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

Optimal. Leaf size=272 \[ \frac {\log \left (2^{2/3}-\frac {1-x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}-\frac {\log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3\ 2^{2/3}}+\frac {1}{3} \sqrt [3]{2} \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )-\frac {\log \left (\frac {(1-x)^2}{\left (1-x^3\right )^{2/3}}+\frac {2^{2/3} (1-x)}{\sqrt [3]{1-x^3}}+2 \sqrt [3]{2}\right )}{6\ 2^{2/3}}+\frac {\sqrt [3]{2} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \]

[Out]

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

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Rubi [C]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 0.08, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {429} \[ x F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-x^3\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[(1 - x^3)^(1/3)/(1 + x^3),x]

[Out]

x*AppellF1[1/3, -1/3, 1, 4/3, x^3, -x^3]

Rule 429

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])

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{1-x^3}}{1+x^3} \, dx &=x F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-x^3\right )\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 109, normalized size = 0.40 \[ -\frac {4 x \sqrt [3]{1-x^3} F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-x^3\right )}{\left (x^3+1\right ) \left (x^3 \left (3 F_1\left (\frac {4}{3};-\frac {1}{3},2;\frac {7}{3};x^3,-x^3\right )+F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};x^3,-x^3\right )\right )-4 F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-x^3\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*x*(1 - x^3)^(1/3)*AppellF1[1/3, -1/3, 1, 4/3, x^3, -x^3])/((1 + x^3)*(-4*AppellF1[1/3, -1/3, 1, 4/3, x^3,
-x^3] + x^3*(3*AppellF1[4/3, -1/3, 2, 7/3, x^3, -x^3] + AppellF1[4/3, 2/3, 1, 7/3, x^3, -x^3])))

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fricas [A]  time = 3.53, size = 341, normalized size = 1.25 \[ \frac {1}{18} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (-\frac {6 \, \sqrt {3} 2^{\frac {2}{3}} {\left (x^{16} - 33 \, x^{13} + 110 \, x^{10} - 110 \, x^{7} + 33 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 24 \, \sqrt {3} 2^{\frac {1}{3}} {\left (x^{14} - 2 \, x^{11} - 6 \, x^{8} - 2 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - \sqrt {3} {\left (x^{18} + 42 \, x^{15} - 417 \, x^{12} + 812 \, x^{9} - 417 \, x^{6} + 42 \, x^{3} + 1\right )}}{3 \, {\left (x^{18} - 102 \, x^{15} + 447 \, x^{12} - 628 \, x^{9} + 447 \, x^{6} - 102 \, x^{3} + 1\right )}}\right ) + \frac {1}{18} \cdot 2^{\frac {1}{3}} \log \left (-\frac {12 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x^{2} + 2^{\frac {2}{3}} {\left (x^{6} + 2 \, x^{3} + 1\right )} - 6 \cdot 2^{\frac {1}{3}} {\left (x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) - \frac {1}{36} \cdot 2^{\frac {1}{3}} \log \left (\frac {12 \cdot 2^{\frac {2}{3}} {\left (x^{8} - 4 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{12} - 32 \, x^{9} + 78 \, x^{6} - 32 \, x^{3} + 1\right )} + 6 \, {\left (x^{10} - 11 \, x^{7} + 11 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{12} + 4 \, x^{9} + 6 \, x^{6} + 4 \, x^{3} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*sqrt(3)*2^(1/3)*arctan(-1/3*(6*sqrt(3)*2^(2/3)*(x^16 - 33*x^13 + 110*x^10 - 110*x^7 + 33*x^4 - x)*(-x^3 +
1)^(1/3) - 24*sqrt(3)*2^(1/3)*(x^14 - 2*x^11 - 6*x^8 - 2*x^5 + x^2)*(-x^3 + 1)^(2/3) - sqrt(3)*(x^18 + 42*x^1
5 - 417*x^12 + 812*x^9 - 417*x^6 + 42*x^3 + 1))/(x^18 - 102*x^15 + 447*x^12 - 628*x^9 + 447*x^6 - 102*x^3 + 1)
) + 1/18*2^(1/3)*log(-(12*(-x^3 + 1)^(2/3)*x^2 + 2^(2/3)*(x^6 + 2*x^3 + 1) - 6*2^(1/3)*(x^4 - x)*(-x^3 + 1)^(1
/3))/(x^6 + 2*x^3 + 1)) - 1/36*2^(1/3)*log((12*2^(2/3)*(x^8 - 4*x^5 + x^2)*(-x^3 + 1)^(2/3) + 2^(1/3)*(x^12 -
32*x^9 + 78*x^6 - 32*x^3 + 1) + 6*(x^10 - 11*x^7 + 11*x^4 - x)*(-x^3 + 1)^(1/3))/(x^12 + 4*x^9 + 6*x^6 + 4*x^3
+ 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 6.55, size = 1214, normalized size = 4.46 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*ln(-(-6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*
_Z^2)*RootOf(_Z^3-2)^4*x^3-36*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)^2*RootOf(_Z^3-2)^3*x^3+x^6*R
ootOf(_Z^3-2)^2+6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)*x^6+18*RootOf(RootOf(_Z^3
-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^2*(-x^3+1)^(2/3)*x^2+18*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^
2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*x^4-6*RootOf(_Z^3-2)^2*x^3-36*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2
)*RootOf(_Z^3-2)*x^3-18*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*x+RootOf(_Z^3-2)^2+
6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2))/(x+1)^2/(x^2-x+1)^2)-1/6*ln((6*RootOf(Ro
otOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^4*x^3+36*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)
+9*_Z^2)^2*RootOf(_Z^3-2)^3*x^3+x^6*RootOf(_Z^3-2)^2+6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*Roo
tOf(_Z^3-2)*x^6+18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^2*(-x^3+1)^(2/3)*x^2+6*R
ootOf(_Z^3-2)*(-x^3+1)^(1/3)*x^4+18*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*x^4-2*R
ootOf(_Z^3-2)^2*x^3-12*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)*x^3+12*(-x^3+1)^(2/3
)*x^2-6*RootOf(_Z^3-2)*(-x^3+1)^(1/3)*x-18*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*
x+RootOf(_Z^3-2)^2+6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2))/(x+1)^2/(x^2-x+1)^2)*
RootOf(_Z^3-2)-1/2*ln((6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^4*x^3+36*RootOf(Ro
otOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)^2*RootOf(_Z^3-2)^3*x^3+x^6*RootOf(_Z^3-2)^2+6*RootOf(RootOf(_Z^3-2)
^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)*x^6+18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootO
f(_Z^3-2)^2*(-x^3+1)^(2/3)*x^2+6*RootOf(_Z^3-2)*(-x^3+1)^(1/3)*x^4+18*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+3
*_Z*RootOf(_Z^3-2)+9*_Z^2)*x^4-2*RootOf(_Z^3-2)^2*x^3-12*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*R
ootOf(_Z^3-2)*x^3+12*(-x^3+1)^(2/3)*x^2-6*RootOf(_Z^3-2)*(-x^3+1)^(1/3)*x-18*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3
-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*x+RootOf(_Z^3-2)^2+6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*Roo
tOf(_Z^3-2))/(x+1)^2/(x^2-x+1)^2)*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x^3)^(1/3)/(x^3 + 1),x)

[Out]

int((1 - x^3)^(1/3)/(x^3 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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