3.175 \(\int \frac{1}{x^3 (a^4+x^4)^3} \, dx\)

Optimal. Leaf size=64 \[ \frac{5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac{15}{16 a^{12} x^2}+\frac{1}{8 a^4 x^2 \left (a^4+x^4\right )^2}-\frac{15 \tan ^{-1}\left (\frac{x^2}{a^2}\right )}{16 a^{14}} \]

[Out]

-15/(16*a^12*x^2) + 1/(8*a^4*x^2*(a^4 + x^4)^2) + 5/(16*a^8*x^2*(a^4 + x^4)) - (15*ArcTan[x^2/a^2])/(16*a^14)

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Rubi [A]  time = 0.0302262, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {275, 290, 325, 203} \[ \frac{5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac{15}{16 a^{12} x^2}+\frac{1}{8 a^4 x^2 \left (a^4+x^4\right )^2}-\frac{15 \tan ^{-1}\left (\frac{x^2}{a^2}\right )}{16 a^{14}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a^4 + x^4)^3),x]

[Out]

-15/(16*a^12*x^2) + 1/(8*a^4*x^2*(a^4 + x^4)^2) + 5/(16*a^8*x^2*(a^4 + x^4)) - (15*ArcTan[x^2/a^2])/(16*a^14)

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 290

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

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]

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])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (a^4+x^4\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a^4+x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac{1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a^4+x^2\right )^2} \, dx,x,x^2\right )}{8 a^4}\\ &=\frac{1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac{5}{16 a^8 x^2 \left (a^4+x^4\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a^4+x^2\right )} \, dx,x,x^2\right )}{16 a^8}\\ &=-\frac{15}{16 a^{12} x^2}+\frac{1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac{5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{a^4+x^2} \, dx,x,x^2\right )}{16 a^{12}}\\ &=-\frac{15}{16 a^{12} x^2}+\frac{1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac{5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac{15 \tan ^{-1}\left (\frac{x^2}{a^2}\right )}{16 a^{14}}\\ \end{align*}

Mathematica [A]  time = 0.0478722, size = 75, normalized size = 1.17 \[ \frac{-\frac{a^2 \left (25 a^4 x^4+8 a^8+15 x^8\right )}{x^2 \left (a^4+x^4\right )^2}+15 \tan ^{-1}\left (1-\frac{\sqrt{2} x}{a}\right )+15 \tan ^{-1}\left (\frac{\sqrt{2} x}{a}+1\right )}{16 a^{14}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a^4 + x^4)^3),x]

[Out]

(-((a^2*(8*a^8 + 25*a^4*x^4 + 15*x^8))/(x^2*(a^4 + x^4)^2)) + 15*ArcTan[1 - (Sqrt[2]*x)/a] + 15*ArcTan[1 + (Sq
rt[2]*x)/a])/(16*a^14)

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Maple [A]  time = 0.013, size = 57, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{a}^{12}{x}^{2}}}-{\frac{9\,{x}^{2}}{16\,{a}^{8} \left ({a}^{4}+{x}^{4} \right ) ^{2}}}-{\frac{7\,{x}^{6}}{16\,{a}^{12} \left ({a}^{4}+{x}^{4} \right ) ^{2}}}-{\frac{15}{16\,{a}^{14}}\arctan \left ({\frac{{x}^{2}}{{a}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a^4+x^4)^3,x)

[Out]

-1/2/a^12/x^2-9/16/a^8/(a^4+x^4)^2*x^2-7/16/a^12/(a^4+x^4)^2*x^6-15/16*arctan(x^2/a^2)/a^14

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Maxima [A]  time = 1.41359, size = 81, normalized size = 1.27 \begin{align*} -\frac{8 \, a^{8} + 25 \, a^{4} x^{4} + 15 \, x^{8}}{16 \,{\left (a^{20} x^{2} + 2 \, a^{16} x^{6} + a^{12} x^{10}\right )}} - \frac{15 \, \arctan \left (\frac{x^{2}}{a^{2}}\right )}{16 \, a^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a^4+x^4)^3,x, algorithm="maxima")

[Out]

-1/16*(8*a^8 + 25*a^4*x^4 + 15*x^8)/(a^20*x^2 + 2*a^16*x^6 + a^12*x^10) - 15/16*arctan(x^2/a^2)/a^14

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Fricas [A]  time = 2.01592, size = 173, normalized size = 2.7 \begin{align*} -\frac{8 \, a^{10} + 25 \, a^{6} x^{4} + 15 \, a^{2} x^{8} + 15 \,{\left (a^{8} x^{2} + 2 \, a^{4} x^{6} + x^{10}\right )} \arctan \left (\frac{x^{2}}{a^{2}}\right )}{16 \,{\left (a^{22} x^{2} + 2 \, a^{18} x^{6} + a^{14} x^{10}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a^4+x^4)^3,x, algorithm="fricas")

[Out]

-1/16*(8*a^10 + 25*a^6*x^4 + 15*a^2*x^8 + 15*(a^8*x^2 + 2*a^4*x^6 + x^10)*arctan(x^2/a^2))/(a^22*x^2 + 2*a^18*
x^6 + a^14*x^10)

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Sympy [C]  time = 6.35304, size = 76, normalized size = 1.19 \begin{align*} - \frac{8 a^{8} + 25 a^{4} x^{4} + 15 x^{8}}{16 a^{20} x^{2} + 32 a^{16} x^{6} + 16 a^{12} x^{10}} + \frac{\frac{15 i \log{\left (- i a^{2} + x^{2} \right )}}{32} - \frac{15 i \log{\left (i a^{2} + x^{2} \right )}}{32}}{a^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a**4+x**4)**3,x)

[Out]

-(8*a**8 + 25*a**4*x**4 + 15*x**8)/(16*a**20*x**2 + 32*a**16*x**6 + 16*a**12*x**10) + (15*I*log(-I*a**2 + x**2
)/32 - 15*I*log(I*a**2 + x**2)/32)/a**14

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Giac [A]  time = 1.06292, size = 68, normalized size = 1.06 \begin{align*} -\frac{9 \, a^{4} x^{2} + 7 \, x^{6}}{16 \,{\left (a^{4} + x^{4}\right )}^{2} a^{12}} - \frac{15 \, \arctan \left (\frac{x^{2}}{a^{2}}\right )}{16 \, a^{14}} - \frac{1}{2 \, a^{12} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a^4+x^4)^3,x, algorithm="giac")

[Out]

-1/16*(9*a^4*x^2 + 7*x^6)/((a^4 + x^4)^2*a^12) - 15/16*arctan(x^2/a^2)/a^14 - 1/2/(a^12*x^2)