3.405 \(\int \frac {x^3}{2+3 x^4+x^8} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{4} \log \left (x^4+1\right )-\frac {1}{4} \log \left (x^4+2\right ) \]

[Out]

1/4*ln(x^4+1)-1/4*ln(x^4+2)

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1352, 616, 31} \[ \frac {1}{4} \log \left (x^4+1\right )-\frac {1}{4} \log \left (x^4+2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3/(2 + 3*x^4 + x^8),x]

[Out]

Log[1 + x^4]/4 - Log[2 + x^4]/4

Rule 31

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 1352

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int \frac {x^3}{2+3 x^4+x^8} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{2+3 x+x^2} \, dx,x,x^4\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^4\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,x^4\right )\\ &=\frac {1}{4} \log \left (1+x^4\right )-\frac {1}{4} \log \left (2+x^4\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \[ \frac {1}{4} \log \left (x^4+1\right )-\frac {1}{4} \log \left (x^4+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(2 + 3*x^4 + x^8),x]

[Out]

Log[1 + x^4]/4 - Log[2 + x^4]/4

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fricas [A]  time = 0.88, size = 17, normalized size = 0.81 \[ -\frac {1}{4} \, \log \left (x^{4} + 2\right ) + \frac {1}{4} \, \log \left (x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^8+3*x^4+2),x, algorithm="fricas")

[Out]

-1/4*log(x^4 + 2) + 1/4*log(x^4 + 1)

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giac [A]  time = 0.28, size = 17, normalized size = 0.81 \[ -\frac {1}{4} \, \log \left (x^{4} + 2\right ) + \frac {1}{4} \, \log \left (x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^8+3*x^4+2),x, algorithm="giac")

[Out]

-1/4*log(x^4 + 2) + 1/4*log(x^4 + 1)

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maple [A]  time = 0.00, size = 18, normalized size = 0.86 \[ \frac {\ln \left (x^{4}+1\right )}{4}-\frac {\ln \left (x^{4}+2\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^8+3*x^4+2),x)

[Out]

1/4*ln(x^4+1)-1/4*ln(x^4+2)

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maxima [A]  time = 0.59, size = 17, normalized size = 0.81 \[ -\frac {1}{4} \, \log \left (x^{4} + 2\right ) + \frac {1}{4} \, \log \left (x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^8+3*x^4+2),x, algorithm="maxima")

[Out]

-1/4*log(x^4 + 2) + 1/4*log(x^4 + 1)

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mupad [B]  time = 0.06, size = 16, normalized size = 0.76 \[ -\frac {\mathrm {atanh}\left (\frac {256}{9\,\left (144\,x^4+160\right )}-\frac {7}{9}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(3*x^4 + x^8 + 2),x)

[Out]

-atanh(256/(9*(144*x^4 + 160)) - 7/9)/2

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sympy [A]  time = 0.12, size = 15, normalized size = 0.71 \[ \frac {\log {\left (x^{4} + 1 \right )}}{4} - \frac {\log {\left (x^{4} + 2 \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(x**8+3*x**4+2),x)

[Out]

log(x**4 + 1)/4 - log(x**4 + 2)/4

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