3.89.8 \(\int \frac {e^x (e^{5-x}-4 x^2+4 x^3)}{8 x^2 \log (4)} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^x \left (-\frac {e^{5-x}}{4}-2 x+x^2\right )}{2 x \log (4)} \]

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Rubi [A]  time = 0.17, antiderivative size = 42, normalized size of antiderivative = 1.31, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6688, 2176, 2194} \begin {gather*} -\frac {e^x (1-x)}{2 \log (4)}-\frac {e^x}{2 \log (4)}-\frac {e^5}{8 x \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(E^(5 - x) - 4*x^2 + 4*x^3))/(8*x^2*Log[4]),x]

[Out]

-1/2*E^x/Log[4] - (E^x*(1 - x))/(2*Log[4]) - E^5/(8*x*Log[4])

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^x \left (e^{5-x}-4 x^2+4 x^3\right )}{x^2} \, dx}{8 \log (4)}\\ &=\frac {\int \left (4 e^x (-1+x)+\frac {e^5}{x^2}\right ) \, dx}{8 \log (4)}\\ &=-\frac {e^5}{8 x \log (4)}+\frac {\int e^x (-1+x) \, dx}{2 \log (4)}\\ &=-\frac {e^x (1-x)}{2 \log (4)}-\frac {e^5}{8 x \log (4)}-\frac {\int e^x \, dx}{2 \log (4)}\\ &=-\frac {e^x}{2 \log (4)}-\frac {e^x (1-x)}{2 \log (4)}-\frac {e^5}{8 x \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 26, normalized size = 0.81 \begin {gather*} \frac {-\frac {e^5}{x}+e^x (-8+4 x)}{8 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(E^(5 - x) - 4*x^2 + 4*x^3))/(8*x^2*Log[4]),x]

[Out]

(-(E^5/x) + E^x*(-8 + 4*x))/(8*Log[4])

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fricas [A]  time = 0.53, size = 25, normalized size = 0.78 \begin {gather*} \frac {4 \, {\left (x^{2} - 2 \, x\right )} e^{x} - e^{5}}{16 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(exp(5-x)+4*x^3-4*x^2)*exp(x)/x^2/log(2),x, algorithm="fricas")

[Out]

1/16*(4*(x^2 - 2*x)*e^x - e^5)/(x*log(2))

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giac [A]  time = 0.16, size = 26, normalized size = 0.81 \begin {gather*} \frac {4 \, x^{2} e^{x} - 8 \, x e^{x} - e^{5}}{16 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(exp(5-x)+4*x^3-4*x^2)*exp(x)/x^2/log(2),x, algorithm="giac")

[Out]

1/16*(4*x^2*e^x - 8*x*e^x - e^5)/(x*log(2))

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maple [A]  time = 0.06, size = 24, normalized size = 0.75

method	result	size
default	\(\frac {-\frac {{\mathrm e}^{5}}{x}+4 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}}{16 \ln \relax (2)}\)	\(24\)
risch	\(-\frac {{\mathrm e}^{5}}{16 \ln \relax (2) x}+\frac {\left (4 x -8\right ) {\mathrm e}^{x}}{16 \ln \relax (2)}\)	\(26\)
norman	\(\frac {\left (-\frac {x \,{\mathrm e}^{2 x}}{2 \ln \relax (2)}+\frac {x^{2} {\mathrm e}^{2 x}}{4 \ln \relax (2)}-\frac {{\mathrm e}^{5} {\mathrm e}^{x}}{16 \ln \relax (2)}\right ) {\mathrm e}^{-x}}{x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(exp(5-x)+4*x^3-4*x^2)*exp(x)/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/16/ln(2)*(-exp(5)/x+4*exp(x)*x-8*exp(x))

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maxima [A]  time = 0.36, size = 25, normalized size = 0.78 \begin {gather*} \frac {4 \, {\left (x - 1\right )} e^{x} - \frac {e^{5}}{x} - 4 \, e^{x}}{16 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(exp(5-x)+4*x^3-4*x^2)*exp(x)/x^2/log(2),x, algorithm="maxima")

[Out]

1/16*(4*(x - 1)*e^x - e^5/x - 4*e^x)/log(2)

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mupad [B]  time = 0.08, size = 23, normalized size = 0.72 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (x-2\right )}{4\,\ln \relax (2)}-\frac {{\mathrm {e}}^5}{16\,x\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(exp(5 - x) - 4*x^2 + 4*x^3))/(16*x^2*log(2)),x)

[Out]

(exp(x)*(x - 2))/(4*log(2)) - exp(5)/(16*x*log(2))

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sympy [A]  time = 0.13, size = 20, normalized size = 0.62 \begin {gather*} \frac {\left (x - 2\right ) e^{x}}{4 \log {\relax (2 )}} - \frac {e^{5}}{16 x \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(exp(5-x)+4*x**3-4*x**2)*exp(x)/x**2/ln(2),x)

[Out]

(x - 2)*exp(x)/(4*log(2)) - exp(5)/(16*x*log(2))

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