3.2929 \(\int \frac{1}{1+(x^2)^{3/2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac{x \log \left (x^2-\sqrt{x^2}+1\right )}{6 \sqrt{x^2}}+\frac{x \log \left (\sqrt{x^2}+1\right )}{3 \sqrt{x^2}}-\frac{x \tan ^{-1}\left (\frac{1-2 \sqrt{x^2}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt{x^2}} \]

[Out]

-((x*ArcTan[(1 - 2*Sqrt[x^2])/Sqrt[3]])/(Sqrt[3]*Sqrt[x^2])) - (x*Log[1 + x^2 - Sqrt[x^2]])/(6*Sqrt[x^2]) + (x
*Log[1 + Sqrt[x^2]])/(3*Sqrt[x^2])

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Rubi [A]  time = 0.0385793, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {254, 200, 31, 634, 618, 204, 628} \[ -\frac{x \log \left (x^2-\sqrt{x^2}+1\right )}{6 \sqrt{x^2}}+\frac{x \log \left (\sqrt{x^2}+1\right )}{3 \sqrt{x^2}}-\frac{x \tan ^{-1}\left (\frac{1-2 \sqrt{x^2}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + (x^2)^(3/2))^(-1),x]

[Out]

-((x*ArcTan[(1 - 2*Sqrt[x^2])/Sqrt[3]])/(Sqrt[3]*Sqrt[x^2])) - (x*Log[1 + x^2 - Sqrt[x^2]])/(6*Sqrt[x^2]) + (x
*Log[1 + Sqrt[x^2]])/(3*Sqrt[x^2])

Rule 254

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> Dist[x/(c*x^q)^(1/q), Subst[Int[(a + b*x^(n*q))
^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, n, p, q}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{1+\left (x^2\right )^{3/2}} \, dx &=\frac{x \operatorname{Subst}\left (\int \frac{1}{1+x^3} \, dx,x,\sqrt{x^2}\right )}{\sqrt{x^2}}\\ &=\frac{x \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\sqrt{x^2}\right )}{3 \sqrt{x^2}}+\frac{x \operatorname{Subst}\left (\int \frac{2-x}{1-x+x^2} \, dx,x,\sqrt{x^2}\right )}{3 \sqrt{x^2}}\\ &=\frac{x \log \left (1+\sqrt{x^2}\right )}{3 \sqrt{x^2}}-\frac{x \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\sqrt{x^2}\right )}{6 \sqrt{x^2}}+\frac{x \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\sqrt{x^2}\right )}{2 \sqrt{x^2}}\\ &=-\frac{x \log \left (1+x^2-\sqrt{x^2}\right )}{6 \sqrt{x^2}}+\frac{x \log \left (1+\sqrt{x^2}\right )}{3 \sqrt{x^2}}-\frac{x \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \sqrt{x^2}\right )}{\sqrt{x^2}}\\ &=-\frac{x \tan ^{-1}\left (\frac{1-2 \sqrt{x^2}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt{x^2}}-\frac{x \log \left (1+x^2-\sqrt{x^2}\right )}{6 \sqrt{x^2}}+\frac{x \log \left (1+\sqrt{x^2}\right )}{3 \sqrt{x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0217577, size = 67, normalized size = 0.81 \[ \frac{x \left (-\log \left (x^2-\sqrt{x^2}+1\right )+2 \log \left (\sqrt{x^2}+1\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt{x^2}-1}{\sqrt{3}}\right )\right )}{6 \sqrt{x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + (x^2)^(3/2))^(-1),x]

[Out]

(x*(2*Sqrt[3]*ArcTan[(-1 + 2*Sqrt[x^2])/Sqrt[3]] - Log[1 + x^2 - Sqrt[x^2]] + 2*Log[1 + Sqrt[x^2]]))/(6*Sqrt[x
^2])

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Maple [A]  time = 0.008, size = 108, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}}{6} \left ( -2\,\sqrt{3}\arctan \left ( 1/3\,{\sqrt{3} \left ( -2\,x+\sqrt [3]{{\frac{{x}^{3}}{ \left ({x}^{2} \right ) ^{3/2}}}} \right ){\frac{1}{\sqrt [3]{{\frac{{x}^{3}}{ \left ({x}^{2} \right ) ^{3/2}}}}}}} \right ) +2\,\ln \left ( x+\sqrt [3]{{\frac{{x}^{3}}{ \left ({x}^{2} \right ) ^{3/2}}}} \right ) -\ln \left ({x}^{2}-x\sqrt [3]{{{x}^{3} \left ({x}^{2} \right ) ^{-{\frac{3}{2}}}}}+ \left ({{x}^{3} \left ({x}^{2} \right ) ^{-{\frac{3}{2}}}} \right ) ^{{\frac{2}{3}}} \right ) \right ) \left ({x}^{2} \right ) ^{-{\frac{3}{2}}} \left ({{x}^{3} \left ({x}^{2} \right ) ^{-{\frac{3}{2}}}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+(x^2)^(3/2)),x)

[Out]

1/6*x^3*(-2*3^(1/2)*arctan(1/3*3^(1/2)*(-2*x+(1/(x^2)^(3/2)*x^3)^(1/3))/(1/(x^2)^(3/2)*x^3)^(1/3))+2*ln(x+(1/(
x^2)^(3/2)*x^3)^(1/3))-ln(x^2-x*(1/(x^2)^(3/2)*x^3)^(1/3)+(1/(x^2)^(3/2)*x^3)^(2/3)))/(x^2)^(3/2)/(1/(x^2)^(3/
2)*x^3)^(2/3)

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Maxima [A]  time = 1.41972, size = 46, normalized size = 0.55 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac{1}{3} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+(x^2)^(3/2)),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) + 1/3*log(x + 1)

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Fricas [A]  time = 1.30044, size = 153, normalized size = 1.84 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{2}{3} \, \sqrt{3} \sqrt{x^{2}} - \frac{1}{3} \, \sqrt{3}\right ) - \frac{1}{6} \, \log \left (x^{2} - \sqrt{x^{2}} + 1\right ) + \frac{1}{3} \, \log \left (\sqrt{x^{2}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+(x^2)^(3/2)),x, algorithm="fricas")

[Out]

1/3*sqrt(3)*arctan(2/3*sqrt(3)*sqrt(x^2) - 1/3*sqrt(3)) - 1/6*log(x^2 - sqrt(x^2) + 1) + 1/3*log(sqrt(x^2) + 1
)

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Sympy [A]  time = 0.130622, size = 41, normalized size = 0.49 \begin{align*} \frac{\log{\left (x + 1 \right )}}{3} - \frac{\log{\left (x^{2} - x + 1 \right )}}{6} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+(x**2)**(3/2)),x)

[Out]

log(x + 1)/3 - log(x**2 - x + 1)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/3

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Giac [C]  time = 1.20624, size = 146, normalized size = 1.76 \begin{align*} -\frac{\sqrt{3}{\left (-i \, \sqrt{3} - 1\right )} \arctan \left (\frac{\sqrt{3}{\left (2 \, x + \left (-\frac{1}{\mathrm{sgn}\left (x\right )}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{1}{\mathrm{sgn}\left (x\right )}\right )^{\frac{1}{3}}}\right )}{6 \, \mathrm{sgn}\left (x\right )^{\frac{1}{3}}} - \frac{1}{9} i \, \pi \mathrm{sgn}\left (x\right ) - \frac{{\left (-i \, \sqrt{3} - 1\right )} \log \left (x^{2} + x \left (-\frac{1}{\mathrm{sgn}\left (x\right )}\right )^{\frac{1}{3}} + \left (-\frac{1}{\mathrm{sgn}\left (x\right )}\right )^{\frac{2}{3}}\right )}{12 \, \mathrm{sgn}\left (x\right )^{\frac{1}{3}}} - \frac{1}{3} \, \left (-\frac{1}{\mathrm{sgn}\left (x\right )}\right )^{\frac{1}{3}} \log \left ({\left | x - \left (-\frac{1}{\mathrm{sgn}\left (x\right )}\right )^{\frac{1}{3}} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+(x^2)^(3/2)),x, algorithm="giac")

[Out]

-1/6*sqrt(3)*(-I*sqrt(3) - 1)*arctan(1/3*sqrt(3)*(2*x + (-1/sgn(x))^(1/3))/(-1/sgn(x))^(1/3))/sgn(x)^(1/3) - 1
/9*I*pi*sgn(x) - 1/12*(-I*sqrt(3) - 1)*log(x^2 + x*(-1/sgn(x))^(1/3) + (-1/sgn(x))^(2/3))/sgn(x)^(1/3) - 1/3*(
-1/sgn(x))^(1/3)*log(abs(x - (-1/sgn(x))^(1/3)))