\(\int \frac {e^{3 x+9 e^{4+x} x} (2+6 x+3 x^2+e^{4+x} (18 x+27 x^2+9 x^3))}{4+4 x+x^2} \, dx\) [4162]

Optimal result

Integrand size = 55, antiderivative size = 23 \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=3+\frac {e^{3 \left (x+3 e^{4+x} x\right )} x}{2+x} \]

[Out]

3+exp(9*x*exp(4)*exp(x)+3*x)/(2+x)*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(23)=46\).

Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {27, 2326} \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\frac {e^{9 e^{x+4} x+3 x} \left (x^2+3 e^{x+4} \left (x^3+3 x^2+2 x\right )+2 x\right )}{(x+2)^2 \left (3 e^{x+4} x+3 e^{x+4}+1\right )} \]

[In]

Int[(E^(3*x + 9*E^(4 + x)*x)*(2 + 6*x + 3*x^2 + E^(4 + x)*(18*x + 27*x^2 + 9*x^3)))/(4 + 4*x + x^2),x]

[Out]

(E^(3*x + 9*E^(4 + x)*x)*(2*x + x^2 + 3*E^(4 + x)*(2*x + 3*x^2 + x^3)))/((2 + x)^2*(1 + 3*E^(4 + x) + 3*E^(4 +
x)*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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{(2+x)^2} \, dx \\ & = \frac {e^{3 x+9 e^{4+x} x} \left (2 x+x^2+3 e^{4+x} \left (2 x+3 x^2+x^3\right )\right )}{(2+x)^2 \left (1+3 e^{4+x}+3 e^{4+x} x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\frac {e^{3 x+9 e^{4+x} x} x}{2+x} \]

[In]

Integrate[(E^(3*x + 9*E^(4 + x)*x)*(2 + 6*x + 3*x^2 + E^(4 + x)*(18*x + 27*x^2 + 9*x^3)))/(4 + 4*x + x^2),x]

[Out]

(E^(3*x + 9*E^(4 + x)*x)*x)/(2 + x)

Maple [A] (verified)

Time = 3.62 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
norman	\(\frac {{\mathrm e}^{9 x \,{\mathrm e}^{4} {\mathrm e}^{x}+3 x} x}{2+x}\)	\(20\)
risch	\(\frac {{\mathrm e}^{3 x \left (3 \,{\mathrm e}^{4+x}+1\right )} x}{2+x}\)	\(20\)
parallelrisch	\(\frac {{\mathrm e}^{9 x \,{\mathrm e}^{4} {\mathrm e}^{x}+3 x} x}{2+x}\)	\(20\)

[In]

int(((9*x^3+27*x^2+18*x)*exp(4)*exp(x)+3*x^2+6*x+2)*exp(9*x*exp(4)*exp(x)+3*x)/(x^2+4*x+4),x,method=_RETURNVER
BOSE)

[Out]

exp(9*x*exp(4)*exp(x)+3*x)/(2+x)*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\frac {x e^{\left (9 \, x e^{\left (x + 4\right )} + 3 \, x\right )}}{x + 2} \]

[In]

integrate(((9*x^3+27*x^2+18*x)*exp(4)*exp(x)+3*x^2+6*x+2)*exp(9*x*exp(4)*exp(x)+3*x)/(x^2+4*x+4),x, algorithm=
"fricas")

[Out]

x*e^(9*x*e^(x + 4) + 3*x)/(x + 2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\frac {x e^{9 x e^{4} e^{x} + 3 x}}{x + 2} \]

[In]

integrate(((9*x**3+27*x**2+18*x)*exp(4)*exp(x)+3*x**2+6*x+2)*exp(9*x*exp(4)*exp(x)+3*x)/(x**2+4*x+4),x)

[Out]

x*exp(9*x*exp(4)*exp(x) + 3*x)/(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\frac {x e^{\left (9 \, x e^{\left (x + 4\right )} + 3 \, x\right )}}{x + 2} \]

[In]

integrate(((9*x^3+27*x^2+18*x)*exp(4)*exp(x)+3*x^2+6*x+2)*exp(9*x*exp(4)*exp(x)+3*x)/(x^2+4*x+4),x, algorithm=
"maxima")

[Out]

x*e^(9*x*e^(x + 4) + 3*x)/(x + 2)

Giac [F(-1)]

Timed out. \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\text {Timed out} \]

[In]

integrate(((9*x^3+27*x^2+18*x)*exp(4)*exp(x)+3*x^2+6*x+2)*exp(9*x*exp(4)*exp(x)+3*x)/(x^2+4*x+4),x, algorithm=
"giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.85 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3 x+9 e^{4+x} x} \left (2+6 x+3 x^2+e^{4+x} \left (18 x+27 x^2+9 x^3\right )\right )}{4+4 x+x^2} \, dx=\frac {x\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{9\,x\,{\mathrm {e}}^4\,{\mathrm {e}}^x}}{x+2} \]

[In]

int((exp(3*x + 9*x*exp(4)*exp(x))*(6*x + 3*x^2 + exp(4)*exp(x)*(18*x + 27*x^2 + 9*x^3) + 2))/(4*x + x^2 + 4),x
)

[Out]

(x*exp(3*x)*exp(9*x*exp(4)*exp(x)))/(x + 2)