\(\int \frac {(1+x^4)^{2/3}}{x} \, dx\) [1181]

Optimal result

Integrand size = 13, antiderivative size = 87 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {3}{8} \left (1+x^4\right )^{2/3}+\frac {1}{4} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )+\frac {1}{4} \log \left (-1+\sqrt [3]{1+x^4}\right )-\frac {1}{8} \log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right ) \]

[Out]

3/8*(x^4+1)^(2/3)+1/4*3^(1/2)*arctan(1/3*3^(1/2)+2/3*(x^4+1)^(1/3)*3^(1/2))+1/4*ln(-1+(x^4+1)^(1/3))-1/8*ln(1+
(x^4+1)^(1/3)+(x^4+1)^(2/3))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 52, 57, 632, 210, 31} \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x^4+1}+1}{\sqrt {3}}\right )+\frac {3}{8} \left (x^4+1\right )^{2/3}+\frac {3}{8} \log \left (1-\sqrt [3]{x^4+1}\right )-\frac {\log (x)}{2} \]

[In]

Int[(1 + x^4)^(2/3)/x,x]

[Out]

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

Rule 31

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

Rule 52

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

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 210

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

Rule 272

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

Rule 632

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*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {(1+x)^{2/3}}{x} \, dx,x,x^4\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,x^4\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}-\frac {\log (x)}{2}-\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^4}\right )+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^4}\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}-\frac {\log (x)}{2}+\frac {3}{8} \log \left (1-\sqrt [3]{1+x^4}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^4}\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}+\frac {1}{4} \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )-\frac {\log (x)}{2}+\frac {3}{8} \log \left (1-\sqrt [3]{1+x^4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{8} \left (3 \left (1+x^4\right )^{2/3}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^4}\right )-\log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right )\right ) \]

[In]

Integrate[(1 + x^4)^(2/3)/x,x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 5.38 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+4 \ln \left (x \right )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{12 \pi }\)	\(64\)
pseudoelliptic	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}+\frac {\ln \left (-1+\left (x^{4}+1\right )^{\frac {1}{3}}\right )}{4}-\frac {\ln \left (1+\left (x^{4}+1\right )^{\frac {1}{3}}+\left (x^{4}+1\right )^{\frac {2}{3}}\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{4}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{4}\)	\(64\)
trager	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}+\frac {\ln \left (\frac {-333 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}-393 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-60 x^{4}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}-48 \left (x^{4}+1\right )^{\frac {2}{3}}+144 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+333 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+165 \left (x^{4}+1\right )^{\frac {1}{3}}-384 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-80}{x^{4}}\right )}{4}-\frac {\ln \left (-\frac {-153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+162 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-40 x^{4}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}+165 \left (x^{4}+1\right )^{\frac {2}{3}}-495 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-48 \left (x^{4}+1\right )^{\frac {1}{3}}+195 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-100}{x^{4}}\right )}{4}-\frac {3 \ln \left (-\frac {-153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+162 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-40 x^{4}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}+165 \left (x^{4}+1\right )^{\frac {2}{3}}-495 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-48 \left (x^{4}+1\right )^{\frac {1}{3}}+195 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-100}{x^{4}}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{4}\)	\(426\)

[In]

int((x^4+1)^(2/3)/x,x,method=_RETURNVERBOSE)

[Out]

-1/12/Pi*3^(1/2)*GAMMA(2/3)*(-2/3*Pi*3^(1/2)/GAMMA(2/3)*x^4*hypergeom([1/3,1,1],[2,2],-x^4)-(3/2-1/6*Pi*3^(1/2
)-3/2*ln(3)+4*ln(x))*Pi*3^(1/2)/GAMMA(2/3))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.75 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{4} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + \frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} - \frac {1}{8} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.43 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=- \frac {x^{\frac {8}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 \Gamma \left (\frac {1}{3}\right )} \]

[In]

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

[Out]

-x**(8/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), exp_polar(I*pi)/x**4)/(4*gamma(1/3))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} - \frac {1}{8} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} - \frac {1}{8} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {\ln \left (\frac {9\,{\left (x^4+1\right )}^{1/3}}{16}-\frac {9}{16}\right )}{4}+\ln \left (\frac {9\,{\left (x^4+1\right )}^{1/3}}{16}-9\,{\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )-\ln \left (\frac {9\,{\left (x^4+1\right )}^{1/3}}{16}-9\,{\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )+\frac {3\,{\left (x^4+1\right )}^{2/3}}{8} \]

[In]

int((x^4 + 1)^(2/3)/x,x)

[Out]

log((9*(x^4 + 1)^(1/3))/16 - 9/16)/4 + log((9*(x^4 + 1)^(1/3))/16 - 9*((3^(1/2)*1i)/8 - 1/8)^2)*((3^(1/2)*1i)/
8 - 1/8) - log((9*(x^4 + 1)^(1/3))/16 - 9*((3^(1/2)*1i)/8 + 1/8)^2)*((3^(1/2)*1i)/8 + 1/8) + (3*(x^4 + 1)^(2/3
))/8