3.93.13 \(\int (1-e^x+e^{5+x} (3+3 x)) \, dx\)

Optimal. Leaf size=20 \[ 2-e^x+\left (1+3 e^{5+x}\right ) x+\log (5) \]

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.20, 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*} x-e^x-3 e^{x+5}+3 e^{x+5} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - E^x + E^(5 + x)*(3 + 3*x),x]

[Out]

-E^x - 3*E^(5 + x) + x + 3*E^(5 + x)*(1 + 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} &=x-\int e^x \, dx+\int e^{5+x} (3+3 x) \, dx\\ &=-e^x+x+3 e^{5+x} (1+x)-3 \int e^{5+x} \, dx\\ &=-e^x-3 e^{5+x}+x+3 e^{5+x} (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.75 \begin {gather*} -e^x+x+3 e^{5+x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - E^x + E^(5 + x)*(3 + 3*x),x]

[Out]

-E^x + x + 3*E^(5 + x)*x

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fricas [A]  time = 0.90, size = 20, normalized size = 1.00 \begin {gather*} {\left (x e^{5} + {\left (3 \, x e^{5} - 1\right )} e^{\left (x + 5\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(5+x)+1-exp(x),x, algorithm="fricas")

[Out]

(x*e^5 + (3*x*e^5 - 1)*e^(x + 5))*e^(-5)

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giac [A]  time = 0.12, size = 13, normalized size = 0.65 \begin {gather*} 3 \, x e^{\left (x + 5\right )} + x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(5+x)+1-exp(x),x, algorithm="giac")

[Out]

3*x*e^(x + 5) + x - e^x

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maple [A]  time = 0.03, size = 14, normalized size = 0.70

method	result	size
norman	\(x +3 x \,{\mathrm e}^{5} {\mathrm e}^{x}-{\mathrm e}^{x}\)	\(14\)
risch	\(3 x \,{\mathrm e}^{5+x}+x -{\mathrm e}^{x}\)	\(14\)
default	\(x +3 \,{\mathrm e}^{5+x} \left (5+x \right )-15 \,{\mathrm e}^{5+x}-{\mathrm e}^{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+3)*exp(5+x)+1-exp(x),x,method=_RETURNVERBOSE)

[Out]

x+3*x*exp(5)*exp(x)-exp(x)

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maxima [A]  time = 0.37, size = 13, normalized size = 0.65 \begin {gather*} 3 \, x e^{\left (x + 5\right )} + x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(5+x)+1-exp(x),x, algorithm="maxima")

[Out]

3*x*e^(x + 5) + x - e^x

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mupad [B]  time = 0.05, size = 13, normalized size = 0.65 \begin {gather*} x-{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^5\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x + 5)*(3*x + 3) - exp(x) + 1,x)

[Out]

x - exp(x) + 3*x*exp(5)*exp(x)

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sympy [A]  time = 0.14, size = 12, normalized size = 0.60 \begin {gather*} x + \left (3 x e^{5} - 1\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(5+x)+1-exp(x),x)

[Out]

x + (3*x*exp(5) - 1)*exp(x)

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