3.13.57 \(\int \frac {-1+2 x^8}{\sqrt [4]{1+x^4} (-1+x^8)} \, dx\)

Optimal. Leaf size=91 \[ -\frac {x}{2 \sqrt [4]{x^4+1}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}} \]

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Rubi [A]  time = 0.20, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6725, 240, 212, 206, 203, 1404, 382, 377} \begin {gather*} -\frac {x}{2 \sqrt [4]{x^4+1}}+\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 )}{4 \sqrt [4]{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 )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 2*x^8)/((1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

-1/2*x/(1 + x^4)^(1/4) + ArcTan[x/(1 + x^4)^(1/4)] - ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(4*2^(1/4)) + ArcTanh
[x/(1 + x^4)^(1/4)] - ArcTanh[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(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 382

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

Rule 1404

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {-1+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx &=\int \left (\frac {2}{\sqrt [4]{1+x^4}}+\frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt [4]{1+x^4}} \, dx+\int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\int \frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}+\frac {1}{2} \int \frac {1}{\left (-1+x^4\right ) \sqrt [4]{1+x^4}} \, dx+\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+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-1+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}+\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 )}{4 \sqrt [4]{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 )}{4 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 111, normalized size = 1.22 \begin {gather*} -\frac {2}{5} x^5 F_1\left (\frac {5}{4};\frac {1}{4},1;\frac {9}{4};-x^4,x^4\right )-\frac {x}{2 \sqrt [4]{x^4+1}}+\frac {3 \left (-\log \left (1-\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )+\log \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}+1\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )\right )}{8 \sqrt [4]{2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(-1 + 2*x^8)/((1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

-1/2*x/(1 + x^4)^(1/4) - (2*x^5*AppellF1[5/4, 1/4, 1, 9/4, -x^4, x^4])/5 + (3*(2*ArcTan[(2^(1/4)*x)/(1 + x^4)^
(1/4)] - Log[1 - (2^(1/4)*x)/(1 + x^4)^(1/4)] + Log[1 + (2^(1/4)*x)/(1 + x^4)^(1/4)]))/(8*2^(1/4))

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IntegrateAlgebraic [A]  time = 0.40, size = 91, normalized size = 1.00 \begin {gather*} -\frac {x}{2 \sqrt [4]{1+x^4}}+\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 )}{4 \sqrt [4]{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 )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + 2*x^8)/((1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

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

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fricas [B]  time = 0.45, size = 197, normalized size = 2.16 \begin {gather*} -\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \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 ) + 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 16 \, {\left (x^{4} + 1\right )} \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 8 \, {\left (x^{4} + 1\right )} \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 8 \, {\left (x^{4} + 1\right )} \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 8 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{16 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-1)/(x^4+1)^(1/4)/(x^8-1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{8} - 1}{{\left (x^{8} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-1)/(x^4+1)^(1/4)/(x^8-1),x, algorithm="giac")

[Out]

integrate((2*x^8 - 1)/((x^8 - 1)*(x^4 + 1)^(1/4)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {2 x^{8}-1}{\left (x^{4}+1\right )^{\frac {1}{4}} \left (x^{8}-1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8-1)/(x^4+1)^(1/4)/(x^8-1),x)

[Out]

int((2*x^8-1)/(x^4+1)^(1/4)/(x^8-1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{8} - 1}{{\left (x^{8} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-1)/(x^4+1)^(1/4)/(x^8-1),x, algorithm="maxima")

[Out]

integrate((2*x^8 - 1)/((x^8 - 1)*(x^4 + 1)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {2\,x^8-1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8 - 1)/((x^4 + 1)^(1/4)*(x^8 - 1)),x)

[Out]

int((2*x^8 - 1)/((x^4 + 1)^(1/4)*(x^8 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 x^{8} - 1}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**8-1)/(x**4+1)**(1/4)/(x**8-1),x)

[Out]

Integral((2*x**8 - 1)/((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)**(5/4)), x)

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