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

Optimal. Leaf size=125 \[ -\frac {\log \left (2^{2/3} \sqrt [3]{x^2+2 x-1}+2\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{2} \left (x^2+2 x-1\right )^{2/3}+2^{2/3} \sqrt [3]{x^2+2 x-1}-2\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{x^2+2 x-1}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {694, 266, 56, 617, 204, 31} \begin {gather*} \frac {\log (x+1)}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{(x+1)^2-2}+\sqrt [3]{2}\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-2^{2/3} \sqrt [3]{(x+1)^2-2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 - 2^(2/3)*(-2 + (1 + x)^2)^(1/3))/Sqrt[3]])/2^(1/3) + Log[1 + x]/(2*2^(1/3)) - (3*Log[
2^(1/3) + (-2 + (1 + 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.29 \begin {gather*} \frac {3}{8} \left ((x+1)^2-2\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {1}{2} \left (2-(x+1)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-2 + (1 + x)^2)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (2 - (1 + x)^2)/2])/8

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IntegrateAlgebraic [A]  time = 0.21, size = 125, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{-1+2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{-1+2 x+x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{-1+2 x+x^2}-\sqrt [3]{2} \left (-1+2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 116, normalized size = 0.93 \begin {gather*} \frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 2 \, x - 1\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} + 2 \, x - 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (x^{2} + 2 \, x - 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (x^{2} + 2 \, x - 1\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*2^(2/3)*(-1)^(1/3)*arctan(1/6*sqrt(3)*2^(1/6)*(2*sqrt(2)*(-1)^(1/3)*(x^2 + 2*x - 1)^(1/3) + 2^(5/6
))) - 1/8*2^(2/3)*(-1)^(1/3)*log(-2^(1/3)*(-1)^(2/3)*(x^2 + 2*x - 1)^(1/3) - 2^(2/3)*(-1)^(1/3) + (x^2 + 2*x -
1)^(2/3)) + 1/4*2^(2/3)*(-1)^(1/3)*log(2^(1/3)*(-1)^(2/3) + (x^2 + 2*x - 1)^(1/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 8.21, size = 1212, normalized size = 9.70

method	result	size
trager	\(\text {Expression too large to display}\)	\(1212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*ln((RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*Root
Of(_Z^3+4)+4*_Z^2)*x^2+10*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x^2+2*RootOf(
_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+20*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*
_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x-24*(x^2+2*x-1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(
_Z^3+4)^2-9*(x^2+2*x-1)^(1/3)*RootOf(_Z^3+4)^2-48*(x^2+2*x-1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4
)+4*_Z^2)*RootOf(_Z^3+4)+RootOf(_Z^3+4)*x^2+10*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+2*RootO
f(_Z^3+4)*x+20*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+18*(x^2+2*x-1)^(2/3)-7*RootOf(_Z^3+4)-70*
RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(1+x)^2)-1/4*ln((4*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)
^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+10*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x
^2+8*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+20*RootOf(_Z^3+4)^2*RootOf(RootOf(
_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x+24*(x^2+2*x-1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_
Z^2)*RootOf(_Z^3+4)^2+15*(x^2+2*x-1)^(1/3)*RootOf(_Z^3+4)^2+48*(x^2+2*x-1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*
RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)-12*RootOf(_Z^3+4)*x^2-30*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_
Z^2)*x^2-24*RootOf(_Z^3+4)*x-60*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x-30*(x^2+2*x-1)^(2/3)+28*
RootOf(_Z^3+4)+70*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(1+x)^2)*RootOf(_Z^3+4)-1/2*ln((4*RootO
f(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+10*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^
2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x^2+8*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+2
0*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x+24*(x^2+2*x-1)^(2/3)*RootOf(RootOf(
_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2+15*(x^2+2*x-1)^(1/3)*RootOf(_Z^3+4)^2+48*(x^2+2*x-1)^(1
/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)-12*RootOf(_Z^3+4)*x^2-30*RootOf(RootOf(
_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2-24*RootOf(_Z^3+4)*x-60*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4
*_Z^2)*x-30*(x^2+2*x-1)^(2/3)+28*RootOf(_Z^3+4)+70*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(1+x)^
2)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 [3]{x^{2} + 2 x - 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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