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

Optimal. Leaf size=63 \[ -\frac {4}{3} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+1}+x^2-x+1}\right )-\frac {1}{3} \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}+x^2+2 x+1}\right ) \]

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Rubi [C]  time = 1.04, antiderivative size = 380, normalized size of antiderivative = 6.03, number of steps used = 29, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {6725, 220, 2074, 1725, 1211, 1699, 206, 1248, 725, 6728, 1217, 1707} \begin {gather*} -\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {x^4+1}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{3 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x^2+1}{\sqrt {2} \sqrt {x^4+1}}\right )}{3 \sqrt {2}}-\frac {\left (1+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {2-\left (1-i \sqrt {3}\right ) x^2}{\sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {x^4+1}}\right )}{3 \sqrt {2 \left (1-i \sqrt {3}\right )}}-\frac {\left (1-i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {2-\left (1+i \sqrt {3}\right ) x^2}{\sqrt {2 \left (1+i \sqrt {3}\right )} \sqrt {x^4+1}}\right )}{3 \sqrt {2 \left (1+i \sqrt {3}\right )}}-\frac {\left (1+i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{6 \sqrt {x^4+1}}-\frac {\left (1-i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{6 \sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^3)/((1 + x^3)*Sqrt[1 + x^4]),x]

[Out]

(-2*ArcTan[x/Sqrt[1 + x^4]])/3 - ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(3*Sqrt[2]) + ArcTanh[(1 + x^2)/(Sqrt[2]*S
qrt[1 + x^4])]/(3*Sqrt[2]) - ((1 + I*Sqrt[3])*ArcTanh[(2 - (1 - I*Sqrt[3])*x^2)/(Sqrt[2*(1 - I*Sqrt[3])]*Sqrt[
1 + x^4])])/(3*Sqrt[2*(1 - I*Sqrt[3])]) - ((1 - I*Sqrt[3])*ArcTanh[(2 - (1 + I*Sqrt[3])*x^2)/(Sqrt[2*(1 + I*Sq
rt[3])]*Sqrt[1 + x^4])])/(3*Sqrt[2*(1 + I*Sqrt[3])]) + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcT
an[x], 1/2])/(3*Sqrt[1 + x^4]) - ((1 - I*Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x],
1/2])/(6*Sqrt[1 + x^4]) - ((1 + I*Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])
/(6*Sqrt[1 + x^4])

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 1211

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1217

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 1725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*Sqrt[a + c*
x^4]), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

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]

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+x^3}{\left (1+x^3\right ) \sqrt {1+x^4}} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}-\frac {2}{\left (1+x^3\right ) \sqrt {1+x^4}}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (1+x^3\right ) \sqrt {1+x^4}} \, dx\right )+\int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-2 \int \left (\frac {1}{3 (1+x) \sqrt {1+x^4}}+\frac {2-x}{3 \left (1-x+x^2\right ) \sqrt {1+x^4}}\right ) \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {2}{3} \int \frac {1}{(1+x) \sqrt {1+x^4}} \, dx-\frac {2}{3} \int \frac {2-x}{\left (1-x+x^2\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {2}{3} \int \frac {1}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx+\frac {2}{3} \int \frac {x}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx-\frac {2}{3} \int \left (\frac {-1-i \sqrt {3}}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt {1+x^4}}+\frac {-1+i \sqrt {3}}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt {1+x^4}}\right ) \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{3} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{3} \int \frac {1+x^2}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {1+x^2}} \, dx,x,x^2\right )+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {1}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {1}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\frac {-1-x^2}{\sqrt {1+x^4}}\right )+\frac {1}{3} \left (4 \left (1-i \sqrt {3}\right )\right ) \int \frac {1}{\left (\left (-1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \left (4 \left (1-i \sqrt {3}\right )\right ) \int \frac {x}{\left (\left (-1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{3} \left (4 \left (1+i \sqrt {3}\right )\right ) \int \frac {x}{\left (\left (-1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (4 \left (1+i \sqrt {3}\right )\right ) \int \frac {1}{\left (\left (-1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {1+x^2}{\sqrt {2} \sqrt {1+x^4}}\right )}{3 \sqrt {2}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}-\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (-1+i \sqrt {3}\right )^2-4 x\right ) \sqrt {1+x^2}} \, dx,x,x^2\right )-\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (-1-i \sqrt {3}\right )^2-4 x\right ) \sqrt {1+x^2}} \, dx,x,x^2\right )+\frac {\left (2 \left (i-\sqrt {3}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (i+\sqrt {3}\right )}+\frac {\left (8 \left (i-\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (\left (-1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (i+\sqrt {3}\right )}+\frac {\left (2 \left (i+\sqrt {3}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (i-\sqrt {3}\right )}+\frac {\left (8 \left (i+\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (\left (-1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (i-\sqrt {3}\right )}\\ &=-\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {1+x^2}{\sqrt {2} \sqrt {1+x^4}}\right )}{3 \sqrt {2}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}+\frac {\left (i-\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (i+\sqrt {3}\right ) \sqrt {1+x^4}}+\frac {\left (i+\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (i-\sqrt {3}\right ) \sqrt {1+x^4}}+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{16+\left (-1+i \sqrt {3}\right )^4-x^2} \, dx,x,\frac {-4-\left (-1+i \sqrt {3}\right )^2 x^2}{\sqrt {1+x^4}}\right )+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{16+\left (-1-i \sqrt {3}\right )^4-x^2} \, dx,x,\frac {-4-\left (-1-i \sqrt {3}\right )^2 x^2}{\sqrt {1+x^4}}\right )\\ &=-\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {1+x^2}{\sqrt {2} \sqrt {1+x^4}}\right )}{3 \sqrt {2}}-\frac {\left (1+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {2-\left (1-i \sqrt {3}\right ) x^2}{\sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{3 \sqrt {2 \left (1-i \sqrt {3}\right )}}-\frac {\left (1-i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {2-\left (1+i \sqrt {3}\right ) x^2}{\sqrt {2 \left (1+i \sqrt {3}\right )} \sqrt {1+x^4}}\right )}{3 \sqrt {2 \left (1+i \sqrt {3}\right )}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}+\frac {\left (i-\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (i+\sqrt {3}\right ) \sqrt {1+x^4}}+\frac {\left (i+\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (i-\sqrt {3}\right ) \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.72, size = 525, normalized size = 8.33 \begin {gather*} \frac {1}{18} \left (-4 (-1)^{2/3} \sqrt {1-\sqrt [3]{-1}} \tanh ^{-1}\left (\frac {\sqrt [3]{-1}-x^2}{\sqrt {1+(-1)^{2/3}} \sqrt {x^4+1}}\right )-2 \sqrt [3]{-1} \sqrt {1-\sqrt [3]{-1}} \tanh ^{-1}\left (\frac {\sqrt [3]{-1}-x^2}{\sqrt {1+(-1)^{2/3}} \sqrt {x^4+1}}\right )+2 \sqrt {1-\sqrt [3]{-1}} \tanh ^{-1}\left (\frac {\sqrt [3]{-1}-x^2}{\sqrt {1+(-1)^{2/3}} \sqrt {x^4+1}}\right )-(-1)^{2/3} \sqrt {2} \tanh ^{-1}\left (\frac {x^2+1}{\sqrt {2} \sqrt {x^4+1}}\right )+\sqrt [3]{-1} \sqrt {2} \tanh ^{-1}\left (\frac {x^2+1}{\sqrt {2} \sqrt {x^4+1}}\right )+2 \sqrt {2} \tanh ^{-1}\left (\frac {x^2+1}{\sqrt {2} \sqrt {x^4+1}}\right )-2 (-1)^{2/3} \sqrt {1+(-1)^{2/3}} \tanh ^{-1}\left (\frac {x^2+(-1)^{2/3}}{\sqrt {1-\sqrt [3]{-1}} \sqrt {x^4+1}}\right )-4 \sqrt [3]{-1} \sqrt {1+(-1)^{2/3}} \tanh ^{-1}\left (\frac {x^2+(-1)^{2/3}}{\sqrt {1-\sqrt [3]{-1}} \sqrt {x^4+1}}\right )-2 \sqrt {1+(-1)^{2/3}} \tanh ^{-1}\left (\frac {x^2+(-1)^{2/3}}{\sqrt {1-\sqrt [3]{-1}} \sqrt {x^4+1}}\right )-18 \sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+12 \sqrt [4]{-1} \Pi \left (i;\left .\sin ^{-1}\left ((-1)^{3/4} x\right )\right |-1\right )-12 (-1)^{11/12} \Pi \left (-\sqrt [6]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+12 (-1)^{7/12} \Pi \left (-\sqrt [6]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+12 \sqrt [4]{-1} \Pi \left (-(-1)^{5/6};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[1 - (-1)^(1/3)]*ArcTanh[((-1)^(1/3) - x^2)/(Sqrt[1 + (-1)^(2/3)]*Sqrt[1 + x^4])] - 2*(-1)^(1/3)*Sqrt[1
- (-1)^(1/3)]*ArcTanh[((-1)^(1/3) - x^2)/(Sqrt[1 + (-1)^(2/3)]*Sqrt[1 + x^4])] - 4*(-1)^(2/3)*Sqrt[1 - (-1)^(
1/3)]*ArcTanh[((-1)^(1/3) - x^2)/(Sqrt[1 + (-1)^(2/3)]*Sqrt[1 + x^4])] + 2*Sqrt[2]*ArcTanh[(1 + x^2)/(Sqrt[2]*
Sqrt[1 + x^4])] + (-1)^(1/3)*Sqrt[2]*ArcTanh[(1 + x^2)/(Sqrt[2]*Sqrt[1 + x^4])] - (-1)^(2/3)*Sqrt[2]*ArcTanh[(
1 + x^2)/(Sqrt[2]*Sqrt[1 + x^4])] - 2*Sqrt[1 + (-1)^(2/3)]*ArcTanh[((-1)^(2/3) + x^2)/(Sqrt[1 - (-1)^(1/3)]*Sq
rt[1 + x^4])] - 4*(-1)^(1/3)*Sqrt[1 + (-1)^(2/3)]*ArcTanh[((-1)^(2/3) + x^2)/(Sqrt[1 - (-1)^(1/3)]*Sqrt[1 + x^
4])] - 2*(-1)^(2/3)*Sqrt[1 + (-1)^(2/3)]*ArcTanh[((-1)^(2/3) + x^2)/(Sqrt[1 - (-1)^(1/3)]*Sqrt[1 + x^4])] - 18
*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1] + 12*(-1)^(1/4)*EllipticPi[I, ArcSin[(-1)^(3/4)*x], -1] + 1
2*(-1)^(7/12)*EllipticPi[-(-1)^(1/6), I*ArcSinh[(-1)^(1/4)*x], -1] - 12*(-1)^(11/12)*EllipticPi[-(-1)^(1/6), I
*ArcSinh[(-1)^(1/4)*x], -1] + 12*(-1)^(1/4)*EllipticPi[-(-1)^(5/6), I*ArcSinh[(-1)^(1/4)*x], -1])/18

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IntegrateAlgebraic [A]  time = 1.05, size = 63, normalized size = 1.00 \begin {gather*} -\frac {4}{3} \tan ^{-1}\left (\frac {x}{1-x+x^2+\sqrt {1+x^4}}\right )-\frac {1}{3} \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{1+2 x+x^2+\sqrt {1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 88, normalized size = 1.40 \begin {gather*} \frac {1}{12} \, \sqrt {2} \log \left (-\frac {3 \, x^{4} + 4 \, x^{3} + 2 \, \sqrt {2} \sqrt {x^{4} + 1} {\left (x^{2} + x + 1\right )} + 6 \, x^{2} + 4 \, x + 3}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1}\right ) - \frac {2}{3} \, \arctan \left (\frac {\sqrt {x^{4} + 1}}{x^{2} - 2 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)/(x^3+1)/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(2)*log(-(3*x^4 + 4*x^3 + 2*sqrt(2)*sqrt(x^4 + 1)*(x^2 + x + 1) + 6*x^2 + 4*x + 3)/(x^4 + 4*x^3 + 6*x
^2 + 4*x + 1)) - 2/3*arctan(sqrt(x^4 + 1)/(x^2 - 2*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)/(x^3+1)/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.39, size = 244, normalized size = 3.87

method	result	size
elliptic	\(\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4 \sqrt {\left (x^{2}-1\right )^{2}+2 x^{2}}}\right )}{6}-\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x^{2}+1\right )^{2}}{\left (-x^{2}+1\right )^{2}}+2}\, \arctan \left (\frac {\sqrt {\frac {2 \left (x^{2}+1\right )^{2}}{\left (-x^{2}+1\right )^{2}}+2}\, \left (x^{2}+1\right )}{\left (\frac {\left (x^{2}+1\right )^{2}}{\left (-x^{2}+1\right )^{2}}+1\right ) \left (-x^{2}+1\right )}\right )}{3 \sqrt {\frac {\frac {\left (x^{2}+1\right )^{2}}{\left (-x^{2}+1\right )^{2}}+1}{\left (\frac {x^{2}+1}{-x^{2}+1}+1\right )^{2}}}\, \left (\frac {x^{2}+1}{-x^{2}+1}+1\right )}+\frac {\left (\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{6}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{6}+\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {x^{4}+1}}{x}\right )}{3}\right ) \sqrt {2}}{2}\)	\(244\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4 \sqrt {x^{4}+1}}\right )}{6}+\frac {2 \left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{3 \sqrt {x^{4}+1}}+\frac {2 \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {\arctanh \left (\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{2}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+1}}\right )}{2 \sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}}+\frac {\left (-1\right )^{\frac {3}{4}} \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ), i\right )}{\sqrt {x^{4}+1}}\right )}{3}+\frac {2 \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (-\frac {\arctanh \left (\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x^{2}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{\sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+1}}\right )}{2 \sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}}+\frac {\left (-1\right )^{\frac {3}{4}} \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ), i\right )}{\sqrt {x^{4}+1}}\right )}{3}\)	\(374\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)/(x^3+1)/(x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*2^(1/2)*arctanh(1/4*(2*x^2+2)*2^(1/2)/((x^2-1)^2+2*x^2)^(1/2))-1/3*2^(1/2)/(((x^2+1)^2/(-x^2+1)^2+1)/((x^2
+1)/(-x^2+1)+1)^2)^(1/2)/((x^2+1)/(-x^2+1)+1)*(2*(x^2+1)^2/(-x^2+1)^2+2)^(1/2)*arctan(1/((x^2+1)^2/(-x^2+1)^2+
1)*(2*(x^2+1)^2/(-x^2+1)^2+2)^(1/2)*(x^2+1)/(-x^2+1))+1/2*(1/6*ln(-1+1/2*2^(1/2)/x*(x^4+1)^(1/2))-1/6*ln(1+1/2
*2^(1/2)/x*(x^4+1)^(1/2))+2/3*2^(1/2)*arctan((x^4+1)^(1/2)/x))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)/(x^3+1)/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^3-1}{\left (x^3+1\right )\,\sqrt {x^4+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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