3.8.20 \(\int \frac {1}{(1+x) \sqrt [6]{1+x^2}} \, dx\) [720]

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

[Out]

x*AppellF1(1/2,1,1/6,3/2,x^2,-x^2)-1/2*arctanh(1/2*(x^2+1)^(1/6)*2^(5/6))*2^(5/6)+1/8*ln(2^(1/3)-2^(1/6)*(x^2+
1)^(1/6)+(x^2+1)^(1/3))*2^(5/6)-1/8*ln(2^(1/3)+2^(1/6)*(x^2+1)^(1/6)+(x^2+1)^(1/3))*2^(5/6)-1/4*arctan(1/3*(1-
(x^2+1)^(1/6)*2^(5/6))*3^(1/2))*3^(1/2)*2^(5/6)+1/4*arctan(1/3*(1+(x^2+1)^(1/6)*2^(5/6))*3^(1/2))*3^(1/2)*2^(5
/6)

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Rubi [A]
time = 0.25, antiderivative size = 203, normalized size of antiderivative = 1.00, 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 = {771, 440, 455, 65, 302, 648, 632, 210, 642, 212} \begin {gather*} x F_1\left (\frac {1}{2};1,\frac {1}{6};\frac {3}{2};x^2,-x^2\right )-\frac {\sqrt {3} \text {ArcTan}\left (\frac {1-2^{5/6} \sqrt [6]{x^2+1}}{\sqrt {3}}\right )}{2 \sqrt [6]{2}}+\frac {\sqrt {3} \text {ArcTan}\left (\frac {2^{5/6} \sqrt [6]{x^2+1}+1}{\sqrt {3}}\right )}{2 \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 {\log \left (\sqrt [3]{x^2+1}+\sqrt [6]{2} \sqrt [6]{x^2+1}+\sqrt [3]{2}\right )}{4 \sqrt [6]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{x^2+1}}{\sqrt [6]{2}}\right )}{\sqrt [6]{2}} \end {gather*}

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 65

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 210

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

Rule 212

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

Rule 302

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)/(a*n*s^m))*Int[1/(r^2 - s^2*x^2), x] + 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 440

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 455

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 632

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 642

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 648

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 771

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]

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} \text {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 \text {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 {\text {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 {\text {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}}-\text {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} \text {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} \text {Subst}\left (\int \frac {1}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx,x,\sqrt [6]{1+x^2}\right )+\frac {\text {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 {\text {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 \text {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 \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 15.29, size = 72, normalized size = 0.35 \begin {gather*} -\frac {3 \sqrt [6]{\frac {-i+x}{1+x}} \sqrt [6]{\frac {i+x}{1+x}} F_1\left (\frac {1}{3};\frac {1}{6},\frac {1}{6};\frac {4}{3};\frac {1-i}{1+x},\frac {1+i}{1+x}\right )}{\sqrt [6]{1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.08, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (x +1\right ) \left (x^{2}+1\right )^{\frac {1}{6}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   Not integrable (provided residues have n
o relations)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (x + 1\right ) \sqrt [6]{x^{2} + 1}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (x^2+1\right )}^{1/6}\,\left (x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^2 + 1)^(1/6)*(x + 1)),x)

[Out]

int(1/((x^2 + 1)^(1/6)*(x + 1)), x)

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