\(\int \frac {(1+x^4)^{2/3}}{x^5} \, dx\) [1233]

Optimal result

Integrand size = 13, antiderivative size = 90 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=-\frac {\left (1+x^4\right )^{2/3}}{4 x^4}+\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{6} \log \left (-1+\sqrt [3]{1+x^4}\right )-\frac {1}{12} \log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right ) \]

[Out]

-1/4*(x^4+1)^(2/3)/x^4+1/6*3^(1/2)*arctan(1/3*3^(1/2)+2/3*(x^4+1)^(1/3)*3^(1/2))+1/6*ln(-1+(x^4+1)^(1/3))-1/12
*ln(1+(x^4+1)^(1/3)+(x^4+1)^(2/3))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 43, 57, 632, 210, 31} \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=\frac {\arctan \left (\frac {2 \sqrt [3]{x^4+1}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\left (x^4+1\right )^{2/3}}{4 x^4}+\frac {1}{4} \log \left (1-\sqrt [3]{x^4+1}\right )-\frac {\log (x)}{3} \]

[In]

Int[(1 + x^4)^(2/3)/x^5,x]

[Out]

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

Rule 31

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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^2} \, dx,x,x^4\right ) \\ & = -\frac {\left (1+x^4\right )^{2/3}}{4 x^4}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,x^4\right ) \\ & = -\frac {\left (1+x^4\right )^{2/3}}{4 x^4}-\frac {\log (x)}{3}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^4}\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^4}\right ) \\ & = -\frac {\left (1+x^4\right )^{2/3}}{4 x^4}-\frac {\log (x)}{3}+\frac {1}{4} \log \left (1-\sqrt [3]{1+x^4}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^4}\right ) \\ & = -\frac {\left (1+x^4\right )^{2/3}}{4 x^4}+\frac {\arctan \left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\log (x)}{3}+\frac {1}{4} \log \left (1-\sqrt [3]{1+x^4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=\frac {1}{12} \left (-\frac {3 \left (1+x^4\right )^{2/3}}{x^4}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^4}\right )-\log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right )\right ) \]

[In]

Integrate[(1 + x^4)^(2/3)/x^5,x]

[Out]

((-3*(1 + x^4)^(2/3))/x^4 + 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)])/12

Maple [C] (verified)

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

Time = 4.93 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {\pi \sqrt {3}\, x^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], -x^{4}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+4 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{4}}\right )}{12 \pi }\)	\(76\)
risch	\(-\frac {\left (x^{4}+1\right )^{\frac {2}{3}}}{4 x^{4}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], -x^{4}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+4 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{12 \pi }\)	\(76\)
pseudoelliptic	\(\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{4}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right ) x^{4}+2 \ln \left (-1+\left (x^{4}+1\right )^{\frac {1}{3}}\right ) x^{4}-\ln \left (1+\left (x^{4}+1\right )^{\frac {1}{3}}+\left (x^{4}+1\right )^{\frac {2}{3}}\right ) x^{4}-3 \left (x^{4}+1\right )^{\frac {2}{3}}}{12 \left (-1+\left (x^{4}+1\right )^{\frac {1}{3}}\right ) \left (1+\left (x^{4}+1\right )^{\frac {1}{3}}+\left (x^{4}+1\right )^{\frac {2}{3}}\right )}\)	\(104\)
trager	\(-\frac {\left (x^{4}+1\right )^{\frac {2}{3}}}{4 x^{4}}-\frac {\ln \left (\frac {180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+129 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-51 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}}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-48 \left (x^{4}+1\right )^{\frac {1}{3}}+291 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-68}{x^{4}}\right )}{6}-\frac {\ln \left (\frac {180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+129 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-51 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}}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}-180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-48 \left (x^{4}+1\right )^{\frac {1}{3}}+291 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-68}{x^{4}}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{2}+\frac {\operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (-\frac {-180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}+74 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}}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+165 \left (x^{4}+1\right )^{\frac {1}{3}}+411 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+185}{x^{4}}\right )}{2}\)	\(439\)

[In]

int((x^4+1)^(2/3)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=\frac {2 \, \sqrt {3} x^{4} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{4} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - x^{4} \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, x^{4} \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) - 3 \, {\left (x^{4} + 1\right )}^{\frac {2}{3}}}{12 \, x^{4}} \]

[In]

integrate((x^4+1)^(2/3)/x^5,x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.38 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 x^{\frac {4}{3}} \Gamma \left (\frac {4}{3}\right )} \]

[In]

integrate((x**4+1)**(2/3)/x**5,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{4} + 1\right )}^{\frac {2}{3}}}{4 \, x^{4}} - \frac {1}{12} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

integrate((x^4+1)^(2/3)/x^5,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{4} + 1\right )}^{\frac {2}{3}}}{4 \, x^{4}} - \frac {1}{12} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

integrate((x^4+1)^(2/3)/x^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x^5} \, dx=\frac {\ln \left (\frac {{\left (x^4+1\right )}^{1/3}}{4}-\frac {1}{4}\right )}{6}+\ln \left (\frac {{\left (x^4+1\right )}^{1/3}}{4}-9\,{\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )}^2\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\ln \left (\frac {{\left (x^4+1\right )}^{1/3}}{4}-9\,{\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )}^2\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\frac {{\left (x^4+1\right )}^{2/3}}{4\,x^4} \]

[In]

int((x^4 + 1)^(2/3)/x^5,x)

[Out]

log((x^4 + 1)^(1/3)/4 - 1/4)/6 + log((x^4 + 1)^(1/3)/4 - 9*((3^(1/2)*1i)/12 - 1/12)^2)*((3^(1/2)*1i)/12 - 1/12
) - log((x^4 + 1)^(1/3)/4 - 9*((3^(1/2)*1i)/12 + 1/12)^2)*((3^(1/2)*1i)/12 + 1/12) - (x^4 + 1)^(2/3)/(4*x^4)