3.19.89 \(\int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx\) [1889]

Optimal. Leaf size=131 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{-1+3 x^2}}{2 \sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left (-\sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}\right )}{3 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2+\sqrt [3]{2} x \sqrt [3]{-1+3 x^2}+\left (-1+3 x^2\right )^{2/3}\right )}{6 \sqrt [3]{2}} \]

[Out]

1/6*arctan(3^(1/2)*(3*x^2-1)^(1/3)/(2*2^(1/3)*x+(3*x^2-1)^(1/3)))*2^(2/3)*3^(1/2)+1/6*ln(-2^(1/3)*x+(3*x^2-1)^
(1/3))*2^(2/3)-1/12*ln(2^(2/3)*x^2+2^(1/3)*x*(3*x^2-1)^(1/3)+(3*x^2-1)^(2/3))*2^(2/3)

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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.40, antiderivative size = 265, normalized size of antiderivative = 2.02, number of steps used = 20, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6874, 771, 441, 440, 455, 57, 631, 210, 31, 58} \begin {gather*} -\frac {2 x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};3 x^2,x^2\right )}{3 \sqrt [3]{3 x^2-1}}-\frac {x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,3 x^2\right )}{3 \sqrt [3]{3 x^2-1}}+\frac {\text {ArcTan}\left (\frac {1-2\ 2^{2/3} \sqrt [3]{3 x^2-1}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{3 x^2-1}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{3 x^2-1}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (2^{2/3} \sqrt [3]{3 x^2-1}+1\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(1 - 3*x^2)^(1/3)*AppellF1[1/2, 1/3, 1, 3/2, 3*x^2, x^2])/(3*(-1 + 3*x^2)^(1/3)) - (x*(1 - 3*x^2)^(1/3)*
AppellF1[1/2, 1, 1/3, 3/2, 4*x^2, 3*x^2])/(3*(-1 + 3*x^2)^(1/3)) + ArcTan[(1 - 2*2^(2/3)*(-1 + 3*x^2)^(1/3))/S
qrt[3]]/(2*2^(1/3)*Sqrt[3]) + ArcTan[(1 + 2^(2/3)*(-1 + 3*x^2)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) - Log[1 - 4*x
^2]/(12*2^(1/3)) - Log[1 - x^2]/(6*2^(1/3)) + Log[2^(1/3) - (-1 + 3*x^2)^(1/3)]/(2*2^(1/3)) + Log[1 + 2^(2/3)*
(-1 + 3*x^2)^(1/3)]/(4*2^(1/3))

Rule 31

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

Rule 57

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 58

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] && NegQ
[(b*c - a*d)/b]

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 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 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 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 631

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 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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx &=\int \left (\frac {2}{3 (-1+x) \sqrt [3]{-1+3 x^2}}-\frac {1}{3 (1+2 x) \sqrt [3]{-1+3 x^2}}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{(1+2 x) \sqrt [3]{-1+3 x^2}} \, dx\right )+\frac {2}{3} \int \frac {1}{(-1+x) \sqrt [3]{-1+3 x^2}} \, dx\\ &=-\left (\frac {1}{3} \int \left (\frac {1}{\left (1-4 x^2\right ) \sqrt [3]{-1+3 x^2}}+\frac {2 x}{\sqrt [3]{-1+3 x^2} \left (-1+4 x^2\right )}\right ) \, dx\right )+\frac {2}{3} \int \left (\frac {1}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}}+\frac {x}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{\left (1-4 x^2\right ) \sqrt [3]{-1+3 x^2}} \, dx\right )+\frac {2}{3} \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}} \, dx+\frac {2}{3} \int \frac {x}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}} \, dx-\frac {2}{3} \int \frac {x}{\sqrt [3]{-1+3 x^2} \left (-1+4 x^2\right )} \, dx\\ &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt [3]{-1+3 x}} \, dx,x,x^2\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+3 x} (-1+4 x)} \, dx,x,x^2\right )-\frac {\sqrt [3]{1-3 x^2} \int \frac {1}{\left (1-4 x^2\right ) \sqrt [3]{1-3 x^2}} \, dx}{3 \sqrt [3]{-1+3 x^2}}+\frac {\left (2 \sqrt [3]{1-3 x^2}\right ) \int \frac {1}{\sqrt [3]{1-3 x^2} \left (-1+x^2\right )} \, dx}{3 \sqrt [3]{-1+3 x^2}}\\ &=-\frac {2 x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};3 x^2,x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,3 x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{\frac {1}{2 \sqrt [3]{2}}-\frac {x}{2^{2/3}}+x^2} \, dx,x,\sqrt [3]{-1+3 x^2}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{-1+3 x^2}\right )+\frac {\text {Subst}\left (\int \frac {1}{\frac {1}{2^{2/3}}+x} \, dx,x,\sqrt [3]{-1+3 x^2}\right )}{4 \sqrt [3]{2}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}\\ &=-\frac {2 x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};3 x^2,x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,3 x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (1+2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{4 \sqrt [3]{2}}-\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2\ 2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}-\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{\sqrt [3]{2}}\\ &=-\frac {2 x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};3 x^2,x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,3 x^2\right )}{3 \sqrt [3]{-1+3 x^2}}+\frac {\tan ^{-1}\left (\frac {1-2\ 2^{2/3} \sqrt [3]{-1+3 x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{-1+3 x^2}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (1+2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{4 \sqrt [3]{2}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 117, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{-1+3 x^2}}{2 \sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}}\right )+2 \log \left (-\sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}\right )-\log \left (2^{2/3} x^2+\left (-1+3 x^2\right )^{2/3}+x \sqrt [3]{-2+6 x^2}\right )}{6 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 18.65, size = 651, normalized size = 4.97

method	result	size
trager	\(\text {Expression too large to display}\)	\(651\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*ln((3*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z
^2)*RootOf(_Z^3-4)^3*x^3+9*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-3*RootOf
(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*RootOf(_Z^3-4)^2*(3*x^2-1)^(2/3)*x+2*RootOf(_Z^3-4)^2*(3*x^2-1)^
(1/3)*x^2+12*RootOf(_Z^3-4)*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*(3*x^2-1)^(1/3)*x^2-6*RootOf(_
Z^3-4)*x^2-18*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*x^2+4*(3*x^2-1)^(2/3)*x+2*RootOf(_Z^3-4)+6*R
ootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2))/(-1+x)^2/(1+2*x))+1/6*RootOf(_Z^3-4)*ln(-(3*RootOf(RootOf(
_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*RootOf(_Z^3-4)^3*x^3+9*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^
2)^2*RootOf(_Z^3-4)^2*x^3-3*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*RootOf(_Z^3-4)^2*(3*x^2-1)^(2/
3)*x+2*RootOf(_Z^3-4)^2*(3*x^2-1)^(1/3)*x^2-6*RootOf(_Z^3-4)*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^
2)*(3*x^2-1)^(1/3)*x^2+4*RootOf(_Z^3-4)*x^3+12*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*x^3+6*RootO
f(_Z^3-4)*x^2+18*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*x^2-8*(3*x^2-1)^(2/3)*x-2*RootOf(_Z^3-4)-
6*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2))/(-1+x)^2/(1+2*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+x)/(-1+x)/(1+2*x)/(3*x^2-1)^(1/3),x, algorithm="maxima")

[Out]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x+1}{\left (2\,x+1\right )\,{\left (3\,x^2-1\right )}^{1/3}\,\left (x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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