3.130 \(\int \frac {1}{x^2 (a^4-x^4)} \, dx\)

Optimal. Leaf size=35 \[ -\frac {\tan ^{-1}\left (\frac {x}{a}\right )}{2 a^5}+\frac {\tanh ^{-1}\left (\frac {x}{a}\right )}{2 a^5}-\frac {1}{a^4 x} \]

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {325, 298, 203, 206} \[ -\frac {1}{a^4 x}-\frac {\tan ^{-1}\left (\frac {x}{a}\right )}{2 a^5}+\frac {\tanh ^{-1}\left (\frac {x}{a}\right )}{2 a^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^4 - x^4)),x]

[Out]

-(1/(a^4*x)) - ArcTan[x/a]/(2*a^5) + ArcTanh[x/a]/(2*a^5)

Rule 203

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

Rule 206

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (a^4-x^4\right )} \, dx &=-\frac {1}{a^4 x}+\frac {\int \frac {x^2}{a^4-x^4} \, dx}{a^4}\\ &=-\frac {1}{a^4 x}+\frac {\int \frac {1}{a^2-x^2} \, dx}{2 a^4}-\frac {\int \frac {1}{a^2+x^2} \, dx}{2 a^4}\\ &=-\frac {1}{a^4 x}-\frac {\tan ^{-1}\left (\frac {x}{a}\right )}{2 a^5}+\frac {\tanh ^{-1}\left (\frac {x}{a}\right )}{2 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.31 \[ -\frac {\log (a-x)}{4 a^5}+\frac {\log (a+x)}{4 a^5}-\frac {\tan ^{-1}\left (\frac {x}{a}\right )}{2 a^5}-\frac {1}{a^4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^4 - x^4)),x]

[Out]

-(1/(a^4*x)) - ArcTan[x/a]/(2*a^5) - Log[a - x]/(4*a^5) + Log[a + x]/(4*a^5)

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IntegrateAlgebraic [A]  time = 0.01, size = 35, normalized size = 1.00 \[ -\frac {\tan ^{-1}\left (\frac {x}{a}\right )}{2 a^5}+\frac {\tanh ^{-1}\left (\frac {x}{a}\right )}{2 a^5}-\frac {1}{a^4 x} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(a^4 - x^4)),x]

[Out]

-(1/(a^4*x)) - ArcTan[x/a]/(2*a^5) + ArcTanh[x/a]/(2*a^5)

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fricas [A]  time = 0.97, size = 36, normalized size = 1.03 \[ -\frac {2 \, x \arctan \left (\frac {x}{a}\right ) - x \log \left (a + x\right ) + x \log \left (-a + x\right ) + 4 \, a}{4 \, a^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a^4-x^4),x, algorithm="fricas")

[Out]

-1/4*(2*x*arctan(x/a) - x*log(a + x) + x*log(-a + x) + 4*a)/(a^5*x)

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giac [A]  time = 0.93, size = 42, normalized size = 1.20 \[ -\frac {\arctan \left (\frac {x}{a}\right )}{2 \, a^{5}} + \frac {\log \left ({\left | a + x \right |}\right )}{4 \, a^{5}} - \frac {\log \left ({\left | -a + x \right |}\right )}{4 \, a^{5}} - \frac {1}{a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a^4-x^4),x, algorithm="giac")

[Out]

-1/2*arctan(x/a)/a^5 + 1/4*log(abs(a + x))/a^5 - 1/4*log(abs(-a + x))/a^5 - 1/(a^4*x)

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maple [A]  time = 0.28, size = 41, normalized size = 1.17

method	result	size
default	\(-\frac {\arctan \left (\frac {x}{a}\right )}{2 a^{5}}-\frac {1}{a^{4} x}+\frac {\ln \left (a +x \right )}{4 a^{5}}-\frac {\ln \left (a -x \right )}{4 a^{5}}\)	\(41\)
risch	\(-\frac {1}{a^{4} x}-\frac {\ln \left (-a +x \right )}{4 a^{5}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{10} \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{20}-4\right ) x +a^{16} \textit {\_R}^{3}\right )\right )}{4}+\frac {\ln \left (a +x \right )}{4 a^{5}}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(a^4-x^4),x,method=_RETURNVERBOSE)

[Out]

-1/2*arctan(x/a)/a^5-1/a^4/x+1/4*ln(a+x)/a^5-1/4/a^5*ln(a-x)

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maxima [A]  time = 0.97, size = 40, normalized size = 1.14 \[ -\frac {\arctan \left (\frac {x}{a}\right )}{2 \, a^{5}} + \frac {\log \left (a + x\right )}{4 \, a^{5}} - \frac {\log \left (-a + x\right )}{4 \, a^{5}} - \frac {1}{a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a^4-x^4),x, algorithm="maxima")

[Out]

-1/2*arctan(x/a)/a^5 + 1/4*log(a + x)/a^5 - 1/4*log(-a + x)/a^5 - 1/(a^4*x)

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mupad [B]  time = 0.22, size = 31, normalized size = 0.89 \[ \frac {\mathrm {atanh}\left (\frac {x}{a}\right )}{2\,a^5}-\frac {\mathrm {atan}\left (\frac {x}{a}\right )}{2\,a^5}-\frac {1}{a^4\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a^4 - x^4)),x)

[Out]

atanh(x/a)/(2*a^5) - atan(x/a)/(2*a^5) - 1/(a^4*x)

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sympy [C]  time = 0.20, size = 44, normalized size = 1.26 \[ - \frac {1}{a^{4} x} - \frac {\frac {\log {\left (- a + x \right )}}{4} - \frac {\log {\left (a + x \right )}}{4} - \frac {i \log {\left (- i a + x \right )}}{4} + \frac {i \log {\left (i a + x \right )}}{4}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a**4-x**4),x)

[Out]

-1/(a**4*x) - (log(-a + x)/4 - log(a + x)/4 - I*log(-I*a + x)/4 + I*log(I*a + x)/4)/a**5

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