\(\int \frac {-1+x^5}{-x+x^3} \, dx\) [372]

Optimal result

Integrand size = 15, antiderivative size = 17 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=x+\frac {x^3}{3}+\log (x)-\log (1+x) \]

[Out]

x+1/3*x^3+ln(x)-ln(1+x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1607, 1816} \[ \int \frac {-1+x^5}{-x+x^3} \, dx=\frac {x^3}{3}+x+\log (x)-\log (x+1) \]

[In]

Int[(-1 + x^5)/(-x + x^3),x]

[Out]

x + x^3/3 + Log[x] - Log[1 + x]

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]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+x^5}{x \left (-1+x^2\right )} \, dx \\ & = \int \left (1+\frac {1}{-1-x}+\frac {1}{x}+x^2\right ) \, dx \\ & = x+\frac {x^3}{3}+\log (x)-\log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=x+\frac {x^3}{3}+\log (x)-\log (1+x) \]

[In]

Integrate[(-1 + x^5)/(-x + x^3),x]

[Out]

x + x^3/3 + Log[x] - Log[1 + x]

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(x +\frac {x^{3}}{3}+\ln \left (x \right )-\ln \left (x +1\right )\)	\(16\)
norman	\(x +\frac {x^{3}}{3}+\ln \left (x \right )-\ln \left (x +1\right )\)	\(16\)
risch	\(x +\frac {x^{3}}{3}+\ln \left (x \right )-\ln \left (x +1\right )\)	\(16\)
parallelrisch	\(x +\frac {x^{3}}{3}+\ln \left (x \right )-\ln \left (x +1\right )\)	\(16\)
meijerg	\(-\frac {\ln \left (-x^{2}+1\right )}{2}+\ln \left (x \right )+\frac {i \pi }{2}+\frac {i \left (-\frac {2 i x \left (5 x^{2}+15\right )}{15}+2 i \operatorname {arctanh}\left (x \right )\right )}{2}\)	\(38\)

[In]

int((x^5-1)/(x^3-x),x,method=_RETURNVERBOSE)

[Out]

x+1/3*x^3+ln(x)-ln(x+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=\frac {1}{3} \, x^{3} + x - \log \left (x + 1\right ) + \log \left (x\right ) \]

[In]

integrate((x^5-1)/(x^3-x),x, algorithm="fricas")

[Out]

1/3*x^3 + x - log(x + 1) + log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=\frac {x^{3}}{3} + x + \log {\left (x \right )} - \log {\left (x + 1 \right )} \]

[In]

integrate((x**5-1)/(x**3-x),x)

[Out]

x**3/3 + x + log(x) - log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=\frac {1}{3} \, x^{3} + x - \log \left (x + 1\right ) + \log \left (x\right ) \]

[In]

integrate((x^5-1)/(x^3-x),x, algorithm="maxima")

[Out]

1/3*x^3 + x - log(x + 1) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=\frac {1}{3} \, x^{3} + x - \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((x^5-1)/(x^3-x),x, algorithm="giac")

[Out]

1/3*x^3 + x - log(abs(x + 1)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^5}{-x+x^3} \, dx=x-2\,\mathrm {atanh}\left (2\,x+1\right )+\frac {x^3}{3} \]

[In]

int(-(x^5 - 1)/(x - x^3),x)

[Out]

x - 2*atanh(2*x + 1) + x^3/3