3.2.36 \(\int \frac {1}{a^5+x^5} \, dx\) [136]

Optimal. Leaf size=201 \[ -\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^4}-\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^4}+\frac {\log (a+x)}{5 a^4}-\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^4}-\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^4} \]

[Out]

1/5*ln(a+x)/a^4-1/20*ln(a^2+x^2-1/2*a*x*(-5^(1/2)+1))*(-5^(1/2)+1)/a^4-1/20*ln(a^2+x^2-1/2*a*x*(5^(1/2)+1))*(5
^(1/2)+1)/a^4-1/10*arctan(1/20*(-4*x+a*(5^(1/2)+1))*(50+10*5^(1/2))^(1/2)/a)*(10-2*5^(1/2))^(1/2)/a^4-1/10*arc
tan((-4*x+a*(-5^(1/2)+1))/a/(10+2*5^(1/2))^(1/2))*(10+2*5^(1/2))^(1/2)/a^4

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Rubi [A]
time = 0.23, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {207, 648, 632, 210, 642, 31} \begin {gather*} -\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \text {ArcTan}\left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^4}-\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \text {ArcTan}\left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^4}+\frac {\log (a+x)}{5 a^4}-\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^4}-\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^5 + x^5)^(-1),x]

[Out]

-1/5*(Sqrt[(5 + Sqrt[5])/2]*ArcTan[((1 - Sqrt[5])*a - 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)])/a^4 - (Sqrt[(5 - Sqrt[5
])/2]*ArcTan[(Sqrt[(5 + Sqrt[5])/10]*((1 + Sqrt[5])*a - 4*x))/(2*a)])/(5*a^4) + Log[a + x]/(5*a^4) - ((1 - Sqr
t[5])*Log[a^2 - ((1 - Sqrt[5])*a*x)/2 + x^2])/(20*a^4) - ((1 + Sqrt[5])*Log[a^2 - ((1 + Sqrt[5])*a*x)/2 + x^2]
)/(20*a^4)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 207

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u}, Simp[u = Int[(r - s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; (r/
(a*n))*Int[1/(r + s*x), x] + Dist[2*(r/(a*n)), Sum[u, {k, 1, (n - 1)/2}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[
(n - 3)/2, 0] && PosQ[a/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

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

Rubi steps

\begin {align*} \int \frac {1}{a^5+x^5} \, dx &=\frac {2 \int \frac {a-\frac {1}{4} \left (1-\sqrt {5}\right ) x}{a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2} \, dx}{5 a^4}+\frac {2 \int \frac {a-\frac {1}{4} \left (1+\sqrt {5}\right ) x}{a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2} \, dx}{5 a^4}+\frac {\int \frac {1}{a+x} \, dx}{5 a^4}\\ &=\frac {\log (a+x)}{5 a^4}-\frac {\left (1-\sqrt {5}\right ) \int \frac {-\frac {1}{2} \left (1-\sqrt {5}\right ) a+2 x}{a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2} \, dx}{20 a^4}-\frac {\left (1+\sqrt {5}\right ) \int \frac {-\frac {1}{2} \left (1+\sqrt {5}\right ) a+2 x}{a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2} \, dx}{20 a^4}+\frac {\left (5-\sqrt {5}\right ) \int \frac {1}{a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2} \, dx}{20 a^3}+\frac {\left (5+\sqrt {5}\right ) \int \frac {1}{a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2} \, dx}{20 a^3}\\ &=\frac {\log (a+x)}{5 a^4}-\frac {\left (1+\sqrt {5}\right ) \log \left (2 a^2-a x-\sqrt {5} a x+2 x^2\right )}{20 a^4}-\frac {\left (1-\sqrt {5}\right ) \log \left (2 a^2-a x+\sqrt {5} a x+2 x^2\right )}{20 a^4}-\frac {\left (5-\sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{2} \left (5-\sqrt {5}\right ) a^2-x^2} \, dx,x,-\frac {1}{2} \left (1+\sqrt {5}\right ) a+2 x\right )}{10 a^3}-\frac {\left (5+\sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{2} \left (5+\sqrt {5}\right ) a^2-x^2} \, dx,x,-\frac {1}{2} \left (1-\sqrt {5}\right ) a+2 x\right )}{10 a^3}\\ &=-\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^4}-\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^4}+\frac {\log (a+x)}{5 a^4}-\frac {\left (1+\sqrt {5}\right ) \log \left (2 a^2-a x-\sqrt {5} a x+2 x^2\right )}{20 a^4}-\frac {\left (1-\sqrt {5}\right ) \log \left (2 a^2-a x+\sqrt {5} a x+2 x^2\right )}{20 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 204, normalized size = 1.01 \begin {gather*} -\frac {-2 \sqrt {2 \left (5+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\left (-1+\sqrt {5}\right ) a+4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )-2 \sqrt {10-2 \sqrt {5}} \tan ^{-1}\left (\frac {-\left (\left (1+\sqrt {5}\right ) a\right )+4 x}{\sqrt {10-2 \sqrt {5}} a}\right )-4 \log (a+x)+\log \left (a^2+\frac {1}{2} \left (-1+\sqrt {5}\right ) a x+x^2\right )-\sqrt {5} \log \left (a^2+\frac {1}{2} \left (-1+\sqrt {5}\right ) a x+x^2\right )+\log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )+\sqrt {5} \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^5 + x^5)^(-1),x]

[Out]

-1/20*(-2*Sqrt[2*(5 + Sqrt[5])]*ArcTan[((-1 + Sqrt[5])*a + 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)] - 2*Sqrt[10 - 2*Sqr
t[5]]*ArcTan[(-((1 + Sqrt[5])*a) + 4*x)/(Sqrt[10 - 2*Sqrt[5]]*a)] - 4*Log[a + x] + Log[a^2 + ((-1 + Sqrt[5])*a
*x)/2 + x^2] - Sqrt[5]*Log[a^2 + ((-1 + Sqrt[5])*a*x)/2 + x^2] + Log[a^2 - ((1 + Sqrt[5])*a*x)/2 + x^2] + Sqrt
[5]*Log[a^2 - ((1 + Sqrt[5])*a*x)/2 + x^2])/a^4

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.05, size = 101, normalized size = 0.50

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{16} \textit {\_Z}^{4}+a^{12} \textit {\_Z}^{3}+a^{8} \textit {\_Z}^{2}+a^{4} \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R} \,a^{5}+x \right )\right )}{5}+\frac {\ln \left (a +x \right )}{5 a^{4}}\)	\(55\)
default	\(\frac {\ln \left (a +x \right )}{5 a^{4}}+\frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}-a \,\textit {\_Z}^{3}+\textit {\_Z}^{2} a^{2}-a^{3} \textit {\_Z} +a^{4}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+2 \textit {\_R}^{2} a -3 a^{2} \textit {\_R} +4 a^{3}\right ) \ln \left (-\textit {\_R} +x \right )}{4 \textit {\_R}^{3}-3 \textit {\_R}^{2} a +2 a^{2} \textit {\_R} -a^{3}}}{5 a^{4}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^5+x^5),x,method=_RETURNVERBOSE)

[Out]

1/5*ln(a+x)/a^4+1/5/a^4*sum((-_R^3+2*_R^2*a-3*_R*a^2+4*a^3)/(4*_R^3-3*_R^2*a+2*_R*a^2-a^3)*ln(-_R+x),_R=RootOf
(_Z^4-_Z^3*a+_Z^2*a^2-_Z*a^3+a^4))

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Maxima [A]
time = 5.06, size = 180, normalized size = 0.90 \begin {gather*} \frac {\sqrt {5} {\left (\sqrt {5} + 1\right )} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{5 \, a^{4} \sqrt {2 \, \sqrt {5} + 10}} + \frac {\sqrt {5} {\left (\sqrt {5} - 1\right )} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{5 \, a^{4} \sqrt {-2 \, \sqrt {5} + 10}} - \frac {{\left (\sqrt {5} + 3\right )} \log \left (-a x {\left (\sqrt {5} + 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{10 \, a^{4} {\left (\sqrt {5} + 1\right )}} - \frac {{\left (\sqrt {5} - 3\right )} \log \left (a x {\left (\sqrt {5} - 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{10 \, a^{4} {\left (\sqrt {5} - 1\right )}} + \frac {\log \left (a + x\right )}{5 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^5+x^5),x, algorithm="maxima")

[Out]

1/5*sqrt(5)*(sqrt(5) + 1)*arctan((a*(sqrt(5) - 1) + 4*x)/(a*sqrt(2*sqrt(5) + 10)))/(a^4*sqrt(2*sqrt(5) + 10))
+ 1/5*sqrt(5)*(sqrt(5) - 1)*arctan(-(a*(sqrt(5) + 1) - 4*x)/(a*sqrt(-2*sqrt(5) + 10)))/(a^4*sqrt(-2*sqrt(5) +
10)) - 1/10*(sqrt(5) + 3)*log(-a*x*(sqrt(5) + 1) + 2*a^2 + 2*x^2)/(a^4*(sqrt(5) + 1)) - 1/10*(sqrt(5) - 3)*log
(a*x*(sqrt(5) - 1) + 2*a^2 + 2*x^2)/(a^4*(sqrt(5) - 1)) + 1/5*log(a + x)/a^4

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Fricas [C] Result contains complex when optimal does not.
time = 1.29, size = 11094, normalized size = 55.19 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^5+x^5),x, algorithm="fricas")

[Out]

-1/4800*(2*(800*(1/2)^(2/3)*(-I*sqrt(3) + 1)*((sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^2/a^8 - 6*(s
qrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4*sqrt((sqrt(5) - 5)/a^8)) + 4)/a^8)/(16000*(sqrt(2)*a^4*sqrt(
(sqrt(5) - 5)/a^8) + sqrt(5) - 3)^3/a^12 - 144000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)*(sqrt(2)
*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4*sqrt((sqrt(5) - 5)/a^8)) + 4)/a^12 + 432000*(sqrt(2)*(a^12*((sqrt(
5) - 5)/a^8)^(3/2) - sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) + 13*a^4*sqrt((sqrt(5) - 5)/a^8)) - 8*sqrt(5) - 8)/a^
12 + 9*sqrt(-2048000000*sqrt(5)*sqrt(2)*a^20*((sqrt(5) - 5)/a^8)^(5/2) - 20480000000/3*(9*sqrt(5)*sqrt(2) + 5*
sqrt(2))*a^12*((sqrt(5) - 5)/a^8)^(3/2) - 102400000000/3*(11*sqrt(5)*sqrt(2) + 7*sqrt(2))*a^4*sqrt((sqrt(5) -
5)/a^8) - 1024000000/3*(9*sqrt(5) - 305)*(sqrt(5) - 5)^2 + 15872000000/3*(sqrt(5) - 5)^3 - 35840000000/3*(sqrt
(5) - 5)*(sqrt(5) - 51) + 778240000000/3*sqrt(5) + 2662400000000/3)/a^12)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*
(16000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^3/a^12 - 144000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8
) + sqrt(5) - 3)*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4*sqrt((sqrt(5) - 5)/a^8)) + 4)/a^12 + 4320
00*(sqrt(2)*(a^12*((sqrt(5) - 5)/a^8)^(3/2) - sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) + 13*a^4*sqrt((sqrt(5) - 5)/
a^8)) - 8*sqrt(5) - 8)/a^12 + 9*sqrt(-2048000000*sqrt(5)*sqrt(2)*a^20*((sqrt(5) - 5)/a^8)^(5/2) - 20480000000/
3*(9*sqrt(5)*sqrt(2) + 5*sqrt(2))*a^12*((sqrt(5) - 5)/a^8)^(3/2) - 102400000000/3*(11*sqrt(5)*sqrt(2) + 7*sqrt
(2))*a^4*sqrt((sqrt(5) - 5)/a^8) - 1024000000/3*(9*sqrt(5) - 305)*(sqrt(5) - 5)^2 + 15872000000/3*(sqrt(5) - 5
)^3 - 35840000000/3*(sqrt(5) - 5)*(sqrt(5) - 51) + 778240000000/3*sqrt(5) + 2662400000000/3)/a^12)^(1/3) - 40*
(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)/a^4)*a^4*log(-1/480*(800*(1/2)^(2/3)*(-I*sqrt(3) + 1)*((sq
rt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^2/a^8 - 6*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4
*sqrt((sqrt(5) - 5)/a^8)) + 4)/a^8)/(16000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^3/a^12 - 144000
*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4*sqrt(
(sqrt(5) - 5)/a^8)) + 4)/a^12 + 432000*(sqrt(2)*(a^12*((sqrt(5) - 5)/a^8)^(3/2) - sqrt(5)*a^4*sqrt((sqrt(5) -
5)/a^8) + 13*a^4*sqrt((sqrt(5) - 5)/a^8)) - 8*sqrt(5) - 8)/a^12 + 9*sqrt(-2048000000*sqrt(5)*sqrt(2)*a^20*((sq
rt(5) - 5)/a^8)^(5/2) - 20480000000/3*(9*sqrt(5)*sqrt(2) + 5*sqrt(2))*a^12*((sqrt(5) - 5)/a^8)^(3/2) - 1024000
00000/3*(11*sqrt(5)*sqrt(2) + 7*sqrt(2))*a^4*sqrt((sqrt(5) - 5)/a^8) - 1024000000/3*(9*sqrt(5) - 305)*(sqrt(5)
- 5)^2 + 15872000000/3*(sqrt(5) - 5)^3 - 35840000000/3*(sqrt(5) - 5)*(sqrt(5) - 51) + 778240000000/3*sqrt(5)
+ 2662400000000/3)/a^12)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(16000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqr
t(5) - 3)^3/a^12 - 144000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt
(5) - 5)/a^8) - a^4*sqrt((sqrt(5) - 5)/a^8)) + 4)/a^12 + 432000*(sqrt(2)*(a^12*((sqrt(5) - 5)/a^8)^(3/2) - sqr
t(5)*a^4*sqrt((sqrt(5) - 5)/a^8) + 13*a^4*sqrt((sqrt(5) - 5)/a^8)) - 8*sqrt(5) - 8)/a^12 + 9*sqrt(-2048000000*
sqrt(5)*sqrt(2)*a^20*((sqrt(5) - 5)/a^8)^(5/2) - 20480000000/3*(9*sqrt(5)*sqrt(2) + 5*sqrt(2))*a^12*((sqrt(5)
- 5)/a^8)^(3/2) - 102400000000/3*(11*sqrt(5)*sqrt(2) + 7*sqrt(2))*a^4*sqrt((sqrt(5) - 5)/a^8) - 1024000000/3*(
9*sqrt(5) - 305)*(sqrt(5) - 5)^2 + 15872000000/3*(sqrt(5) - 5)^3 - 35840000000/3*(sqrt(5) - 5)*(sqrt(5) - 51)
+ 778240000000/3*sqrt(5) + 2662400000000/3)/a^12)^(1/3) - 40*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) -
3)/a^4)*a^5 + x) + 240*a^4*(2*sqrt(1/2)*sqrt(sqrt(5)/a^8 - 5/a^8) + sqrt(5)/a^4 + 1/a^4)*log(-1/4*a^5*(2*sqrt(
1/2)*sqrt(sqrt(5)/a^8 - 5/a^8) + sqrt(5)/a^4 + 1/a^4) + x) - ((800*(1/2)^(2/3)*(-I*sqrt(3) + 1)*((sqrt(2)*a^4*
sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^2/a^8 - 6*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4*sqrt((sqr
t(5) - 5)/a^8)) + 4)/a^8)/(16000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^3/a^12 - 144000*(sqrt(2)*
a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) - a^4*sqrt((sqrt(5) -
5)/a^8)) + 4)/a^12 + 432000*(sqrt(2)*(a^12*((sqrt(5) - 5)/a^8)^(3/2) - sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a^8) +
13*a^4*sqrt((sqrt(5) - 5)/a^8)) - 8*sqrt(5) - 8)/a^12 + 9*sqrt(-2048000000*sqrt(5)*sqrt(2)*a^20*((sqrt(5) - 5)
/a^8)^(5/2) - 20480000000/3*(9*sqrt(5)*sqrt(2) + 5*sqrt(2))*a^12*((sqrt(5) - 5)/a^8)^(3/2) - 102400000000/3*(1
1*sqrt(5)*sqrt(2) + 7*sqrt(2))*a^4*sqrt((sqrt(5) - 5)/a^8) - 1024000000/3*(9*sqrt(5) - 305)*(sqrt(5) - 5)^2 +
15872000000/3*(sqrt(5) - 5)^3 - 35840000000/3*(sqrt(5) - 5)*(sqrt(5) - 51) + 778240000000/3*sqrt(5) + 26624000
00000/3)/a^12)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(16000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)^
3/a^12 - 144000*(sqrt(2)*a^4*sqrt((sqrt(5) - 5)/a^8) + sqrt(5) - 3)*(sqrt(2)*(sqrt(5)*a^4*sqrt((sqrt(5) - 5)/a
^8) - a^4*sqrt((sqrt(5) - 5)/a^8)) + 4)/a^12 + ...

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Sympy [A]
time = 0.05, size = 39, normalized size = 0.19 \begin {gather*} \frac {\frac {\log {\left (a + x \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} + 125 t^{3} + 25 t^{2} + 5 t + 1, \left ( t \mapsto t \log {\left (5 t a + x \right )} \right )\right )}}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a**5+x**5),x)

[Out]

(log(a + x)/5 + RootSum(625*_t**4 + 125*_t**3 + 25*_t**2 + 5*_t + 1, Lambda(_t, _t*log(5*_t*a + x))))/a**4

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Giac [A]
time = 0.57, size = 177, normalized size = 0.88 \begin {gather*} \frac {\sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{10 \, a^{4}} + \frac {\sqrt {-2 \, \sqrt {5} + 10} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{10 \, a^{4}} - \frac {\sqrt {5} \log \left (a^{2} - \frac {1}{2} \, {\left (\sqrt {5} a + a\right )} x + x^{2}\right )}{20 \, a^{4}} + \frac {\sqrt {5} \log \left (a^{2} + \frac {1}{2} \, {\left (\sqrt {5} a - a\right )} x + x^{2}\right )}{20 \, a^{4}} - \frac {\log \left ({\left | a^{4} - a^{3} x + a^{2} x^{2} - a x^{3} + x^{4} \right |}\right )}{20 \, a^{4}} + \frac {\log \left ({\left | a + x \right |}\right )}{5 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^5+x^5),x, algorithm="giac")

[Out]

1/10*sqrt(2*sqrt(5) + 10)*arctan((a*(sqrt(5) - 1) + 4*x)/(a*sqrt(2*sqrt(5) + 10)))/a^4 + 1/10*sqrt(-2*sqrt(5)
+ 10)*arctan(-(a*(sqrt(5) + 1) - 4*x)/(a*sqrt(-2*sqrt(5) + 10)))/a^4 - 1/20*sqrt(5)*log(a^2 - 1/2*(sqrt(5)*a +
a)*x + x^2)/a^4 + 1/20*sqrt(5)*log(a^2 + 1/2*(sqrt(5)*a - a)*x + x^2)/a^4 - 1/20*log(abs(a^4 - a^3*x + a^2*x^
2 - a*x^3 + x^4))/a^4 + 1/5*log(abs(a + x))/a^4

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Mupad [B]
time = 0.59, size = 174, normalized size = 0.87 \begin {gather*} \frac {\ln \left (a+x\right )}{5\,a^4}-\frac {\ln \left (x-\frac {a\,\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}{4}\right )\,\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^4}-\frac {\ln \left (x-\frac {a\,\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}{4}\right )\,\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}{20\,a^4}+\frac {\ln \left (x+\frac {a\,\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}{4}\right )\,\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}{20\,a^4}-\frac {\ln \left (x-\frac {a\,\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}{4}\right )\,\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^5 + x^5),x)

[Out]

log(a + x)/(5*a^4) - (log(x - (a*(5^(1/2) - (2*5^(1/2) - 10)^(1/2) + 1))/4)*(5^(1/2) - (2*5^(1/2) - 10)^(1/2)
+ 1))/(20*a^4) - (log(x - (a*((- 2*5^(1/2) - 10)^(1/2) - 5^(1/2) + 1))/4)*((- 2*5^(1/2) - 10)^(1/2) - 5^(1/2)
+ 1))/(20*a^4) + (log(x + (a*(5^(1/2) + (- 2*5^(1/2) - 10)^(1/2) - 1))/4)*(5^(1/2) + (- 2*5^(1/2) - 10)^(1/2)
- 1))/(20*a^4) - (log(x - (a*(5^(1/2) + (2*5^(1/2) - 10)^(1/2) + 1))/4)*(5^(1/2) + (2*5^(1/2) - 10)^(1/2) + 1)
)/(20*a^4)

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