\(\int \frac {2+2 x+x^4}{x^4+x^5} \, dx\) [284]

Optimal result

Integrand size = 18, antiderivative size = 12 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=-\frac {2}{3 x^3}+\log (1+x) \]

[Out]

-2/3/x^3+ln(1+x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1607, 1634} \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=\log (x+1)-\frac {2}{3 x^3} \]

[In]

Int[(2 + 2*x + x^4)/(x^4 + x^5),x]

[Out]

-2/(3*x^3) + 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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=-\frac {2}{3 x^3}+\log (1+x) \]

[In]

Integrate[(2 + 2*x + x^4)/(x^4 + x^5),x]

[Out]

-2/(3*x^3) + Log[1 + x]

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {2}{3 x^{3}}+\ln \left (x +1\right )\)	\(11\)
norman	\(-\frac {2}{3 x^{3}}+\ln \left (x +1\right )\)	\(11\)
meijerg	\(-\frac {2}{3 x^{3}}+\ln \left (x +1\right )\)	\(11\)
risch	\(-\frac {2}{3 x^{3}}+\ln \left (x +1\right )\)	\(11\)
parallelrisch	\(\frac {3 \ln \left (x +1\right ) x^{3}-2}{3 x^{3}}\)	\(17\)

[In]

int((x^4+2*x+2)/(x^5+x^4),x,method=_RETURNVERBOSE)

[Out]

-2/3/x^3+ln(x+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=\frac {3 \, x^{3} \log \left (x + 1\right ) - 2}{3 \, x^{3}} \]

[In]

integrate((x^4+2*x+2)/(x^5+x^4),x, algorithm="fricas")

[Out]

1/3*(3*x^3*log(x + 1) - 2)/x^3

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=\log {\left (x + 1 \right )} - \frac {2}{3 x^{3}} \]

[In]

integrate((x**4+2*x+2)/(x**5+x**4),x)

[Out]

log(x + 1) - 2/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=-\frac {2}{3 \, x^{3}} + \log \left (x + 1\right ) \]

[In]

integrate((x^4+2*x+2)/(x^5+x^4),x, algorithm="maxima")

[Out]

-2/3/x^3 + log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=-\frac {2}{3 \, x^{3}} + \log \left ({\left | x + 1 \right |}\right ) \]

[In]

integrate((x^4+2*x+2)/(x^5+x^4),x, algorithm="giac")

[Out]

-2/3/x^3 + log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {2+2 x+x^4}{x^4+x^5} \, dx=\ln \left (x+1\right )-\frac {2}{3\,x^3} \]

[In]

int((2*x + x^4 + 2)/(x^4 + x^5),x)

[Out]

log(x + 1) - 2/(3*x^3)