\(\int \frac {x^7}{1+x^{12}} \, dx\) [52]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 49 \[ \int \frac {x^7}{1+x^{12}} \, dx=-\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{12} \log \left (1+x^4\right )+\frac {1}{24} \log \left (1-x^4+x^8\right ) \]

[Out]

-1/12*ln(x^4+1)+1/24*ln(x^8-x^4+1)-1/12*arctan(1/3*(-2*x^4+1)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {281, 298, 31, 648, 632, 210, 642} \[ \int \frac {x^7}{1+x^{12}} \, dx=-\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{12} \log \left (x^4+1\right )+\frac {1}{24} \log \left (x^8-x^4+1\right ) \]

[In]

Int[x^7/(1 + x^12),x]

[Out]

-1/4*ArcTan[(1 - 2*x^4)/Sqrt[3]]/Sqrt[3] - Log[1 + x^4]/12 + Log[1 - x^4 + x^8]/24

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, 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}{4} \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,x^4\right ) \\ & = -\left (\frac {1}{12} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^4\right )\right )+\frac {1}{12} \text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,x^4\right ) \\ & = -\frac {1}{12} \log \left (1+x^4\right )+\frac {1}{24} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^4\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^4\right ) \\ & = -\frac {1}{12} \log \left (1+x^4\right )+\frac {1}{24} \log \left (1-x^4+x^8\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^4\right ) \\ & = -\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{12} \log \left (1+x^4\right )+\frac {1}{24} \log \left (1-x^4+x^8\right ) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(260\) vs. \(2(49)=98\).

Time = 0.07 (sec) , antiderivative size = 260, normalized size of antiderivative = 5.31 \[ \int \frac {x^7}{1+x^{12}} \, dx=\frac {1}{24} \left (2 \sqrt {3} \arctan \left (\frac {1+\sqrt {3}-2 \sqrt {2} x}{1-\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1-\sqrt {3}+2 \sqrt {2} x}{1+\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {-1+\sqrt {3}+2 \sqrt {2} x}{1+\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+\sqrt {3}+2 \sqrt {2} x}{-1+\sqrt {3}}\right )-2 \log \left (1-\sqrt {2} x+x^2\right )-2 \log \left (1+\sqrt {2} x+x^2\right )+\log \left (2+\sqrt {2} x-\sqrt {6} x+2 x^2\right )+\log \left (2+\sqrt {2} \left (-1+\sqrt {3}\right ) x+2 x^2\right )+\log \left (2-\left (\sqrt {2}+\sqrt {6}\right ) x+2 x^2\right )+\log \left (2+\left (\sqrt {2}+\sqrt {6}\right ) x+2 x^2\right )\right ) \]

[In]

Integrate[x^7/(1 + x^12),x]

[Out]

(2*Sqrt[3]*ArcTan[(1 + Sqrt[3] - 2*Sqrt[2]*x)/(1 - Sqrt[3])] - 2*Sqrt[3]*ArcTan[(1 - Sqrt[3] + 2*Sqrt[2]*x)/(1
 + Sqrt[3])] + 2*Sqrt[3]*ArcTan[(-1 + Sqrt[3] + 2*Sqrt[2]*x)/(1 + Sqrt[3])] - 2*Sqrt[3]*ArcTan[(1 + Sqrt[3] +
2*Sqrt[2]*x)/(-1 + Sqrt[3])] - 2*Log[1 - Sqrt[2]*x + x^2] - 2*Log[1 + Sqrt[2]*x + x^2] + Log[2 + Sqrt[2]*x - S
qrt[6]*x + 2*x^2] + Log[2 + Sqrt[2]*(-1 + Sqrt[3])*x + 2*x^2] + Log[2 - (Sqrt[2] + Sqrt[6])*x + 2*x^2] + Log[2
 + (Sqrt[2] + Sqrt[6])*x + 2*x^2])/24

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84

method result size
default \(\frac {\ln \left (x^{8}-x^{4}+1\right )}{24}+\frac {\arctan \left (\frac {\left (2 x^{4}-1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (x^{4}+1\right )}{12}\) \(41\)
risch \(\frac {\ln \left (4 x^{8}-4 x^{4}+4\right )}{24}+\frac {\arctan \left (\frac {\left (2 x^{4}-1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (x^{4}+1\right )}{12}\) \(43\)
meijerg \(-\frac {x^{8} \ln \left (1+\left (x^{12}\right )^{\frac {1}{3}}\right )}{12 \left (x^{12}\right )^{\frac {2}{3}}}+\frac {x^{8} \ln \left (1-\left (x^{12}\right )^{\frac {1}{3}}+\left (x^{12}\right )^{\frac {2}{3}}\right )}{24 \left (x^{12}\right )^{\frac {2}{3}}}+\frac {x^{8} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{12}\right )^{\frac {1}{3}}}{2-\left (x^{12}\right )^{\frac {1}{3}}}\right )}{12 \left (x^{12}\right )^{\frac {2}{3}}}\) \(80\)

[In]

int(x^7/(x^12+1),x,method=_RETURNVERBOSE)

[Out]

1/24*ln(x^8-x^4+1)+1/12*arctan(1/3*(2*x^4-1)*3^(1/2))*3^(1/2)-1/12*ln(x^4+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {x^7}{1+x^{12}} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + \frac {1}{24} \, \log \left (x^{8} - x^{4} + 1\right ) - \frac {1}{12} \, \log \left (x^{4} + 1\right ) \]

[In]

integrate(x^7/(x^12+1),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 - 1)) + 1/24*log(x^8 - x^4 + 1) - 1/12*log(x^4 + 1)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \frac {x^7}{1+x^{12}} \, dx=- \frac {\log {\left (x^{4} + 1 \right )}}{12} + \frac {\log {\left (x^{8} - x^{4} + 1 \right )}}{24} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} \]

[In]

integrate(x**7/(x**12+1),x)

[Out]

-log(x**4 + 1)/12 + log(x**8 - x**4 + 1)/24 + sqrt(3)*atan(2*sqrt(3)*x**4/3 - sqrt(3)/3)/12

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {x^7}{1+x^{12}} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + \frac {1}{24} \, \log \left (x^{8} - x^{4} + 1\right ) - \frac {1}{12} \, \log \left (x^{4} + 1\right ) \]

[In]

integrate(x^7/(x^12+1),x, algorithm="maxima")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 - 1)) + 1/24*log(x^8 - x^4 + 1) - 1/12*log(x^4 + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {x^7}{1+x^{12}} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + \frac {1}{24} \, \log \left (x^{8} - x^{4} + 1\right ) - \frac {1}{12} \, \log \left (x^{4} + 1\right ) \]

[In]

integrate(x^7/(x^12+1),x, algorithm="giac")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 - 1)) + 1/24*log(x^8 - x^4 + 1) - 1/12*log(x^4 + 1)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {x^7}{1+x^{12}} \, dx=-\frac {\ln \left (x^4+1\right )}{12}-\ln \left (x^4-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )+\ln \left (x^4+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right ) \]

[In]

int(x^7/(x^12 + 1),x)

[Out]

log((3^(1/2)*1i)/2 + x^4 - 1/2)*((3^(1/2)*1i)/24 + 1/24) - log(x^4 - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/24 -
1/24) - log(x^4 + 1)/12