3.472 \(\int \frac {1}{x^7 (1+x^2)^3} \, dx\)

Optimal. Leaf size=52 \[ -\frac {1}{6 x^6}+\frac {3}{4 x^4}-\frac {2}{x^2+1}-\frac {3}{x^2}-\frac {1}{4 \left (x^2+1\right )^2}+5 \log \left (x^2+1\right )-10 \log (x) \]

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Rubi [A]  time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {266, 44} \[ -\frac {2}{x^2+1}-\frac {3}{x^2}-\frac {1}{4 \left (x^2+1\right )^2}+\frac {3}{4 x^4}-\frac {1}{6 x^6}+5 \log \left (x^2+1\right )-10 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 + x^2)^3),x]

[Out]

-1/(6*x^6) + 3/(4*x^4) - 3/x^2 - 1/(4*(1 + x^2)^2) - 2/(1 + x^2) - 10*Log[x] + 5*Log[1 + x^2]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 266

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

Rubi steps

\begin {align*} \int \frac {1}{x^7 \left (1+x^2\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 (1+x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{x^4}-\frac {3}{x^3}+\frac {6}{x^2}-\frac {10}{x}+\frac {1}{(1+x)^3}+\frac {4}{(1+x)^2}+\frac {10}{1+x}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}+\frac {3}{4 x^4}-\frac {3}{x^2}-\frac {1}{4 \left (1+x^2\right )^2}-\frac {2}{1+x^2}-10 \log (x)+5 \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.94 \[ 5 \log \left (x^2+1\right )-\frac {60 x^8+90 x^6+20 x^4-5 x^2+2}{12 x^6 \left (x^2+1\right )^2}-10 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 + x^2)^3),x]

[Out]

-1/12*(2 - 5*x^2 + 20*x^4 + 90*x^6 + 60*x^8)/(x^6*(1 + x^2)^2) - 10*Log[x] + 5*Log[1 + x^2]

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IntegrateAlgebraic [A]  time = 0.03, size = 49, normalized size = 0.94 \[ 5 \log \left (x^2+1\right )+\frac {-60 x^8-90 x^6-20 x^4+5 x^2-2}{12 x^6 \left (x^2+1\right )^2}-10 \log (x) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^7*(1 + x^2)^3),x]

[Out]

(-2 + 5*x^2 - 20*x^4 - 90*x^6 - 60*x^8)/(12*x^6*(1 + x^2)^2) - 10*Log[x] + 5*Log[1 + x^2]

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fricas [A]  time = 1.11, size = 74, normalized size = 1.42 \[ -\frac {60 \, x^{8} + 90 \, x^{6} + 20 \, x^{4} - 5 \, x^{2} - 60 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )} \log \left (x^{2} + 1\right ) + 120 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )} \log \relax (x) + 2}{12 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^2+1)^3,x, algorithm="fricas")

[Out]

-1/12*(60*x^8 + 90*x^6 + 20*x^4 - 5*x^2 - 60*(x^10 + 2*x^8 + x^6)*log(x^2 + 1) + 120*(x^10 + 2*x^8 + x^6)*log(
x) + 2)/(x^10 + 2*x^8 + x^6)

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giac [A]  time = 0.62, size = 58, normalized size = 1.12 \[ -\frac {30 \, x^{4} + 68 \, x^{2} + 39}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {110 \, x^{6} - 36 \, x^{4} + 9 \, x^{2} - 2}{12 \, x^{6}} + 5 \, \log \left (x^{2} + 1\right ) - 5 \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^2+1)^3,x, algorithm="giac")

[Out]

-1/4*(30*x^4 + 68*x^2 + 39)/(x^2 + 1)^2 + 1/12*(110*x^6 - 36*x^4 + 9*x^2 - 2)/x^6 + 5*log(x^2 + 1) - 5*log(x^2
)

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maple [A]  time = 0.33, size = 47, normalized size = 0.90

method	result	size
default	\(-\frac {1}{6 x^{6}}+\frac {3}{4 x^{4}}-\frac {3}{x^{2}}-\frac {1}{4 \left (x^{2}+1\right )^{2}}-\frac {2}{x^{2}+1}-10 \ln \relax (x )+5 \ln \left (x^{2}+1\right )\)	\(47\)
norman	\(\frac {-\frac {1}{6}-5 x^{8}-\frac {15}{2} x^{6}+\frac {5}{12} x^{2}-\frac {5}{3} x^{4}}{x^{6} \left (x^{2}+1\right )^{2}}-10 \ln \relax (x )+5 \ln \left (x^{2}+1\right )\)	\(47\)
risch	\(\frac {-\frac {1}{6}-5 x^{8}-\frac {15}{2} x^{6}+\frac {5}{12} x^{2}-\frac {5}{3} x^{4}}{x^{6} \left (x^{2}+1\right )^{2}}-10 \ln \relax (x )+5 \ln \left (x^{2}+1\right )\)	\(47\)
meijerg	\(\frac {x^{2} \left (9 x^{2}+10\right )}{4 \left (x^{2}+1\right )^{2}}+5 \ln \left (x^{2}+1\right )-\frac {9}{4}-10 \ln \relax (x )-\frac {1}{6 x^{6}}+\frac {3}{4 x^{4}}-\frac {3}{x^{2}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

-1/6/x^6+3/4/x^4-3/x^2-1/4/(x^2+1)^2-2/(x^2+1)-10*ln(x)+5*ln(x^2+1)

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maxima [A]  time = 0.45, size = 53, normalized size = 1.02 \[ -\frac {60 \, x^{8} + 90 \, x^{6} + 20 \, x^{4} - 5 \, x^{2} + 2}{12 \, {\left (x^{10} + 2 \, x^{8} + x^{6}\right )}} + 5 \, \log \left (x^{2} + 1\right ) - 5 \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^2+1)^3,x, algorithm="maxima")

[Out]

-1/12*(60*x^8 + 90*x^6 + 20*x^4 - 5*x^2 + 2)/(x^10 + 2*x^8 + x^6) + 5*log(x^2 + 1) - 5*log(x^2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.98 \[ 5\,\ln \left (x^2+1\right )-10\,\ln \relax (x)-\frac {5\,x^8+\frac {15\,x^6}{2}+\frac {5\,x^4}{3}-\frac {5\,x^2}{12}+\frac {1}{6}}{x^{10}+2\,x^8+x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(x^2 + 1)^3),x)

[Out]

5*log(x^2 + 1) - 10*log(x) - ((5*x^4)/3 - (5*x^2)/12 + (15*x^6)/2 + 5*x^8 + 1/6)/(x^6 + 2*x^8 + x^10)

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sympy [A]  time = 0.17, size = 49, normalized size = 0.94 \[ - 10 \log {\relax (x )} + 5 \log {\left (x^{2} + 1 \right )} + \frac {- 60 x^{8} - 90 x^{6} - 20 x^{4} + 5 x^{2} - 2}{12 x^{10} + 24 x^{8} + 12 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**2+1)**3,x)

[Out]

-10*log(x) + 5*log(x**2 + 1) + (-60*x**8 - 90*x**6 - 20*x**4 + 5*x**2 - 2)/(12*x**10 + 24*x**8 + 12*x**6)

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