\(\int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx\) [3200]

Optimal result

Integrand size = 23, antiderivative size = 24 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=2+4 (-1+2 x-\log (3))-\frac {3+\log (3)}{2+x} \]

[Out]

-2+8*x-4*ln(3)-(3+ln(3))/(2+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {27, 697} \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=8 x-\frac {3+\log (3)}{x+2} \]

[In]

Int[(35 + 32*x + 8*x^2 + Log[3])/(4 + 4*x + x^2),x]

[Out]

8*x - (3 + Log[3])/(2 + x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {35+32 x+8 x^2+\log (3)}{(2+x)^2} \, dx \\ & = \int \left (8+\frac {3+\log (3)}{(2+x)^2}\right ) \, dx \\ & = 8 x-\frac {3+\log (3)}{2+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=8 (2+x)+\frac {-3-\log (3)}{2+x} \]

[In]

Integrate[(35 + 32*x + 8*x^2 + Log[3])/(4 + 4*x + x^2),x]

[Out]

8*(2 + x) + (-3 - Log[3])/(2 + x)

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
default	\(8 x -\frac {3+\ln \left (3\right )}{2+x}\)	\(16\)
gosper	\(-\frac {-8 x^{2}+\ln \left (3\right )+35}{2+x}\)	\(17\)
parallelrisch	\(-\frac {-8 x^{2}+\ln \left (3\right )+35}{2+x}\)	\(17\)
norman	\(\frac {8 x^{2}-35-\ln \left (3\right )}{2+x}\)	\(18\)
risch	\(8 x -\frac {3}{2+x}-\frac {\ln \left (3\right )}{2+x}\)	\(21\)
meijerg	\(-\frac {29 x}{4 \left (1+\frac {x}{2}\right )}+\frac {x \ln \left (3\right )}{4+2 x}+\frac {8 x \left (\frac {3 x}{2}+6\right )}{3 \left (1+\frac {x}{2}\right )}\)	\(39\)

[In]

int((ln(3)+8*x^2+32*x+35)/(x^2+4*x+4),x,method=_RETURNVERBOSE)

[Out]

8*x-(3+ln(3))/(2+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=\frac {8 \, x^{2} + 16 \, x - \log \left (3\right ) - 3}{x + 2} \]

[In]

integrate((log(3)+8*x^2+32*x+35)/(x^2+4*x+4),x, algorithm="fricas")

[Out]

(8*x^2 + 16*x - log(3) - 3)/(x + 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.50 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=8 x + \frac {-3 - \log {\left (3 \right )}}{x + 2} \]

[In]

integrate((ln(3)+8*x**2+32*x+35)/(x**2+4*x+4),x)

[Out]

8*x + (-3 - log(3))/(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=8 \, x - \frac {\log \left (3\right ) + 3}{x + 2} \]

[In]

integrate((log(3)+8*x^2+32*x+35)/(x^2+4*x+4),x, algorithm="maxima")

[Out]

8*x - (log(3) + 3)/(x + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=8 \, x - \frac {\log \left (3\right ) + 3}{x + 2} \]

[In]

integrate((log(3)+8*x^2+32*x+35)/(x^2+4*x+4),x, algorithm="giac")

[Out]

8*x - (log(3) + 3)/(x + 2)

Mupad [B] (verification not implemented)

Time = 8.92 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx=8\,x-\frac {\ln \left (3\right )+3}{x+2} \]

[In]

int((32*x + log(3) + 8*x^2 + 35)/(4*x + x^2 + 4),x)

[Out]

8*x - (log(3) + 3)/(x + 2)