\(\int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx\) [6488]

Optimal result

Integrand size = 27, antiderivative size = 23 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=x+\log \left (\frac {e^{\frac {4 \left (4+4 x^2\right )}{x}}}{(4+x)^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1607, 1634} \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=17 x+\frac {16}{x}-2 \log (x+4) \]

[In]

Int[(-64 - 16*x + 66*x^2 + 17*x^3)/(4*x^2 + x^3),x]

[Out]

16/x + 17*x - 2*Log[4 + 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 {-64-16 x+66 x^2+17 x^3}{x^2 (4+x)} \, dx \\ & = \int \left (17-\frac {16}{x^2}-\frac {2}{4+x}\right ) \, dx \\ & = \frac {16}{x}+17 x-2 \log (4+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=\frac {16}{x}+17 x-2 \log (4+x) \]

[In]

Integrate[(-64 - 16*x + 66*x^2 + 17*x^3)/(4*x^2 + x^3),x]

[Out]

16/x + 17*x - 2*Log[4 + x]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method	result	size
default	\(17 x +\frac {16}{x}-2 \ln \left (4+x \right )\)	\(16\)
risch	\(17 x +\frac {16}{x}-2 \ln \left (4+x \right )\)	\(16\)
meijerg	\(\frac {16}{x}-2 \ln \left (1+\frac {x}{4}\right )+17 x\)	\(18\)
norman	\(\frac {17 x^{2}+16}{x}-2 \ln \left (4+x \right )\)	\(19\)
parallelrisch	\(-\frac {2 x \ln \left (4+x \right )-17 x^{2}-16}{x}\)	\(20\)

[In]

int((17*x^3+66*x^2-16*x-64)/(x^3+4*x^2),x,method=_RETURNVERBOSE)

[Out]

17*x+16/x-2*ln(4+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=\frac {17 \, x^{2} - 2 \, x \log \left (x + 4\right ) + 16}{x} \]

[In]

integrate((17*x^3+66*x^2-16*x-64)/(x^3+4*x^2),x, algorithm="fricas")

[Out]

(17*x^2 - 2*x*log(x + 4) + 16)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=17 x - 2 \log {\left (x + 4 \right )} + \frac {16}{x} \]

[In]

integrate((17*x**3+66*x**2-16*x-64)/(x**3+4*x**2),x)

[Out]

17*x - 2*log(x + 4) + 16/x

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=17 \, x + \frac {16}{x} - 2 \, \log \left (x + 4\right ) \]

[In]

integrate((17*x^3+66*x^2-16*x-64)/(x^3+4*x^2),x, algorithm="maxima")

[Out]

17*x + 16/x - 2*log(x + 4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=17 \, x + \frac {16}{x} - 2 \, \log \left ({\left | x + 4 \right |}\right ) \]

[In]

integrate((17*x^3+66*x^2-16*x-64)/(x^3+4*x^2),x, algorithm="giac")

[Out]

17*x + 16/x - 2*log(abs(x + 4))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-64-16 x+66 x^2+17 x^3}{4 x^2+x^3} \, dx=17\,x-2\,\ln \left (x+4\right )+\frac {16}{x} \]

[In]

int(-(16*x - 66*x^2 - 17*x^3 + 64)/(4*x^2 + x^3),x)

[Out]

17*x - 2*log(x + 4) + 16/x