3.64.50 \(\int \frac {1}{5} (e^8+e^{4+x} (-2-2 x)-2 x+e^{2 x} (1+2 x)) \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{5} \left (5-x \left (-\left (e^4-e^x\right )^2+x\right )\right ) \]

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Rubi [B]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 2.50, number of steps used = 6, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2176, 2194} \begin {gather*} -\frac {x^2}{5}+\frac {e^8 x}{5}-\frac {e^{2 x}}{10}+\frac {2 e^{x+4}}{5}-\frac {2}{5} e^{x+4} (x+1)+\frac {1}{10} e^{2 x} (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8 + E^(4 + x)*(-2 - 2*x) - 2*x + E^(2*x)*(1 + 2*x))/5,x]

[Out]

-1/10*E^(2*x) + (2*E^(4 + x))/5 + (E^8*x)/5 - x^2/5 - (2*E^(4 + x)*(1 + x))/5 + (E^(2*x)*(1 + 2*x))/10

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{5} \int \left (e^8+e^{4+x} (-2-2 x)-2 x+e^{2 x} (1+2 x)\right ) \, dx\\ &=\frac {e^8 x}{5}-\frac {x^2}{5}+\frac {1}{5} \int e^{4+x} (-2-2 x) \, dx+\frac {1}{5} \int e^{2 x} (1+2 x) \, dx\\ &=\frac {e^8 x}{5}-\frac {x^2}{5}-\frac {2}{5} e^{4+x} (1+x)+\frac {1}{10} e^{2 x} (1+2 x)-\frac {1}{5} \int e^{2 x} \, dx+\frac {2}{5} \int e^{4+x} \, dx\\ &=-\frac {e^{2 x}}{10}+\frac {2 e^{4+x}}{5}+\frac {e^8 x}{5}-\frac {x^2}{5}-\frac {2}{5} e^{4+x} (1+x)+\frac {1}{10} e^{2 x} (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.25 \begin {gather*} \frac {1}{5} \left (e^8 x+e^{2 x} x-2 e^{4+x} x-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8 + E^(4 + x)*(-2 - 2*x) - 2*x + E^(2*x)*(1 + 2*x))/5,x]

[Out]

(E^8*x + E^(2*x)*x - 2*E^(4 + x)*x - x^2)/5

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fricas [A]  time = 0.53, size = 32, normalized size = 1.33 \begin {gather*} -\frac {1}{5} \, {\left (x^{2} e^{8} - x e^{16} - x e^{\left (2 \, x + 8\right )} + 2 \, x e^{\left (x + 12\right )}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x+1)*exp(x)^2+1/5*(-2*x-2)*exp(4)*exp(x)+1/5*exp(4)^2-2/5*x,x, algorithm="fricas")

[Out]

-1/5*(x^2*e^8 - x*e^16 - x*e^(2*x + 8) + 2*x*e^(x + 12))*e^(-8)

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giac [A]  time = 0.22, size = 25, normalized size = 1.04 \begin {gather*} -\frac {1}{5} \, x^{2} + \frac {1}{5} \, x e^{8} + \frac {1}{5} \, x e^{\left (2 \, x\right )} - \frac {2}{5} \, x e^{\left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x+1)*exp(x)^2+1/5*(-2*x-2)*exp(4)*exp(x)+1/5*exp(4)^2-2/5*x,x, algorithm="giac")

[Out]

-1/5*x^2 + 1/5*x*e^8 + 1/5*x*e^(2*x) - 2/5*x*e^(x + 4)

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maple [A]  time = 0.06, size = 26, normalized size = 1.08

method	result	size
risch	\(-\frac {x^{2}}{5}+\frac {x \,{\mathrm e}^{8}}{5}+\frac {x \,{\mathrm e}^{2 x}}{5}-\frac {2 x \,{\mathrm e}^{4+x}}{5}\)	\(26\)
default	\(-\frac {x^{2}}{5}+\frac {x \,{\mathrm e}^{8}}{5}+\frac {x \,{\mathrm e}^{2 x}}{5}-\frac {2 x \,{\mathrm e}^{4} {\mathrm e}^{x}}{5}\)	\(28\)
norman	\(-\frac {x^{2}}{5}+\frac {x \,{\mathrm e}^{8}}{5}+\frac {x \,{\mathrm e}^{2 x}}{5}-\frac {2 x \,{\mathrm e}^{4} {\mathrm e}^{x}}{5}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(2*x+1)*exp(x)^2+1/5*(-2*x-2)*exp(4)*exp(x)+1/5*exp(4)^2-2/5*x,x,method=_RETURNVERBOSE)

[Out]

-1/5*x^2+1/5*x*exp(8)+1/5*x*exp(2*x)-2/5*x*exp(4+x)

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maxima [A]  time = 0.37, size = 25, normalized size = 1.04 \begin {gather*} -\frac {1}{5} \, x^{2} + \frac {1}{5} \, x e^{8} + \frac {1}{5} \, x e^{\left (2 \, x\right )} - \frac {2}{5} \, x e^{\left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x+1)*exp(x)^2+1/5*(-2*x-2)*exp(4)*exp(x)+1/5*exp(4)^2-2/5*x,x, algorithm="maxima")

[Out]

-1/5*x^2 + 1/5*x*e^8 + 1/5*x*e^(2*x) - 2/5*x*e^(x + 4)

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mupad [B]  time = 0.06, size = 21, normalized size = 0.88 \begin {gather*} -\frac {x\,\left (x-{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^{x+4}-{\mathrm {e}}^8\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(8)/5 - (2*x)/5 + (exp(2*x)*(2*x + 1))/5 - (exp(4)*exp(x)*(2*x + 2))/5,x)

[Out]

-(x*(x - exp(2*x) + 2*exp(x + 4) - exp(8)))/5

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sympy [B]  time = 0.13, size = 31, normalized size = 1.29 \begin {gather*} - \frac {x^{2}}{5} + \frac {x e^{2 x}}{5} - \frac {2 x e^{4} e^{x}}{5} + \frac {x e^{8}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*x+1)*exp(x)**2+1/5*(-2*x-2)*exp(4)*exp(x)+1/5*exp(4)**2-2/5*x,x)

[Out]

-x**2/5 + x*exp(2*x)/5 - 2*x*exp(4)*exp(x)/5 + x*exp(8)/5

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