3.17.66 \(\int \frac {(1+x^3)^{2/3} (4+x^3)}{x^6 (8-4 x^3+x^6)} \, dx\) [1666]

Optimal. Leaf size=112 \[ \frac {\left (-8-23 x^3\right ) \left (1+x^3\right )^{2/3}}{80 x^5}+\frac {1}{96} \text {RootSum}\left [13-20 \text {$\#$1}^3+8 \text {$\#$1}^6\& ,\frac {-39 \log (x)+39 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+32 \log (x) \text {$\#$1}^3-32 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}+4 \text {$\#$1}^4}\& \right ] \]

[Out]

Unintegrable

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Rubi [F]
time = 1.40, 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 {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((1 + x^3)^(2/3)*(4 + x^3))/(x^6*(8 - 4*x^3 + x^6)),x]

[Out]

(1 + x^3)^(2/3)/96 - (3*(1 + x^3)^(2/3))/(16*x^2) - (1 + x^3)^(5/3)/(10*x^5) - (7/384 + (3*I)/128)*x^2*AppellF
1[2/3, 1/3, 1, 5/3, -x^3, (1/4 - I/4)*x^3] - (7/384 - (3*I)/128)*x^2*AppellF1[2/3, 1/3, 1, 5/3, -x^3, (1/4 + I
/4)*x^3] - ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]]/(8*Sqrt[3]) + (Sqrt[3]*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3
))/Sqrt[3]])/8 + ((1/32 + I/96)*(5 - I)^(2/3)*ArcTan[(1 + ((10 - 2*I)^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(2^(
1/3)*Sqrt[3]) + ((1/32 - I/96)*(5 + I)^(2/3)*ArcTan[(1 + ((10 + 2*I)^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(2^(1
/3)*Sqrt[3]) + ((1/96 + I/32)*(3 - 2*I)^(2/3)*ArcTan[((3 - 2*I)^(1/3) + 2*(1 + x^3)^(1/3))/(Sqrt[3]*(3 - 2*I)^
(1/3))])/Sqrt[3] + ((1/96 - I/32)*(3 + 2*I)^(2/3)*ArcTan[((3 + 2*I)^(1/3) + 2*(1 + x^3)^(1/3))/(Sqrt[3]*(3 + 2
*I)^(1/3))])/Sqrt[3] + (x^2*Hypergeometric2F1[1/3, 2/3, 5/3, -x^3])/48 - (1/576 + I/192)*(3 - 2*I)^(2/3)*Log[(
2 - 2*I) - x^3] - (1/576 - I/192)*(3 + 2*I)^(2/3)*Log[(2 + 2*I) - x^3] + ((1/36 + I/288)*Log[(-16 - 16*I) + 8*
x^3])/(10 - 2*I)^(1/3) + ((1/36 - I/288)*Log[(-16 + 16*I) + 8*x^3])/(10 + 2*I)^(1/3) + (1/192 + I/64)*(3 - 2*I
)^(2/3)*Log[(3 - 2*I)^(1/3) - (1 + x^3)^(1/3)] + (1/192 - I/64)*(3 + 2*I)^(2/3)*Log[(3 + 2*I)^(1/3) - (1 + x^3
)^(1/3)] - ((1/12 + I/96)*Log[((5 - I)^(1/3)*x)/2^(2/3) - (1 + x^3)^(1/3)])/(10 - 2*I)^(1/3) - ((1/12 - I/96)*
Log[((5 + I)^(1/3)*x)/2^(2/3) - (1 + x^3)^(1/3)])/(10 + 2*I)^(1/3) - Log[-x + (1 + x^3)^(1/3)]/8 + Defer[Int][
(1 + x^3)^(2/3)/(4 - 4*x + 2*x^2 - 2*x^3 + x^4), x]/3 - (5*Defer[Int][(x*(1 + x^3)^(2/3))/(4 - 4*x + 2*x^2 - 2
*x^3 + x^4), x])/24 - Defer[Int][(x^3*(1 + x^3)^(2/3))/(4 - 4*x + 2*x^2 - 2*x^3 + x^4), x]/48

Rubi steps

\begin {align*} \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx &=\int \left (\frac {\left (1+x^3\right )^{2/3}}{2 x^6}+\frac {3 \left (1+x^3\right )^{2/3}}{8 x^3}+\frac {(4+x) \left (1+x^3\right )^{2/3}}{48 \left (2+2 x+x^2\right )}+\frac {\left (16-10 x-x^3\right ) \left (1+x^3\right )^{2/3}}{48 \left (4-4 x+2 x^2-2 x^3+x^4\right )}\right ) \, dx\\ &=\frac {1}{48} \int \frac {(4+x) \left (1+x^3\right )^{2/3}}{2+2 x+x^2} \, dx+\frac {1}{48} \int \frac {\left (16-10 x-x^3\right ) \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {3}{8} \int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx+\frac {1}{2} \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx\\ &=-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{48} \int \left (\frac {(1-3 i) \left (1+x^3\right )^{2/3}}{(2-2 i)+2 x}+\frac {(1+3 i) \left (1+x^3\right )^{2/3}}{(2+2 i)+2 x}\right ) \, dx+\frac {1}{48} \int \left (\frac {16 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4}-\frac {10 x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4}-\frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4}\right ) \, dx+\frac {3}{8} \int \frac {1}{\sqrt [3]{1+x^3}} \, dx\\ &=-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{8} \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )-\frac {3}{16} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {1}{48}-\frac {i}{16}\right ) \int \frac {\left (1+x^3\right )^{2/3}}{(2-2 i)+2 x} \, dx+\left (\frac {1}{48}+\frac {i}{16}\right ) \int \frac {\left (1+x^3\right )^{2/3}}{(2+2 i)+2 x} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 112, normalized size = 1.00 \begin {gather*} \frac {\left (-8-23 x^3\right ) \left (1+x^3\right )^{2/3}}{80 x^5}+\frac {1}{96} \text {RootSum}\left [13-20 \text {$\#$1}^3+8 \text {$\#$1}^6\&,\frac {-39 \log (x)+39 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+32 \log (x) \text {$\#$1}^3-32 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}+4 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^3)^(2/3)*(4 + x^3))/(x^6*(8 - 4*x^3 + x^6)),x]

[Out]

((-8 - 23*x^3)*(1 + x^3)^(2/3))/(80*x^5) + RootSum[13 - 20*#1^3 + 8*#1^6 & , (-39*Log[x] + 39*Log[(1 + x^3)^(1
/3) - x*#1] + 32*Log[x]*#1^3 - 32*Log[(1 + x^3)^(1/3) - x*#1]*#1^3)/(-5*#1 + 4*#1^4) & ]/96

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 1.
time = 274.71, size = 6957, normalized size = 62.12

method	result	size
risch	\(\text {Expression too large to display}\)	\(6957\)
trager	\(\text {Expression too large to display}\)	\(10757\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x, algorithm="maxima")

[Out]

integrate((x^3 + 4)*(x^3 + 1)^(2/3)/((x^6 - 4*x^3 + 8)*x^6), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (x + 1\right ) \left (x^{2} - x + 1\right )\right )^{\frac {2}{3}} \left (x^{3} + 4\right )}{x^{6} \left (x^{2} + 2 x + 2\right ) \left (x^{4} - 2 x^{3} + 2 x^{2} - 4 x + 4\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)**(2/3)*(x**3+4)/x**6/(x**6-4*x**3+8),x)

[Out]

Integral(((x + 1)*(x**2 - x + 1))**(2/3)*(x**3 + 4)/(x**6*(x**2 + 2*x + 2)*(x**4 - 2*x**3 + 2*x**2 - 4*x + 4))
, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x, algorithm="giac")

[Out]

integrate((x^3 + 4)*(x^3 + 1)^(2/3)/((x^6 - 4*x^3 + 8)*x^6), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^3+1\right )}^{2/3}\,\left (x^3+4\right )}{x^6\,\left (x^6-4\,x^3+8\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 + 1)^(2/3)*(x^3 + 4))/(x^6*(x^6 - 4*x^3 + 8)),x)

[Out]

int(((x^3 + 1)^(2/3)*(x^3 + 4))/(x^6*(x^6 - 4*x^3 + 8)), x)

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