3.1628 \(\int \frac {\sqrt [3]{-1+x}}{\sqrt [3]{1+x}} \, dx\)

Optimal. Leaf size=77 \[ \sqrt [3]{x-1} (x+1)^{2/3}+\frac {1}{3} \log (x-1)+\log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{x-1}}-1\right )+\frac {2 \tan ^{-1}\left (\frac {2 \sqrt [3]{x+1}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {50, 59} \[ \sqrt [3]{x-1} (x+1)^{2/3}+\frac {1}{3} \log (x-1)+\log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{x-1}}-1\right )+\frac {2 \tan ^{-1}\left (\frac {2 \sqrt [3]{x+1}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x)^(1/3)/(1 + x)^(1/3),x]

[Out]

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

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 59

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

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 48, normalized size = 0.62 \[ \frac {3 \left (\frac {x-1}{x+1}\right )^{4/3} (x+1)^{4/3} \, _2F_1\left (\frac {1}{3},\frac {4}{3};\frac {7}{3};\frac {1-x}{2}\right )}{4 \sqrt [3]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*((-1 + x)/(1 + x))^(4/3)*(1 + x)^(4/3)*Hypergeometric2F1[1/3, 4/3, 7/3, (1 - x)/2])/(4*2^(1/3))

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fricas [A]  time = 0.45, size = 107, normalized size = 1.39 \[ -\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x + 1\right )} + 2 \, \sqrt {3} {\left (x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )}^{\frac {1}{3}}}{3 \, {\left (x + 1\right )}}\right ) + {\left (x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )}^{\frac {1}{3}} - \frac {1}{3} \, \log \left (\frac {{\left (x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )}^{\frac {1}{3}} + {\left (x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )}^{\frac {2}{3}} + x + 1}{x + 1}\right ) + \frac {2}{3} \, \log \left (\frac {{\left (x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )}^{\frac {1}{3}} - x - 1}{x + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.38, size = 573, normalized size = 7.44 \[ \left (x -1\right )^{\frac {1}{3}} \left (x +1\right )^{\frac {2}{3}}+\frac {\left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {-2 x^{2} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+x^{2} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2 x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+x^{2}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x -2 x +3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+1}{x -1}\right )}{3}+\frac {2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {2 x^{2} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+5 x^{2} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2 x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+2 x^{2}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-4 x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2}{x -1}\right )}{3}-\frac {2 \ln \left (\frac {-2 x^{2} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+x^{2} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2 x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+x^{2}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x -2 x +3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+1}{x -1}\right )}{3}\right ) \left (\left (x -1\right )^{2} \left (x +1\right )\right )^{\frac {1}{3}}}{\left (x -1\right )^{\frac {2}{3}} \left (x +1\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x-1)^(1/3)*(x+1)^(2/3)+(2/3*RootOf(_Z^2+_Z+1)*ln(-(2*RootOf(_Z^2+_Z+1)^2*x^2+3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1
)^(2/3)+3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(1/3)*x-2*RootOf(_Z^2+_Z+1)^2*x+5*RootOf(_Z^2+_Z+1)*x^2-3*RootOf(_Z^
2+_Z+1)*(x^3-x^2-x+1)^(1/3)-4*RootOf(_Z^2+_Z+1)*x+2*x^2-RootOf(_Z^2+_Z+1)-2)/(x-1))-2/3*ln((-2*RootOf(_Z^2+_Z+
1)^2*x^2+3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(2/3)+3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(1/3)*x+2*RootOf(_Z^2+_Z+1)
^2*x+RootOf(_Z^2+_Z+1)*x^2+3*(x^3-x^2-x+1)^(2/3)-3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(1/3)+3*(x^3-x^2-x+1)^(1/3)
*x+x^2-3*(x^3-x^2-x+1)^(1/3)-RootOf(_Z^2+_Z+1)-2*x+1)/(x-1))*RootOf(_Z^2+_Z+1)-2/3*ln((-2*RootOf(_Z^2+_Z+1)^2*
x^2+3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(2/3)+3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(1/3)*x+2*RootOf(_Z^2+_Z+1)^2*x+
RootOf(_Z^2+_Z+1)*x^2+3*(x^3-x^2-x+1)^(2/3)-3*RootOf(_Z^2+_Z+1)*(x^3-x^2-x+1)^(1/3)+3*(x^3-x^2-x+1)^(1/3)*x+x^
2-3*(x^3-x^2-x+1)^(1/3)-RootOf(_Z^2+_Z+1)-2*x+1)/(x-1)))/(x-1)^(2/3)*((x-1)^2*(x+1))^(1/3)/(x+1)^(1/3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)^(1/3)/(x + 1)^(1/3),x)

[Out]

int((x - 1)^(1/3)/(x + 1)^(1/3), x)

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sympy [C]  time = 2.55, size = 39, normalized size = 0.51 \[ \frac {2^{\frac {2}{3}} \left (x - 1\right )^{\frac {4}{3}} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {\left (x - 1\right ) e^{i \pi }}{2}} \right )}}{2 \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(2/3)*(x - 1)**(4/3)*gamma(4/3)*hyper((1/3, 4/3), (7/3,), (x - 1)*exp_polar(I*pi)/2)/(2*gamma(7/3))

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