\(\int e^{12+16 x+7 x^2+x^3} (64+56 x+12 x^2) \, dx\) [6033]

Optimal result

Integrand size = 26, antiderivative size = 15 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=-36+4 e^{(2+x)^2 (3+x)} \]

[Out]

4*exp((3+x)*(2+x)^2)-36

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6838} \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4 e^{x^3+7 x^2+16 x+12} \]

[In]

Int[E^(12 + 16*x + 7*x^2 + x^3)*(64 + 56*x + 12*x^2),x]

[Out]

4*E^(12 + 16*x + 7*x^2 + x^3)

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}& = 4 e^{12+16 x+7 x^2+x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4 e^{(2+x)^2 (3+x)} \]

[In]

Integrate[E^(12 + 16*x + 7*x^2 + x^3)*(64 + 56*x + 12*x^2),x]

[Out]

4*E^((2 + x)^2*(3 + x))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
risch	\(4 \,{\mathrm e}^{\left (3+x \right ) \left (2+x \right )^{2}}\)	\(13\)
gosper	\(4 \,{\mathrm e}^{x^{3}+7 x^{2}+16 x +12}\)	\(17\)
default	\(4 \,{\mathrm e}^{x^{3}+7 x^{2}+16 x +12}\)	\(17\)
norman	\(4 \,{\mathrm e}^{x^{3}+7 x^{2}+16 x +12}\)	\(17\)
parallelrisch	\(4 \,{\mathrm e}^{x^{3}+7 x^{2}+16 x +12}\)	\(17\)

[In]

int((12*x^2+56*x+64)*exp(x^3+7*x^2+16*x+12),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4 \, e^{\left (x^{3} + 7 \, x^{2} + 16 \, x + 12\right )} \]

[In]

integrate((12*x^2+56*x+64)*exp(x^3+7*x^2+16*x+12),x, algorithm="fricas")

[Out]

4*e^(x^3 + 7*x^2 + 16*x + 12)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4 e^{x^{3} + 7 x^{2} + 16 x + 12} \]

[In]

integrate((12*x**2+56*x+64)*exp(x**3+7*x**2+16*x+12),x)

[Out]

4*exp(x**3 + 7*x**2 + 16*x + 12)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4 \, e^{\left (x^{3} + 7 \, x^{2} + 16 \, x + 12\right )} \]

[In]

integrate((12*x^2+56*x+64)*exp(x^3+7*x^2+16*x+12),x, algorithm="maxima")

[Out]

4*e^(x^3 + 7*x^2 + 16*x + 12)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4 \, e^{\left (x^{3} + 7 \, x^{2} + 16 \, x + 12\right )} \]

[In]

integrate((12*x^2+56*x+64)*exp(x^3+7*x^2+16*x+12),x, algorithm="giac")

[Out]

4*e^(x^3 + 7*x^2 + 16*x + 12)

Mupad [B] (verification not implemented)

Time = 11.44 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int e^{12+16 x+7 x^2+x^3} \left (64+56 x+12 x^2\right ) \, dx=4\,{\mathrm {e}}^{16\,x}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{7\,x^2} \]

[In]

int(exp(16*x + 7*x^2 + x^3 + 12)*(56*x + 12*x^2 + 64),x)

[Out]

4*exp(16*x)*exp(x^3)*exp(12)*exp(7*x^2)