\(\int \frac {1}{x^9 (1+x^8)} \, dx\) [1499]

Optimal result

Integrand size = 11, antiderivative size = 22 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=-\frac {1}{8 x^8}-\log (x)+\frac {1}{8} \log \left (1+x^8\right ) \]

[Out]

-1/8/x^8-ln(x)+1/8*ln(x^8+1)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, 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} \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=-\frac {1}{8 x^8}+\frac {1}{8} \log \left (x^8+1\right )-\log (x) \]

[In]

Int[1/(x^9*(1 + x^8)),x]

[Out]

-1/8*1/x^8 - Log[x] + Log[1 + x^8]/8

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}{8} \text {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,x^8\right ) \\ & = \frac {1}{8} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^8\right ) \\ & = -\frac {1}{8 x^8}-\log (x)+\frac {1}{8} \log \left (1+x^8\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=-\frac {1}{8 x^8}-\log (x)+\frac {1}{8} \log \left (1+x^8\right ) \]

[In]

Integrate[1/(x^9*(1 + x^8)),x]

[Out]

-1/8*1/x^8 - Log[x] + Log[1 + x^8]/8

Maple [A] (verified)

Time = 3.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {1}{8 x^{8}}-\ln \left (x \right )+\frac {\ln \left (x^{8}+1\right )}{8}\)	\(19\)
norman	\(-\frac {1}{8 x^{8}}-\ln \left (x \right )+\frac {\ln \left (x^{8}+1\right )}{8}\)	\(19\)
meijerg	\(-\frac {1}{8 x^{8}}-\ln \left (x \right )+\frac {\ln \left (x^{8}+1\right )}{8}\)	\(19\)
risch	\(-\frac {1}{8 x^{8}}-\ln \left (x \right )+\frac {\ln \left (x^{8}+1\right )}{8}\)	\(19\)
parallelrisch	\(-\frac {8 \ln \left (x \right ) x^{8}-\ln \left (x^{8}+1\right ) x^{8}+1}{8 x^{8}}\)	\(26\)

[In]

int(1/x^9/(x^8+1),x,method=_RETURNVERBOSE)

[Out]

-1/8/x^8-ln(x)+1/8*ln(x^8+1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=\frac {x^{8} \log \left (x^{8} + 1\right ) - 8 \, x^{8} \log \left (x\right ) - 1}{8 \, x^{8}} \]

[In]

integrate(1/x^9/(x^8+1),x, algorithm="fricas")

[Out]

1/8*(x^8*log(x^8 + 1) - 8*x^8*log(x) - 1)/x^8

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=- \log {\left (x \right )} + \frac {\log {\left (x^{8} + 1 \right )}}{8} - \frac {1}{8 x^{8}} \]

[In]

integrate(1/x**9/(x**8+1),x)

[Out]

-log(x) + log(x**8 + 1)/8 - 1/(8*x**8)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=-\frac {1}{8 \, x^{8}} + \frac {1}{8} \, \log \left (x^{8} + 1\right ) - \frac {1}{8} \, \log \left (x^{8}\right ) \]

[In]

integrate(1/x^9/(x^8+1),x, algorithm="maxima")

[Out]

-1/8/x^8 + 1/8*log(x^8 + 1) - 1/8*log(x^8)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=\frac {x^{8} - 1}{8 \, x^{8}} + \frac {1}{8} \, \log \left (x^{8} + 1\right ) - \frac {1}{8} \, \log \left (x^{8}\right ) \]

[In]

integrate(1/x^9/(x^8+1),x, algorithm="giac")

[Out]

1/8*(x^8 - 1)/x^8 + 1/8*log(x^8 + 1) - 1/8*log(x^8)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^9 \left (1+x^8\right )} \, dx=\frac {\ln \left (x^8+1\right )}{8}-\ln \left (x\right )-\frac {1}{8\,x^8} \]

[In]

int(1/(x^9*(x^8 + 1)),x)

[Out]

log(x^8 + 1)/8 - log(x) - 1/(8*x^8)