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

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

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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} $$\begin{array}{c}-\frac{e^x(1-x)}{2\log (4)}-\frac{e^x}{2\log (4)}-\frac{e^5}{8x\log (4)}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{e^x(e^{5-x}-4x^2+4x^3)}{x^2}dx}{8\log (4)}\\ & =\frac{\int (4e^x(-1+x)+\frac{e^5}{x^2})dx}{8\log (4)}\\ & =-\frac{e^5}{8x\log (4)}+\frac{\int e^x(-1+x)dx}{2\log (4)}\\ & =-\frac{e^x(1-x)}{2\log (4)}-\frac{e^5}{8x\log (4)}-\frac{\int e^{x}dx}{2\log (4)}\\ & =-\frac{e^x}{2\log (4)}-\frac{e^x(1-x)}{2\log (4)}-\frac{e^5}{8x\log (4)}\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{4(x^2-2x)e^x-e^5}{16x\log (2)}\end{array}$$

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 $$\begin{array}{c}\frac{4x^2{e}^x-8x{e}^x-e^5}{16x\log (2)}\end{array}$$

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	$\frac{-\frac{{\mathrm{e}}^5}{x}+4{\mathrm{e}}^{x}x-8{\mathrm{e}}^x}{16\ln (2)}$	$24$
risch	$-\frac{{\mathrm{e}}^5}{16\ln (2)x}+\frac{(4x-8){\mathrm{e}}^x}{16\ln (2)}$	$26$
norman	$\frac{(-\frac{x{\mathrm{e}}^{2x}}{2\ln (2)}+\frac{x^2{\mathrm{e}}^{2x}}{4\ln (2)}-\frac{{\mathrm{e}}^5{\mathrm{e}}^x}{16\ln (2)}){\mathrm{e}}^{-x}}{x}$	$44$

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 $$\begin{array}{c}\frac{4(x-1)e^x-\frac{e^5}{x}-4e^x}{16\log (2)}\end{array}$$

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 $$\begin{array}{c}\frac{{\mathrm{e}}^x(x-2)}{4\ln (2)}-\frac{{\mathrm{e}}^5}{16x\ln (2)}\end{array}$$

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 $$\begin{array}{c}\frac{(x-2)e^x}{4\log (2)}-\frac{e^5}{16x\log (2)}\end{array}$$

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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