3.51 \(\int \frac {1}{A^4-A^2 B^2+(-A^2+B^2) x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {\tanh ^{-1}\left (\frac {x}{A}\right )}{A \left (A^2-B^2\right )} \]

[Out]

arctanh(x/A)/A/(A^2-B^2)

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {208} \[ \frac {\tanh ^{-1}\left (\frac {x}{A}\right )}{A \left (A^2-B^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A^4 - A^2*B^2 + (-A^2 + B^2)*x^2)^(-1),x]

[Out]

ArcTanh[x/A]/(A*(A^2 - B^2))

Rule 208

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

Rubi steps

\begin {align*} \int \frac {1}{A^4-A^2 B^2+\left (-A^2+B^2\right ) x^2} \, dx &=\frac {\tanh ^{-1}\left (\frac {x}{A}\right )}{A \left (A^2-B^2\right )}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[(A^4 - A^2*B^2 + (-A^2 + B^2)*x^2)^(-1),x]

[Out]

ArcTanh[x/A]/(A*(A^2 - B^2))

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fricas [A]  time = 0.40, size = 27, normalized size = 1.29 \[ \frac {\log \left (A + x\right ) - \log \left (-A + x\right )}{2 \, {\left (A^{3} - A B^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(A^4-A^2*B^2+(-A^2+B^2)*x^2),x, algorithm="fricas")

[Out]

1/2*(log(A + x) - log(-A + x))/(A^3 - A*B^2)

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giac [A]  time = 1.14, size = 41, normalized size = 1.95 \[ \frac {\log \left ({\left | A + x \right |}\right )}{2 \, {\left (A^{3} - A B^{2}\right )}} - \frac {\log \left ({\left | -A + x \right |}\right )}{2 \, {\left (A^{3} - A B^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(A^4-A^2*B^2+(-A^2+B^2)*x^2),x, algorithm="giac")

[Out]

1/2*log(abs(A + x))/(A^3 - A*B^2) - 1/2*log(abs(-A + x))/(A^3 - A*B^2)

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maple [B]  time = 0.01, size = 44, normalized size = 2.10 \[ -\frac {\ln \left (A -x \right )}{2 \left (A^{2}-B^{2}\right ) A}+\frac {\ln \left (A +x \right )}{2 \left (A^{2}-B^{2}\right ) A} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(A^4-A^2*B^2+(-A^2+B^2)*x^2),x)

[Out]

1/2/(A^2-B^2)/A*ln(A+x)-1/2/(A^2-B^2)/A*ln(A-x)

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maxima [A]  time = 0.51, size = 39, normalized size = 1.86 \[ \frac {\log \left (A + x\right )}{2 \, {\left (A^{3} - A B^{2}\right )}} - \frac {\log \left (-A + x\right )}{2 \, {\left (A^{3} - A B^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(A^4-A^2*B^2+(-A^2+B^2)*x^2),x, algorithm="maxima")

[Out]

1/2*log(A + x)/(A^3 - A*B^2) - 1/2*log(-A + x)/(A^3 - A*B^2)

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mupad [B]  time = 0.24, size = 21, normalized size = 1.00 \[ -\frac {\mathrm {atanh}\left (\frac {x}{A}\right )}{A\,B^2-A^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^2*(A^2 - B^2) - A^4 + A^2*B^2),x)

[Out]

-atanh(x/A)/(A*B^2 - A^3)

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sympy [B]  time = 0.30, size = 70, normalized size = 3.33 \[ - \frac {\log {\left (- \frac {A^{3}}{\left (A - B\right ) \left (A + B\right )} + \frac {A B^{2}}{\left (A - B\right ) \left (A + B\right )} + x \right )}}{2 A \left (A - B\right ) \left (A + B\right )} + \frac {\log {\left (\frac {A^{3}}{\left (A - B\right ) \left (A + B\right )} - \frac {A B^{2}}{\left (A - B\right ) \left (A + B\right )} + x \right )}}{2 A \left (A - B\right ) \left (A + B\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(A**4-A**2*B**2+(-A**2+B**2)*x**2),x)

[Out]

-log(-A**3/((A - B)*(A + B)) + A*B**2/((A - B)*(A + B)) + x)/(2*A*(A - B)*(A + B)) + log(A**3/((A - B)*(A + B)
) - A*B**2/((A - B)*(A + B)) + x)/(2*A*(A - B)*(A + B))

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