\(\int \frac {1}{x^7 (1+2 x^4+x^8)} \, dx\) [280]

Optimal result

Integrand size = 16, antiderivative size = 37 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=-\frac {5}{12 x^6}+\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1+x^4\right )}+\frac {5 \arctan \left (x^2\right )}{4} \]

[Out]

-5/12/x^6+5/4/x^2+1/4/x^6/(x^4+1)+5/4*arctan(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {28, 281, 296, 331, 209} \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=\frac {5 \arctan \left (x^2\right )}{4}-\frac {5}{12 x^6}+\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (x^4+1\right )} \]

[In]

Int[1/(x^7*(1 + 2*x^4 + x^8)),x]

[Out]

-5/(12*x^6) + 5/(4*x^2) + 1/(4*x^6*(1 + x^4)) + (5*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^7 \left (1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^4 \left (1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{4 x^6 \left (1+x^4\right )}+\frac {5}{4} \text {Subst}\left (\int \frac {1}{x^4 \left (1+x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {5}{12 x^6}+\frac {1}{4 x^6 \left (1+x^4\right )}-\frac {5}{4} \text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {5}{12 x^6}+\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1+x^4\right )}+\frac {5}{4} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,x^2\right ) \\ & = -\frac {5}{12 x^6}+\frac {5}{4 x^2}+\frac {1}{4 x^6 \left (1+x^4\right )}+\frac {5}{4} \tan ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}+\frac {1}{x^2}+\frac {x^2}{4 \left (1+x^4\right )}-\frac {5}{4} \arctan \left (\frac {1}{x^2}\right ) \]

[In]

Integrate[1/(x^7*(1 + 2*x^4 + x^8)),x]

[Out]

-1/6*1/x^6 + x^(-2) + x^2/(4*(1 + x^4)) - (5*ArcTan[x^(-2)])/4

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {1}{6 x^{6}}+\frac {1}{x^{2}}+\frac {x^{2}}{4 x^{4}+4}+\frac {5 \arctan \left (x^{2}\right )}{4}\)	\(28\)
risch	\(\frac {\frac {5}{4} x^{8}+\frac {5}{6} x^{4}-\frac {1}{6}}{x^{6} \left (x^{4}+1\right )}+\frac {5 \arctan \left (x^{2}\right )}{4}\)	\(31\)
parallelrisch	\(-\frac {15 i \ln \left (x^{2}-i\right ) x^{10}-15 i \ln \left (x^{2}+i\right ) x^{10}+4+15 i \ln \left (x^{2}-i\right ) x^{6}-15 i \ln \left (x^{2}+i\right ) x^{6}-30 x^{8}-20 x^{4}}{24 x^{6} \left (x^{4}+1\right )}\)	\(77\)

[In]

int(1/x^7/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/6/x^6+1/x^2+1/4*x^2/(x^4+1)+5/4*arctan(x^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=\frac {15 \, x^{8} + 10 \, x^{4} + 15 \, {\left (x^{10} + x^{6}\right )} \arctan \left (x^{2}\right ) - 2}{12 \, {\left (x^{10} + x^{6}\right )}} \]

[In]

integrate(1/x^7/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

1/12*(15*x^8 + 10*x^4 + 15*(x^10 + x^6)*arctan(x^2) - 2)/(x^10 + x^6)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=\frac {5 \operatorname {atan}{\left (x^{2} \right )}}{4} + \frac {15 x^{8} + 10 x^{4} - 2}{12 x^{10} + 12 x^{6}} \]

[In]

integrate(1/x**7/(x**8+2*x**4+1),x)

[Out]

5*atan(x**2)/4 + (15*x**8 + 10*x**4 - 2)/(12*x**10 + 12*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=\frac {15 \, x^{8} + 10 \, x^{4} - 2}{12 \, {\left (x^{10} + x^{6}\right )}} + \frac {5}{4} \, \arctan \left (x^{2}\right ) \]

[In]

integrate(1/x^7/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

1/12*(15*x^8 + 10*x^4 - 2)/(x^10 + x^6) + 5/4*arctan(x^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=\frac {x^{2}}{4 \, {\left (x^{4} + 1\right )}} + \frac {6 \, x^{4} - 1}{6 \, x^{6}} + \frac {5}{4} \, \arctan \left (x^{2}\right ) \]

[In]

integrate(1/x^7/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

1/4*x^2/(x^4 + 1) + 1/6*(6*x^4 - 1)/x^6 + 5/4*arctan(x^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^7 \left (1+2 x^4+x^8\right )} \, dx=\frac {5\,\mathrm {atan}\left (x^2\right )}{4}+\frac {\frac {5\,x^8}{4}+\frac {5\,x^4}{6}-\frac {1}{6}}{x^6\,\left (x^4+1\right )} \]

[In]

int(1/(x^7*(2*x^4 + x^8 + 1)),x)

[Out]

(5*atan(x^2))/4 + ((5*x^4)/6 + (5*x^8)/4 - 1/6)/(x^6*(x^4 + 1))