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

Optimal. Leaf size=19 \[ e^{2 x^2}+e^{16} \left (29-e^x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2194, 2209} \begin {gather*} e^{2 x^2}-e^{x+16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-E^(16 + x) + 4*E^(2*x^2)*x,x]

[Out]

E^(2*x^2) - E^(16 + x)

Rule 2194

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 2209

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.79 \begin {gather*} e^{2 x^2}-e^{16+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*x^2) - E^(16 + x)

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fricas [A]  time = 1.02, size = 13, normalized size = 0.68 \begin {gather*} e^{\left (2 \, x^{2}\right )} - e^{\left (x + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x^2) - e^(x + 16)

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giac [A]  time = 0.13, size = 13, normalized size = 0.68 \begin {gather*} e^{\left (2 \, x^{2}\right )} - e^{\left (x + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x^2) - e^(x + 16)

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maple [A]  time = 0.03, size = 14, normalized size = 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	\({\mathrm e}^{2 x^{2}}-{\mathrm e}^{x +16}\)	\(14\)
meijerg	\(-1+{\mathrm e}^{2 x^{2}}+{\mathrm e}^{16} \left (1-{\mathrm e}^{x}\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 13, normalized size = 0.68 \begin {gather*} e^{\left (2 \, x^{2}\right )} - e^{\left (x + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x^2) - e^(x + 16)

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mupad [B]  time = 4.53, size = 13, normalized size = 0.68 \begin {gather*} {\mathrm {e}}^{2\,x^2}-{\mathrm {e}}^{16}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*x^2) - exp(16)*exp(x)

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sympy [A]  time = 0.13, size = 12, normalized size = 0.63 \begin {gather*} - e^{16} e^{x} + e^{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(16)*exp(x) + exp(2*x**2)

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