\(\int \frac {1}{128} (e^{2 x}-1152 x^2) \, dx\) [4820]

Optimal result

Integrand size = 15, antiderivative size = 28 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=2-e^2+\frac {e^{2 x}}{256}-3 x^2 \left (\frac {e}{x^2}+x\right ) \]

[Out]

2+1/256*exp(x)^2-exp(2)-3*x^2*(x+exp(1)/x^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 2225} \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=\frac {e^{2 x}}{256}-3 x^3 \]

[In]

Int[(E^(2*x) - 1152*x^2)/128,x]

[Out]

E^(2*x)/256 - 3*x^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{128} \int \left (e^{2 x}-1152 x^2\right ) \, dx \\ & = -3 x^3+\frac {1}{128} \int e^{2 x} \, dx \\ & = \frac {e^{2 x}}{256}-3 x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=\frac {e^{2 x}}{256}-3 x^3 \]

[In]

Integrate[(E^(2*x) - 1152*x^2)/128,x]

[Out]

E^(2*x)/256 - 3*x^3

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.46

method	result	size
default	\(\frac {{\mathrm e}^{2 x}}{256}-3 x^{3}\)	\(13\)
norman	\(\frac {{\mathrm e}^{2 x}}{256}-3 x^{3}\)	\(13\)
risch	\(\frac {{\mathrm e}^{2 x}}{256}-3 x^{3}\)	\(13\)
parallelrisch	\(\frac {{\mathrm e}^{2 x}}{256}-3 x^{3}\)	\(13\)
parts	\(\frac {{\mathrm e}^{2 x}}{256}-3 x^{3}\)	\(13\)

[In]

int(1/128*exp(x)^2-9*x^2,x,method=_RETURNVERBOSE)

[Out]

1/256*exp(x)^2-3*x^3

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.43 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=-3 \, x^{3} + \frac {1}{256} \, e^{\left (2 \, x\right )} \]

[In]

integrate(1/128*exp(x)^2-9*x^2,x, algorithm="fricas")

[Out]

-3*x^3 + 1/256*e^(2*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.36 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=- 3 x^{3} + \frac {e^{2 x}}{256} \]

[In]

integrate(1/128*exp(x)**2-9*x**2,x)

[Out]

-3*x**3 + exp(2*x)/256

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.43 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=-3 \, x^{3} + \frac {1}{256} \, e^{\left (2 \, x\right )} \]

[In]

integrate(1/128*exp(x)^2-9*x^2,x, algorithm="maxima")

[Out]

-3*x^3 + 1/256*e^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.43 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=-3 \, x^{3} + \frac {1}{256} \, e^{\left (2 \, x\right )} \]

[In]

integrate(1/128*exp(x)^2-9*x^2,x, algorithm="giac")

[Out]

-3*x^3 + 1/256*e^(2*x)

Mupad [B] (verification not implemented)

Time = 10.84 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.43 \[ \int \frac {1}{128} \left (e^{2 x}-1152 x^2\right ) \, dx=\frac {{\mathrm {e}}^{2\,x}}{256}-3\,x^3 \]

[In]

int(exp(2*x)/128 - 9*x^2,x)

[Out]

exp(2*x)/256 - 3*x^3