\(\int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx\) [280]

Optimal result

Integrand size = 19, antiderivative size = 17 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=4 \log (x)-\frac {9}{10} \log \left (2-x^{10}\right ) \]

[Out]

4*ln(x)-9/10*ln(-x^10+2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1607, 457, 78} \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=4 \log (x)-\frac {9}{10} \log \left (2-x^{10}\right ) \]

[In]

Int[(8 + 5*x^10)/(2*x - x^11),x]

[Out]

4*Log[x] - (9*Log[2 - x^10])/10

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {8+5 x^{10}}{x \left (2-x^{10}\right )} \, dx \\ & = \frac {1}{10} \text {Subst}\left (\int \frac {8+5 x}{(2-x) x} \, dx,x,x^{10}\right ) \\ & = \frac {1}{10} \text {Subst}\left (\int \left (-\frac {9}{-2+x}+\frac {4}{x}\right ) \, dx,x,x^{10}\right ) \\ & = 4 \log (x)-\frac {9}{10} \log \left (2-x^{10}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=4 \log (x)-\frac {9}{10} \log \left (2-x^{10}\right ) \]

[In]

Integrate[(8 + 5*x^10)/(2*x - x^11),x]

[Out]

4*Log[x] - (9*Log[2 - x^10])/10

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
default	\(4 \ln \left (x \right )-\frac {9 \ln \left (x^{10}-2\right )}{10}\)	\(14\)
norman	\(4 \ln \left (x \right )-\frac {9 \ln \left (x^{10}-2\right )}{10}\)	\(14\)
risch	\(4 \ln \left (x \right )-\frac {9 \ln \left (x^{10}-2\right )}{10}\)	\(14\)
parallelrisch	\(4 \ln \left (x \right )-\frac {9 \ln \left (x^{10}-2\right )}{10}\)	\(14\)
meijerg	\(4 \ln \left (x \right )-\frac {2 \ln \left (2\right )}{5}+\frac {2 i \pi }{5}-\frac {9 \ln \left (1-\frac {x^{10}}{2}\right )}{10}\)	\(24\)

[In]

int((5*x^10+8)/(-x^11+2*x),x,method=_RETURNVERBOSE)

[Out]

4*ln(x)-9/10*ln(x^10-2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=-\frac {9}{10} \, \log \left (x^{10} - 2\right ) + 4 \, \log \left (x\right ) \]

[In]

integrate((5*x^10+8)/(-x^11+2*x),x, algorithm="fricas")

[Out]

-9/10*log(x^10 - 2) + 4*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=4 \log {\left (x \right )} - \frac {9 \log {\left (x^{10} - 2 \right )}}{10} \]

[In]

integrate((5*x**10+8)/(-x**11+2*x),x)

[Out]

4*log(x) - 9*log(x**10 - 2)/10

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=-\frac {9}{10} \, \log \left (x^{10} - 2\right ) + 4 \, \log \left (x\right ) \]

[In]

integrate((5*x^10+8)/(-x^11+2*x),x, algorithm="maxima")

[Out]

-9/10*log(x^10 - 2) + 4*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=\frac {2}{5} \, \log \left (x^{10}\right ) - \frac {9}{10} \, \log \left ({\left | x^{10} - 2 \right |}\right ) \]

[In]

integrate((5*x^10+8)/(-x^11+2*x),x, algorithm="giac")

[Out]

2/5*log(x^10) - 9/10*log(abs(x^10 - 2))

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {8+5 x^{10}}{2 x-x^{11}} \, dx=4\,\ln \left (x\right )-\frac {9\,\ln \left (x^{10}-2\right )}{10} \]

[In]

int((5*x^10 + 8)/(2*x - x^11),x)

[Out]

4*log(x) - (9*log(x^10 - 2))/10