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

Optimal. Leaf size=499 \[ \frac {(-1)^{2/3} \sqrt [3]{-54+35 i \sqrt {3}} \log \left (6 \sqrt [3]{x^2-1}-i 2^{2/3} \sqrt {3} x+3\ 2^{2/3}\right )}{6\ 6^{2/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{-54-35 i \sqrt {3}} \log \left (6 \sqrt [3]{x^2-1}+i 2^{2/3} \sqrt {3} x+3\ 2^{2/3}\right )}{6\ 6^{2/3}}+\frac {\sqrt [3]{54+35 i \sqrt {3}} \log \left (-\sqrt [3]{2} x^2-i 2^{2/3} \sqrt {3} \sqrt [3]{x^2-1} x+6 \left (x^2-1\right )^{2/3}-3\ 2^{2/3} \sqrt [3]{x^2-1}+2 i \sqrt [3]{2} \sqrt {3} x+3 \sqrt [3]{2}\right )}{12\ 6^{2/3}}+\frac {\sqrt [3]{54-35 i \sqrt {3}} \log \left (-\sqrt [3]{2} x^2+i 2^{2/3} \sqrt {3} \sqrt [3]{x^2-1} x+6 \left (x^2-1\right )^{2/3}-3\ 2^{2/3} \sqrt [3]{x^2-1}-2 i \sqrt [3]{2} \sqrt {3} x+3 \sqrt [3]{2}\right )}{12\ 6^{2/3}}-\frac {\sqrt [6]{-253+1260 i \sqrt {3}} \tan ^{-1}\left (\frac {3 \sqrt [3]{x^2-1}}{\sqrt {3} \sqrt [3]{x^2-1}-i 2^{2/3} x-2^{2/3} \sqrt {3}}\right )}{6\ 2^{2/3}}-\frac {\sqrt [6]{-253-1260 i \sqrt {3}} \tan ^{-1}\left (\frac {3 \sqrt [3]{x^2-1}}{\sqrt {3} \sqrt [3]{x^2-1}+i 2^{2/3} x-2^{2/3} \sqrt {3}}\right )}{6\ 2^{2/3}} \]

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Rubi [A]  time = 0.09, antiderivative size = 213, normalized size of antiderivative = 0.43, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1010, 393, 444, 56, 617, 204, 31} \begin {gather*} \frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}-\frac {3 \log \left (\sqrt [3]{x^2-1}+2^{2/3}\right )}{2\ 2^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\sqrt [3]{2} \sqrt [3]{x^2-1}}{\sqrt {3}}\right )}{2^{2/3}}-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left ((-1)^{2/3} \sqrt [3]{2} \sqrt [3]{x^2-1}+1\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{2} \sqrt [3]{x^2-1}+\sqrt [3]{-1}}\right )-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-1)^(2/3)*ArcTan[Sqrt[3]/x])/(2^(2/3)*Sqrt[3]) - (Sqrt[3]*ArcTan[(1 - 2^(1/3)*(-1 + x^2)^(1/3))/Sqrt[3]
])/2^(2/3) - ((-1)^(2/3)*ArcTan[(Sqrt[3]*(1 + (-1)^(2/3)*2^(1/3)*(-1 + x^2)^(1/3)))/x])/(2*2^(2/3)*Sqrt[3]) +
((-1/2)^(2/3)*ArcTanh[x])/6 - ((-1/2)^(2/3)*ArcTanh[((-1)^(1/3)*x)/((-1)^(1/3) + 2^(1/3)*(-1 + x^2)^(1/3))])/2
+ Log[3 + x^2]/(2*2^(2/3)) - (3*Log[2^(2/3) + (-1 + x^2)^(1/3)])/(2*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 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[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] && Ne
gQ[(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 393

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b/a), 2]}, Simp[(q*ArcT
an[Sqrt[3]/(q*x)])/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d), x] + (Simp[(q*ArcTanh[(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*ArcTanh[q*x])/(6*2^(2/3)*a^(1/3)*d), x] + Simp[(q*ArcTan[(
Sqrt[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] && NegQ[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+2 x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx &=2 \int \frac {x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx+\int \frac {1}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx\\ &=-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\frac {\log \left (3+x^2\right )}{2\ 2^{2/3}}+\frac {3}{2} \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 )}{2\ 2^{2/3}}\\ &=-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\frac {\log \left (3+x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}+\sqrt [3]{-1+x^2}\right )}{2\ 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^{2/3}}\\ &=-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\sqrt [3]{2} \sqrt [3]{-1+x^2}}{\sqrt {3}}\right )}{2^{2/3}}-\frac {(-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\frac {\log \left (3+x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}+\sqrt [3]{-1+x^2}\right )}{2\ 2^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 151, normalized size = 0.30 \begin {gather*} \frac {x \left (x \sqrt [3]{1-x^2} F_1\left (1;\frac {1}{3},1;2;x^2,-\frac {x^2}{3}\right )-\frac {27 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 ) \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 )}\right )}{3 \sqrt [3]{x^2-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(x*(1 - x^2)^(1/3)*AppellF1[1, 1/3, 1, 2, x^2, -1/3*x^2] - (27*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2])/(
(3 + x^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2] + 2*x^2*(AppellF1[3/2, 1/3, 2, 5/2, x^2, -1/3*x^2] - A
ppellF1[3/2, 4/3, 1, 5/2, x^2, -1/3*x^2])))))/(3*(-1 + x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 7.98, size = 499, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [6]{-253+1260 i \sqrt {3}} \tan ^{-1}\left (\frac {3 \sqrt [3]{-1+x^2}}{-2^{2/3} \sqrt {3}-i 2^{2/3} x+\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{6\ 2^{2/3}}-\frac {\sqrt [6]{-253-1260 i \sqrt {3}} \tan ^{-1}\left (\frac {3 \sqrt [3]{-1+x^2}}{-2^{2/3} \sqrt {3}+i 2^{2/3} x+\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{6\ 2^{2/3}}+\frac {(-1)^{2/3} \sqrt [3]{-54+35 i \sqrt {3}} \log \left (3\ 2^{2/3}-i 2^{2/3} \sqrt {3} x+6 \sqrt [3]{-1+x^2}\right )}{6\ 6^{2/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{-54-35 i \sqrt {3}} \log \left (3\ 2^{2/3}+i 2^{2/3} \sqrt {3} x+6 \sqrt [3]{-1+x^2}\right )}{6\ 6^{2/3}}+\frac {\sqrt [3]{54+35 i \sqrt {3}} \log \left (3 \sqrt [3]{2}+2 i \sqrt [3]{2} \sqrt {3} x-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{-1+x^2}-i 2^{2/3} \sqrt {3} x \sqrt [3]{-1+x^2}+6 \left (-1+x^2\right )^{2/3}\right )}{12\ 6^{2/3}}+\frac {\sqrt [3]{54-35 i \sqrt {3}} \log \left (3 \sqrt [3]{2}-2 i \sqrt [3]{2} \sqrt {3} x-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{-1+x^2}+i 2^{2/3} \sqrt {3} x \sqrt [3]{-1+x^2}+6 \left (-1+x^2\right )^{2/3}\right )}{12\ 6^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((-253 + (1260*I)*Sqrt[3])^(1/6)*ArcTan[(3*(-1 + x^2)^(1/3))/(-(2^(2/3)*Sqrt[3]) - I*2^(2/3)*x + Sqrt[3]*
(-1 + x^2)^(1/3))])/2^(2/3) - ((-253 - (1260*I)*Sqrt[3])^(1/6)*ArcTan[(3*(-1 + x^2)^(1/3))/(-(2^(2/3)*Sqrt[3])
+ I*2^(2/3)*x + Sqrt[3]*(-1 + x^2)^(1/3))])/(6*2^(2/3)) + ((-1)^(2/3)*(-54 + (35*I)*Sqrt[3])^(1/3)*Log[3*2^(2
/3) - I*2^(2/3)*Sqrt[3]*x + 6*(-1 + x^2)^(1/3)])/(6*6^(2/3)) - ((-1)^(1/3)*(-54 - (35*I)*Sqrt[3])^(1/3)*Log[3*
2^(2/3) + I*2^(2/3)*Sqrt[3]*x + 6*(-1 + x^2)^(1/3)])/(6*6^(2/3)) + ((54 + (35*I)*Sqrt[3])^(1/3)*Log[3*2^(1/3)
+ (2*I)*2^(1/3)*Sqrt[3]*x - 2^(1/3)*x^2 - 3*2^(2/3)*(-1 + x^2)^(1/3) - I*2^(2/3)*Sqrt[3]*x*(-1 + x^2)^(1/3) +
6*(-1 + x^2)^(2/3)])/(12*6^(2/3)) + ((54 - (35*I)*Sqrt[3])^(1/3)*Log[3*2^(1/3) - (2*I)*2^(1/3)*Sqrt[3]*x - 2^(
1/3)*x^2 - 3*2^(2/3)*(-1 + x^2)^(1/3) + I*2^(2/3)*Sqrt[3]*x*(-1 + x^2)^(1/3) + 6*(-1 + x^2)^(2/3)])/(12*6^(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+2*x)/(x^2-1)^(1/3)/(x^2+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 {2 \, x + 1}{{\left (x^{2} + 3\right )} {\left (x^{2} - 1\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 hanged

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

int((2*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 {2 x + 1}{\sqrt [3]{\left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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