3.95.7 \(\int (2-15 e^{3 x}+e^{260+e^4-32 x+x^2} (-32+2 x)) \, dx\)

Optimal. Leaf size=26 \[ 8+e^{4+e^4+(16-x)^2}-5 e^{3 x}+2 x \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2194, 2236} \begin {gather*} e^{x^2-32 x+e^4+260}+2 x-5 e^{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2 - 15*E^(3*x) + E^(260 + E^4 - 32*x + x^2)*(-32 + 2*x),x]

[Out]

-5*E^(3*x) + E^(260 + E^4 - 32*x + x^2) + 2*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 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 x-15 \int e^{3 x} \, dx+\int e^{260+e^4-32 x+x^2} (-32+2 x) \, dx\\ &=-5 e^{3 x}+e^{260+e^4-32 x+x^2}+2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 0.92 \begin {gather*} -5 e^{3 x}+e^{260+e^4-32 x+x^2}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2 - 15*E^(3*x) + E^(260 + E^4 - 32*x + x^2)*(-32 + 2*x),x]

[Out]

-5*E^(3*x) + E^(260 + E^4 - 32*x + x^2) + 2*x

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fricas [A]  time = 1.00, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + e^{\left (x^{2} - 32 \, x + e^{4} + 260\right )} - 5 \, e^{\left (3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-32)*exp(exp(4)+x^2-32*x+260)-15*exp(3*x)+2,x, algorithm="fricas")

[Out]

2*x + e^(x^2 - 32*x + e^4 + 260) - 5*e^(3*x)

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giac [A]  time = 0.12, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + e^{\left (x^{2} - 32 \, x + e^{4} + 260\right )} - 5 \, e^{\left (3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-32)*exp(exp(4)+x^2-32*x+260)-15*exp(3*x)+2,x, algorithm="giac")

[Out]

2*x + e^(x^2 - 32*x + e^4 + 260) - 5*e^(3*x)

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maple [A]  time = 0.03, size = 22, normalized size = 0.85

method	result	size
default	\(2 x +{\mathrm e}^{{\mathrm e}^{4}+x^{2}-32 x +260}-5 \,{\mathrm e}^{3 x}\)	\(22\)
norman	\(2 x +{\mathrm e}^{{\mathrm e}^{4}+x^{2}-32 x +260}-5 \,{\mathrm e}^{3 x}\)	\(22\)
risch	\(2 x +{\mathrm e}^{{\mathrm e}^{4}+x^{2}-32 x +260}-5 \,{\mathrm e}^{3 x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x-32)*exp(exp(4)+x^2-32*x+260)-15*exp(3*x)+2,x,method=_RETURNVERBOSE)

[Out]

2*x+exp(exp(4)+x^2-32*x+260)-5*exp(3*x)

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maxima [A]  time = 0.35, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + e^{\left (x^{2} - 32 \, x + e^{4} + 260\right )} - 5 \, e^{\left (3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-32)*exp(exp(4)+x^2-32*x+260)-15*exp(3*x)+2,x, algorithm="maxima")

[Out]

2*x + e^(x^2 - 32*x + e^4 + 260) - 5*e^(3*x)

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mupad [B]  time = 0.19, size = 24, normalized size = 0.92 \begin {gather*} 2\,x-5\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{-32\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{260}\,{\mathrm {e}}^{{\mathrm {e}}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(4) - 32*x + x^2 + 260)*(2*x - 32) - 15*exp(3*x) + 2,x)

[Out]

2*x - 5*exp(3*x) + exp(-32*x)*exp(x^2)*exp(260)*exp(exp(4))

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sympy [A]  time = 0.14, size = 22, normalized size = 0.85 \begin {gather*} 2 x - 5 e^{3 x} + e^{x^{2} - 32 x + e^{4} + 260} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-32)*exp(exp(4)+x**2-32*x+260)-15*exp(3*x)+2,x)

[Out]

2*x - 5*exp(3*x) + exp(x**2 - 32*x + exp(4) + 260)

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