3.6.95 \(\int \frac {(-1+x^4) \sqrt [4]{1+x^4}}{x} \, dx\)

Optimal. Leaf size=47 \[ \frac {1}{5} \sqrt [4]{x^4+1} \left (x^4-4\right )+\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{x^4+1}\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{x^4+1}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {446, 80, 50, 63, 212, 206, 203} \begin {gather*} \frac {1}{5} \left (x^4+1\right )^{5/4}-\sqrt [4]{x^4+1}+\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{x^4+1}\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{x^4+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^4)*(1 + x^4)^(1/4))/x,x]

[Out]

-(1 + x^4)^(1/4) + (1 + x^4)^(5/4)/5 + ArcTan[(1 + x^4)^(1/4)]/2 + ArcTanh[(1 + x^4)^(1/4)]/2

Rule 50

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

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

Rule 203

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

Rule 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/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]]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {(-1+x) \sqrt [4]{1+x}}{x} \, dx,x,x^4\right )\\ &=\frac {1}{5} \left (1+x^4\right )^{5/4}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt [4]{1+x}}{x} \, dx,x,x^4\right )\\ &=-\sqrt [4]{1+x^4}+\frac {1}{5} \left (1+x^4\right )^{5/4}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^4\right )\\ &=-\sqrt [4]{1+x^4}+\frac {1}{5} \left (1+x^4\right )^{5/4}-\operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^4}\right )\\ &=-\sqrt [4]{1+x^4}+\frac {1}{5} \left (1+x^4\right )^{5/4}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^4}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^4}\right )\\ &=-\sqrt [4]{1+x^4}+\frac {1}{5} \left (1+x^4\right )^{5/4}+\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{1+x^4}\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{1+x^4}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 0.96 \begin {gather*} \frac {1}{10} \left (2 \sqrt [4]{x^4+1} \left (x^4-4\right )+5 \tan ^{-1}\left (\sqrt [4]{x^4+1}\right )+5 \tanh ^{-1}\left (\sqrt [4]{x^4+1}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-4 + x^4)*(1 + x^4)^(1/4) + 5*ArcTan[(1 + x^4)^(1/4)] + 5*ArcTanh[(1 + x^4)^(1/4)])/10

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IntegrateAlgebraic [A]  time = 0.04, size = 47, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (-4+x^4\right ) \sqrt [4]{1+x^4}+\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{1+x^4}\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{1+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4 + x^4)*(1 + x^4)^(1/4))/5 + ArcTan[(1 + x^4)^(1/4)]/2 + ArcTanh[(1 + x^4)^(1/4)]/2

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fricas [A]  time = 0.47, size = 49, normalized size = 1.04 \begin {gather*} \frac {1}{5} \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} {\left (x^{4} - 4\right )} + \frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^4 + 1)^(1/4)*(x^4 - 4) + 1/2*arctan((x^4 + 1)^(1/4)) + 1/4*log((x^4 + 1)^(1/4) + 1) - 1/4*log((x^4 + 1)
^(1/4) - 1)

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giac [A]  time = 0.29, size = 53, normalized size = 1.13 \begin {gather*} \frac {1}{5} \, {\left (x^{4} + 1\right )}^{\frac {5}{4}} - {\left (x^{4} + 1\right )}^{\frac {1}{4}} + \frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^4 + 1)^(5/4) - (x^4 + 1)^(1/4) + 1/2*arctan((x^4 + 1)^(1/4)) + 1/4*log((x^4 + 1)^(1/4) + 1) - 1/4*log((
x^4 + 1)^(1/4) - 1)

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maple [C]  time = 4.36, size = 62, normalized size = 1.32

method	result	size
meijerg	\(\frac {-4 \left (4-3 \ln \relax (2)+\frac {\pi }{2}+4 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )-\hypergeom \left (\left [\frac {3}{4}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right ) \Gamma \left (\frac {3}{4}\right ) x^{4}}{16 \Gamma \left (\frac {3}{4}\right )}+\frac {\hypergeom \left (\left [-\frac {1}{4}, 1\right ], \relax [2], -x^{4}\right ) x^{4}}{4}\)	\(62\)
trager	\(\left (\frac {x^{4}}{5}-\frac {4}{5}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}+2 \left (x^{4}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}+1\right )^{\frac {1}{4}}}{x^{4}}\right )}{4}-\frac {\ln \left (\frac {-x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}}-2 \sqrt {x^{4}+1}+2 \left (x^{4}+1\right )^{\frac {1}{4}}-2}{x^{4}}\right )}{4}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/GAMMA(3/4)*(-4*(4-3*ln(2)+1/2*Pi+4*ln(x))*GAMMA(3/4)-hypergeom([3/4,1,1],[2,2],-x^4)*GAMMA(3/4)*x^4)+1/4*
hypergeom([-1/4,1],[2],-x^4)*x^4

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maxima [A]  time = 0.49, size = 53, normalized size = 1.13 \begin {gather*} \frac {1}{5} \, {\left (x^{4} + 1\right )}^{\frac {5}{4}} - {\left (x^{4} + 1\right )}^{\frac {1}{4}} + \frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^4 + 1)^(5/4) - (x^4 + 1)^(1/4) + 1/2*arctan((x^4 + 1)^(1/4)) + 1/4*log((x^4 + 1)^(1/4) + 1) - 1/4*log((
x^4 + 1)^(1/4) - 1)

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mupad [B]  time = 0.73, size = 39, normalized size = 0.83 \begin {gather*} \frac {\mathrm {atan}\left ({\left (x^4+1\right )}^{1/4}\right )}{2}+\frac {\mathrm {atanh}\left ({\left (x^4+1\right )}^{1/4}\right )}{2}-{\left (x^4+1\right )}^{1/4}+\frac {{\left (x^4+1\right )}^{5/4}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)*(x^4 + 1)^(1/4))/x,x)

[Out]

atan((x^4 + 1)^(1/4))/2 + atanh((x^4 + 1)^(1/4))/2 - (x^4 + 1)^(1/4) + (x^4 + 1)^(5/4)/5

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sympy [A]  time = 37.10, size = 56, normalized size = 1.19 \begin {gather*} \frac {\left (x^{4} + 1\right )^{\frac {5}{4}}}{5} - \sqrt [4]{x^{4} + 1} - \frac {\log {\left (\sqrt [4]{x^{4} + 1} - 1 \right )}}{4} + \frac {\log {\left (\sqrt [4]{x^{4} + 1} + 1 \right )}}{4} + \frac {\operatorname {atan}{\left (\sqrt [4]{x^{4} + 1} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-1)*(x**4+1)**(1/4)/x,x)

[Out]

(x**4 + 1)**(5/4)/5 - (x**4 + 1)**(1/4) - log((x**4 + 1)**(1/4) - 1)/4 + log((x**4 + 1)**(1/4) + 1)/4 + atan((
x**4 + 1)**(1/4))/2

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