\(\int \frac {x^9}{1+x^5} \, dx\) [1296]

Optimal result

Integrand size = 11, antiderivative size = 18 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {x^5}{5}-\frac {1}{5} \log \left (1+x^5\right ) \]

[Out]

1/5*x^5-1/5*ln(x^5+1)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, 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, 45} \[ \int \frac {x^9}{1+x^5} \, dx=\frac {x^5}{5}-\frac {1}{5} \log \left (x^5+1\right ) \]

[In]

Int[x^9/(1 + x^5),x]

[Out]

x^5/5 - Log[1 + x^5]/5

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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}{5} \text {Subst}\left (\int \frac {x}{1+x} \, dx,x,x^5\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,x^5\right ) \\ & = \frac {x^5}{5}-\frac {1}{5} \log \left (1+x^5\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {x^5}{5}-\frac {1}{5} \log \left (1+x^5\right ) \]

[In]

Integrate[x^9/(1 + x^5),x]

[Out]

x^5/5 - Log[1 + x^5]/5

Maple [A] (verified)

Time = 4.35 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {x^{5}}{5}-\frac {\ln \left (x^{5}+1\right )}{5}\)	\(15\)
meijerg	\(\frac {x^{5}}{5}-\frac {\ln \left (x^{5}+1\right )}{5}\)	\(15\)
risch	\(\frac {x^{5}}{5}-\frac {\ln \left (x^{5}+1\right )}{5}\)	\(15\)
norman	\(\frac {x^{5}}{5}-\frac {\ln \left (1+x \right )}{5}-\frac {\ln \left (x^{4}-x^{3}+x^{2}-x +1\right )}{5}\)	\(32\)
parallelrisch	\(\frac {x^{5}}{5}-\frac {\ln \left (1+x \right )}{5}-\frac {\ln \left (x^{4}-x^{3}+x^{2}-x +1\right )}{5}\)	\(32\)

[In]

int(x^9/(x^5+1),x,method=_RETURNVERBOSE)

[Out]

1/5*x^5-1/5*ln(x^5+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {1}{5} \, x^{5} - \frac {1}{5} \, \log \left (x^{5} + 1\right ) \]

[In]

integrate(x^9/(x^5+1),x, algorithm="fricas")

[Out]

1/5*x^5 - 1/5*log(x^5 + 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {x^{5}}{5} - \frac {\log {\left (x^{5} + 1 \right )}}{5} \]

[In]

integrate(x**9/(x**5+1),x)

[Out]

x**5/5 - log(x**5 + 1)/5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {1}{5} \, x^{5} - \frac {1}{5} \, \log \left (x^{5} + 1\right ) \]

[In]

integrate(x^9/(x^5+1),x, algorithm="maxima")

[Out]

1/5*x^5 - 1/5*log(x^5 + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {1}{5} \, x^{5} - \frac {1}{5} \, \log \left ({\left | x^{5} + 1 \right |}\right ) \]

[In]

integrate(x^9/(x^5+1),x, algorithm="giac")

[Out]

1/5*x^5 - 1/5*log(abs(x^5 + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^9}{1+x^5} \, dx=\frac {x^5}{5}-\frac {\ln \left (x^5+1\right )}{5} \]

[In]

int(x^9/(x^5 + 1),x)

[Out]

x^5/5 - log(x^5 + 1)/5