3.117 \(\int \frac {x}{(1+x^2) (2+x^2) (3+x^2) (4+x^2)} \, dx\)

Optimal. Leaf size=41 \[ \frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6694, 180} \[ \frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 + x^2)*(2 + x^2)*(3 + x^2)*(4 + x^2)),x]

[Out]

Log[1 + x^2]/12 - Log[2 + x^2]/4 + Log[3 + x^2]/4 - Log[4 + x^2]/12

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 6694

Int[(u_)*((c_.) + (d_.)*(v_))^(n_.)*((e_.) + (f_.)*(w_))^(p_.)*((a_.) + (b_.)*(y_))^(m_.)*((g_.) + (h_.)*(z_))
^(q_.), x_Symbol] :> With[{r = DerivativeDivides[y, u, x]}, Dist[r, Subst[Int[(a + b*x)^m*(c + d*x)^n*(e + f*x
)^p*(g + h*x)^q, x], x, y], x] /;  !FalseQ[r]] /; FreeQ[{a, b, c, d, e, f, g, h, m, n, p, q}, x] && EqQ[v, y]
&& EqQ[w, y] && EqQ[z, y]

Rubi steps

\begin {align*} \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1+x) (2+x) (3+x) (4+x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{6 (1+x)}-\frac {1}{2 (2+x)}+\frac {1}{2 (3+x)}-\frac {1}{6 (4+x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{12} \log \left (1+x^2\right )-\frac {1}{4} \log \left (2+x^2\right )+\frac {1}{4} \log \left (3+x^2\right )-\frac {1}{12} \log \left (4+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 1.00 \[ \frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/((1 + x^2)*(2 + x^2)*(3 + x^2)*(4 + x^2)),x]

[Out]

Log[1 + x^2]/12 - Log[2 + x^2]/4 + Log[3 + x^2]/4 - Log[4 + x^2]/12

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IntegrateAlgebraic [A]  time = 0.01, size = 29, normalized size = 0.71 \[ \frac {1}{2} \tanh ^{-1}\left (2 x^2+5\right )-\frac {1}{6} \tanh ^{-1}\left (\frac {2 x^2}{3}+\frac {5}{3}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/((1 + x^2)*(2 + x^2)*(3 + x^2)*(4 + x^2)),x]

[Out]

-1/6*ArcTanh[5/3 + (2*x^2)/3] + ArcTanh[5 + 2*x^2]/2

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fricas [A]  time = 0.86, size = 33, normalized size = 0.80 \[ -\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+1)/(x^2+2)/(x^2+3)/(x^2+4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.97, size = 33, normalized size = 0.80 \[ -\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+1)/(x^2+2)/(x^2+3)/(x^2+4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.28, size = 34, normalized size = 0.83

method	result	size
default	\(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\)	\(34\)
norman	\(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\)	\(34\)
risch	\(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2+1)/(x^2+2)/(x^2+3)/(x^2+4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 33, normalized size = 0.80 \[ -\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+1)/(x^2+2)/(x^2+3)/(x^2+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 33, normalized size = 0.80 \[ \frac {\mathrm {atanh}\left (\frac {3072}{5\,\left (1280\,x^2+3072\right )}-\frac {1}{5}\right )}{2}-\frac {\mathrm {atanh}\left (\frac {1024}{405\,\left (\frac {640\,x^2}{243}+\frac {1024}{243}\right )}-\frac {3}{5}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((x^2 + 1)*(x^2 + 2)*(x^2 + 3)*(x^2 + 4)),x)

[Out]

atanh(3072/(5*(1280*x^2 + 3072)) - 1/5)/2 - atanh(1024/(405*((640*x^2)/243 + 1024/243)) - 3/5)/6

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sympy [A]  time = 0.19, size = 32, normalized size = 0.78 \[ \frac {\log {\left (x^{2} + 1 \right )}}{12} - \frac {\log {\left (x^{2} + 2 \right )}}{4} + \frac {\log {\left (x^{2} + 3 \right )}}{4} - \frac {\log {\left (x^{2} + 4 \right )}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**2+1)/(x**2+2)/(x**2+3)/(x**2+4),x)

[Out]

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

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