\(\int \frac {1}{x^3 (1+2 x^4+x^8)} \, dx\) [278]

Optimal result

Integrand size = 16, antiderivative size = 30 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1+x^4\right )}-\frac {3 \arctan \left (x^2\right )}{4} \]

[Out]

-3/4/x^2+1/4/x^2/(x^4+1)-3/4*arctan(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {28, 281, 296, 331, 209} \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {3}{4} \arctan \left (x^2\right )-\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (x^4+1\right )} \]

[In]

Int[1/(x^3*(1 + 2*x^4 + x^8)),x]

[Out]

-3/(4*x^2) + 1/(4*x^2*(1 + x^4)) - (3*ArcTan[x^2])/4

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 281

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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*} \text {integral}& = \int \frac {1}{x^3 \left (1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{4 x^2 \left (1+x^4\right )}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1+x^4\right )}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,x^2\right ) \\ & = -\frac {3}{4 x^2}+\frac {1}{4 x^2 \left (1+x^4\right )}-\frac {3}{4} \tan ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {x^2}{4 \left (1+x^4\right )}+\frac {3}{4} \arctan \left (\frac {1}{x^2}\right ) \]

[In]

Integrate[1/(x^3*(1 + 2*x^4 + x^8)),x]

[Out]

-1/2*1/x^2 - x^2/(4*(1 + x^4)) + (3*ArcTan[x^(-2)])/4

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {1}{2 x^{2}}-\frac {x^{2}}{4 \left (x^{4}+1\right )}-\frac {3 \arctan \left (x^{2}\right )}{4}\)	\(25\)
risch	\(\frac {-\frac {3 x^{4}}{4}-\frac {1}{2}}{x^{2} \left (x^{4}+1\right )}-\frac {3 \arctan \left (x^{2}\right )}{4}\)	\(26\)
parallelrisch	\(\frac {3 i \ln \left (x^{2}-i\right ) x^{6}-3 i \ln \left (x^{2}+i\right ) x^{6}-4+3 i \ln \left (x^{2}-i\right ) x^{2}-3 i \ln \left (x^{2}+i\right ) x^{2}-6 x^{4}}{8 x^{2} \left (x^{4}+1\right )}\)	\(72\)

[In]

int(1/x^3/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/2/x^2-1/4*x^2/(x^4+1)-3/4*arctan(x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {3 \, x^{4} + 3 \, {\left (x^{6} + x^{2}\right )} \arctan \left (x^{2}\right ) + 2}{4 \, {\left (x^{6} + x^{2}\right )}} \]

[In]

integrate(1/x^3/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

-1/4*(3*x^4 + 3*(x^6 + x^2)*arctan(x^2) + 2)/(x^6 + x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 3 x^{4} - 2}{4 x^{6} + 4 x^{2}} - \frac {3 \operatorname {atan}{\left (x^{2} \right )}}{4} \]

[In]

integrate(1/x**3/(x**8+2*x**4+1),x)

[Out]

(-3*x**4 - 2)/(4*x**6 + 4*x**2) - 3*atan(x**2)/4

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {3 \, x^{4} + 2}{4 \, {\left (x^{6} + x^{2}\right )}} - \frac {3}{4} \, \arctan \left (x^{2}\right ) \]

[In]

integrate(1/x^3/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

-1/4*(3*x^4 + 2)/(x^6 + x^2) - 3/4*arctan(x^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {3 \, x^{4} + 2}{4 \, {\left (x^{6} + x^{2}\right )}} - \frac {3}{4} \, \arctan \left (x^{2}\right ) \]

[In]

integrate(1/x^3/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

-1/4*(3*x^4 + 2)/(x^6 + x^2) - 3/4*arctan(x^2)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (1+2 x^4+x^8\right )} \, dx=-\frac {3\,\mathrm {atan}\left (x^2\right )}{4}-\frac {\frac {3\,x^4}{4}+\frac {1}{2}}{x^6+x^2} \]

[In]

int(1/(x^3*(2*x^4 + x^8 + 1)),x)

[Out]

- (3*atan(x^2))/4 - ((3*x^4)/4 + 1/2)/(x^2 + x^6)