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

Optimal. Leaf size=84 \[ 3 \sqrt [3]{1-x}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x}\right )}{2^{2/3}}-\frac {\log (x+1)}{2^{2/3}}-\sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x}+1}{\sqrt {3}}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {50, 57, 617, 204, 31} \begin {gather*} 3 \sqrt [3]{1-x}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x}\right )}{2^{2/3}}-\frac {\log (x+1)}{2^{2/3}}-\sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x}+1}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(1/3)/(1 + x),x]

[Out]

3*(1 - x)^(1/3) - 2^(1/3)*Sqrt[3]*ArcTan[(1 + 2^(2/3)*(1 - x)^(1/3))/Sqrt[3]] + (3*Log[2^(1/3) - (1 - x)^(1/3)
])/2^(2/3) - Log[1 + x]/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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 57

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 104, normalized size = 1.24 \begin {gather*} 3 \sqrt [3]{1-x}+\sqrt [3]{2} \log \left (\sqrt [3]{2}-\sqrt [3]{1-x}\right )-\frac {\log \left ((1-x)^{2/3}+\sqrt [3]{2-2 x}+2^{2/3}\right )}{2^{2/3}}-\sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x}+1}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^(1/3)/(1 + x),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 115, normalized size = 1.37 \begin {gather*} 3 \sqrt [3]{1-x}+\sqrt [3]{2} \log \left (2^{2/3} \sqrt [3]{1-x}-2\right )-\frac {\log \left (\sqrt [3]{2} (1-x)^{2/3}+2^{2/3} \sqrt [3]{1-x}+2\right )}{2^{2/3}}-\sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.33, size = 86, normalized size = 1.02 \begin {gather*} -\sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} 2^{\frac {2}{3}} {\left (-x + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{2} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}} + {\left (-x + 1\right )}^{\frac {2}{3}}\right ) + 2^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x + 1\right )}^{\frac {1}{3}}\right ) + 3 \, {\left (-x + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*2^(1/3)*arctan(1/3*sqrt(3)*2^(2/3)*(-x + 1)^(1/3) + 1/3*sqrt(3)) - 1/2*2^(1/3)*log(2^(2/3) + 2^(1/3)*
(-x + 1)^(1/3) + (-x + 1)^(2/3)) + 2^(1/3)*log(-2^(1/3) + (-x + 1)^(1/3)) + 3*(-x + 1)^(1/3)

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giac [A]  time = 1.05, size = 87, normalized size = 1.04 \begin {gather*} -\sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{2} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}} + {\left (-x + 1\right )}^{\frac {2}{3}}\right ) + 2^{\frac {1}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x + 1\right )}^{\frac {1}{3}} \right |}\right ) + 3 \, {\left (-x + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x + 1)^(1/3))) - 1/2*2^(1/3)*log(2^(2/3) + 2^(1/3)*
(-x + 1)^(1/3) + (-x + 1)^(2/3)) + 2^(1/3)*log(abs(-2^(1/3) + (-x + 1)^(1/3))) + 3*(-x + 1)^(1/3)

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maple [A]  time = 0.01, size = 84, normalized size = 1.00 \begin {gather*} -2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\left (1+2^{\frac {2}{3}} \left (-x +1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )+2^{\frac {1}{3}} \ln \left (\left (-x +1\right )^{\frac {1}{3}}-2^{\frac {1}{3}}\right )-\frac {2^{\frac {1}{3}} \ln \left (\left (-x +1\right )^{\frac {2}{3}}+2^{\frac {1}{3}} \left (-x +1\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )}{2}+3 \left (-x +1\right )^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 86, normalized size = 1.02 \begin {gather*} -\sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{2} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}} + {\left (-x + 1\right )}^{\frac {2}{3}}\right ) + 2^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x + 1\right )}^{\frac {1}{3}}\right ) + 3 \, {\left (-x + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 104, normalized size = 1.24 \begin {gather*} 2^{1/3}\,\ln \left (18\,{\left (1-x\right )}^{1/3}-18\,2^{1/3}\right )+3\,{\left (1-x\right )}^{1/3}+\frac {2^{1/3}\,\ln \left (18\,{\left (1-x\right )}^{1/3}-9\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-\frac {2^{1/3}\,\ln \left (18\,{\left (1-x\right )}^{1/3}+9\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x)^(1/3)/(x + 1),x)

[Out]

2^(1/3)*log(18*(1 - x)^(1/3) - 18*2^(1/3)) + 3*(1 - x)^(1/3) + (2^(1/3)*log(18*(1 - x)^(1/3) - 9*2^(1/3)*(3^(1
/2)*1i - 1))*(3^(1/2)*1i - 1))/2 - (2^(1/3)*log(18*(1 - x)^(1/3) + 9*2^(1/3)*(3^(1/2)*1i + 1))*(3^(1/2)*1i + 1
))/2

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sympy [C]  time = 2.26, size = 170, normalized size = 2.02 \begin {gather*} \frac {4 \sqrt [3]{-1} \sqrt [3]{x - 1} \Gamma \left (\frac {4}{3}\right )}{\Gamma \left (\frac {7}{3}\right )} + \frac {4 \sqrt [3]{-2} e^{- \frac {i \pi }{3}} \log {\left (- \frac {2^{\frac {2}{3}} \sqrt [3]{x - 1} e^{\frac {i \pi }{3}}}{2} + 1 \right )} \Gamma \left (\frac {4}{3}\right )}{3 \Gamma \left (\frac {7}{3}\right )} - \frac {4 \sqrt [3]{-2} \log {\left (- \frac {2^{\frac {2}{3}} \sqrt [3]{x - 1} e^{i \pi }}{2} + 1 \right )} \Gamma \left (\frac {4}{3}\right )}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {4 \sqrt [3]{-2} e^{\frac {i \pi }{3}} \log {\left (- \frac {2^{\frac {2}{3}} \sqrt [3]{x - 1} e^{\frac {5 i \pi }{3}}}{2} + 1 \right )} \Gamma \left (\frac {4}{3}\right )}{3 \Gamma \left (\frac {7}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(-1)**(1/3)*(x - 1)**(1/3)*gamma(4/3)/gamma(7/3) + 4*(-2)**(1/3)*exp(-I*pi/3)*log(-2**(2/3)*(x - 1)**(1/3)*e
xp_polar(I*pi/3)/2 + 1)*gamma(4/3)/(3*gamma(7/3)) - 4*(-2)**(1/3)*log(-2**(2/3)*(x - 1)**(1/3)*exp_polar(I*pi)
/2 + 1)*gamma(4/3)/(3*gamma(7/3)) + 4*(-2)**(1/3)*exp(I*pi/3)*log(-2**(2/3)*(x - 1)**(1/3)*exp_polar(5*I*pi/3)
/2 + 1)*gamma(4/3)/(3*gamma(7/3))

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