\(\int \frac {1+x+x^2+x^3}{1-x^4} \, dx\) [167]

Optimal result

Integrand size = 19, antiderivative size = 8 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\log (1-x) \]

[Out]

-ln(1-x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1600, 31} \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\log (1-x) \]

[In]

Int[(1 + x + x^2 + x^3)/(1 - x^4),x]

[Out]

-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 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{1-x} \, dx \\ & = -\log (1-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\log (1-x) \]

[In]

Integrate[(1 + x + x^2 + x^3)/(1 - x^4),x]

[Out]

-Log[1 - x]

Maple [A] (verified)

Time = 1.48 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(-\ln \left (-1+x \right )\)	\(7\)
norman	\(-\ln \left (-1+x \right )\)	\(7\)
risch	\(-\ln \left (-1+x \right )\)	\(7\)
parallelrisch	\(-\ln \left (-1+x \right )\)	\(7\)
meijerg	\(-\frac {\ln \left (-x^{4}+1\right )}{4}-\frac {x^{3} \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {\operatorname {arctanh}\left (x^{2}\right )}{2}-\frac {x \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(94\)

[In]

int((x^3+x^2+x+1)/(-x^4+1),x,method=_RETURNVERBOSE)

[Out]

-ln(-1+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\log \left (x - 1\right ) \]

[In]

integrate((x^3+x^2+x+1)/(-x^4+1),x, algorithm="fricas")

[Out]

-log(x - 1)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=- \log {\left (x - 1 \right )} \]

[In]

integrate((x**3+x**2+x+1)/(-x**4+1),x)

[Out]

-log(x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\log \left (x - 1\right ) \]

[In]

integrate((x^3+x^2+x+1)/(-x^4+1),x, algorithm="maxima")

[Out]

-log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate((x^3+x^2+x+1)/(-x^4+1),x, algorithm="giac")

[Out]

-log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1+x+x^2+x^3}{1-x^4} \, dx=-\ln \left (x-1\right ) \]

[In]

int(-(x + x^2 + x^3 + 1)/(x^4 - 1),x)

[Out]

-log(x - 1)