\(\int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx\) [7646]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 19 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=e^2+x+x^2+\log \left (3 x^2 (3+2 x)\right ) \]

[Out]

x^2+ln(3*x^2*(3+2*x))+exp(2)+x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1607, 1634} \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x^2+x+2 \log (x)+\log (2 x+3) \]

[In]

Int[(6 + 9*x + 8*x^2 + 4*x^3)/(3*x + 2*x^2),x]

[Out]

x + x^2 + 2*Log[x] + Log[3 + 2*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 {6+9 x+8 x^2+4 x^3}{x (3+2 x)} \, dx \\ & = \int \left (1+\frac {2}{x}+2 x+\frac {2}{3+2 x}\right ) \, dx \\ & = x+x^2+2 \log (x)+\log (3+2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x+x^2+2 \log (x)+\log (3+2 x) \]

[In]

Integrate[(6 + 9*x + 8*x^2 + 4*x^3)/(3*x + 2*x^2),x]

[Out]

x + x^2 + 2*Log[x] + Log[3 + 2*x]

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method result size
parallelrisch \(x^{2}+x +2 \ln \left (x \right )+\ln \left (x +\frac {3}{2}\right )\) \(14\)
default \(x +x^{2}+2 \ln \left (x \right )+\ln \left (3+2 x \right )\) \(16\)
norman \(x +x^{2}+2 \ln \left (x \right )+\ln \left (3+2 x \right )\) \(16\)
risch \(x +x^{2}+2 \ln \left (x \right )+\ln \left (3+2 x \right )\) \(16\)
meijerg \(2 \ln \left (x \right )+2 \ln \left (2\right )-2 \ln \left (3\right )+\ln \left (1+\frac {2 x}{3}\right )-\frac {x \left (6-2 x \right )}{2}+4 x\) \(31\)

[In]

int((4*x^3+8*x^2+9*x+6)/(2*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

x^2+x+2*ln(x)+ln(x+3/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x^{2} + x + \log \left (2 \, x + 3\right ) + 2 \, \log \left (x\right ) \]

[In]

integrate((4*x^3+8*x^2+9*x+6)/(2*x^2+3*x),x, algorithm="fricas")

[Out]

x^2 + x + log(2*x + 3) + 2*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x^{2} + x + 2 \log {\left (x \right )} + \log {\left (x + \frac {3}{2} \right )} \]

[In]

integrate((4*x**3+8*x**2+9*x+6)/(2*x**2+3*x),x)

[Out]

x**2 + x + 2*log(x) + log(x + 3/2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x^{2} + x + \log \left (2 \, x + 3\right ) + 2 \, \log \left (x\right ) \]

[In]

integrate((4*x^3+8*x^2+9*x+6)/(2*x^2+3*x),x, algorithm="maxima")

[Out]

x^2 + x + log(2*x + 3) + 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x^{2} + x + \log \left ({\left | 2 \, x + 3 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((4*x^3+8*x^2+9*x+6)/(2*x^2+3*x),x, algorithm="giac")

[Out]

x^2 + x + log(abs(2*x + 3)) + 2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {6+9 x+8 x^2+4 x^3}{3 x+2 x^2} \, dx=x+\ln \left (x+\frac {3}{2}\right )+2\,\ln \left (x\right )+x^2 \]

[In]

int((9*x + 8*x^2 + 4*x^3 + 6)/(3*x + 2*x^2),x)

[Out]

x + log(x + 3/2) + 2*log(x) + x^2