\(\int \frac {1}{x^5 (5+x^2)} \, dx\) [458]

Optimal result

Integrand size = 11, antiderivative size = 31 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=-\frac {1}{20 x^4}+\frac {1}{50 x^2}+\frac {\log (x)}{125}-\frac {1}{250} \log \left (5+x^2\right ) \]

[Out]

-1/20/x^4+1/50/x^2+1/125*ln(x)-1/250*ln(x^2+5)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {272, 46} \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=-\frac {1}{20 x^4}+\frac {1}{50 x^2}-\frac {1}{250} \log \left (x^2+5\right )+\frac {\log (x)}{125} \]

[In]

Int[1/(x^5*(5 + x^2)),x]

[Out]

-1/20*1/x^4 + 1/(50*x^2) + Log[x]/125 - Log[5 + x^2]/250

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^3 (5+x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{5 x^3}-\frac {1}{25 x^2}+\frac {1}{125 x}-\frac {1}{125 (5+x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {1}{20 x^4}+\frac {1}{50 x^2}+\frac {\log (x)}{125}-\frac {1}{250} \log \left (5+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=-\frac {1}{20 x^4}+\frac {1}{50 x^2}+\frac {\log (x)}{125}-\frac {1}{250} \log \left (5+x^2\right ) \]

[In]

Integrate[1/(x^5*(5 + x^2)),x]

[Out]

-1/20*1/x^4 + 1/(50*x^2) + Log[x]/125 - Log[5 + x^2]/250

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {1}{20 x^{4}}+\frac {1}{50 x^{2}}+\frac {\ln \left (x \right )}{125}-\frac {\ln \left (x^{2}+5\right )}{250}\)	\(24\)
norman	\(\frac {-\frac {1}{20}+\frac {x^{2}}{50}}{x^{4}}+\frac {\ln \left (x \right )}{125}-\frac {\ln \left (x^{2}+5\right )}{250}\)	\(25\)
risch	\(\frac {-\frac {1}{20}+\frac {x^{2}}{50}}{x^{4}}+\frac {\ln \left (x \right )}{125}-\frac {\ln \left (x^{2}+5\right )}{250}\)	\(25\)
meijerg	\(-\frac {1}{20 x^{4}}+\frac {1}{50 x^{2}}+\frac {\ln \left (x \right )}{125}-\frac {\ln \left (5\right )}{250}-\frac {\ln \left (1+\frac {x^{2}}{5}\right )}{250}\)	\(30\)
parallelrisch	\(\frac {4 x^{4} \ln \left (x \right )-2 \ln \left (x^{2}+5\right ) x^{4}-25+10 x^{2}}{500 x^{4}}\)	\(31\)

[In]

int(1/x^5/(x^2+5),x,method=_RETURNVERBOSE)

[Out]

-1/20/x^4+1/50/x^2+1/125*ln(x)-1/250*ln(x^2+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=-\frac {2 \, x^{4} \log \left (x^{2} + 5\right ) - 4 \, x^{4} \log \left (x\right ) - 10 \, x^{2} + 25}{500 \, x^{4}} \]

[In]

integrate(1/x^5/(x^2+5),x, algorithm="fricas")

[Out]

-1/500*(2*x^4*log(x^2 + 5) - 4*x^4*log(x) - 10*x^2 + 25)/x^4

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=\frac {\log {\left (x \right )}}{125} - \frac {\log {\left (x^{2} + 5 \right )}}{250} + \frac {2 x^{2} - 5}{100 x^{4}} \]

[In]

integrate(1/x**5/(x**2+5),x)

[Out]

log(x)/125 - log(x**2 + 5)/250 + (2*x**2 - 5)/(100*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=\frac {2 \, x^{2} - 5}{100 \, x^{4}} - \frac {1}{250} \, \log \left (x^{2} + 5\right ) + \frac {1}{250} \, \log \left (x^{2}\right ) \]

[In]

integrate(1/x^5/(x^2+5),x, algorithm="maxima")

[Out]

1/100*(2*x^2 - 5)/x^4 - 1/250*log(x^2 + 5) + 1/250*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=-\frac {3 \, x^{4} - 10 \, x^{2} + 25}{500 \, x^{4}} - \frac {1}{250} \, \log \left (x^{2} + 5\right ) + \frac {1}{250} \, \log \left (x^{2}\right ) \]

[In]

integrate(1/x^5/(x^2+5),x, algorithm="giac")

[Out]

-1/500*(3*x^4 - 10*x^2 + 25)/x^4 - 1/250*log(x^2 + 5) + 1/250*log(x^2)

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^5 \left (5+x^2\right )} \, dx=\frac {\ln \left (x\right )}{125}-\frac {\ln \left (x^2+5\right )}{250}+\frac {\frac {x^2}{50}-\frac {1}{20}}{x^4} \]

[In]

int(1/(x^5*(x^2 + 5)),x)

[Out]

log(x)/125 - log(x^2 + 5)/250 + (x^2/50 - 1/20)/x^4