3.720 \(\int \frac{1}{(1+x) \sqrt [6]{1+x^2}} \, dx\)

Optimal. Leaf size=203 \[ x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )+\frac{\log \left (\sqrt [3]{x^2+1}-\sqrt [6]{2} \sqrt [6]{x^2+1}+\sqrt [3]{2}\right )}{4 \sqrt [6]{2}}-\frac{\log \left (\sqrt [3]{x^2+1}+\sqrt [6]{2} \sqrt [6]{x^2+1}+\sqrt [3]{2}\right )}{4 \sqrt [6]{2}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-2^{5/6} \sqrt [6]{x^2+1}}{\sqrt{3}}\right )}{2 \sqrt [6]{2}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{2^{5/6} \sqrt [6]{x^2+1}+1}{\sqrt{3}}\right )}{2 \sqrt [6]{2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{x^2+1}}{\sqrt [6]{2}}\right )}{\sqrt [6]{2}} \]

[Out]

x*AppellF1[1/2, 1, 1/6, 3/2, x^2, -x^2] - (Sqrt[3]*ArcTan[(1 - 2^(5/6)*(1 + x^2)^(1/6))/Sqrt[3]])/(2*2^(1/6))
+ (Sqrt[3]*ArcTan[(1 + 2^(5/6)*(1 + x^2)^(1/6))/Sqrt[3]])/(2*2^(1/6)) - ArcTanh[(1 + x^2)^(1/6)/2^(1/6)]/2^(1/
6) + Log[2^(1/3) - 2^(1/6)*(1 + x^2)^(1/6) + (1 + x^2)^(1/3)]/(4*2^(1/6)) - Log[2^(1/3) + 2^(1/6)*(1 + x^2)^(1
/6) + (1 + x^2)^(1/3)]/(4*2^(1/6))

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Rubi [A]  time = 0.367675, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {757, 429, 444, 63, 296, 634, 618, 204, 628, 206} \[ x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )+\frac{\log \left (\sqrt [3]{x^2+1}-\sqrt [6]{2} \sqrt [6]{x^2+1}+\sqrt [3]{2}\right )}{4 \sqrt [6]{2}}-\frac{\log \left (\sqrt [3]{x^2+1}+\sqrt [6]{2} \sqrt [6]{x^2+1}+\sqrt [3]{2}\right )}{4 \sqrt [6]{2}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-2^{5/6} \sqrt [6]{x^2+1}}{\sqrt{3}}\right )}{2 \sqrt [6]{2}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{2^{5/6} \sqrt [6]{x^2+1}+1}{\sqrt{3}}\right )}{2 \sqrt [6]{2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{x^2+1}}{\sqrt [6]{2}}\right )}{\sqrt [6]{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + x)*(1 + x^2)^(1/6)),x]

[Out]

x*AppellF1[1/2, 1, 1/6, 3/2, x^2, -x^2] - (Sqrt[3]*ArcTan[(1 - 2^(5/6)*(1 + x^2)^(1/6))/Sqrt[3]])/(2*2^(1/6))
+ (Sqrt[3]*ArcTan[(1 + 2^(5/6)*(1 + x^2)^(1/6))/Sqrt[3]])/(2*2^(1/6)) - ArcTanh[(1 + x^2)^(1/6)/2^(1/6)]/2^(1/
6) + Log[2^(1/3) - 2^(1/6)*(1 + x^2)^(1/6) + (1 + x^2)^(1/3)]/(4*2^(1/6)) - Log[2^(1/3) + 2^(1/6)*(1 + x^2)^(1
/6) + (1 + x^2)^(1/3)]/(4*2^(1/6))

Rule 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - (e*x)/(d^2 - e^2*x^2))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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]

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{1}{(1+x) \sqrt [6]{1+x^2}} \, dx &=\int \left (\frac{1}{\left (1-x^2\right ) \sqrt [6]{1+x^2}}+\frac{x}{\left (-1+x^2\right ) \sqrt [6]{1+x^2}}\right ) \, dx\\ &=\int \frac{1}{\left (1-x^2\right ) \sqrt [6]{1+x^2}} \, dx+\int \frac{x}{\left (-1+x^2\right ) \sqrt [6]{1+x^2}} \, dx\\ &=x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt [6]{1+x}} \, dx,x,x^2\right )\\ &=x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )+3 \operatorname{Subst}\left (\int \frac{x^4}{-2+x^6} \, dx,x,\sqrt [6]{1+x^2}\right )\\ &=x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2^{5/6}}-\frac{x}{2}}{\sqrt [3]{2}-\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )}{\sqrt [6]{2}}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2^{5/6}}+\frac{x}{2}}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )}{\sqrt [6]{2}}-\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2}-x^2} \, dx,x,\sqrt [6]{1+x^2}\right )\\ &=x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{1+x^2}}{\sqrt [6]{2}}\right )}{\sqrt [6]{2}}+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2}-\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )+\frac{\operatorname{Subst}\left (\int \frac{-\sqrt [6]{2}+2 x}{\sqrt [3]{2}-\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )}{4 \sqrt [6]{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt [6]{2}+2 x}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )}{4 \sqrt [6]{2}}\\ &=x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{1+x^2}}{\sqrt [6]{2}}\right )}{\sqrt [6]{2}}+\frac{\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt [6]{1+x^2}+\sqrt [3]{1+x^2}\right )}{4 \sqrt [6]{2}}-\frac{\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt [6]{1+x^2}+\sqrt [3]{1+x^2}\right )}{4 \sqrt [6]{2}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-2^{5/6} \sqrt [6]{1+x^2}\right )}{2 \sqrt [6]{2}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2^{5/6} \sqrt [6]{1+x^2}\right )}{2 \sqrt [6]{2}}\\ &=x F_1\left (\frac{1}{2};1,\frac{1}{6};\frac{3}{2};x^2,-x^2\right )-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-2^{5/6} \sqrt [6]{1+x^2}}{\sqrt{3}}\right )}{2 \sqrt [6]{2}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{1+2^{5/6} \sqrt [6]{1+x^2}}{\sqrt{3}}\right )}{2 \sqrt [6]{2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{1+x^2}}{\sqrt [6]{2}}\right )}{\sqrt [6]{2}}+\frac{\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt [6]{1+x^2}+\sqrt [3]{1+x^2}\right )}{4 \sqrt [6]{2}}-\frac{\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt [6]{1+x^2}+\sqrt [3]{1+x^2}\right )}{4 \sqrt [6]{2}}\\ \end{align*}

Mathematica [C]  time = 0.0385877, size = 72, normalized size = 0.35 \[ -\frac{3 \sqrt [6]{\frac{x-i}{x+1}} \sqrt [6]{\frac{x+i}{x+1}} F_1\left (\frac{1}{3};\frac{1}{6},\frac{1}{6};\frac{4}{3};\frac{1-i}{x+1},\frac{1+i}{x+1}\right )}{\sqrt [6]{x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((1 + x)*(1 + x^2)^(1/6)),x]

[Out]

(-3*((-I + x)/(1 + x))^(1/6)*((I + x)/(1 + x))^(1/6)*AppellF1[1/3, 1/6, 1/6, 4/3, (1 - I)/(1 + x), (1 + I)/(1
+ x)])/(1 + x^2)^(1/6)

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Maple [F]  time = 0.366, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{1+x}{\frac{1}{\sqrt [6]{{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+x)/(x^2+1)^(1/6),x)

[Out]

int(1/(1+x)/(x^2+1)^(1/6),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{2} + 1\right )}^{\frac{1}{6}}{\left (x + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)/(x^2+1)^(1/6),x, algorithm="maxima")

[Out]

integrate(1/((x^2 + 1)^(1/6)*(x + 1)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)/(x^2+1)^(1/6),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (x + 1\right ) \sqrt [6]{x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)/(x**2+1)**(1/6),x)

[Out]

Integral(1/((x + 1)*(x**2 + 1)**(1/6)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{2} + 1\right )}^{\frac{1}{6}}{\left (x + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)/(x^2+1)^(1/6),x, algorithm="giac")

[Out]

integrate(1/((x^2 + 1)^(1/6)*(x + 1)), x)