\(\int (-e^{16+x}+4 e^{2 x^2} x) \, dx\) [7519]

Optimal result

Integrand size = 18, antiderivative size = 19 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=e^{2 x^2}+e^{16} \left (29-e^x\right ) \]

[Out]

exp(2*x^2)+(29-exp(x))*exp(16)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2225, 2240} \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=e^{2 x^2}-e^{x+16} \]

[In]

Int[-E^(16 + x) + 4*E^(2*x^2)*x,x]

[Out]

E^(2*x^2) - E^(16 + 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]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = 4 \int e^{2 x^2} x \, dx-\int e^{16+x} \, dx \\ & = e^{2 x^2}-e^{16+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=e^{2 x^2}-e^{16+x} \]

[In]

Integrate[-E^(16 + x) + 4*E^(2*x^2)*x,x]

[Out]

E^(2*x^2) - E^(16 + x)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
default	\({\mathrm e}^{2 x^{2}}-{\mathrm e}^{16} {\mathrm e}^{x}\)	\(14\)
norman	\({\mathrm e}^{2 x^{2}}-{\mathrm e}^{16} {\mathrm e}^{x}\)	\(14\)
risch	\({\mathrm e}^{2 x^{2}}-{\mathrm e}^{x +16}\)	\(14\)
parallelrisch	\({\mathrm e}^{2 x^{2}}-{\mathrm e}^{16} {\mathrm e}^{x}\)	\(14\)
parts	\({\mathrm e}^{2 x^{2}}-{\mathrm e}^{16} {\mathrm e}^{x}\)	\(14\)
meijerg	\(-1+{\mathrm e}^{2 x^{2}}+{\mathrm e}^{16} \left (1-{\mathrm e}^{x}\right )\)	\(18\)

[In]

int(4*x*exp(2*x^2)-exp(16)*exp(x),x,method=_RETURNVERBOSE)

[Out]

exp(2*x^2)-exp(16)*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=e^{\left (2 \, x^{2}\right )} - e^{\left (x + 16\right )} \]

[In]

integrate(4*x*exp(2*x^2)-exp(16)*exp(x),x, algorithm="fricas")

[Out]

e^(2*x^2) - e^(x + 16)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=- e^{16} e^{x} + e^{2 x^{2}} \]

[In]

integrate(4*x*exp(2*x**2)-exp(16)*exp(x),x)

[Out]

-exp(16)*exp(x) + exp(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=e^{\left (2 \, x^{2}\right )} - e^{\left (x + 16\right )} \]

[In]

integrate(4*x*exp(2*x^2)-exp(16)*exp(x),x, algorithm="maxima")

[Out]

e^(2*x^2) - e^(x + 16)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx=e^{\left (2 \, x^{2}\right )} - e^{\left (x + 16\right )} \]

[In]

integrate(4*x*exp(2*x^2)-exp(16)*exp(x),x, algorithm="giac")

[Out]

e^(2*x^2) - e^(x + 16)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (-e^{16+x}+4 e^{2 x^2} x\right ) \, dx={\mathrm {e}}^{2\,x^2}-{\mathrm {e}}^{16}\,{\mathrm {e}}^x \]

[In]

int(4*x*exp(2*x^2) - exp(16)*exp(x),x)

[Out]

exp(2*x^2) - exp(16)*exp(x)