\(\int \frac {-1+x^5}{-1+x^2} \, dx\) [291]

Optimal result

Integrand size = 13, antiderivative size = 19 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {x^2}{2}+\frac {x^4}{4}+\log (1+x) \]

[Out]

1/2*x^2+1/4*x^4+ln(1+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1824, 641, 31} \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {x^4}{4}+\frac {x^2}{2}+\log (x+1) \]

[In]

Int[(-1 + x^5)/(-1 + x^2),x]

[Out]

x^2/2 + x^4/4 + Log[1 + x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 1824

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

Rubi steps \begin{align*} \text {integral}& = \int \left (x+x^3-\frac {1-x}{-1+x^2}\right ) \, dx \\ & = \frac {x^2}{2}+\frac {x^4}{4}-\int \frac {1-x}{-1+x^2} \, dx \\ & = \frac {x^2}{2}+\frac {x^4}{4}-\int \frac {1}{-1-x} \, dx \\ & = \frac {x^2}{2}+\frac {x^4}{4}+\log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {x^2}{2}+\frac {x^4}{4}+\log (1+x) \]

[In]

Integrate[(-1 + x^5)/(-1 + x^2),x]

[Out]

x^2/2 + x^4/4 + Log[1 + x]

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {x^{2}}{2}+\frac {x^{4}}{4}+\ln \left (x +1\right )\)	\(16\)
norman	\(\frac {x^{2}}{2}+\frac {x^{4}}{4}+\ln \left (x +1\right )\)	\(16\)
parallelrisch	\(\frac {x^{2}}{2}+\frac {x^{4}}{4}+\ln \left (x +1\right )\)	\(16\)
risch	\(\frac {x^{4}}{4}+\frac {x^{2}}{2}+\frac {1}{4}+\ln \left (x +1\right )\)	\(17\)
meijerg	\(\operatorname {arctanh}\left (x \right )+\frac {x^{2} \left (3 x^{2}+6\right )}{12}+\frac {\ln \left (-x^{2}+1\right )}{2}\)	\(26\)

[In]

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

[Out]

1/2*x^2+1/4*x^4+ln(x+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {1}{4} \, x^{4} + \frac {1}{2} \, x^{2} + \log \left (x + 1\right ) \]

[In]

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

[Out]

1/4*x^4 + 1/2*x^2 + log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {x^{4}}{4} + \frac {x^{2}}{2} + \log {\left (x + 1 \right )} \]

[In]

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

[Out]

x**4/4 + x**2/2 + log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {1}{4} \, x^{4} + \frac {1}{2} \, x^{2} + \log \left (x + 1\right ) \]

[In]

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

[Out]

1/4*x^4 + 1/2*x^2 + log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\frac {1}{4} \, x^{4} + \frac {1}{2} \, x^{2} + \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

1/4*x^4 + 1/2*x^2 + log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 9.67 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-1+x^5}{-1+x^2} \, dx=\ln \left (x+1\right )+\frac {x^2}{2}+\frac {x^4}{4} \]

[In]

int((x^5 - 1)/(x^2 - 1),x)

[Out]

log(x + 1) + x^2/2 + x^4/4