ[a, 0]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 725

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\sqrt{x^2+\sqrt{1+x^4}}}{(1+x)^2 \sqrt{1+x^4}} \, dx &=\left (\frac{1}{2}-\frac{i}{2}\right ) \int \frac{1}{(1+x)^2 \sqrt{1-i x^2}} \, dx+\left (\frac{1}{2}+\frac{i}{2}\right ) \int \frac{1}{(1+x)^2 \sqrt{1+i x^2}} \, dx\\ &=-\frac{\sqrt{1-i x^2}}{2 (1+x)}-\frac{\sqrt{1+i x^2}}{2 (1+x)}-\frac{1}{2} i \int \frac{1}{(1+x) \sqrt{1-i x^2}} \, dx+\frac{1}{2} i \int \frac{1}{(1+x) \sqrt{1+i x^2}} \, dx\\ &=-\frac{\sqrt{1-i x^2}}{2 (1+x)}-\frac{\sqrt{1+i x^2}}{2 (1+x)}+\frac{1}{2} i \operatorname{Subst}\left (\int \frac{1}{(1-i)-x^2} \, dx,x,\frac{1+i x}{\sqrt{1-i x^2}}\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{1}{(1+i)-x^2} \, dx,x,\frac{1-i x}{\sqrt{1+i x^2}}\right )\\ &=-\frac{\sqrt{1-i x^2}}{2 (1+x)}-\frac{\sqrt{1+i x^2}}{2 (1+x)}-\frac{1}{4} (1-i)^{3/2} \tanh ^{-1}\left (\frac{1+i x}{\sqrt{1-i} \sqrt{1-i x^2}}\right )-\frac{1}{4} (1+i)^{3/2} \tanh ^{-1}\left (\frac{1-i x}{\sqrt{1+i} \sqrt{1+i x^2}}\right )\\ \end{align*}

Mathematica [F]  time = 0.239471, size = 0, normalized size = 0. \[ \int \frac{\sqrt{x^2+\sqrt{1+x^4}}}{(1+x)^2 \sqrt{1+x^4}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sqrt[x^2 + Sqrt[1 + x^4]]/((1 + x)^2*Sqrt[1 + x^4]),x]

[Out]

Integrate[Sqrt[x^2 + Sqrt[1 + x^4]]/((1 + x)^2*Sqrt[1 + x^4]), x]

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( 1+x \right ) ^{2}}\sqrt{{x}^{2}+\sqrt{{x}^{4}+1}}{\frac{1}{\sqrt{{x}^{4}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x)

[Out]

int((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + \sqrt{x^{4} + 1}}}{\sqrt{x^{4} + 1}{\left (x + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 + sqrt(x^4 + 1))/(sqrt(x^4 + 1)*(x + 1)^2), x)

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Fricas [B]  time = 19.3436, size = 1068, normalized size = 8.54 \begin{align*} \frac{4 \,{\left (x + 1\right )} \sqrt{\sqrt{2} + 1} \arctan \left (\frac{2 \,{\left (x^{3} + x^{2} - \sqrt{2}{\left (x^{3} + 1\right )} + \sqrt{x^{4} + 1}{\left (\sqrt{2} x - x - 1\right )} - x + 1\right )} \sqrt{x^{2} + \sqrt{x^{4} + 1}} \sqrt{\sqrt{2} + 1} +{\left (2 \, x^{2} - \sqrt{2}{\left (x^{2} + 1\right )} + 2 \, \sqrt{x^{4} + 1}{\left (\sqrt{2} - 1\right )} + 2\right )} \sqrt{2 \, \sqrt{2} + 2} \sqrt{\sqrt{2} + 1}}{2 \,{\left (x^{2} - 2 \, x + 1\right )}}\right ) +{\left (x + 1\right )} \sqrt{\sqrt{2} - 1} \log \left (-\frac{{\left (2 \, x^{3} - \sqrt{2}{\left (x^{3} - x^{2} - x - 1\right )} + \sqrt{x^{4} + 1}{\left (\sqrt{2}{\left (x - 1\right )} - 2 \, x\right )} - 2\right )} \sqrt{x^{2} + \sqrt{x^{4} + 1}} +{\left (\sqrt{2}{\left (x^{2} + 1\right )} + 2 \, \sqrt{x^{4} + 1}\right )} \sqrt{\sqrt{2} - 1}}{x^{2} + 2 \, x + 1}\right ) -{\left (x + 1\right )} \sqrt{\sqrt{2} - 1} \log \left (-\frac{{\left (2 \, x^{3} - \sqrt{2}{\left (x^{3} - x^{2} - x - 1\right )} + \sqrt{x^{4} + 1}{\left (\sqrt{2}{\left (x - 1\right )} - 2 \, x\right )} - 2\right )} \sqrt{x^{2} + \sqrt{x^{4} + 1}} -{\left (\sqrt{2}{\left (x^{2} + 1\right )} + 2 \, \sqrt{x^{4} + 1}\right )} \sqrt{\sqrt{2} - 1}}{x^{2} + 2 \, x + 1}\right ) + 4 \, \sqrt{x^{2} + \sqrt{x^{4} + 1}}{\left (x^{2} - \sqrt{x^{4} + 1} - 1\right )}}{8 \,{\left (x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + \sqrt{x^{4} + 1}}}{\left (x + 1\right )^{2} \sqrt{x^{4} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+(x**4+1)**(1/2))**(1/2)/(1+x)**2/(x**4+1)**(1/2),x)

[Out]

Integral(sqrt(x**2 + sqrt(x**4 + 1))/((x + 1)**2*sqrt(x**4 + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + \sqrt{x^{4} + 1}}}{\sqrt{x^{4} + 1}{\left (x + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(x^2 + sqrt(x^4 + 1))/(sqrt(x^4 + 1)*(x + 1)^2), x)