3.13 \(\int \frac{x}{(a^2+x^2) (b^2+x^2)} \, dx\)

Optimal. Leaf size=47 \[ \frac{\log \left (b^2+x^2\right )}{2 \left (a^2-b^2\right )}-\frac{\log \left (a^2+x^2\right )}{2 \left (a^2-b^2\right )} \]

[Out]

-Log[a^2 + x^2]/(2*(a^2 - b^2)) + Log[b^2 + x^2]/(2*(a^2 - b^2))

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Rubi [A]  time = 0.0197179, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {444, 36, 31} \[ \frac{\log \left (b^2+x^2\right )}{2 \left (a^2-b^2\right )}-\frac{\log \left (a^2+x^2\right )}{2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/((a^2 + x^2)*(b^2 + x^2)),x]

[Out]

-Log[a^2 + x^2]/(2*(a^2 - b^2)) + Log[b^2 + x^2]/(2*(a^2 - b^2))

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]

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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0105171, size = 34, normalized size = 0.72 \[ \frac{\log \left (b^2+x^2\right )-\log \left (a^2+x^2\right )}{2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((a^2 + x^2)*(b^2 + x^2)),x]

[Out]

(-Log[a^2 + x^2] + Log[b^2 + x^2])/(2*(a^2 - b^2))

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Maple [A]  time = 0.007, size = 44, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ({a}^{2}+{x}^{2} \right ) }{2\,{a}^{2}-2\,{b}^{2}}}+{\frac{\ln \left ({b}^{2}+{x}^{2} \right ) }{2\,{a}^{2}-2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a^2+x^2)/(b^2+x^2),x)

[Out]

-1/2*ln(a^2+x^2)/(a^2-b^2)+1/2*ln(b^2+x^2)/(a^2-b^2)

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Maxima [A]  time = 0.948073, size = 58, normalized size = 1.23 \begin{align*} -\frac{\log \left (a^{2} + x^{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )}} + \frac{\log \left (b^{2} + x^{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a^2+x^2)/(b^2+x^2),x, algorithm="maxima")

[Out]

-1/2*log(a^2 + x^2)/(a^2 - b^2) + 1/2*log(b^2 + x^2)/(a^2 - b^2)

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Fricas [A]  time = 2.01894, size = 70, normalized size = 1.49 \begin{align*} -\frac{\log \left (a^{2} + x^{2}\right ) - \log \left (b^{2} + x^{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a^2+x^2)/(b^2+x^2),x, algorithm="fricas")

[Out]

-1/2*(log(a^2 + x^2) - log(b^2 + x^2))/(a^2 - b^2)

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Sympy [B]  time = 0.540466, size = 121, normalized size = 2.57 \begin{align*} \frac{\log{\left (- \frac{a^{4}}{2 \left (a - b\right ) \left (a + b\right )} + \frac{a^{2} b^{2}}{\left (a - b\right ) \left (a + b\right )} + \frac{a^{2}}{2} - \frac{b^{4}}{2 \left (a - b\right ) \left (a + b\right )} + \frac{b^{2}}{2} + x^{2} \right )}}{2 \left (a - b\right ) \left (a + b\right )} - \frac{\log{\left (\frac{a^{4}}{2 \left (a - b\right ) \left (a + b\right )} - \frac{a^{2} b^{2}}{\left (a - b\right ) \left (a + b\right )} + \frac{a^{2}}{2} + \frac{b^{4}}{2 \left (a - b\right ) \left (a + b\right )} + \frac{b^{2}}{2} + x^{2} \right )}}{2 \left (a - b\right ) \left (a + b\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a**2+x**2)/(b**2+x**2),x)

[Out]

log(-a**4/(2*(a - b)*(a + b)) + a**2*b**2/((a - b)*(a + b)) + a**2/2 - b**4/(2*(a - b)*(a + b)) + b**2/2 + x**
2)/(2*(a - b)*(a + b)) - log(a**4/(2*(a - b)*(a + b)) - a**2*b**2/((a - b)*(a + b)) + a**2/2 + b**4/(2*(a - b)
*(a + b)) + b**2/2 + x**2)/(2*(a - b)*(a + b))

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Giac [A]  time = 1.14083, size = 88, normalized size = 1.87 \begin{align*} \frac{\log \left (\frac{{\left | a^{2} + b^{2} + 2 \, x^{2} -{\left | a^{2} - b^{2} \right |} \right |}}{a^{2} + b^{2} + 2 \, x^{2} +{\left | a^{2} - b^{2} \right |}}\right )}{2 \,{\left | a^{2} - b^{2} \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a^2+x^2)/(b^2+x^2),x, algorithm="giac")

[Out]

1/2*log(abs(a^2 + b^2 + 2*x^2 - abs(a^2 - b^2))/(a^2 + b^2 + 2*x^2 + abs(a^2 - b^2)))/abs(a^2 - b^2)