\(\int \frac {1}{x^2 (a^5+x^5)} \, dx\) [142]

Optimal result

Integrand size = 13, antiderivative size = 209 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=-\frac {1}{a^5 x}+\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^6}+\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^6}+\frac {\log (a+x)}{5 a^6}-\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^6}-\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^6} \]

[Out]

-1/a^5/x+1/5*ln(a+x)/a^6-1/20*ln(a^2+x^2-1/2*a*x*(-5^(1/2)+1))*(-5^(1/2)+1)/a^6-1/20*ln(a^2+x^2-1/2*a*x*(5^(1/
2)+1))*(5^(1/2)+1)/a^6+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^6
+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^6

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 209, 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} \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=\frac {\sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {\left (1-\sqrt {5}\right ) a-4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )}{5 a^6}+\frac {\sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{10} \left (5+\sqrt {5}\right )} \left (\left (1+\sqrt {5}\right ) a-4 x\right )}{2 a}\right )}{5 a^6}+\frac {\log (a+x)}{5 a^6}-\frac {1}{a^5 x}-\frac {\left (1-\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1-\sqrt {5}\right ) a x+x^2\right )}{20 a^6}-\frac {\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^6} \]

[In]

Int[1/(x^2*(a^5 + x^5)),x]

[Out]

-(1/(a^5*x)) + (Sqrt[(5 + Sqrt[5])/2]*ArcTan[((1 - Sqrt[5])*a - 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)])/(5*a^6) + (Sq
rt[(5 - Sqrt[5])/2]*ArcTan[(Sqrt[(5 + Sqrt[5])/10]*((1 + Sqrt[5])*a - 4*x))/(2*a)])/(5*a^6) + Log[a + x]/(5*a^
6) - ((1 - Sqrt[5])*Log[a^2 - ((1 - Sqrt[5])*a*x)/2 + x^2])/(20*a^6) - ((1 + Sqrt[5])*Log[a^2 - ((1 + Sqrt[5])
*a*x)/2 + x^2])/(20*a^6)

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}{a^5 x}-\frac {\int \frac {x^3}{a^5+x^5} \, dx}{a^5} \\ & = -\frac {1}{a^5 x}+\frac {\int \frac {1}{a+x} \, dx}{5 a^6}-\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^6}-\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^6} \\ & = -\frac {1}{a^5 x}+\frac {\log (a+x)}{5 a^6}-\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^6}-\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^6}-\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^5}-\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^5} \\ & = -\frac {1}{a^5 x}+\frac {\log (a+x)}{5 a^6}-\frac {\left (1+\sqrt {5}\right ) \log \left (2 a^2-a x-\sqrt {5} a x+2 x^2\right )}{20 a^6}-\frac {\left (1-\sqrt {5}\right ) \log \left (2 a^2-a x+\sqrt {5} a x+2 x^2\right )}{20 a^6}+\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^5}+\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^5} \\ & = -\frac {1}{a^5 x}+\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^6}+\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^6}+\frac {\log (a+x)}{5 a^6}-\frac {\left (1+\sqrt {5}\right ) \log \left (2 a^2-a x-\sqrt {5} a x+2 x^2\right )}{20 a^6}-\frac {\left (1-\sqrt {5}\right ) \log \left (2 a^2-a x+\sqrt {5} a x+2 x^2\right )}{20 a^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=-\frac {\frac {20 a}{x}+2 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {\left (-1+\sqrt {5}\right ) a+4 x}{\sqrt {2 \left (5+\sqrt {5}\right )} a}\right )+2 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {-\left (\left (1+\sqrt {5}\right ) a\right )+4 x}{\sqrt {10-2 \sqrt {5}} a}\right )-4 \log (a+x)-\left (-1+\sqrt {5}\right ) \log \left (a^2+\frac {1}{2} \left (-1+\sqrt {5}\right ) a x+x^2\right )+\left (1+\sqrt {5}\right ) \log \left (a^2-\frac {1}{2} \left (1+\sqrt {5}\right ) a x+x^2\right )}{20 a^6} \]

[In]

Integrate[1/(x^2*(a^5 + x^5)),x]

[Out]

-1/20*((20*a)/x + 2*Sqrt[2*(5 + Sqrt[5])]*ArcTan[((-1 + Sqrt[5])*a + 4*x)/(Sqrt[2*(5 + Sqrt[5])]*a)] + 2*Sqrt[
10 - 2*Sqrt[5]]*ArcTan[(-((1 + Sqrt[5])*a) + 4*x)/(Sqrt[10 - 2*Sqrt[5]]*a)] - 4*Log[a + x] - (-1 + Sqrt[5])*Lo
g[a^2 + ((-1 + Sqrt[5])*a*x)/2 + x^2] + (1 + Sqrt[5])*Log[a^2 - ((1 + Sqrt[5])*a*x)/2 + x^2])/a^6

Maple [C] (verified)

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

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.36

method	result	size
risch	\(-\frac {1}{a^{5} x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{24} \textit {\_Z}^{4}+a^{18} \textit {\_Z}^{3}+a^{12} \textit {\_Z}^{2}+a^{6} \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\left (6 \textit {\_R}^{5} a^{30}-5\right ) x +a^{25} \textit {\_R}^{4}\right )\right )}{5}+\frac {\ln \left (a +x \right )}{5 a^{6}}\)	\(76\)
default	\(-\frac {1}{a^{5} x}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-a \,\textit {\_Z}^{3}+\textit {\_Z}^{2} a^{2}-a^{3} \textit {\_Z} +a^{4}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}-3 \textit {\_R}^{2} a +2 a^{2} \textit {\_R} -a^{3}\right ) \ln \left (x -\textit {\_R} \right )}{4 \textit {\_R}^{3}-3 \textit {\_R}^{2} a +2 a^{2} \textit {\_R} -a^{3}}}{5 a^{6}}+\frac {\ln \left (a +x \right )}{5 a^{6}}\)	\(109\)

[In]

int(1/x^2/(a^5+x^5),x,method=_RETURNVERBOSE)

[Out]

-1/a^5/x+1/5*sum(_R*ln((6*_R^5*a^30-5)*x+a^25*_R^4),_R=RootOf(_Z^4*a^24+_Z^3*a^18+_Z^2*a^12+_Z*a^6+1))+1/5*ln(
a+x)/a^6

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 15275, normalized size of antiderivative = 73.09 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=\text {Too large to display} \]

[In]

integrate(1/x^2/(a^5+x^5),x, algorithm="fricas")

[Out]

Too large to include

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=- \frac {1}{a^{5} x} + \frac {\frac {\log {\left (a + x \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} + 125 t^{3} + 25 t^{2} + 5 t + 1, \left ( t \mapsto t \log {\left (625 t^{4} a + x \right )} \right )\right )}}{a^{6}} \]

[In]

integrate(1/x**2/(a**5+x**5),x)

[Out]

-1/(a**5*x) + (log(a + x)/5 + RootSum(625*_t**4 + 125*_t**3 + 25*_t**2 + 5*_t + 1, Lambda(_t, _t*log(625*_t**4
*a + x))))/a**6

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=-\frac {\frac {2 \, \sqrt {5} {\left (\sqrt {5} + 1\right )} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{a \sqrt {2 \, \sqrt {5} + 10}} + \frac {2 \, \sqrt {5} {\left (\sqrt {5} - 1\right )} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{a \sqrt {-2 \, \sqrt {5} + 10}} + \frac {{\left (\sqrt {5} + 3\right )} \log \left (-a x {\left (\sqrt {5} + 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{a {\left (\sqrt {5} + 1\right )}} + \frac {{\left (\sqrt {5} - 3\right )} \log \left (a x {\left (\sqrt {5} - 1\right )} + 2 \, a^{2} + 2 \, x^{2}\right )}{a {\left (\sqrt {5} - 1\right )}} - \frac {2 \, \log \left (a + x\right )}{a}}{10 \, a^{5}} - \frac {1}{a^{5} x} \]

[In]

integrate(1/x^2/(a^5+x^5),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=-\frac {\sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {a {\left (\sqrt {5} - 1\right )} + 4 \, x}{a \sqrt {2 \, \sqrt {5} + 10}}\right )}{10 \, a^{6}} - \frac {\sqrt {-2 \, \sqrt {5} + 10} \arctan \left (-\frac {a {\left (\sqrt {5} + 1\right )} - 4 \, x}{a \sqrt {-2 \, \sqrt {5} + 10}}\right )}{10 \, a^{6}} - \frac {\sqrt {5} \log \left (a^{2} - \frac {1}{2} \, {\left (\sqrt {5} a + a\right )} x + x^{2}\right )}{20 \, a^{6}} + \frac {\sqrt {5} \log \left (a^{2} + \frac {1}{2} \, {\left (\sqrt {5} a - a\right )} x + x^{2}\right )}{20 \, a^{6}} - \frac {\log \left ({\left | a^{4} - a^{3} x + a^{2} x^{2} - a x^{3} + x^{4} \right |}\right )}{20 \, a^{6}} + \frac {\log \left ({\left | a + x \right |}\right )}{5 \, a^{6}} - \frac {1}{a^{5} x} \]

[In]

integrate(1/x^2/(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^6 - 1/10*sqrt(-2*sqrt(5)
+ 10)*arctan(-(a*(sqrt(5) + 1) - 4*x)/(a*sqrt(-2*sqrt(5) + 10)))/a^6 - 1/20*sqrt(5)*log(a^2 - 1/2*(sqrt(5)*a
+ a)*x + x^2)/a^6 + 1/20*sqrt(5)*log(a^2 + 1/2*(sqrt(5)*a - a)*x + x^2)/a^6 - 1/20*log(abs(a^4 - a^3*x + a^2*x
^2 - a*x^3 + x^4))/a^6 + 1/5*log(abs(a + x))/a^6 - 1/(a^5*x)

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (a^5+x^5\right )} \, dx=\frac {\ln \left (a+x\right )}{5\,a^6}-\frac {1}{a^5\,x}+\frac {\ln \left (5\,a^{30}+\frac {5\,x\,\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )\,a^{29}}{4}\right )\,\left (\sqrt {5}+\sqrt {-2\,\sqrt {5}-10}-1\right )}{20\,a^6}-\frac {\ln \left (5\,a^{30}-\frac {5\,a^{29}\,x\,\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}{4}\right )\,\left (\sqrt {5}+\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^6}-\frac {\ln \left (5\,a^{30}-\frac {5\,a^{29}\,x\,\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}{4}\right )\,\left (\sqrt {5}-\sqrt {2\,\sqrt {5}-10}+1\right )}{20\,a^6}-\frac {\ln \left (5\,a^{30}-\frac {5\,a^{29}\,x\,\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}{4}\right )\,\left (\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}+1\right )}{20\,a^6} \]

[In]

int(1/(x^2*(a^5 + x^5)),x)

[Out]

log(a + x)/(5*a^6) - 1/(a^5*x) + (log(5*a^30 + (5*a^29*x*(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^6) - (log(5*a^30 - (5*a^29*x*(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^6) - (log(5*a^30 - (5*a^29*x*(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^6) - (log(5*a^30 - (5*a^29*x*((- 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^6)