3.25.4 \(\int \frac {-2+x^4}{\sqrt [4]{1+x^4} (-1+x^4+x^8)} \, dx\)
Optimal. Leaf size=193 \[ \frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4+1}}\right ) \]
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Rubi [A] time = 0.31, antiderivative size = 189, normalized size of antiderivative = 0.98,
number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used
= {6728, 377, 212, 206, 203} \begin {gather*} \frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-2 + x^4)/((1 + x^4)^(1/4)*(-1 + x^4 + x^8)),x]
[Out]
(((3 + Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)])/2 + (((3 - Sqrt[5])/2)^(1/4)*Arc
Tan[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/2 + (((3 + Sqrt[5])/2)^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4
)*x)/(1 + x^4)^(1/4)])/2 + (((3 - Sqrt[5])/2)^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/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 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 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 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 {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx &=\int \left (\frac {1-\sqrt {5}}{\sqrt [4]{1+x^4} \left (1-\sqrt {5}+2 x^4\right )}+\frac {1+\sqrt {5}}{\sqrt [4]{1+x^4} \left (1+\sqrt {5}+2 x^4\right )}\right ) \, dx\\ &=\left (1-\sqrt {5}\right ) \int \frac {1}{\sqrt [4]{1+x^4} \left (1-\sqrt {5}+2 x^4\right )} \, dx+\left (1+\sqrt {5}\right ) \int \frac {1}{\sqrt [4]{1+x^4} \left (1+\sqrt {5}+2 x^4\right )} \, dx\\ &=\left (1-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {5}-\left (-1-\sqrt {5}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\left (1+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {5}-\left (-1+\sqrt {5}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {\left (1-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}-\frac {\left (1-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}+\frac {\left (1+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}+\frac {\left (1+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {2}}\\ &=\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )\\ \end {align*}
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Mathematica [A] time = 0.12, size = 170, normalized size = 0.88 \begin {gather*} \frac {\sqrt [4]{3+\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )+\sqrt [4]{3-\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )+\sqrt [4]{3+\sqrt {5}} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )+\sqrt [4]{3-\sqrt {5}} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-2 + x^4)/((1 + x^4)^(1/4)*(-1 + x^4 + x^8)),x]
[Out]
((3 + Sqrt[5])^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)] + (3 - Sqrt[5])^(1/4)*ArcTan[(((3 + S
qrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)] + (3 + Sqrt[5])^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)
] + (3 - Sqrt[5])^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(2*2^(1/4))
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IntegrateAlgebraic [A] time = 0.67, size = 193, normalized size = 1.00 \begin {gather*} \frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(-2 + x^4)/((1 + x^4)^(1/4)*(-1 + x^4 + x^8)),x]
[Out]
(Sqrt[(1 + Sqrt[5])/2]*ArcTan[(Sqrt[-1/2 + Sqrt[5]/2]*x)/(1 + x^4)^(1/4)])/2 + (Sqrt[(-1 + Sqrt[5])/2]*ArcTan[
(Sqrt[1/2 + Sqrt[5]/2]*x)/(1 + x^4)^(1/4)])/2 + (Sqrt[(1 + Sqrt[5])/2]*ArcTanh[(Sqrt[-1/2 + Sqrt[5]/2]*x)/(1 +
x^4)^(1/4)])/2 + (Sqrt[(-1 + Sqrt[5])/2]*ArcTanh[(Sqrt[1/2 + Sqrt[5]/2]*x)/(1 + x^4)^(1/4)])/2
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fricas [B] time = 24.25, size = 985, normalized size = 5.10
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((x^4-2)/(x^4+1)^(1/4)/(x^8+x^4-1),x, algorithm="fricas")
[Out]
-1/4*sqrt(2)*sqrt(sqrt(5) - 1)*arctan(1/4*(sqrt(2)*(sqrt(5)*sqrt(2)*(x^8 + x^4) + sqrt(2)*(2*x^4 + 1) + sqrt(x
^4 + 1)*(sqrt(5)*sqrt(2)*x^2 + sqrt(2)*(2*x^6 + x^2)))*sqrt(sqrt(5) + 1)*sqrt(sqrt(5) - 1) + 2*((x^4 + 1)^(3/4
)*(sqrt(5)*sqrt(2)*x + sqrt(2)*(2*x^5 + x)) + (x^4 + 1)^(1/4)*(sqrt(5)*sqrt(2)*(x^7 + x^3) + sqrt(2)*(x^7 + 3*
x^3)))*sqrt(sqrt(5) - 1))/(x^8 + x^4 - 1)) + 1/4*sqrt(2)*sqrt(sqrt(5) + 1)*arctan(-1/4*(sqrt(2)*(sqrt(5)*sqrt(
2)*(x^8 + x^4) - sqrt(2)*(2*x^4 + 1) - sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*x^2 - sqrt(2)*(2*x^6 + x^2)))*sqrt(sqrt(
5) + 1)*sqrt(sqrt(5) - 1) + 2*((x^4 + 1)^(3/4)*(sqrt(5)*sqrt(2)*x - sqrt(2)*(2*x^5 + x)) - (x^4 + 1)^(1/4)*(sq
rt(5)*sqrt(2)*(x^7 + x^3) - sqrt(2)*(x^7 + 3*x^3)))*sqrt(sqrt(5) + 1))/(x^8 + x^4 - 1)) + 1/16*sqrt(2)*sqrt(sq
rt(5) - 1)*log((4*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) + (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) + sqrt(2)*(5*x^
8 + 7*x^4 + 1) + 2*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x^6 + x^2) + sqrt(2)*(x^6 + 3*x^2)))*sqrt(sqrt(5) - 1) + 4*
(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) - 1/16*sqrt(2)*sqrt(sqrt(5) - 1)*log((4*
(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) - (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) + sqrt(2)*(5*x^8 + 7*x^4 + 1) + 2
*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x^6 + x^2) + sqrt(2)*(x^6 + 3*x^2)))*sqrt(sqrt(5) - 1) + 4*(x^7 + 3*x^3 + sqr
t(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) - 1/16*sqrt(2)*sqrt(sqrt(5) + 1)*log((4*(2*x^5 - sqrt(5)*x
+ x)*(x^4 + 1)^(3/4) + (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) - sqrt(2)*(5*x^8 + 7*x^4 + 1) - 2*sqrt(x^4 + 1)*(sq
rt(5)*sqrt(2)*(x^6 + x^2) - sqrt(2)*(x^6 + 3*x^2)))*sqrt(sqrt(5) + 1) - 4*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*
(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/16*sqrt(2)*sqrt(sqrt(5) + 1)*log((4*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3
/4) - (sqrt(5)*sqrt(2)*(x^8 + 3*x^4 + 1) - sqrt(2)*(5*x^8 + 7*x^4 + 1) - 2*sqrt(x^4 + 1)*(sqrt(5)*sqrt(2)*(x^6
+ x^2) - sqrt(2)*(x^6 + 3*x^2)))*sqrt(sqrt(5) + 1) - 4*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(
x^8 + x^4 - 1))
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((x^4-2)/(x^4+1)^(1/4)/(x^8+x^4-1),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to convert to real 1/4 Error: Bad Argument ValueUnable to convert to real 1/4 Error: Bad Argument Value
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maple [C] time = 10.98, size = 1696, normalized size = 8.79
| | |
| method |
result |
size |
| | |
| trager |
\(\text {Expression too large to display}\) |
\(1696\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^4-2)/(x^4+1)^(1/4)/(x^8+x^4-1),x,method=_RETURNVERBOSE)
[Out]
-RootOf(256*_Z^4+16*_Z^2-1)*ln(-(1536*RootOf(256*_Z^4+16*_Z^2-1)^5*x^4+256*RootOf(256*_Z^4+16*_Z^2-1)^3*(x^4+1
)^(1/2)*x^2+416*x^4*RootOf(256*_Z^4+16*_Z^2-1)^3-48*RootOf(256*_Z^4+16*_Z^2-1)^2*(x^4+1)^(3/4)*x-64*(x^4+1)^(1
/4)*RootOf(256*_Z^4+16*_Z^2-1)^2*x^3+28*RootOf(256*_Z^4+16*_Z^2-1)*(x^4+1)^(1/2)*x^2+24*x^4*RootOf(256*_Z^4+16
*_Z^2-1)-4*(x^4+1)^(3/4)*x-7*x^3*(x^4+1)^(1/4)+96*RootOf(256*_Z^4+16*_Z^2-1)^3+8*RootOf(256*_Z^4+16*_Z^2-1))/(
64*RootOf(256*_Z^4+16*_Z^2-1)^3*x^2+4*RootOf(256*_Z^4+16*_Z^2-1)*x^2-1)/(64*RootOf(256*_Z^4+16*_Z^2-1)^3*x^2+4
*RootOf(256*_Z^4+16*_Z^2-1)*x^2+1))+1/4*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*ln((1024*RootOf(_Z^2+16
*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*RootOf(256*_Z^4+16*_Z^2-1)^4*x^4-224*(x^4+1)^(1/2)*RootOf(256*_Z^4+16*_Z^2-1)
^2*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*x^2-176*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*RootO
f(256*_Z^4+16*_Z^2-1)^2*x^4-128*RootOf(256*_Z^4+16*_Z^2-1)^2*(x^4+1)^(3/4)*x-224*(x^4+1)^(1/4)*RootOf(256*_Z^4
+16*_Z^2-1)^2*x^3+8*(x^4+1)^(1/2)*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*x^2+6*RootOf(_Z^2+16*RootOf(2
56*_Z^4+16*_Z^2-1)^2+1)*x^4+6*(x^4+1)^(3/4)*x+8*x^3*(x^4+1)^(1/4)-64*RootOf(256*_Z^4+16*_Z^2-1)^2*RootOf(_Z^2+
16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)+3*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1))/(16*RootOf(256*_Z^4+16*_Z
^2-1)^2*x^4+1))-4*RootOf(256*_Z^4+16*_Z^2-1)^2*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*ln(-(384*RootOf(
_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*RootOf(256*_Z^4+16*_Z^2-1)^4*x^4-64*(x^4+1)^(1/2)*RootOf(256*_Z^4+16*_
Z^2-1)^2*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*x^2+104*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)
*RootOf(256*_Z^4+16*_Z^2-1)^2*x^4-64*RootOf(256*_Z^4+16*_Z^2-1)^2*(x^4+1)^(3/4)*x+112*(x^4+1)^(1/4)*RootOf(256
*_Z^4+16*_Z^2-1)^2*x^3-7*(x^4+1)^(1/2)*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)*x^2+6*RootOf(_Z^2+16*Roo
tOf(256*_Z^4+16*_Z^2-1)^2+1)*x^4-7*(x^4+1)^(3/4)*x+11*x^3*(x^4+1)^(1/4)+24*RootOf(256*_Z^4+16*_Z^2-1)^2*RootOf
(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1)+2*RootOf(_Z^2+16*RootOf(256*_Z^4+16*_Z^2-1)^2+1))/(64*RootOf(256*_Z^4
+16*_Z^2-1)^3*x^2+4*RootOf(256*_Z^4+16*_Z^2-1)*x^2-1)/(64*RootOf(256*_Z^4+16*_Z^2-1)^3*x^2+4*RootOf(256*_Z^4+1
6*_Z^2-1)*x^2+1))+16*RootOf(256*_Z^4+16*_Z^2-1)^3*ln(-(512*RootOf(256*_Z^4+16*_Z^2-1)^5*x^4+192*RootOf(256*_Z^
4+16*_Z^2-1)^3*(x^4+1)^(1/2)*x^2-128*x^4*RootOf(256*_Z^4+16*_Z^2-1)^3+48*RootOf(256*_Z^4+16*_Z^2-1)^2*(x^4+1)^
(3/4)*x-64*(x^4+1)^(1/4)*RootOf(256*_Z^4+16*_Z^2-1)^2*x^3-4*RootOf(256*_Z^4+16*_Z^2-1)*(x^4+1)^(1/2)*x^2+8*x^4
*RootOf(256*_Z^4+16*_Z^2-1)-(x^4+1)^(3/4)*x+3*x^3*(x^4+1)^(1/4)-32*RootOf(256*_Z^4+16*_Z^2-1)^3+4*RootOf(256*_
Z^4+16*_Z^2-1))/(16*RootOf(256*_Z^4+16*_Z^2-1)^2*x^4+1))+RootOf(256*_Z^4+16*_Z^2-1)*ln(-(512*RootOf(256*_Z^4+1
6*_Z^2-1)^5*x^4+192*RootOf(256*_Z^4+16*_Z^2-1)^3*(x^4+1)^(1/2)*x^2-128*x^4*RootOf(256*_Z^4+16*_Z^2-1)^3+48*Roo
tOf(256*_Z^4+16*_Z^2-1)^2*(x^4+1)^(3/4)*x-64*(x^4+1)^(1/4)*RootOf(256*_Z^4+16*_Z^2-1)^2*x^3-4*RootOf(256*_Z^4+
16*_Z^2-1)*(x^4+1)^(1/2)*x^2+8*x^4*RootOf(256*_Z^4+16*_Z^2-1)-(x^4+1)^(3/4)*x+3*x^3*(x^4+1)^(1/4)-32*RootOf(25
6*_Z^4+16*_Z^2-1)^3+4*RootOf(256*_Z^4+16*_Z^2-1))/(16*RootOf(256*_Z^4+16*_Z^2-1)^2*x^4+1))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} - 2}{{\left (x^{8} + x^{4} - 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((x^4-2)/(x^4+1)^(1/4)/(x^8+x^4-1),x, algorithm="maxima")
[Out]
integrate((x^4 - 2)/((x^8 + x^4 - 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 {x^4-2}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+x^4-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^4 - 2)/((x^4 + 1)^(1/4)*(x^4 + x^8 - 1)),x)
[Out]
int((x^4 - 2)/((x^4 + 1)^(1/4)*(x^4 + x^8 - 1)), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((x**4-2)/(x**4+1)**(1/4)/(x**8+x**4-1),x)
[Out]
Timed out
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