3.14.92 \(\int \frac {\sqrt [3]{1+x^4} (3+x^4)}{x^{13}} \, dx\)

Optimal. Leaf size=100 \[ \frac {1}{54} \log \left (\sqrt [3]{x^4+1}-1\right )-\frac {1}{108} \log \left (\left (x^4+1\right )^{2/3}+\sqrt [3]{x^4+1}+1\right )-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^4+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{18 \sqrt {3}}+\frac {\sqrt [3]{x^4+1} \left (x^8-6 x^4-9\right )}{36 x^{12}} \]

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Rubi [A]  time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {446, 78, 47, 51, 57, 618, 204, 31} \begin {gather*} \frac {\sqrt [3]{x^4+1}}{36 x^4}+\frac {1}{36} \log \left (1-\sqrt [3]{x^4+1}\right )-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^4+1}+1}{\sqrt {3}}\right )}{18 \sqrt {3}}-\frac {\left (x^4+1\right )^{4/3}}{4 x^{12}}+\frac {\sqrt [3]{x^4+1}}{12 x^8}-\frac {\log (x)}{27} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^4)^(1/3)*(3 + x^4))/x^13,x]

[Out]

(1 + x^4)^(1/3)/(12*x^8) + (1 + x^4)^(1/3)/(36*x^4) - (1 + x^4)^(4/3)/(4*x^12) - ArcTan[(1 + 2*(1 + x^4)^(1/3)
)/Sqrt[3]]/(18*Sqrt[3]) - Log[x]/27 + Log[1 - (1 + x^4)^(1/3)]/36

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 446

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{1+x^4} \left (3+x^4\right )}{x^{13}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt [3]{1+x} (3+x)}{x^4} \, dx,x,x^4\right )\\ &=-\frac {\left (1+x^4\right )^{4/3}}{4 x^{12}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\sqrt [3]{1+x}}{x^3} \, dx,x,x^4\right )\\ &=\frac {\sqrt [3]{1+x^4}}{12 x^8}-\frac {\left (1+x^4\right )^{4/3}}{4 x^{12}}-\frac {1}{36} \operatorname {Subst}\left (\int \frac {1}{x^2 (1+x)^{2/3}} \, dx,x,x^4\right )\\ &=\frac {\sqrt [3]{1+x^4}}{12 x^8}+\frac {\sqrt [3]{1+x^4}}{36 x^4}-\frac {\left (1+x^4\right )^{4/3}}{4 x^{12}}+\frac {1}{54} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{2/3}} \, dx,x,x^4\right )\\ &=\frac {\sqrt [3]{1+x^4}}{12 x^8}+\frac {\sqrt [3]{1+x^4}}{36 x^4}-\frac {\left (1+x^4\right )^{4/3}}{4 x^{12}}-\frac {\log (x)}{27}-\frac {1}{36} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^4}\right )-\frac {1}{36} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^4}\right )\\ &=\frac {\sqrt [3]{1+x^4}}{12 x^8}+\frac {\sqrt [3]{1+x^4}}{36 x^4}-\frac {\left (1+x^4\right )^{4/3}}{4 x^{12}}-\frac {\log (x)}{27}+\frac {1}{36} \log \left (1-\sqrt [3]{1+x^4}\right )+\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^4}\right )\\ &=\frac {\sqrt [3]{1+x^4}}{12 x^8}+\frac {\sqrt [3]{1+x^4}}{36 x^4}-\frac {\left (1+x^4\right )^{4/3}}{4 x^{12}}-\frac {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )}{18 \sqrt {3}}-\frac {\log (x)}{27}+\frac {1}{36} \log \left (1-\sqrt [3]{1+x^4}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.35 \begin {gather*} \frac {\left (x^4+1\right )^{4/3} \left (x^{12} \, _2F_1\left (\frac {4}{3},3;\frac {7}{3};x^4+1\right )-2\right )}{8 x^{12}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^4)^(1/3)*(3 + x^4))/x^13,x]

[Out]

((1 + x^4)^(4/3)*(-2 + x^12*Hypergeometric2F1[4/3, 3, 7/3, 1 + x^4]))/(8*x^12)

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IntegrateAlgebraic [A]  time = 0.15, size = 100, normalized size = 1.00 \begin {gather*} \frac {\sqrt [3]{1+x^4} \left (-9-6 x^4+x^8\right )}{36 x^{12}}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )}{18 \sqrt {3}}+\frac {1}{54} \log \left (-1+\sqrt [3]{1+x^4}\right )-\frac {1}{108} \log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^4)^(1/3)*(3 + x^4))/x^13,x]

[Out]

((1 + x^4)^(1/3)*(-9 - 6*x^4 + x^8))/(36*x^12) - ArcTan[1/Sqrt[3] + (2*(1 + x^4)^(1/3))/Sqrt[3]]/(18*Sqrt[3])
+ Log[-1 + (1 + x^4)^(1/3)]/54 - Log[1 + (1 + x^4)^(1/3) + (1 + x^4)^(2/3)]/108

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fricas [A]  time = 0.45, size = 88, normalized size = 0.88 \begin {gather*} -\frac {2 \, \sqrt {3} x^{12} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{4} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + x^{12} \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) - 2 \, x^{12} \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) - 3 \, {\left (x^{8} - 6 \, x^{4} - 9\right )} {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{108 \, x^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(1/3)*(x^4+3)/x^13,x, algorithm="fricas")

[Out]

-1/108*(2*sqrt(3)*x^12*arctan(2/3*sqrt(3)*(x^4 + 1)^(1/3) + 1/3*sqrt(3)) + x^12*log((x^4 + 1)^(2/3) + (x^4 + 1
)^(1/3) + 1) - 2*x^12*log((x^4 + 1)^(1/3) - 1) - 3*(x^8 - 6*x^4 - 9)*(x^4 + 1)^(1/3))/x^12

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giac [A]  time = 0.16, size = 85, normalized size = 0.85 \begin {gather*} -\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {{\left (x^{4} + 1\right )}^{\frac {7}{3}} - 8 \, {\left (x^{4} + 1\right )}^{\frac {4}{3}} - 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{36 \, x^{12}} - \frac {1}{108} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{54} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(1/3)*(x^4+3)/x^13,x, algorithm="giac")

[Out]

-1/54*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^4 + 1)^(1/3) + 1)) + 1/36*((x^4 + 1)^(7/3) - 8*(x^4 + 1)^(4/3) - 2*(x^4
+ 1)^(1/3))/x^12 - 1/108*log((x^4 + 1)^(2/3) + (x^4 + 1)^(1/3) + 1) + 1/54*log((x^4 + 1)^(1/3) - 1)

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maple [C]  time = 3.99, size = 74, normalized size = 0.74

method	result	size
risch	\(\frac {x^{12}-5 x^{8}-15 x^{4}-9}{36 x^{12} \left (x^{4}+1\right )^{\frac {2}{3}}}+\frac {-\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{4} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], -x^{4}\right )}{3}+\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+4 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{54 \Gamma \left (\frac {2}{3}\right )}\)	\(74\)
meijerg	\(-\frac {-\frac {5 \Gamma \left (\frac {2}{3}\right ) x^{4} \hypergeom \left (\left [1, 1, \frac {8}{3}\right ], \left [2, 4\right ], -x^{4}\right )}{27}+\frac {\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+4 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{3}+\frac {3 \Gamma \left (\frac {2}{3}\right )}{2 x^{8}}+\frac {\Gamma \left (\frac {2}{3}\right )}{x^{4}}}{12 \Gamma \left (\frac {2}{3}\right )}-\frac {\frac {10 \Gamma \left (\frac {2}{3}\right ) x^{4} \hypergeom \left (\left [1, 1, \frac {11}{3}\right ], \left [2, 5\right ], -x^{4}\right )}{81}-\frac {5 \left (\frac {4}{15}+\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+4 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{27}+\frac {\Gamma \left (\frac {2}{3}\right )}{x^{12}}+\frac {\Gamma \left (\frac {2}{3}\right )}{2 x^{8}}-\frac {\Gamma \left (\frac {2}{3}\right )}{3 x^{4}}}{4 \Gamma \left (\frac {2}{3}\right )}\)	\(128\)
trager	\(\frac {\left (x^{4}+1\right )^{\frac {1}{3}} \left (x^{8}-6 x^{4}-9\right )}{36 x^{12}}+\frac {\RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (\frac {180 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+129 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-51 x^{4}+351 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}-180 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+351 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}-48 \left (x^{4}+1\right )^{\frac {2}{3}}+291 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-48 \left (x^{4}+1\right )^{\frac {1}{3}}-68}{x^{4}}\right )}{18}-\frac {\ln \left (\frac {180 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}-9 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-74 x^{4}-351 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}-180 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-351 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}-165 \left (x^{4}+1\right )^{\frac {2}{3}}-411 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-165 \left (x^{4}+1\right )^{\frac {1}{3}}-185}{x^{4}}\right )}{54}-\frac {\ln \left (\frac {180 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}-9 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-74 x^{4}-351 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}-180 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-351 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}-165 \left (x^{4}+1\right )^{\frac {2}{3}}-411 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-165 \left (x^{4}+1\right )^{\frac {1}{3}}-185}{x^{4}}\right ) \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{18}\)	\(448\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)^(1/3)*(x^4+3)/x^13,x,method=_RETURNVERBOSE)

[Out]

1/36*(x^12-5*x^8-15*x^4-9)/x^12/(x^4+1)^(2/3)+1/54/GAMMA(2/3)*(-2/3*GAMMA(2/3)*x^4*hypergeom([1,1,5/3],[2,2],-
x^4)+(1/6*Pi*3^(1/2)-3/2*ln(3)+4*ln(x))*GAMMA(2/3))

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maxima [A]  time = 0.46, size = 146, normalized size = 1.46 \begin {gather*} -\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {5 \, {\left (x^{4} + 1\right )}^{\frac {7}{3}} - 13 \, {\left (x^{4} + 1\right )}^{\frac {4}{3}} - 10 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{72 \, {\left (3 \, x^{4} + {\left (x^{4} + 1\right )}^{3} - 3 \, {\left (x^{4} + 1\right )}^{2} + 2\right )}} + \frac {{\left (x^{4} + 1\right )}^{\frac {4}{3}} + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{24 \, {\left (2 \, x^{4} - {\left (x^{4} + 1\right )}^{2} + 1\right )}} - \frac {1}{108} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{54} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(1/3)*(x^4+3)/x^13,x, algorithm="maxima")

[Out]

-1/54*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^4 + 1)^(1/3) + 1)) + 1/72*(5*(x^4 + 1)^(7/3) - 13*(x^4 + 1)^(4/3) - 10*
(x^4 + 1)^(1/3))/(3*x^4 + (x^4 + 1)^3 - 3*(x^4 + 1)^2 + 2) + 1/24*((x^4 + 1)^(4/3) + 2*(x^4 + 1)^(1/3))/(2*x^4
- (x^4 + 1)^2 + 1) - 1/108*log((x^4 + 1)^(2/3) + (x^4 + 1)^(1/3) + 1) + 1/54*log((x^4 + 1)^(1/3) - 1)

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mupad [B]  time = 1.53, size = 232, normalized size = 2.32 \begin {gather*} \frac {5\,\ln \left (\frac {25\,{\left (x^4+1\right )}^{1/3}}{1296}-\frac {25}{1296}\right )}{108}-\frac {\ln \left (\frac {{\left (x^4+1\right )}^{1/3}}{144}-\frac {1}{144}\right )}{36}-\frac {\frac {5\,{\left (x^4+1\right )}^{1/3}}{36}+\frac {13\,{\left (x^4+1\right )}^{4/3}}{72}-\frac {5\,{\left (x^4+1\right )}^{7/3}}{72}}{{\left (x^4+1\right )}^3-3\,{\left (x^4+1\right )}^2+3\,x^4+2}+\frac {\frac {{\left (x^4+1\right )}^{1/3}}{12}+\frac {{\left (x^4+1\right )}^{4/3}}{24}}{2\,x^4-{\left (x^4+1\right )}^2+1}-\ln \left (\frac {{\left (x^4+1\right )}^{1/3}}{4}+\frac {1}{8}-\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )\,\left (-\frac {1}{72}+\frac {\sqrt {3}\,1{}\mathrm {i}}{72}\right )+\ln \left (\frac {{\left (x^4+1\right )}^{1/3}}{4}+\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )\,\left (\frac {1}{72}+\frac {\sqrt {3}\,1{}\mathrm {i}}{72}\right )+\ln \left (\frac {5\,{\left (x^4+1\right )}^{1/3}}{12}+\frac {5}{24}-\frac {\sqrt {3}\,5{}\mathrm {i}}{24}\right )\,\left (-\frac {5}{216}+\frac {\sqrt {3}\,5{}\mathrm {i}}{216}\right )-\ln \left (\frac {5\,{\left (x^4+1\right )}^{1/3}}{12}+\frac {5}{24}+\frac {\sqrt {3}\,5{}\mathrm {i}}{24}\right )\,\left (\frac {5}{216}+\frac {\sqrt {3}\,5{}\mathrm {i}}{216}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 + 1)^(1/3)*(x^4 + 3))/x^13,x)

[Out]

(5*log((25*(x^4 + 1)^(1/3))/1296 - 25/1296))/108 - log((x^4 + 1)^(1/3)/144 - 1/144)/36 - ((5*(x^4 + 1)^(1/3))/
36 + (13*(x^4 + 1)^(4/3))/72 - (5*(x^4 + 1)^(7/3))/72)/((x^4 + 1)^3 - 3*(x^4 + 1)^2 + 3*x^4 + 2) + ((x^4 + 1)^
(1/3)/12 + (x^4 + 1)^(4/3)/24)/(2*x^4 - (x^4 + 1)^2 + 1) - log((x^4 + 1)^(1/3)/4 - (3^(1/2)*1i)/8 + 1/8)*((3^(
1/2)*1i)/72 - 1/72) + log((3^(1/2)*1i)/8 + (x^4 + 1)^(1/3)/4 + 1/8)*((3^(1/2)*1i)/72 + 1/72) + log((5*(x^4 + 1
)^(1/3))/12 - (3^(1/2)*5i)/24 + 5/24)*((3^(1/2)*5i)/216 - 5/216) - log((3^(1/2)*5i)/24 + (5*(x^4 + 1)^(1/3))/1
2 + 5/24)*((3^(1/2)*5i)/216 + 5/216)

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sympy [F(-1)]  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**4+1)**(1/3)*(x**4+3)/x**13,x)

[Out]

Timed out

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