3.97.73 \(\int (11+24 e^{2 x}+e^x (-48+12 x)) \, dx\)

Optimal. Leaf size=20 \[ -x+12 \left (e^{2 x}+e^x (-5+x)+x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.30, number of steps used = 4, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2194, 2176} \begin {gather*} -12 e^x (4-x)-12 e^x+12 e^{2 x}+11 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[11 + 24*E^(2*x) + E^x*(-48 + 12*x),x]

[Out]

-12*E^x + 12*E^(2*x) - 12*E^x*(4 - x) + 11*x

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=11 x+24 \int e^{2 x} \, dx+\int e^x (-48+12 x) \, dx\\ &=12 e^{2 x}-12 e^x (4-x)+11 x-12 \int e^x \, dx\\ &=-12 e^x+12 e^{2 x}-12 e^x (4-x)+11 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.95 \begin {gather*} 12 e^{2 x}+12 e^x (-5+x)+11 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[11 + 24*E^(2*x) + E^x*(-48 + 12*x),x]

[Out]

12*E^(2*x) + 12*E^x*(-5 + x) + 11*x

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fricas [A]  time = 0.81, size = 17, normalized size = 0.85 \begin {gather*} 12 \, {\left (x - 5\right )} e^{x} + 11 \, x + 12 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(24*exp(x)^2+(12*x-48)*exp(x)+11,x, algorithm="fricas")

[Out]

12*(x - 5)*e^x + 11*x + 12*e^(2*x)

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giac [A]  time = 0.13, size = 17, normalized size = 0.85 \begin {gather*} 12 \, {\left (x - 5\right )} e^{x} + 11 \, x + 12 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(24*exp(x)^2+(12*x-48)*exp(x)+11,x, algorithm="giac")

[Out]

12*(x - 5)*e^x + 11*x + 12*e^(2*x)

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maple [A]  time = 0.02, size = 19, normalized size = 0.95

method	result	size
risch	\(12 \,{\mathrm e}^{2 x}+\left (12 x -60\right ) {\mathrm e}^{x}+11 x\)	\(19\)
default	\(11 x +12 \,{\mathrm e}^{x} x -60 \,{\mathrm e}^{x}+12 \,{\mathrm e}^{2 x}\)	\(20\)
norman	\(11 x +12 \,{\mathrm e}^{x} x -60 \,{\mathrm e}^{x}+12 \,{\mathrm e}^{2 x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(24*exp(x)^2+(12*x-48)*exp(x)+11,x,method=_RETURNVERBOSE)

[Out]

12*exp(2*x)+(12*x-60)*exp(x)+11*x

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maxima [A]  time = 0.45, size = 21, normalized size = 1.05 \begin {gather*} 12 \, {\left (x - 1\right )} e^{x} + 11 \, x + 12 \, e^{\left (2 \, x\right )} - 48 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(24*exp(x)^2+(12*x-48)*exp(x)+11,x, algorithm="maxima")

[Out]

12*(x - 1)*e^x + 11*x + 12*e^(2*x) - 48*e^x

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mupad [B]  time = 7.44, size = 19, normalized size = 0.95 \begin {gather*} 11\,x+12\,{\mathrm {e}}^{2\,x}-60\,{\mathrm {e}}^x+12\,x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(24*exp(2*x) + exp(x)*(12*x - 48) + 11,x)

[Out]

11*x + 12*exp(2*x) - 60*exp(x) + 12*x*exp(x)

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sympy [A]  time = 0.09, size = 17, normalized size = 0.85 \begin {gather*} 11 x + \left (12 x - 60\right ) e^{x} + 12 e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(24*exp(x)**2+(12*x-48)*exp(x)+11,x)

[Out]

11*x + (12*x - 60)*exp(x) + 12*exp(2*x)

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