3.52 \(\int \frac {x^7}{1+x^{12}} \, dx\)

Optimal. Leaf size=49 \[ -\frac {1}{12} \log \left (x^4+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{24} \log \left (x^8-x^4+1\right ) \]

[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)

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Rubi [A]  time = 0.04, antiderivative size = 49, 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 = {275, 292, 31, 634, 618, 204, 628} \[ -\frac {1}{12} \log \left (x^4+1\right )+\frac {1}{24} \log \left (x^8-x^4+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(1 + x^12),x]

[Out]

-ArcTan[(1 - 2*x^4)/Sqrt[3]]/(4*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 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 275

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 292

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 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 {x^7}{1+x^{12}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,x^4\right )\\ &=-\left (\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^4\right )\right )+\frac {1}{12} \operatorname {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} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^4\right )+\frac {1}{8} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^4\right )\\ &=-\frac {\tan ^{-1}\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*}

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Mathematica [B]  time = 0.11, size = 260, normalized size = 5.31 \[ \frac {1}{24} \left (-2 \log \left (x^2-\sqrt {2} x+1\right )-2 \log \left (x^2+\sqrt {2} x+1\right )+\log \left (2 x^2-\sqrt {6} x+\sqrt {2} x+2\right )+\log \left (2 x^2+\sqrt {2} \left (\sqrt {3}-1\right ) x+2\right )+\log \left (2 x^2-\left (\sqrt {2}+\sqrt {6}\right ) x+2\right )+\log \left (2 x^2+\left (\sqrt {2}+\sqrt {6}\right ) x+2\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {-2 \sqrt {2} x+\sqrt {3}+1}{1-\sqrt {3}}\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt {2} x-\sqrt {3}+1}{1+\sqrt {3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt {2} x+\sqrt {3}-1}{1+\sqrt {3}}\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt {2} x+\sqrt {3}+1}{\sqrt {3}-1}\right )\right ) \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.43, size = 40, normalized size = 0.82 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + \frac {1}{24} \, \log \left (x^{8} - x^{4} + 1\right ) - \frac {1}{12} \, \log \left (x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.13, size = 40, normalized size = 0.82 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + \frac {1}{24} \, \log \left (x^{8} - x^{4} + 1\right ) - \frac {1}{12} \, \log \left (x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 41, normalized size = 0.84 \[ \frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{4}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{4}+1\right )}{12}+\frac {\ln \left (x^{8}-x^{4}+1\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(x^12+1),x)

[Out]

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

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maxima [A]  time = 0.96, size = 40, normalized size = 0.82 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) + \frac {1}{24} \, \log \left (x^{8} - x^{4} + 1\right ) - \frac {1}{12} \, \log \left (x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.22, size = 52, normalized size = 1.06 \[ -\frac {\ln \left (x^4+1\right )}{12}-\ln \left (x^4-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )+\ln \left (x^4+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.17, size = 46, normalized size = 0.94 \[ - \frac {\log {\left (x^{4} + 1 \right )}}{12} + \frac {\log {\left (x^{8} - x^{4} + 1 \right )}}{24} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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