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

Optimal. Leaf size=231 \[ \frac {\log \left (5 \sqrt [3]{1-x^3}+\sqrt [3]{2} 5^{2/3} x-\sqrt [3]{2} 5^{2/3}\right )}{\sqrt [3]{2} 5^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {5 \sqrt {3} \sqrt [3]{1-x^3}}{5 \sqrt [3]{1-x^3}-2 \sqrt [3]{2} 5^{2/3} x+2 \sqrt [3]{2} 5^{2/3}}\right )}{\sqrt [3]{2} 5^{2/3}}-\frac {\log \left (5 \left (1-x^3\right )^{2/3}+\left (\sqrt [3]{2} 5^{2/3}-\sqrt [3]{2} 5^{2/3} x\right ) \sqrt [3]{1-x^3}+2^{2/3} \sqrt [3]{5} x^2-2\ 2^{2/3} \sqrt [3]{5} x+2^{2/3} \sqrt [3]{5}\right )}{2 \sqrt [3]{2} 5^{2/3}} \]

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Rubi [F]  time = 0.31, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

((5 - Sqrt[5])*Defer[Int][1/((3 - Sqrt[5] + 2*x)*(1 - x^3)^(1/3)), x])/5 + ((5 + Sqrt[5])*Defer[Int][1/((3 + S
qrt[5] + 2*x)*(1 - x^3)^(1/3)), x])/5

Rubi steps

\begin {align*} \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx &=\int \left (\frac {1-\frac {1}{\sqrt {5}}}{\left (3-\sqrt {5}+2 x\right ) \sqrt [3]{1-x^3}}+\frac {1+\frac {1}{\sqrt {5}}}{\left (3+\sqrt {5}+2 x\right ) \sqrt [3]{1-x^3}}\right ) \, dx\\ &=\frac {1}{5} \left (5-\sqrt {5}\right ) \int \frac {1}{\left (3-\sqrt {5}+2 x\right ) \sqrt [3]{1-x^3}} \, dx+\frac {1}{5} \left (5+\sqrt {5}\right ) \int \frac {1}{\left (3+\sqrt {5}+2 x\right ) \sqrt [3]{1-x^3}} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.20, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.23, size = 231, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {5 \sqrt {3} \sqrt [3]{1-x^3}}{2 \sqrt [3]{2} 5^{2/3}-2 \sqrt [3]{2} 5^{2/3} x+5 \sqrt [3]{1-x^3}}\right )}{\sqrt [3]{2} 5^{2/3}}+\frac {\log \left (-\sqrt [3]{2} 5^{2/3}+\sqrt [3]{2} 5^{2/3} x+5 \sqrt [3]{1-x^3}\right )}{\sqrt [3]{2} 5^{2/3}}-\frac {\log \left (2^{2/3} \sqrt [3]{5}-2\ 2^{2/3} \sqrt [3]{5} x+2^{2/3} \sqrt [3]{5} x^2+\left (\sqrt [3]{2} 5^{2/3}-\sqrt [3]{2} 5^{2/3} x\right ) \sqrt [3]{1-x^3}+5 \left (1-x^3\right )^{2/3}\right )}{2 \sqrt [3]{2} 5^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(5*Sqrt[3]*(1 - x^3)^(1/3))/(2*2^(1/3)*5^(2/3) - 2*2^(1/3)*5^(2/3)*x + 5*(1 - x^3)^(1/3))])/(2
^(1/3)*5^(2/3)) + Log[-(2^(1/3)*5^(2/3)) + 2^(1/3)*5^(2/3)*x + 5*(1 - x^3)^(1/3)]/(2^(1/3)*5^(2/3)) - Log[2^(2
/3)*5^(1/3) - 2*2^(2/3)*5^(1/3)*x + 2^(2/3)*5^(1/3)*x^2 + (2^(1/3)*5^(2/3) - 2^(1/3)*5^(2/3)*x)*(1 - x^3)^(1/3
) + 5*(1 - x^3)^(2/3)]/(2*2^(1/3)*5^(2/3))

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fricas [A]  time = 12.66, size = 324, normalized size = 1.40 \begin {gather*} \frac {1}{30} \cdot 50^{\frac {1}{6}} \sqrt {3} \sqrt {2} \arctan \left (\frac {50^{\frac {1}{6}} \sqrt {3} {\left (2 \cdot 50^{\frac {2}{3}} \sqrt {2} {\left (3 \, x^{4} + 8 \, x^{3} + 3 \, x^{2} + 8 \, x + 3\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 50^{\frac {1}{3}} \sqrt {2} {\left (41 \, x^{6} - 11 \, x^{5} + 50 \, x^{4} - 35 \, x^{3} + 50 \, x^{2} - 11 \, x + 41\right )} - 20 \, \sqrt {2} {\left (11 \, x^{5} - 15 \, x^{4} + 15 \, x^{3} - 15 \, x^{2} + 15 \, x - 11\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{30 \, {\left (19 \, x^{6} - 69 \, x^{5} + 30 \, x^{4} - 85 \, x^{3} + 30 \, x^{2} - 69 \, x + 19\right )}}\right ) - \frac {1}{300} \cdot 50^{\frac {2}{3}} \log \left (\frac {50^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} {\left (3 \, x^{2} - x + 3\right )} + 50^{\frac {1}{3}} {\left (11 \, x^{4} - 4 \, x^{3} + 11 \, x^{2} - 4 \, x + 11\right )} - 20 \, {\left (2 \, x^{3} - x^{2} + x - 2\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{4} + 6 \, x^{3} + 11 \, x^{2} + 6 \, x + 1}\right ) + \frac {1}{150} \cdot 50^{\frac {2}{3}} \log \left (\frac {50^{\frac {2}{3}} {\left (x^{2} + 3 \, x + 1\right )} - 10 \cdot 50^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 50 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} + 3 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*50^(1/6)*sqrt(3)*sqrt(2)*arctan(1/30*50^(1/6)*sqrt(3)*(2*50^(2/3)*sqrt(2)*(3*x^4 + 8*x^3 + 3*x^2 + 8*x +
3)*(-x^3 + 1)^(2/3) + 50^(1/3)*sqrt(2)*(41*x^6 - 11*x^5 + 50*x^4 - 35*x^3 + 50*x^2 - 11*x + 41) - 20*sqrt(2)*(
11*x^5 - 15*x^4 + 15*x^3 - 15*x^2 + 15*x - 11)*(-x^3 + 1)^(1/3))/(19*x^6 - 69*x^5 + 30*x^4 - 85*x^3 + 30*x^2 -
69*x + 19)) - 1/300*50^(2/3)*log((50^(2/3)*(-x^3 + 1)^(2/3)*(3*x^2 - x + 3) + 50^(1/3)*(11*x^4 - 4*x^3 + 11*x
^2 - 4*x + 11) - 20*(2*x^3 - x^2 + x - 2)*(-x^3 + 1)^(1/3))/(x^4 + 6*x^3 + 11*x^2 + 6*x + 1)) + 1/150*50^(2/3)
*log((50^(2/3)*(x^2 + 3*x + 1) - 10*50^(1/3)*(-x^3 + 1)^(1/3)*(x - 1) - 50*(-x^3 + 1)^(2/3))/(x^2 + 3*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 10.68, size = 1175, normalized size = 5.09

method	result	size
trager	\(\text {Expression too large to display}\)	\(1175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*ln(-(15*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)^3*x+25*(-x^3+1)^(2/3)*R
ootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)^2+100*RootOf(RootOf(_Z^3-20)^2+10*_Z*R
ootOf(_Z^3-20)+100*_Z^2)^2*RootOf(_Z^3-20)^2*x-11*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2*x-50*(-x^3+1)^(1/3)*RootOf(
RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)*x+11*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2+50*(-x
^3+1)^(1/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)+39*RootOf(_Z^3-20)*x^2+26
0*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x^2+27*RootOf(_Z^3-20)*x+110*(-x^3+1)^(2/3)+180*Roo
tOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x+39*RootOf(_Z^3-20)+260*RootOf(RootOf(_Z^3-20)^2+10*_Z*
RootOf(_Z^3-20)+100*_Z^2))/(x^2+3*x+1))*RootOf(_Z^3-20)-ln(-(15*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)
+100*_Z^2)*RootOf(_Z^3-20)^3*x+25*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*Root
Of(_Z^3-20)^2+100*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)^2*RootOf(_Z^3-20)^2*x-11*(-x^3+1)^(
1/3)*RootOf(_Z^3-20)^2*x-50*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^
3-20)*x+11*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2+50*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+1
00*_Z^2)*RootOf(_Z^3-20)+39*RootOf(_Z^3-20)*x^2+260*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x
^2+27*RootOf(_Z^3-20)*x+110*(-x^3+1)^(2/3)+180*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x+39*R
ootOf(_Z^3-20)+260*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2))/(x^2+3*x+1))*RootOf(RootOf(_Z^3-2
0)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)+RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*ln((5*RootOf(Roo
tOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)^3*x+25*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-20)^2+
10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)^2-100*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)
^2*RootOf(_Z^3-20)^2*x+6*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2*x-50*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*R
ootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)*x-6*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2+50*(-x^3+1)^(1/3)*RootOf(RootOf(
_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)-13*RootOf(_Z^3-20)*x^2+260*RootOf(RootOf(_Z^3-20)^2
+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x^2+RootOf(_Z^3-20)*x-60*(-x^3+1)^(2/3)-20*RootOf(RootOf(_Z^3-20)^2+10*_Z*Roo
tOf(_Z^3-20)+100*_Z^2)*x-13*RootOf(_Z^3-20)+260*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2))/(x^2
+3*x+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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