\(\int (3+e^{4 x+4 x^2+x^3} (12+24 x+9 x^2)) \, dx\) [6450]

Optimal result

Integrand size = 27, antiderivative size = 20 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=51+3 e^{x+(3+x) \left (x+x^2\right )}+3 x \]

[Out]

3*x+51+3*exp((3+x)*(x^2+x)+x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6838} \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3 e^{x^3+4 x^2+4 x}+3 x \]

[In]

Int[3 + E^(4*x + 4*x^2 + x^3)*(12 + 24*x + 9*x^2),x]

[Out]

3*E^(4*x + 4*x^2 + x^3) + 3*x

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 3 x+\int e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right ) \, dx \\ & = 3 e^{4 x+4 x^2+x^3}+3 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3 e^{x (2+x)^2}+3 x \]

[In]

Integrate[3 + E^(4*x + 4*x^2 + x^3)*(12 + 24*x + 9*x^2),x]

[Out]

3*E^(x*(2 + x)^2) + 3*x

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
risch	\(3 x +3 \,{\mathrm e}^{x \left (2+x \right )^{2}}\)	\(15\)
default	\(3 x +3 \,{\mathrm e}^{x^{3}+4 x^{2}+4 x}\)	\(20\)
norman	\(3 x +3 \,{\mathrm e}^{x^{3}+4 x^{2}+4 x}\)	\(20\)
parallelrisch	\(3 x +3 \,{\mathrm e}^{x^{3}+4 x^{2}+4 x}\)	\(20\)
parts	\(3 x +3 \,{\mathrm e}^{x^{3}+4 x^{2}+4 x}\)	\(20\)

[In]

int((9*x^2+24*x+12)*exp(x^3+4*x^2+4*x)+3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3 \, x + 3 \, e^{\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \]

[In]

integrate((9*x^2+24*x+12)*exp(x^3+4*x^2+4*x)+3,x, algorithm="fricas")

[Out]

3*x + 3*e^(x^3 + 4*x^2 + 4*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3 x + 3 e^{x^{3} + 4 x^{2} + 4 x} \]

[In]

integrate((9*x**2+24*x+12)*exp(x**3+4*x**2+4*x)+3,x)

[Out]

3*x + 3*exp(x**3 + 4*x**2 + 4*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3 \, x + 3 \, e^{\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \]

[In]

integrate((9*x^2+24*x+12)*exp(x^3+4*x^2+4*x)+3,x, algorithm="maxima")

[Out]

3*x + 3*e^(x^3 + 4*x^2 + 4*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3 \, x + 3 \, e^{\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \]

[In]

integrate((9*x^2+24*x+12)*exp(x^3+4*x^2+4*x)+3,x, algorithm="giac")

[Out]

3*x + 3*e^(x^3 + 4*x^2 + 4*x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \left (3+e^{4 x+4 x^2+x^3} \left (12+24 x+9 x^2\right )\right ) \, dx=3\,x+3\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{4\,x^2} \]

[In]

int(exp(4*x + 4*x^2 + x^3)*(24*x + 9*x^2 + 12) + 3,x)

[Out]

3*x + 3*exp(4*x)*exp(x^3)*exp(4*x^2)