3.1.14 \(\int \frac {x^2}{(a^2+x^2) (b^2+x^2)} \, dx\) [14]

Optimal. Leaf size=40 \[ \frac {a \tan ^{-1}\left (\frac {x}{a}\right )}{a^2-b^2}-\frac {b \tan ^{-1}\left (\frac {x}{b}\right )}{a^2-b^2} \]

[Out]

a*arctan(x/a)/(a^2-b^2)-b*arctan(x/b)/(a^2-b^2)

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {492, 209} \begin {gather*} \frac {a \text {ArcTan}\left (\frac {x}{a}\right )}{a^2-b^2}-\frac {b \text {ArcTan}\left (\frac {x}{b}\right )}{a^2-b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((a^2 + x^2)*(b^2 + x^2)),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 492

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(-a)*(e^n/(b*c -
a*d)), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[c*(e^n/(b*c - a*d)), Int[(e*x)^(m - n)/(c + d*x^n), x], x
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 30, normalized size = 0.75 \begin {gather*} \frac {a \tan ^{-1}\left (\frac {x}{a}\right )-b \tan ^{-1}\left (\frac {x}{b}\right )}{a^2-b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 41, normalized size = 1.02

method	result	size
default	\(\frac {a \arctan \left (\frac {x}{a}\right )}{a^{2}-b^{2}}-\frac {b \arctan \left (\frac {x}{b}\right )}{a^{2}-b^{2}}\)	\(41\)
risch	\(\frac {a \arctan \left (\frac {x}{a}\right )}{a^{2}-b^{2}}-\frac {b \arctan \left (\frac {x}{b}\right )}{a^{2}-b^{2}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a^2+x^2)/(b^2+x^2),x,method=_RETURNVERBOSE)

[Out]

a*arctan(x/a)/(a^2-b^2)-b*arctan(x/b)/(a^2-b^2)

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Maxima [A]
time = 2.41, size = 40, normalized size = 1.00 \begin {gather*} \frac {a \arctan \left (\frac {x}{a}\right )}{a^{2} - b^{2}} - \frac {b \arctan \left (\frac {x}{b}\right )}{a^{2} - b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*arctan(x/a)/(a^2 - b^2) - b*arctan(x/b)/(a^2 - b^2)

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Fricas [A]
time = 0.73, size = 30, normalized size = 0.75 \begin {gather*} \frac {a \arctan \left (\frac {x}{a}\right ) - b \arctan \left (\frac {x}{b}\right )}{a^{2} - b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*arctan(x/a) - b*arctan(x/b))/(a^2 - b^2)

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Sympy [C] Result contains complex when optimal does not.
time = 0.67, size = 393, normalized size = 9.82 \begin {gather*} - \frac {i a \log {\left (- \frac {2 i a^{7}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} + \frac {4 i a^{5} b^{2}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} - \frac {2 i a^{3} b^{4}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} + \frac {i a^{3}}{\left (a - b\right ) \left (a + b\right )} + \frac {i a b^{2}}{\left (a - b\right ) \left (a + b\right )} + x \right )}}{2 \left (a - b\right ) \left (a + b\right )} + \frac {i a \log {\left (\frac {2 i a^{7}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} - \frac {4 i a^{5} b^{2}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} + \frac {2 i a^{3} b^{4}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} - \frac {i a^{3}}{\left (a - b\right ) \left (a + b\right )} - \frac {i a b^{2}}{\left (a - b\right ) \left (a + b\right )} + x \right )}}{2 \left (a - b\right ) \left (a + b\right )} - \frac {i b \log {\left (- \frac {2 i a^{4} b^{3}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} + \frac {4 i a^{2} b^{5}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} + \frac {i a^{2} b}{\left (a - b\right ) \left (a + b\right )} - \frac {2 i b^{7}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} + \frac {i b^{3}}{\left (a - b\right ) \left (a + b\right )} + x \right )}}{2 \left (a - b\right ) \left (a + b\right )} + \frac {i b \log {\left (\frac {2 i a^{4} b^{3}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} - \frac {4 i a^{2} b^{5}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} - \frac {i a^{2} b}{\left (a - b\right ) \left (a + b\right )} + \frac {2 i b^{7}}{\left (a - b\right )^{3} \left (a + b\right )^{3}} - \frac {i b^{3}}{\left (a - b\right ) \left (a + b\right )} + x \right )}}{2 \left (a - b\right ) \left (a + b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*log(-2*I*a**7/((a - b)**3*(a + b)**3) + 4*I*a**5*b**2/((a - b)**3*(a + b)**3) - 2*I*a**3*b**4/((a - b)**3
*(a + b)**3) + I*a**3/((a - b)*(a + b)) + I*a*b**2/((a - b)*(a + b)) + x)/(2*(a - b)*(a + b)) + I*a*log(2*I*a*
*7/((a - b)**3*(a + b)**3) - 4*I*a**5*b**2/((a - b)**3*(a + b)**3) + 2*I*a**3*b**4/((a - b)**3*(a + b)**3) - I
*a**3/((a - b)*(a + b)) - I*a*b**2/((a - b)*(a + b)) + x)/(2*(a - b)*(a + b)) - I*b*log(-2*I*a**4*b**3/((a - b
)**3*(a + b)**3) + 4*I*a**2*b**5/((a - b)**3*(a + b)**3) + I*a**2*b/((a - b)*(a + b)) - 2*I*b**7/((a - b)**3*(
a + b)**3) + I*b**3/((a - b)*(a + b)) + x)/(2*(a - b)*(a + b)) + I*b*log(2*I*a**4*b**3/((a - b)**3*(a + b)**3)
- 4*I*a**2*b**5/((a - b)**3*(a + b)**3) - I*a**2*b/((a - b)*(a + b)) + 2*I*b**7/((a - b)**3*(a + b)**3) - I*b
**3/((a - b)*(a + b)) + x)/(2*(a - b)*(a + b))

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Giac [A]
time = 0.45, size = 40, normalized size = 1.00 \begin {gather*} \frac {a \arctan \left (\frac {x}{a}\right )}{a^{2} - b^{2}} - \frac {b \arctan \left (\frac {x}{b}\right )}{a^{2} - b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*arctan(x/a)/(a^2 - b^2) - b*arctan(x/b)/(a^2 - b^2)

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Mupad [B]
time = 0.18, size = 191, normalized size = 4.78 \begin {gather*} -\frac {a\,\mathrm {atan}\left (\frac {x\,\left (2\,a^4+2\,b^4\right )-\frac {a^2\,x\,\left (8\,a^6-8\,a^4\,b^2-8\,a^2\,b^4+8\,b^6\right )}{{\left (2\,a^2-2\,b^2\right )}^2}}{a\,b^2\,\left (2\,a^2-2\,b^2\right )}\right )}{a^2-b^2}-\frac {b\,\mathrm {atan}\left (\frac {x\,\left (2\,a^4+2\,b^4\right )-\frac {b^2\,x\,\left (8\,a^6-8\,a^4\,b^2-8\,a^2\,b^4+8\,b^6\right )}{{\left (2\,a^2-2\,b^2\right )}^2}}{a^2\,b\,\left (2\,a^2-2\,b^2\right )}\right )}{a^2-b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a^2 + x^2)*(b^2 + x^2)),x)

[Out]

- (a*atan((x*(2*a^4 + 2*b^4) - (a^2*x*(8*a^6 + 8*b^6 - 8*a^2*b^4 - 8*a^4*b^2))/(2*a^2 - 2*b^2)^2)/(a*b^2*(2*a^
2 - 2*b^2))))/(a^2 - b^2) - (b*atan((x*(2*a^4 + 2*b^4) - (b^2*x*(8*a^6 + 8*b^6 - 8*a^2*b^4 - 8*a^4*b^2))/(2*a^
2 - 2*b^2)^2)/(a^2*b*(2*a^2 - 2*b^2))))/(a^2 - b^2)

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