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

Optimal. Leaf size=61 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x^3+x}}{x^2+1}\right )}{\sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x^3+x}}{x^2+1}\right )}{\sqrt [4]{2}} \]

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Rubi [C]  time = 1.09, antiderivative size = 386, normalized size of antiderivative = 6.33, number of steps used = 23, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {2056, 1586, 6715, 6725, 406, 220, 409, 1217, 1707} \begin {gather*} -\frac {\sqrt {x} \sqrt {x^2+1} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt {x^2+1}}\right )}{\sqrt [4]{2} \sqrt {x^3+x}}-\frac {\sqrt {x} \sqrt {x^2+1} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt {x^2+1}}\right )}{\sqrt [4]{2} \sqrt {x^3+x}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{4 \sqrt {x^3+x}}+\frac {i \left (\sqrt {2}+(-1+i)\right ) \sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{4 \sqrt {x^3+x}}+\frac {\left ((-1-i)-i \sqrt {2}\right ) \sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{4 \sqrt {x^3+x}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (1+(-1)^{3/4}\right ) \sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^3+x}}+\frac {\sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 406

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

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 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, 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]

Rubi steps

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

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Mathematica [C]  time = 0.80, size = 146, normalized size = 2.39 \begin {gather*} \frac {\sqrt [4]{-1} \sqrt {\frac {1}{x^2}+1} x^{3/2} \left (-2 F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt [4]{-1}}{\sqrt {x}}\right )\right |-1\right )+\Pi \left (-\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\frac {\sqrt [4]{-1}}{\sqrt {x}}\right )\right |-1\right )+\Pi \left (\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\frac {\sqrt [4]{-1}}{\sqrt {x}}\right )\right |-1\right )+\Pi \left (-(-1)^{3/4};\left .i \sinh ^{-1}\left (\frac {\sqrt [4]{-1}}{\sqrt {x}}\right )\right |-1\right )+\Pi \left ((-1)^{3/4};\left .i \sinh ^{-1}\left (\frac {\sqrt [4]{-1}}{\sqrt {x}}\right )\right |-1\right )\right )}{\sqrt {x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*Sqrt[1 + x^(-2)]*x^(3/2)*(-2*EllipticF[I*ArcSinh[(-1)^(1/4)/Sqrt[x]], -1] + EllipticPi[-(-1)^(1/4)
, I*ArcSinh[(-1)^(1/4)/Sqrt[x]], -1] + EllipticPi[(-1)^(1/4), I*ArcSinh[(-1)^(1/4)/Sqrt[x]], -1] + EllipticPi[
-(-1)^(3/4), I*ArcSinh[(-1)^(1/4)/Sqrt[x]], -1] + EllipticPi[(-1)^(3/4), I*ArcSinh[(-1)^(1/4)/Sqrt[x]], -1]))/
Sqrt[x + x^3]

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IntegrateAlgebraic [A]  time = 0.28, size = 61, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x+x^3}}{1+x^2}\right )}{\sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x+x^3}}{1+x^2}\right )}{\sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 141, normalized size = 2.31 \begin {gather*} -\frac {1}{2} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {1}{4}} \sqrt {x^{3} + x}}{x^{2} + 1}\right ) - \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left (\frac {x^{4} + 4 \, x^{2} + 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 2 \, \sqrt {x^{3} + x} {\left (2^{\frac {3}{4}} x + 2^{\frac {1}{4}} {\left (x^{2} + 1\right )}\right )} + 1}{x^{4} + 1}\right ) + \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left (\frac {x^{4} + 4 \, x^{2} + 2 \, \sqrt {2} {\left (x^{3} + x\right )} - 2 \, \sqrt {x^{3} + x} {\left (2^{\frac {3}{4}} x + 2^{\frac {1}{4}} {\left (x^{2} + 1\right )}\right )} + 1}{x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.57, size = 151, normalized size = 2.48

method	result	size
default	\(\frac {i \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticF \left (\sqrt {-i \left (i+x \right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{3}-i \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha +i\right ) \sqrt {-i \left (i+x \right )}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, -\frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha ^{3}-\frac {1}{2} \underline {\hspace {1.25 ex}}\alpha ^{2}+\frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha +\frac {1}{2}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x \left (x^{2}+1\right )}}\right )}{4}\)	\(151\)
elliptic	\(\frac {i \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticF \left (\sqrt {-i \left (i+x \right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{3}-i \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha +i\right ) \sqrt {-i \left (i+x \right )}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, -\frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha ^{3}-\frac {1}{2} \underline {\hspace {1.25 ex}}\alpha ^{2}+\frac {1}{2} i \underline {\hspace {1.25 ex}}\alpha +\frac {1}{2}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x \left (x^{2}+1\right )}}\right )}{4}\)	\(151\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}-8\right )^{5} x^{2}-2 x \RootOf \left (\textit {\_Z}^{4}-8\right )^{5}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{5}-4 \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x -16 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{2}+16 \RootOf \left (\textit {\_Z}^{4}-8\right ) x -16 \RootOf \left (\textit {\_Z}^{4}-8\right )-32 \sqrt {x^{3}+x}}{\RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 x \RootOf \left (\textit {\_Z}^{4}-8\right )^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}+4 x^{2}-4 x +4}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{4} x^{2}-2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{4} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{4}+4 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x -16 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2}+16 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x -16 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )+32 \sqrt {x^{3}+x}}{\RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 x \RootOf \left (\textit {\_Z}^{4}-8\right )^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}-4 x^{2}+4 x -4}\right )}{4}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(-I*(I+x))^(1/2)*2^(1/2)*(I*(-I+x))^(1/2)*(I*x)^(1/2)/(x^3+x)^(1/2)*EllipticF((-I*(I+x))^(1/2),1/2*2^(1/2))+
1/4*I*2^(1/2)*sum(_alpha*(I+_alpha^3-I*_alpha^2-_alpha)*(-I*(I+x))^(1/2)*(I*(-I+x))^(1/2)*(I*x)^(1/2)/(x*(x^2+
1))^(1/2)*EllipticPi((-I*(I+x))^(1/2),-1/2*I*_alpha^3-1/2*_alpha^2+1/2*I*_alpha+1/2,1/2*2^(1/2)),_alpha=RootOf
(_Z^4+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.59, size = 234, normalized size = 3.84 \begin {gather*} -\frac {\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (\sqrt {2}\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right );\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,1{}\mathrm {i}}{\sqrt {x^3+x}}-\frac {\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (\sqrt {2}\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right );\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,1{}\mathrm {i}}{\sqrt {x^3+x}}-\frac {\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (\sqrt {2}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right );\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,1{}\mathrm {i}}{\sqrt {x^3+x}}-\frac {\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (\sqrt {2}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right );\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,1{}\mathrm {i}}{\sqrt {x^3+x}}+\frac {\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,2{}\mathrm {i}}{\sqrt {x^3+x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1 - x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipticF(asin((-x*1i)^(1/2)), -1)*2i)/(x + x^3)^(1/2) - ((1
- x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipticPi(2^(1/2)*(- 1/2 + 1i/2), asin((-x*1i)^(1/2)), -1)*1i)/(
x + x^3)^(1/2) - ((1 - x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipticPi(2^(1/2)*(1/2 - 1i/2), asin((-x*1i
)^(1/2)), -1)*1i)/(x + x^3)^(1/2) - ((1 - x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipticPi(2^(1/2)*(1/2 +
1i/2), asin((-x*1i)^(1/2)), -1)*1i)/(x + x^3)^(1/2) - ((1 - x*1i)^(1/2)*(x*1i + 1)^(1/2)*(-x*1i)^(1/2)*ellipt
icPi(2^(1/2)*(- 1/2 - 1i/2), asin((-x*1i)^(1/2)), -1)*1i)/(x + x^3)^(1/2)

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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 + 1\right ) \left (x^{2} + 1\right )}{\sqrt {x \left (x^{2} + 1\right )} \left (x^{4} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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