3.1307 \(\int \frac {1}{x^4 (1+x^5)} \, dx\)

Optimal. Leaf size=192 \[ -\frac {1}{3 x^3}-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (x^2-\frac {1}{2} \left (1-\sqrt {5}\right ) x+1\right )-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (x^2-\frac {1}{2} \left (1+\sqrt {5}\right ) x+1\right )+\frac {1}{5} \log (x+1)+\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{5+\sqrt {5}}} x+\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}\right )+\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}-\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} x\right ) \]

[Out]

-1/3/x^3+1/5*ln(1+x)-1/20*ln(1+x^2-1/2*x*(5^(1/2)+1))*(-5^(1/2)+1)-1/20*ln(1+x^2-1/2*x*(-5^(1/2)+1))*(5^(1/2)+
1)+1/10*arctan(1/5*(25-10*5^(1/2))^(1/2)+2*x*2^(1/2)/(5+5^(1/2))^(1/2))*(10-2*5^(1/2))^(1/2)-1/10*arctan(1/5*x
*(50+10*5^(1/2))^(1/2)-1/5*(25+10*5^(1/2))^(1/2))*(10+2*5^(1/2))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {325, 293, 634, 618, 204, 628, 31} \[ -\frac {1}{3 x^3}-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (x^2-\frac {1}{2} \left (1-\sqrt {5}\right ) x+1\right )-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (x^2-\frac {1}{2} \left (1+\sqrt {5}\right ) x+1\right )+\frac {1}{5} \log (x+1)+\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{5+\sqrt {5}}} x+\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}\right )+\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}-\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} x\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(1 + x^5)),x]

[Out]

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

Rule 31

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

Rule 204

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

Rule 293

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)*Int[1/(r + s*x), x])/(a*n*s^m)) + 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 325

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 618

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 628

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 634

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

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Mathematica [A]  time = 0.14, size = 150, normalized size = 0.78 \[ \frac {1}{60} \left (-\frac {20}{x^3}-3 \left (1+\sqrt {5}\right ) \log \left (x^2+\frac {1}{2} \left (\sqrt {5}-1\right ) x+1\right )+3 \left (\sqrt {5}-1\right ) \log \left (x^2-\frac {1}{2} \left (1+\sqrt {5}\right ) x+1\right )+12 \log (x+1)+6 \sqrt {2 \left (5+\sqrt {5}\right )} \tan ^{-1}\left (\frac {-4 x+\sqrt {5}+1}{\sqrt {10-2 \sqrt {5}}}\right )+6 \sqrt {10-2 \sqrt {5}} \tan ^{-1}\left (\frac {4 x+\sqrt {5}-1}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^4*(1 + x^5)),x]

[Out]

(-20/x^3 + 6*Sqrt[2*(5 + Sqrt[5])]*ArcTan[(1 + Sqrt[5] - 4*x)/Sqrt[10 - 2*Sqrt[5]]] + 6*Sqrt[10 - 2*Sqrt[5]]*A
rcTan[(-1 + Sqrt[5] + 4*x)/Sqrt[2*(5 + Sqrt[5])]] + 12*Log[1 + x] - 3*(1 + Sqrt[5])*Log[1 + ((-1 + Sqrt[5])*x)
/2 + x^2] + 3*(-1 + Sqrt[5])*Log[1 - ((1 + Sqrt[5])*x)/2 + x^2])/60

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fricas [B]  time = 2.65, size = 1174, normalized size = 6.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(6*x^3*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1)*log(1/64*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) +
1)^3 + 1/64*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1)^2 - 1
/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^2 - 1/64*((2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^2 -
8*sqrt(1/2)*sqrt(sqrt(5) - 5) - 4*sqrt(5) + 12)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) + x + 1/2*sqrt(1
/2)*sqrt(sqrt(5) - 5) + 1/4*sqrt(5) - 3/4) - 6*x^3*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)*log(-1/64*(2*
sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^3 + x) + 24*x^3*log(x + 1) + 3*(x^3*(2*sqrt(1/2)*sqrt(sqrt(5) - 5)
+ sqrt(5) + 1) - x^3*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) - 4*sqrt(-3/16*(2*sqrt(1/2)*sqrt(sqrt(5) -
5) + sqrt(5) + 1)^2 + 1/8*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(
5) - 1) - 3/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1)^2 + sqrt(1/2)*sqrt(sqrt(5) - 5) + 1/2*sqrt(5) - 5
/2)*x^3 - 4*x^3)*log(-1/64*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt
(5) - 1)^2 + 1/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^2 + 1/64*((2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqr
t(5) + 1)^2 - 8*sqrt(1/2)*sqrt(sqrt(5) - 5) - 4*sqrt(5) + 12)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) +
1/16*sqrt(-3/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^2 + 1/8*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5)
- 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) - 3/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1)^2 + s
qrt(1/2)*sqrt(sqrt(5) - 5) + 1/2*sqrt(5) - 5/2)*((2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sq
rt(sqrt(5) - 5) - sqrt(5) - 1) + 8*sqrt(1/2)*sqrt(sqrt(5) - 5) + 4*sqrt(5) + 4) + 2*x - 1/2*sqrt(1/2)*sqrt(sqr
t(5) - 5) - 1/4*sqrt(5) - 1/4) + 3*(x^3*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1) - x^3*(2*sqrt(1/2)*sqrt(
sqrt(5) - 5) - sqrt(5) - 1) + 4*sqrt(-3/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^2 + 1/8*(2*sqrt(1/2)*
sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) - 3/16*(2*sqrt(1/2)*sqrt(sqrt(5
) - 5) - sqrt(5) - 1)^2 + sqrt(1/2)*sqrt(sqrt(5) - 5) + 1/2*sqrt(5) - 5/2)*x^3 - 4*x^3)*log(-1/64*(2*sqrt(1/2)
*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1)^2 + 1/16*(2*sqrt(1/2)*sqrt(sqr
t(5) - 5) + sqrt(5) + 1)^2 + 1/64*((2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) + 1)^2 - 8*sqrt(1/2)*sqrt(sqrt(5)
- 5) - 4*sqrt(5) + 12)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) - 1/16*sqrt(-3/16*(2*sqrt(1/2)*sqrt(sqrt(
5) - 5) + sqrt(5) + 1)^2 + 1/8*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) -
sqrt(5) - 1) - 3/16*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1)^2 + sqrt(1/2)*sqrt(sqrt(5) - 5) + 1/2*sqrt(5
) - 5/2)*((2*sqrt(1/2)*sqrt(sqrt(5) - 5) + sqrt(5) - 3)*(2*sqrt(1/2)*sqrt(sqrt(5) - 5) - sqrt(5) - 1) + 8*sqrt
(1/2)*sqrt(sqrt(5) - 5) + 4*sqrt(5) + 4) + 2*x - 1/2*sqrt(1/2)*sqrt(sqrt(5) - 5) - 1/4*sqrt(5) - 1/4) - 40)/x^
3

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giac [A]  time = 0.18, size = 132, normalized size = 0.69 \[ \frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) - \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) + \frac {1}{20} \, \sqrt {5} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {5} + 1\right )} + 1\right ) - \frac {1}{20} \, \sqrt {5} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {5} - 1\right )} + 1\right ) - \frac {1}{3 \, x^{3}} - \frac {1}{20} \, \log \left (x^{4} - x^{3} + x^{2} - x + 1\right ) + \frac {1}{5} \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 161, normalized size = 0.84 \[ -\frac {2 \sqrt {5}\, \arctan \left (\frac {4 x -\sqrt {5}-1}{\sqrt {10-2 \sqrt {5}}}\right )}{5 \sqrt {10-2 \sqrt {5}}}+\frac {2 \sqrt {5}\, \arctan \left (\frac {4 x +\sqrt {5}-1}{\sqrt {10+2 \sqrt {5}}}\right )}{5 \sqrt {10+2 \sqrt {5}}}+\frac {\ln \left (x +1\right )}{5}+\frac {\sqrt {5}\, \ln \left (2 x^{2}-\sqrt {5}\, x -x +2\right )}{20}-\frac {\ln \left (2 x^{2}-\sqrt {5}\, x -x +2\right )}{20}-\frac {\sqrt {5}\, \ln \left (2 x^{2}+\sqrt {5}\, x -x +2\right )}{20}-\frac {\ln \left (2 x^{2}+\sqrt {5}\, x -x +2\right )}{20}-\frac {1}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(x^5+1),x)

[Out]

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

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maxima [A]  time = 2.29, size = 129, normalized size = 0.67 \[ \frac {2 \, \sqrt {5} \arctan \left (\frac {4 \, x + \sqrt {5} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {2 \, \sqrt {5} + 10}} - \frac {2 \, \sqrt {5} \arctan \left (\frac {4 \, x - \sqrt {5} - 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {-2 \, \sqrt {5} + 10}} + \frac {\log \left (2 \, x^{2} - x {\left (\sqrt {5} + 1\right )} + 2\right )}{5 \, {\left (\sqrt {5} + 1\right )}} - \frac {\log \left (2 \, x^{2} + x {\left (\sqrt {5} - 1\right )} + 2\right )}{5 \, {\left (\sqrt {5} - 1\right )}} - \frac {1}{3 \, x^{3}} + \frac {1}{5} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 207, normalized size = 1.08 \[ \frac {\ln \left (x+1\right )}{5}-\ln \left (x-125\,{\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}-\frac {\sqrt {5}}{20}+\frac {1}{20}\right )}^3\right )\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}-\frac {\sqrt {5}}{20}+\frac {1}{20}\right )+\ln \left (x+125\,{\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}+\frac {\sqrt {5}}{20}-\frac {1}{20}\right )}^3\right )\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{20}+\frac {\sqrt {5}}{20}-\frac {1}{20}\right )-\ln \left (x-125\,{\left (\frac {\sqrt {5}}{20}-\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )}^3\right )\,\left (\frac {\sqrt {5}}{20}-\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )-\ln \left (x-125\,{\left (\frac {\sqrt {5}}{20}+\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )}^3\right )\,\left (\frac {\sqrt {5}}{20}+\frac {\sqrt {2}\,\sqrt {\sqrt {5}-5}}{20}+\frac {1}{20}\right )-\frac {1}{3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(x^5 + 1)),x)

[Out]

log(x + 1)/5 - log(x - 125*((2^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 - 5^(1/2)/20 + 1/20)^3)*((2^(1/2)*(- 5^(1/2) -
5)^(1/2))/20 - 5^(1/2)/20 + 1/20) + log(x + 125*((2^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 + 5^(1/2)/20 - 1/20)^3)*((
2^(1/2)*(- 5^(1/2) - 5)^(1/2))/20 + 5^(1/2)/20 - 1/20) - log(x - 125*(5^(1/2)/20 - (2^(1/2)*(5^(1/2) - 5)^(1/2
))/20 + 1/20)^3)*(5^(1/2)/20 - (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 + 1/20) - log(x - 125*(5^(1/2)/20 + (2^(1/2)*(
5^(1/2) - 5)^(1/2))/20 + 1/20)^3)*(5^(1/2)/20 + (2^(1/2)*(5^(1/2) - 5)^(1/2))/20 + 1/20) - 1/(3*x^3)

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sympy [A]  time = 3.06, size = 42, normalized size = 0.22 \[ \frac {\log {\left (x + 1 \right )}}{5} + \operatorname {RootSum} {\left (625 t^{4} + 125 t^{3} + 25 t^{2} + 5 t + 1, \left (t \mapsto t \log {\left (125 t^{3} + x \right )} \right )\right )} - \frac {1}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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