3.31.56 \(\int \frac {1+x}{(-3+x^2) \sqrt [3]{1+x^2}} \, dx\)

Optimal. Leaf size=468 \[ \frac {\sqrt [3]{\frac {1}{2} \left (9+5 \sqrt {3}\right )} \log \left (-6 \sqrt [3]{x^2+1}+2^{2/3} \sqrt {3} x+3\ 2^{2/3}\right )}{6\ 3^{2/3}}+\frac {\sqrt [3]{\frac {1}{2} \left (9-5 \sqrt {3}\right )} \log \left (6 \sqrt [3]{x^2+1}+2^{2/3} \sqrt {3} x-3\ 2^{2/3}\right )}{6\ 3^{2/3}}-\frac {\sqrt [3]{\frac {1}{2} \left (9-5 \sqrt {3}\right )} \log \left (-\sqrt [3]{2} x^2+2^{2/3} \sqrt {3} \sqrt [3]{x^2+1} x-6 \left (x^2+1\right )^{2/3}-3\ 2^{2/3} \sqrt [3]{x^2+1}+2 \sqrt [3]{2} \sqrt {3} x-3 \sqrt [3]{2}\right )}{12\ 3^{2/3}}-\frac {\sqrt [3]{\frac {1}{2} \left (9+5 \sqrt {3}\right )} \log \left (\sqrt [3]{2} x^2+2^{2/3} \sqrt {3} \sqrt [3]{x^2+1} x+6 \left (x^2+1\right )^{2/3}+3\ 2^{2/3} \sqrt [3]{x^2+1}+2 \sqrt [3]{2} \sqrt {3} x+3 \sqrt [3]{2}\right )}{12\ 3^{2/3}}+\frac {1}{6} \sqrt [3]{\frac {1}{2} \left (3 \sqrt {3}-5\right )} \tan ^{-1}\left (\frac {-\frac {\sqrt [3]{x^2+1}}{\sqrt {3}}+\frac {1}{3} 2^{2/3} x-\frac {2^{2/3}}{\sqrt {3}}}{\sqrt [3]{x^2+1}}\right )-\frac {1}{6} \sqrt [3]{\frac {1}{2} \left (5+3 \sqrt {3}\right )} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+1}}{\sqrt {3}}+\frac {1}{3} 2^{2/3} x+\frac {2^{2/3}}{\sqrt {3}}}{\sqrt [3]{x^2+1}}\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 191, normalized size of antiderivative = 0.41, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1010, 392, 444, 55, 617, 204, 31} \begin {gather*} -\frac {\log \left (3-x^2\right )}{4\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{4\ 2^{2/3}}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2} \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\tan ^{-1}(x)}{6\ 2^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x)/((-3 + x^2)*(1 + x^2)^(1/3)),x]

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

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 392

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b/a, 2]}, Simp[(q*ArcTanh
[Sqrt[3]/(q*x)])/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d), x] + (-Simp[(q*ArcTan[(a^(1/3)*q*x)/(a^(1/3) + 2^(1/3)*(a + b*
x^2)^(1/3))])/(2*2^(2/3)*a^(1/3)*d), x] + Simp[(q*ArcTan[q*x])/(6*2^(2/3)*a^(1/3)*d), x] + Simp[(q*ArcTanh[(Sq
rt[3]*(a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3)))/(a^(1/3)*q*x)])/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d), x])] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 3*a*d, 0] && PosQ[b/a]

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1010

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rubi steps

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

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Mathematica [C]  time = 0.27, size = 151, normalized size = 0.32 \begin {gather*} \frac {9 x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};-x^2,\frac {x^2}{3}\right )}{\left (x^2-3\right ) \sqrt [3]{x^2+1} \left (2 x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};-x^2,\frac {x^2}{3}\right )-F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};-x^2,\frac {x^2}{3}\right )\right )+9 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};-x^2,\frac {x^2}{3}\right )\right )}-\frac {1}{6} x^2 F_1\left (1;\frac {1}{3},1;2;-x^2,\frac {x^2}{3}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + x)/((-3 + x^2)*(1 + x^2)^(1/3)),x]

[Out]

-1/6*(x^2*AppellF1[1, 1/3, 1, 2, -x^2, x^2/3]) + (9*x*AppellF1[1/2, 1/3, 1, 3/2, -x^2, x^2/3])/((-3 + x^2)*(1
+ x^2)^(1/3)*(9*AppellF1[1/2, 1/3, 1, 3/2, -x^2, x^2/3] + 2*x^2*(AppellF1[3/2, 1/3, 2, 5/2, -x^2, x^2/3] - App
ellF1[3/2, 4/3, 1, 5/2, -x^2, x^2/3])))

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IntegrateAlgebraic [A]  time = 6.42, size = 468, normalized size = 1.00 \begin {gather*} \frac {1}{6} \sqrt [3]{\frac {1}{2} \left (-5+3 \sqrt {3}\right )} \tan ^{-1}\left (\frac {-\frac {2^{2/3}}{\sqrt {3}}+\frac {1}{3} 2^{2/3} x-\frac {\sqrt [3]{1+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x^2}}\right )-\frac {1}{6} \sqrt [3]{\frac {1}{2} \left (5+3 \sqrt {3}\right )} \tan ^{-1}\left (\frac {\frac {2^{2/3}}{\sqrt {3}}+\frac {1}{3} 2^{2/3} x+\frac {\sqrt [3]{1+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x^2}}\right )+\frac {\sqrt [3]{\frac {1}{2} \left (9+5 \sqrt {3}\right )} \log \left (3\ 2^{2/3}+2^{2/3} \sqrt {3} x-6 \sqrt [3]{1+x^2}\right )}{6\ 3^{2/3}}+\frac {\sqrt [3]{\frac {1}{2} \left (9-5 \sqrt {3}\right )} \log \left (-3 2^{2/3}+2^{2/3} \sqrt {3} x+6 \sqrt [3]{1+x^2}\right )}{6\ 3^{2/3}}-\frac {\sqrt [3]{\frac {1}{2} \left (9-5 \sqrt {3}\right )} \log \left (-3 \sqrt [3]{2}+2 \sqrt [3]{2} \sqrt {3} x-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{1+x^2}+2^{2/3} \sqrt {3} x \sqrt [3]{1+x^2}-6 \left (1+x^2\right )^{2/3}\right )}{12\ 3^{2/3}}-\frac {\sqrt [3]{\frac {1}{2} \left (9+5 \sqrt {3}\right )} \log \left (3 \sqrt [3]{2}+2 \sqrt [3]{2} \sqrt {3} x+\sqrt [3]{2} x^2+3\ 2^{2/3} \sqrt [3]{1+x^2}+2^{2/3} \sqrt {3} x \sqrt [3]{1+x^2}+6 \left (1+x^2\right )^{2/3}\right )}{12\ 3^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x)/((-3 + x^2)*(1 + x^2)^(1/3)),x]

[Out]

(((-5 + 3*Sqrt[3])/2)^(1/3)*ArcTan[(-(2^(2/3)/Sqrt[3]) + (2^(2/3)*x)/3 - (1 + x^2)^(1/3)/Sqrt[3])/(1 + x^2)^(1
/3)])/6 - (((5 + 3*Sqrt[3])/2)^(1/3)*ArcTan[(2^(2/3)/Sqrt[3] + (2^(2/3)*x)/3 + (1 + x^2)^(1/3)/Sqrt[3])/(1 + x
^2)^(1/3)])/6 + (((9 + 5*Sqrt[3])/2)^(1/3)*Log[3*2^(2/3) + 2^(2/3)*Sqrt[3]*x - 6*(1 + x^2)^(1/3)])/(6*3^(2/3))
+ (((9 - 5*Sqrt[3])/2)^(1/3)*Log[-3*2^(2/3) + 2^(2/3)*Sqrt[3]*x + 6*(1 + x^2)^(1/3)])/(6*3^(2/3)) - (((9 - 5*
Sqrt[3])/2)^(1/3)*Log[-3*2^(1/3) + 2*2^(1/3)*Sqrt[3]*x - 2^(1/3)*x^2 - 3*2^(2/3)*(1 + x^2)^(1/3) + 2^(2/3)*Sqr
t[3]*x*(1 + x^2)^(1/3) - 6*(1 + x^2)^(2/3)])/(12*3^(2/3)) - (((9 + 5*Sqrt[3])/2)^(1/3)*Log[3*2^(1/3) + 2*2^(1/
3)*Sqrt[3]*x + 2^(1/3)*x^2 + 3*2^(2/3)*(1 + x^2)^(1/3) + 2^(2/3)*Sqrt[3]*x*(1 + x^2)^(1/3) + 6*(1 + x^2)^(2/3)
])/(12*3^(2/3))

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 117.77, size = 17124, normalized size = 36.59 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x+1}{{\left (x^2+1\right )}^{1/3}\,\left (x^2-3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{\left (x^{2} - 3\right ) \sqrt [3]{x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 1)/((x**2 - 3)*(x**2 + 1)**(1/3)), x)

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