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

Optimal. Leaf size=105 \[ -\frac {5 x}{3 \sqrt [4]{x^4+1}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \]

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Rubi [A]  time = 0.25, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6728, 240, 212, 206, 203, 1403, 382, 377} \begin {gather*} -\frac {5 x}{3 \sqrt [4]{x^4+1}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x)/(3*(1 + x^4)^(1/4)) + ArcTan[x/(1 + x^4)^(1/4)] - (5*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4
)*3^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4)] - (5*ArcTanh[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(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 1403

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*
x^n)^(p + q)*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx &=\int \left (\frac {2}{\sqrt [4]{1+x^4}}+\frac {5}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt [4]{1+x^4}} \, dx+5 \int \frac {1}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+5 \int \frac {1}{\left (-2+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx\\ &=-\frac {5 x}{3 \sqrt [4]{1+x^4}}+\frac {5}{3} \int \frac {1}{\left (-2+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 {5 x}{3 \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 {5}{3} \operatorname {Subst}\left (\int \frac {1}{-2+3 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {5 x}{3 \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 {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}\\ &=-\frac {5 x}{3 \sqrt [4]{1+x^4}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}\\ \end {align*}

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Mathematica [C]  time = 0.33, size = 132, normalized size = 1.26 \begin {gather*} -\frac {1}{5} x^5 F_1\left (\frac {5}{4};\frac {1}{4},1;\frac {9}{4};-x^4,\frac {x^4}{2}\right )-\frac {5 x}{3 \sqrt [4]{x^4+1}}+\frac {7 \left (-\log \left (2-\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )+\log \left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt [4]{x^4+1}}+2\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )\right )}{12\ 2^{3/4} \sqrt [4]{3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-5*x)/(3*(1 + x^4)^(1/4)) - (x^5*AppellF1[5/4, 1/4, 1, 9/4, -x^4, x^4/2])/5 + (7*(2*ArcTan[((3/2)^(1/4)*x)/(1
+ x^4)^(1/4)] - Log[2 - (2^(3/4)*3^(1/4)*x)/(1 + x^4)^(1/4)] + Log[2 + (2^(3/4)*3^(1/4)*x)/(1 + x^4)^(1/4)]))
/(12*2^(3/4)*3^(1/4))

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IntegrateAlgebraic [A]  time = 0.47, size = 105, normalized size = 1.00 \begin {gather*} -\frac {5 x}{3 \sqrt [4]{1+x^4}}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x)/(3*(1 + x^4)^(1/4)) + ArcTan[x/(1 + x^4)^(1/4)] - (5*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4
)*3^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4)] - (5*ArcTanh[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4))

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fricas [B]  time = 0.62, size = 205, normalized size = 1.95 \begin {gather*} -\frac {20 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {24^{\frac {3}{4}} \sqrt {2} x \sqrt {\frac {\sqrt {6} x^{2} + 2 \, \sqrt {x^{4} + 1}}{x^{2}}} - 2 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{24 \, x}\right ) + 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {24^{\frac {1}{4}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (-\frac {24^{\frac {1}{4}} x - 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 288 \, {\left (x^{4} + 1\right )} \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 144 \, {\left (x^{4} + 1\right )} \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 144 \, {\left (x^{4} + 1\right )} \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 480 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{288 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/288*(20*24^(3/4)*(x^4 + 1)*arctan(1/24*(24^(3/4)*sqrt(2)*x*sqrt((sqrt(6)*x^2 + 2*sqrt(x^4 + 1))/x^2) - 2*24
^(3/4)*(x^4 + 1)^(1/4))/x) + 5*24^(3/4)*(x^4 + 1)*log((24^(1/4)*x + 2*(x^4 + 1)^(1/4))/x) - 5*24^(3/4)*(x^4 +
1)*log(-(24^(1/4)*x - 2*(x^4 + 1)^(1/4))/x) + 288*(x^4 + 1)*arctan((x^4 + 1)^(1/4)/x) - 144*(x^4 + 1)*log((x +
(x^4 + 1)^(1/4))/x) + 144*(x^4 + 1)*log(-(x - (x^4 + 1)^(1/4))/x) + 480*(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} - 2 \, x^{4} + 1}{{\left (x^{8} - x^{4} - 2\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^8-2*x^4+1)/(x^4+1)^(1/4)/(x^8-x^4-2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {2 x^{8}-2 x^{4}+1}{\left (x^{4}+1\right )^{\frac {1}{4}} \left (x^{8}-x^{4}-2\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((2*x^8 - 2*x^4 + 1)/((x^8 - x^4 - 2)*(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^8-2\,x^4+1}{{\left (x^4+1\right )}^{1/4}\,\left (-x^8+x^4+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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