3.9.90 \(\int \frac {1+x^8}{\sqrt {1+x^4} (-1+x^8)} \, dx\)

Optimal. Leaf size=67 \[ -\frac {x}{2 \sqrt {x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {6725, 220, 1404, 414, 523, 409, 1211, 1699, 206, 203} \begin {gather*} -\frac {x}{2 \sqrt {x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 409

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

Rule 414

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, 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 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 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 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+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}+\frac {2}{\sqrt {1+x^4} \left (-1+x^8\right )}\right ) \, dx\\ &=2 \int \frac {1}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx+\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 \frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/2}} \, dx\\ &=-\frac {x}{2 \sqrt {1+x^4}}+\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}{2} \int \frac {3-x^4}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {1+x^4}}+\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}{2} \int \frac {1}{\sqrt {1+x^4}} \, dx+\int \frac {1}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {1+x^4}}+\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 )}{4 \sqrt {1+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {1+x^4}}+\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 )}{4 \sqrt {1+x^4}}-2 \left (\frac {1}{4} \int \frac {1}{\sqrt {1+x^4}} \, dx\right )-\frac {1}{4} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1+x^2}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {1+x^4}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt {1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.28, size = 119, normalized size = 1.78 \begin {gather*} \frac {x \left (\frac {5 \left (x^4+1\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-x^4,x^4\right )}{\left (x^4-1\right ) \left (2 x^4 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};-x^4,x^4\right )+F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-x^4,x^4\right )\right )+5 F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-x^4,x^4\right )\right )}-1\right )}{2 \sqrt {x^4+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.39, size = 67, normalized size = 1.00 \begin {gather*} -\frac {x}{2 \sqrt {1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 90, normalized size = 1.34 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \, \sqrt {x^{4} + 1} x}{16 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.46, size = 78, normalized size = 1.16

method	result	size
elliptic	\(\frac {\left (\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{8}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{8}-\frac {\sqrt {2}\, x}{2 \sqrt {x^{4}+1}}\right ) \sqrt {2}}{2}\)	\(78\)
trager	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x -\sqrt {x^{4}+1}}{\left (-1+x \right ) \left (1+x \right )}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x -\sqrt {x^{4}+1}}{x^{2}+1}\right )}{8}\)	\(86\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{2 \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{2 \sqrt {x^{4}+1}}-\frac {x}{2 \sqrt {x^{4}+1}}\)	\(172\)
risch	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{4 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{2 \sqrt {x^{4}+1}}+\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{4 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{4 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{2 \sqrt {x^{4}+1}}\)	\(375\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8+1)/(x^4+1)^(1/2)/(x^8-1),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/8*ln(-1+1/2*2^(1/2)/x*(x^4+1)^(1/2))+1/4*arctan(1/2*2^(1/2)/x*(x^4+1)^(1/2))-1/8*ln(1+1/2*2^(1/2)/x*(x^
4+1)^(1/2))-1/2*2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} + 1}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**8 + 1)/((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)**(3/2)), x)

________________________________________________________________________________________