\(\int \frac {1}{x^4 (a^5+x^5)} \, dx\) [144]

Optimal result

Integrand size = 13, antiderivative size = 211 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=-\frac {1}{3 a^5 x^3}-\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^8}+\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^8}+\frac {\log (a+x)}{5 a^8}-\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^8}-\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^8} \]

[Out]

-1/3/a^5/x^3+1/5*ln(a+x)/a^8-1/20*ln(a^2+x^2-1/2*a*x*(5^(1/2)+1))*(-5^(1/2)+1)/a^8-1/20*ln(a^2+x^2-1/2*a*x*(-5
^(1/2)+1))*(5^(1/2)+1)/a^8-1/10*arctan((-4*x+a*(-5^(1/2)+1))/a/(10+2*5^(1/2))^(1/2))*(10-2*5^(1/2))^(1/2)/a^8+
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^8

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {331, 299, 648, 632, 210, 642, 31} \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=-\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^8}+\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^8}+\frac {\log (a+x)}{5 a^8}-\frac {1}{3 a^5 x^3}-\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^8}-\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^8} \]

[In]

Int[1/(x^4*(a^5 + x^5)),x]

[Out]

-1/3*1/(a^5*x^3) - (Sqrt[(5 - Sqrt[5])/2]*ArcTan[((1 - Sqrt[5])*a - 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)])/(5*a^8) +
(Sqrt[(5 + Sqrt[5])/2]*ArcTan[(Sqrt[(5 + Sqrt[5])/10]*((1 + Sqrt[5])*a - 4*x))/(2*a)])/(5*a^8) + Log[a + x]/(
5*a^8) - ((1 + Sqrt[5])*Log[a^2 - ((1 - Sqrt[5])*a*x)/2 + x^2])/(20*a^8) - ((1 - Sqrt[5])*Log[a^2 - ((1 + Sqrt
[5])*a*x)/2 + x^2])/(20*a^8)

Rule 31

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

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 299

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

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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*} \text {integral}& = -\frac {1}{3 a^5 x^3}-\frac {\int \frac {x}{a^5+x^5} \, dx}{a^5} \\ & = -\frac {1}{3 a^5 x^3}+\frac {\int \frac {1}{a+x} \, dx}{5 a^8}-\frac {2 \int \frac {\frac {1}{4} \left (1-\sqrt {5}\right ) 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^8}-\frac {2 \int \frac {\frac {1}{4} \left (1+\sqrt {5}\right ) 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^8} \\ & = -\frac {1}{3 a^5 x^3}+\frac {\log (a+x)}{5 a^8}-\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^8}-\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^8}+\frac {\int \frac {1}{a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2} \, dx}{2 \sqrt {5} a^7}-\frac {\int \frac {1}{a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2} \, dx}{2 \sqrt {5} a^7} \\ & = -\frac {1}{3 a^5 x^3}+\frac {\log (a+x)}{5 a^8}-\frac {\left (1-\sqrt {5}\right ) \log \left (2 a^2-a x-\sqrt {5} a x+2 x^2\right )}{20 a^8}-\frac {\left (1+\sqrt {5}\right ) \log \left (2 a^2-a x+\sqrt {5} a x+2 x^2\right )}{20 a^8}+\frac {\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 )}{\sqrt {5} a^7}-\frac {\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 )}{\sqrt {5} a^7} \\ & = -\frac {1}{3 a^5 x^3}-\frac {\sqrt {\frac {2}{5 \left (5+\sqrt {5}\right )}} \arctan \left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{a^8}+\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \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^8}+\frac {\log (a+x)}{5 a^8}-\frac {\left (1-\sqrt {5}\right ) \log \left (2 a^2-a x-\sqrt {5} a x+2 x^2\right )}{20 a^8}-\frac {\left (1+\sqrt {5}\right ) \log \left (2 a^2-a x+\sqrt {5} a x+2 x^2\right )}{20 a^8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=\frac {-\frac {20 a^3}{x^3}+6 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {\left (-1+\sqrt {5}\right ) a+4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )-6 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {-\left (\left (1+\sqrt {5}\right ) a\right )+4 x}{\sqrt {10-2 \sqrt {5}} a}\right )+12 \log (a+x)-3 \left (1+\sqrt {5}\right ) \log \left (a^2+\frac {1}{2} \left (-1+\sqrt {5}\right ) a x+x^2\right )+3 \left (-1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{60 a^8} \]

[In]

Integrate[1/(x^4*(a^5 + x^5)),x]

[Out]

((-20*a^3)/x^3 + 6*Sqrt[10 - 2*Sqrt[5]]*ArcTan[((-1 + Sqrt[5])*a + 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)] - 6*Sqrt[2*
(5 + Sqrt[5])]*ArcTan[(-((1 + Sqrt[5])*a) + 4*x)/(Sqrt[10 - 2*Sqrt[5]]*a)] + 12*Log[a + x] - 3*(1 + Sqrt[5])*L
og[a^2 + ((-1 + Sqrt[5])*a*x)/2 + x^2] + 3*(-1 + Sqrt[5])*Log[a^2 - ((1 + Sqrt[5])*a*x)/2 + x^2])/(60*a^8)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.38

method	result	size
risch	\(-\frac {1}{3 a^{5} x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{32} \textit {\_Z}^{4}+a^{24} \textit {\_Z}^{3}+a^{16} \textit {\_Z}^{2}+a^{8} \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\left (-6 \textit {\_R}^{5} a^{40}+5\right ) x -a^{25} \textit {\_R}^{3}\right )\right )}{5}+\frac {\ln \left (-a -x \right )}{5 a^{8}}\)	\(81\)
default	\(-\frac {1}{3 a^{5} x^{3}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-a \,\textit {\_Z}^{3}+\textit {\_Z}^{2} a^{2}-a^{3} \textit {\_Z} +a^{4}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+2 \textit {\_R}^{2} a -3 a^{2} \textit {\_R} -a^{3}\right ) \ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3}-3 \textit {\_R}^{2} a +2 a^{2} \textit {\_R} -a^{3}}}{5 a^{8}}+\frac {\ln \left (a +x \right )}{5 a^{8}}\)	\(109\)

[In]

int(1/x^4/(a^5+x^5),x,method=_RETURNVERBOSE)

[Out]

-1/3/a^5/x^3+1/5*sum(_R*ln((-6*_R^5*a^40+5)*x-a^25*_R^3),_R=RootOf(_Z^4*a^32+_Z^3*a^24+_Z^2*a^16+_Z*a^8+1))+1/
5/a^8*ln(-a-x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.97 (sec) , antiderivative size = 15501, normalized size of antiderivative = 73.46 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=\text {Too large to display} \]

[In]

integrate(1/x^4/(a^5+x^5),x, algorithm="fricas")

[Out]

Too large to include

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.24 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=- \frac {1}{3 a^{5} x^{3}} + \frac {\frac {\log {\left (a + x \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} + 125 t^{3} + 25 t^{2} + 5 t + 1, \left ( t \mapsto t \log {\left (125 t^{3} a + x \right )} \right )\right )}}{a^{8}} \]

[In]

integrate(1/x**4/(a**5+x**5),x)

[Out]

-1/(3*a**5*x**3) + (log(a + x)/5 + RootSum(625*_t**4 + 125*_t**3 + 25*_t**2 + 5*_t + 1, Lambda(_t, _t*log(125*
_t**3*a + x))))/a**8

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=\frac {\frac {2 \, \sqrt {5} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{a^{3} \sqrt {2 \, \sqrt {5} + 10}} - \frac {2 \, \sqrt {5} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{a^{3} \sqrt {-2 \, \sqrt {5} + 10}} + \frac {\log \left (a + x\right )}{a^{3}} + \frac {\log \left (-a x {\left (\sqrt {5} + 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{a^{3} {\left (\sqrt {5} + 1\right )}} - \frac {\log \left (a x {\left (\sqrt {5} - 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{a^{3} {\left (\sqrt {5} - 1\right )}}}{5 \, a^{5}} - \frac {1}{3 \, a^{5} x^{3}} \]

[In]

integrate(1/x^4/(a^5+x^5),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=\frac {\sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{10 \, a^{8}} - \frac {\sqrt {2 \, \sqrt {5} + 10} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{10 \, a^{8}} + \frac {\sqrt {5} \log \left (a^{2} - \frac {1}{2} \, {\left (\sqrt {5} a + a\right )} x + x^{2}\right )}{20 \, a^{8}} - \frac {\sqrt {5} \log \left (a^{2} + \frac {1}{2} \, {\left (\sqrt {5} a - a\right )} x + x^{2}\right )}{20 \, a^{8}} - \frac {\log \left ({\left | a^{4} - a^{3} x + a^{2} x^{2} - a x^{3} + x^{4} \right |}\right )}{20 \, a^{8}} + \frac {\log \left ({\left | a + x \right |}\right )}{5 \, a^{8}} - \frac {1}{3 \, a^{5} x^{3}} \]

[In]

integrate(1/x^4/(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^8 - 1/10*sqrt(2*sqrt(5)
+ 10)*arctan(-(a*(sqrt(5) + 1) - 4*x)/(a*sqrt(-2*sqrt(5) + 10)))/a^8 + 1/20*sqrt(5)*log(a^2 - 1/2*(sqrt(5)*a +
a)*x + x^2)/a^8 - 1/20*sqrt(5)*log(a^2 + 1/2*(sqrt(5)*a - a)*x + x^2)/a^8 - 1/20*log(abs(a^4 - a^3*x + a^2*x^
2 - a*x^3 + x^4))/a^8 + 1/5*log(abs(a + x))/a^8 - 1/3/(a^5*x^3)

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^4 \left (a^5+x^5\right )} \, dx=\frac {\ln \left (a+x\right )}{5\,a^8}-\frac {1}{3\,a^5\,x^3}-\frac {\ln \left (a^{15}\,x-\frac {a^{16}\,{\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}^3}{64}\right )\,\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^8}-\frac {\ln \left (a^{15}\,x-\frac {a^{16}\,{\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}^3}{64}\right )\,\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}{20\,a^8}+\frac {\ln \left (\frac {{\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}^3\,a^{16}}{64}+x\,a^{15}\right )\,\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}{20\,a^8}-\frac {\ln \left (a^{15}\,x-\frac {a^{16}\,{\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}^3}{64}\right )\,\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^8} \]

[In]

int(1/(x^4*(a^5 + x^5)),x)

[Out]

log(a + x)/(5*a^8) - 1/(3*a^5*x^3) - (log(a^15*x - (a^16*(5^(1/2) - (2*5^(1/2) - 10)^(1/2) + 1)^3)/64)*(5^(1/2
) - (2*5^(1/2) - 10)^(1/2) + 1))/(20*a^8) - (log(a^15*x - (a^16*((- 2*5^(1/2) - 10)^(1/2) - 5^(1/2) + 1)^3)/64
)*((- 2*5^(1/2) - 10)^(1/2) - 5^(1/2) + 1))/(20*a^8) + (log(a^15*x + (a^16*(5^(1/2) + (- 2*5^(1/2) - 10)^(1/2)
- 1)^3)/64)*(5^(1/2) + (- 2*5^(1/2) - 10)^(1/2) - 1))/(20*a^8) - (log(a^15*x - (a^16*(5^(1/2) + (2*5^(1/2) -
10)^(1/2) + 1)^3)/64)*(5^(1/2) + (2*5^(1/2) - 10)^(1/2) + 1))/(20*a^8)