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]

-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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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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 (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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.41, size = 32, normalized size = 0.68 \[ -\frac {\log \left (a^{2} + x^{2}\right ) - \log \left (b^{2} + x^{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}} \]

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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giac [A]  time = 1.05, size = 43, normalized size = 0.91 \[ -\frac {\log \left (a^{2} + x^{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}} + \frac {\log \left (b^{2} + x^{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}} \]

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

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maple [A]  time = 0.01, size = 44, normalized size = 0.94 \[ -\frac {\ln \left (a^{2}+x^{2}\right )}{2 \left (a^{2}-b^{2}\right )}+\frac {\ln \left (b^{2}+x^{2}\right )}{2 a^{2}-2 b^{2}} \]

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 = 0.42, size = 43, normalized size = 0.91 \[ -\frac {\log \left (a^{2} + x^{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}} + \frac {\log \left (b^{2} + x^{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )}} \]

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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mupad [B]  time = 0.30, size = 256, normalized size = 5.45 \[ \frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {x^2\,\left (8\,a^2+8\,b^2\right )+16\,a^2\,b^2}{2\,\left (a^2-b^2\right )}-4\,x^2\right )\,1{}\mathrm {i}}{2\,\left (a^2-b^2\right )}-\frac {\left (\frac {x^2\,\left (8\,a^2+8\,b^2\right )+16\,a^2\,b^2}{2\,\left (a^2-b^2\right )}+4\,x^2\right )\,1{}\mathrm {i}}{2\,\left (a^2-b^2\right )}}{\frac {\frac {x^2\,\left (8\,a^2+8\,b^2\right )+16\,a^2\,b^2}{2\,\left (a^2-b^2\right )}-4\,x^2}{2\,\left (a^2-b^2\right )}+\frac {\frac {x^2\,\left (8\,a^2+8\,b^2\right )+16\,a^2\,b^2}{2\,\left (a^2-b^2\right )}+4\,x^2}{2\,\left (a^2-b^2\right )}}\right )\,1{}\mathrm {i}}{a^2-b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a^2 + x^2)*(b^2 + x^2)),x)

[Out]

(atan(((((x^2*(8*a^2 + 8*b^2) + 16*a^2*b^2)/(2*(a^2 - b^2)) - 4*x^2)*1i)/(2*(a^2 - b^2)) - (((x^2*(8*a^2 + 8*b
^2) + 16*a^2*b^2)/(2*(a^2 - b^2)) + 4*x^2)*1i)/(2*(a^2 - b^2)))/(((x^2*(8*a^2 + 8*b^2) + 16*a^2*b^2)/(2*(a^2 -
b^2)) - 4*x^2)/(2*(a^2 - b^2)) + ((x^2*(8*a^2 + 8*b^2) + 16*a^2*b^2)/(2*(a^2 - b^2)) + 4*x^2)/(2*(a^2 - b^2))
))*1i)/(a^2 - b^2)

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sympy [B]  time = 0.70, size = 121, normalized size = 2.57 \[ \frac {\log {\left (- \frac {a^{4}}{2 \left (a - b\right ) \left (a + b\right )} + \frac {a^{2} b^{2}}{\left (a - b\right ) \left (a + b\right )} + \frac {a^{2}}{2} - \frac {b^{4}}{2 \left (a - b\right ) \left (a + b\right )} + \frac {b^{2}}{2} + x^{2} \right )}}{2 \left (a - b\right ) \left (a + b\right )} - \frac {\log {\left (\frac {a^{4}}{2 \left (a - b\right ) \left (a + b\right )} - \frac {a^{2} b^{2}}{\left (a - b\right ) \left (a + b\right )} + \frac {a^{2}}{2} + \frac {b^{4}}{2 \left (a - b\right ) \left (a + b\right )} + \frac {b^{2}}{2} + x^{2} \right )}}{2 \left (a - b\right ) \left (a + b\right )} \]

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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