3.11.77 \(\int \frac {(-1+x^4)^{3/4}}{1+x^4} \, dx\)

Optimal. Leaf size=81 \[ \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt [4]{2}}+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt [4]{2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {408, 240, 212, 206, 203, 377} \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt [4]{2}}+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^4)^(3/4)/(1 + x^4),x]

[Out]

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

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 377

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

Rule 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rubi steps

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

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Mathematica [C]  time = 0.11, size = 109, normalized size = 1.35 \begin {gather*} -\frac {5 x \left (x^4-1\right )^{3/4} F_1\left (\frac {1}{4};-\frac {3}{4},1;\frac {5}{4};x^4,-x^4\right )}{\left (x^4+1\right ) \left (x^4 \left (4 F_1\left (\frac {5}{4};-\frac {3}{4},2;\frac {9}{4};x^4,-x^4\right )+3 F_1\left (\frac {5}{4};\frac {1}{4},1;\frac {9}{4};x^4,-x^4\right )\right )-5 F_1\left (\frac {1}{4};-\frac {3}{4},1;\frac {5}{4};x^4,-x^4\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-5*x*(-1 + x^4)^(3/4)*AppellF1[1/4, -3/4, 1, 5/4, x^4, -x^4])/((1 + x^4)*(-5*AppellF1[1/4, -3/4, 1, 5/4, x^4,
-x^4] + x^4*(4*AppellF1[5/4, -3/4, 2, 9/4, x^4, -x^4] + 3*AppellF1[5/4, 1/4, 1, 9/4, x^4, -x^4])))

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IntegrateAlgebraic [A]  time = 0.30, size = 81, normalized size = 1.00 \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt [4]{2}}+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.63, size = 149, normalized size = 1.84 \begin {gather*} -2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {3}{4}} x \sqrt {\frac {\sqrt {2} x^{2} + \sqrt {x^{4} - 1}}{x^{2}}} - 2^{\frac {3}{4}} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{2 \, x}\right ) - \frac {1}{4} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - 1\right )}^{\frac {3}{4}}}{x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 - 1)^(3/4)/(x^4 + 1), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{4}-1\right )^{\frac {3}{4}}}{x^{4}+1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)^(3/4)/(x^4+1),x)

[Out]

int((x^4-1)^(3/4)/(x^4+1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - 1\right )}^{\frac {3}{4}}}{x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 - 1)^(3/4)/(x^4 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^4-1\right )}^{3/4}}{x^4+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - 1)^(3/4)/(x^4 + 1),x)

[Out]

int((x^4 - 1)^(3/4)/(x^4 + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )\right )^{\frac {3}{4}}}{x^{4} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((x - 1)*(x + 1)*(x**2 + 1))**(3/4)/(x**4 + 1), x)

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