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

Optimal. Leaf size=61 \[ \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{x^4+1}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{x^4+1}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1833, 240, 212, 206, 203, 266, 63, 298} \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{x^4+1}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\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)/(x*(1 + x^4)^(1/4)),x]

[Out]

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

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 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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 266

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 1833

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

\begin {align*} \int \frac {-1+x}{x \sqrt [4]{1+x^4}} \, dx &=\int \left (\frac {1}{\sqrt [4]{1+x^4}}-\frac {1}{x \sqrt [4]{1+x^4}}\right ) \, dx\\ &=\int \frac {1}{\sqrt [4]{1+x^4}} \, dx-\int \frac {1}{x \sqrt [4]{1+x^4}} \, dx\\ &=-\left (\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{1+x}} \, dx,x,x^4\right )\right )+\operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\operatorname {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt [4]{1+x^4}\right )\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {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 )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^4}\right )\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{1+x^4}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\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 = 61, normalized size = 1.00 \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{x^4+1}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\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]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 4.25, size = 61, normalized size = 1.00 \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{1+x^4}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\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)/(x*(1 + x^4)^(1/4)),x]

[Out]

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

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fricas [B]  time = 4.64, size = 101, normalized size = 1.66 \begin {gather*} \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )} + {\left (x^{4} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + x\right )}}{x^{4} - x^{2} + 1}\right ) + \frac {1}{2} \, \log \left (-\frac {x^{4} + x^{3} + {\left (x^{4} + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )} + \sqrt {x^{4} + 1} {\left (x^{2} + x + 1\right )} + {\left (x^{4} + 1\right )}^{\frac {1}{4}} {\left (x^{3} + x^{2} + x + 1\right )} + x + 1}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{{\left (x^{4} + 1\right )}^{\frac {1}{4}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.99, size = 73, normalized size = 1.20

method	result	size
meijerg	\(-\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{4} \hypergeom \left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{4}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \relax (2)-\frac {\pi }{2}+4 \ln \relax (x )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{8 \pi }+x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -x^{4}\right )\)	\(73\)
trager	\(\frac {\ln \left (-\frac {\left (x^{4}+1\right )^{\frac {3}{4}} x +x^{2} \sqrt {x^{4}+1}+x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+x^{4}+\left (x^{4}+1\right )^{\frac {3}{4}}+\sqrt {x^{4}+1}\, x +x^{2} \left (x^{4}+1\right )^{\frac {1}{4}}+x^{3}+\sqrt {x^{4}+1}+\left (x^{4}+1\right )^{\frac {1}{4}} x +\left (x^{4}+1\right )^{\frac {1}{4}}+x +1}{x^{2}}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x^{2}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+\left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x -x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+\left (x^{4}+1\right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}-x^{2} \left (x^{4}+1\right )^{\frac {1}{4}}-\left (x^{4}+1\right )^{\frac {1}{4}} x +\RootOf \left (\textit {\_Z}^{2}+1\right ) x -\left (x^{4}+1\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{2}}\right )}{2}\)	\(257\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/Pi*2^(1/2)*GAMMA(3/4)*(-1/4*Pi*2^(1/2)/GAMMA(3/4)*x^4*hypergeom([1,1,5/4],[2,2],-x^4)+(-3*ln(2)-1/2*Pi+4*
ln(x))*Pi*2^(1/2)/GAMMA(3/4))+x*hypergeom([1/4,1/4],[5/4],-x^4)

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maxima [A]  time = 0.42, size = 81, normalized size = 1.33 \begin {gather*} -\frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\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 ) + \frac {1}{4} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{4} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.77, size = 33, normalized size = 0.54 \begin {gather*} \frac {\mathrm {atanh}\left ({\left (x^4+1\right )}^{1/4}\right )}{2}-\frac {\mathrm {atan}\left ({\left (x^4+1\right )}^{1/4}\right )}{2}+x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -x^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)/(x*(x^4 + 1)^(1/4)),x)

[Out]

atanh((x^4 + 1)^(1/4))/2 - atan((x^4 + 1)^(1/4))/2 + x*hypergeom([1/4, 1/4], 5/4, -x^4)

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sympy [C]  time = 2.39, size = 56, normalized size = 0.92 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 1/4), (5/4,), x**4*exp_polar(I*pi))/(4*gamma(5/4)) + gamma(1/4)*hyper((1/4, 1/4), (5/
4,), exp_polar(I*pi)/x**4)/(4*x*gamma(5/4))

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