3.261 \(\int \frac {x}{(1+x^2) (4+x^2)} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{6} \log \left (x^2+1\right )-\frac {1}{6} \log \left (x^2+4\right ) \]

[Out]

1/6*ln(x^2+1)-1/6*ln(x^2+4)

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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 = {444, 36, 31} \[ \frac {1}{6} \log \left (x^2+1\right )-\frac {1}{6} \log \left (x^2+4\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 + x^2)*(4 + x^2)),x]

[Out]

Log[1 + x^2]/6 - Log[4 + x^2]/6

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 444

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

Rubi steps

\begin {align*} \int \frac {x}{\left (1+x^2\right ) \left (4+x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1+x) (4+x)} \, dx,x,x^2\right )\\ &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{4+x} \, dx,x,x^2\right )\\ &=\frac {1}{6} \log \left (1+x^2\right )-\frac {1}{6} \log \left (4+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \[ \frac {1}{6} \log \left (x^2+1\right )-\frac {1}{6} \log \left (x^2+4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/((1 + x^2)*(4 + x^2)),x]

[Out]

Log[1 + x^2]/6 - Log[4 + x^2]/6

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fricas [A]  time = 0.47, size = 17, normalized size = 0.81 \[ -\frac {1}{6} \, \log \left (x^{2} + 4\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+1)/(x^2+4),x, algorithm="fricas")

[Out]

-1/6*log(x^2 + 4) + 1/6*log(x^2 + 1)

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giac [A]  time = 0.28, size = 17, normalized size = 0.81 \[ -\frac {1}{6} \, \log \left (x^{2} + 4\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+1)/(x^2+4),x, algorithm="giac")

[Out]

-1/6*log(x^2 + 4) + 1/6*log(x^2 + 1)

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maple [A]  time = 0.01, size = 18, normalized size = 0.86 \[ \frac {\ln \left (x^{2}+1\right )}{6}-\frac {\ln \left (x^{2}+4\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2+1)/(x^2+4),x)

[Out]

1/6*ln(x^2+1)-1/6*ln(x^2+4)

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maxima [A]  time = 0.95, size = 17, normalized size = 0.81 \[ -\frac {1}{6} \, \log \left (x^{2} + 4\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+1)/(x^2+4),x, algorithm="maxima")

[Out]

-1/6*log(x^2 + 4) + 1/6*log(x^2 + 1)

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mupad [B]  time = 0.15, size = 17, normalized size = 0.81 \[ \frac {\mathrm {atanh}\left (\frac {3\,x^2}{5\,x^2+8}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((x^2 + 1)*(x^2 + 4)),x)

[Out]

atanh((3*x^2)/(5*x^2 + 8))/3

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sympy [A]  time = 0.11, size = 15, normalized size = 0.71 \[ \frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\log {\left (x^{2} + 4 \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**2+1)/(x**2+4),x)

[Out]

log(x**2 + 1)/6 - log(x**2 + 4)/6

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