3.17.74 \(\int \frac {(1+x^3)^{2/3} (4+6 x^3+3 x^6)}{x^6 (8+6 x^3+3 x^6)} \, dx\) [1674]

Optimal. Leaf size=112 \[ \frac {\left (-8-23 x^3\right ) \left (1+x^3\right )^{2/3}}{80 x^5}+\frac {1}{16} \text {RootSum}\left [5-10 \text {$\#$1}^3+8 \text {$\#$1}^6\& ,\frac {-5 \log (x)+5 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+6 \log (x) \text {$\#$1}^3-6 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}+8 \text {$\#$1}^4}\& \right ] \]

[Out]

Unintegrable

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Rubi [C] Result contains complex when optimal does not.
time = 0.84, antiderivative size = 615, normalized size of antiderivative = 5.49, number of steps used = 13, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6860, 270, 283, 245, 399, 384} \begin {gather*} -\frac {\left (\sqrt {15}+i\right ) \text {ArcTan}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{16 \sqrt {5}}+\frac {\left (-\sqrt {15}+i\right ) \text {ArcTan}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{16 \sqrt {5}}+\frac {1}{8} \sqrt {3} \text {ArcTan}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )-\frac {\sqrt [3]{3} \left (-\sqrt {15}+i\right ) \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [6]{15} x}{\sqrt [3]{\sqrt {15}-3 i} \sqrt [3]{x^3+1}}}{\sqrt {3}}\right )}{16 \sqrt [6]{5} \left (\sqrt {15}-3 i\right )^{2/3}}+\frac {\sqrt [3]{3} \left (\sqrt {15}+i\right ) \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [6]{15} x}{\sqrt [3]{\sqrt {15}+3 i} \sqrt [3]{x^3+1}}}{\sqrt {3}}\right )}{16 \sqrt [6]{5} \left (\sqrt {15}+3 i\right )^{2/3}}+\frac {\left (\sqrt {15}+i\right ) \log \left (6 x^3+2 \left (3-i \sqrt {15}\right )\right )}{32 \sqrt [6]{15} \left (\sqrt {15}+3 i\right )^{2/3}}-\frac {\left (-\sqrt {15}+i\right ) \log \left (6 x^3+2 \left (3+i \sqrt {15}\right )\right )}{32 \sqrt [6]{15} \left (\sqrt {15}-3 i\right )^{2/3}}+\frac {3^{5/6} \left (-\sqrt {15}+i\right ) \log \left (-\sqrt [3]{x^3+1}+\frac {\sqrt [6]{15} x}{\sqrt [3]{\sqrt {15}-3 i}}\right )}{32 \sqrt [6]{5} \left (\sqrt {15}-3 i\right )^{2/3}}-\frac {3^{5/6} \left (\sqrt {15}+i\right ) \log \left (-\sqrt [3]{x^3+1}+\frac {\sqrt [6]{15} x}{\sqrt [3]{\sqrt {15}+3 i}}\right )}{32 \sqrt [6]{5} \left (\sqrt {15}+3 i\right )^{2/3}}+\frac {1}{160} \left (15+i \sqrt {15}\right ) \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {1}{160} \left (15-i \sqrt {15}\right ) \log \left (\sqrt [3]{x^3+1}-x\right )-\frac {3}{16} \log \left (\sqrt [3]{x^3+1}-x\right )-\frac {\left (x^3+1\right )^{5/3}}{10 x^5}-\frac {3 \left (x^3+1\right )^{2/3}}{16 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 + x^3)^(2/3))/(16*x^2) - (1 + x^3)^(5/3)/(10*x^5) + (Sqrt[3]*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]
])/8 + ((I - Sqrt[15])*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(16*Sqrt[5]) - ((I + Sqrt[15])*ArcTan[(1 +
(2*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(16*Sqrt[5]) - (3^(1/3)*(I - Sqrt[15])*ArcTan[(1 + (2*15^(1/6)*x)/((-3*I + S
qrt[15])^(1/3)*(1 + x^3)^(1/3)))/Sqrt[3]])/(16*5^(1/6)*(-3*I + Sqrt[15])^(2/3)) + (3^(1/3)*(I + Sqrt[15])*ArcT
an[(1 + (2*15^(1/6)*x)/((3*I + Sqrt[15])^(1/3)*(1 + x^3)^(1/3)))/Sqrt[3]])/(16*5^(1/6)*(3*I + Sqrt[15])^(2/3))
+ ((I + Sqrt[15])*Log[2*(3 - I*Sqrt[15]) + 6*x^3])/(32*15^(1/6)*(3*I + Sqrt[15])^(2/3)) - ((I - Sqrt[15])*Log
[2*(3 + I*Sqrt[15]) + 6*x^3])/(32*15^(1/6)*(-3*I + Sqrt[15])^(2/3)) + (3^(5/6)*(I - Sqrt[15])*Log[(15^(1/6)*x)
/(-3*I + Sqrt[15])^(1/3) - (1 + x^3)^(1/3)])/(32*5^(1/6)*(-3*I + Sqrt[15])^(2/3)) - (3^(5/6)*(I + Sqrt[15])*Lo
g[(15^(1/6)*x)/(3*I + Sqrt[15])^(1/3) - (1 + x^3)^(1/3)])/(32*5^(1/6)*(3*I + Sqrt[15])^(2/3)) - (3*Log[-x + (1
+ x^3)^(1/3)])/16 + ((15 - I*Sqrt[15])*Log[-x + (1 + x^3)^(1/3)])/160 + ((15 + I*Sqrt[15])*Log[-x + (1 + x^3)
^(1/3)])/160

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 399

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[b/d, Int[(a + b*x^n)^(p - 1), x]
, x] - Dist[(b*c - a*d)/d, Int[(a + b*x^n)^(p - 1)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c
- a*d, 0] && EqQ[n*(p - 1) + 1, 0] && IntegerQ[n]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (1+x^3\right )^{2/3} \left (4+6 x^3+3 x^6\right )}{x^6 \left (8+6 x^3+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 {3 \left (1+x^3\right )^{2/3} \left (2+3 x^3\right )}{8 \left (8+6 x^3+3 x^6\right )}\right ) \, dx\\ &=\frac {3}{8} \int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx-\frac {3}{8} \int \frac {\left (1+x^3\right )^{2/3} \left (2+3 x^3\right )}{8+6 x^3+3 x^6} \, 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 {3}{8} \int \frac {1}{\sqrt [3]{1+x^3}} \, dx-\frac {3}{8} \int \left (\frac {\left (3+i \sqrt {\frac {3}{5}}\right ) \left (1+x^3\right )^{2/3}}{6-2 i \sqrt {15}+6 x^3}+\frac {\left (3-i \sqrt {\frac {3}{5}}\right ) \left (1+x^3\right )^{2/3}}{6+2 i \sqrt {15}+6 x^3}\right ) \, 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}{40} \left (3 \left (15-i \sqrt {15}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{6+2 i \sqrt {15}+6 x^3} \, dx-\frac {1}{40} \left (3 \left (15+i \sqrt {15}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{6-2 i \sqrt {15}+6 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 {3 \left (15 i-\sqrt {15}\right ) x F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {3 x^3}{3-i \sqrt {15}}\right )}{80 \left (3 i+\sqrt {15}\right )}-\frac {3 \left (15 i+\sqrt {15}\right ) x F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {3 x^3}{3+i \sqrt {15}}\right )}{80 \left (3 i-\sqrt {15}\right )}+\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 )\\ \end {align*}

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Mathematica [A]
time = 0.17, 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}{16} \text {RootSum}\left [5-10 \text {$\#$1}^3+8 \text {$\#$1}^6\&,\frac {-5 \log (x)+5 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+6 \log (x) \text {$\#$1}^3-6 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}+8 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8 - 23*x^3)*(1 + x^3)^(2/3))/(80*x^5) + RootSum[5 - 10*#1^3 + 8*#1^6 & , (-5*Log[x] + 5*Log[(1 + x^3)^(1/3)
- x*#1] + 6*Log[x]*#1^3 - 6*Log[(1 + x^3)^(1/3) - x*#1]*#1^3)/(-5*#1 + 8*#1^4) & ]/16

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 1.
time = 230.80, size = 8003, normalized size = 71.46

method	result	size
risch	\(\text {Expression too large to display}\)	\(8003\)
trager	\(\text {Expression too large to display}\)	\(10918\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((3*x^6 + 6*x^3 + 4)*(x^3 + 1)^(2/3)/((3*x^6 + 6*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)*(3*x^6+6*x^3+4)/x^6/(3*x^6+6*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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integrate((3*x^6 + 6*x^3 + 4)*(x^3 + 1)^(2/3)/((3*x^6 + 6*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 (3\,x^6+6\,x^3+4\right )}{x^6\,\left (3\,x^6+6\,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)*(6*x^3 + 3*x^6 + 4))/(x^6*(6*x^3 + 3*x^6 + 8)),x)

[Out]

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

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