3.32.42 $\int \frac{e^{16-4x^2}(4x^3-8x^5-8x^6)}{(1+x{)}^5}dx$

Optimal. Leaf size=18 $$\frac{e^{16-4x^2}{x}^4}{(1+x{)}^4}$$

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Rubi [A]  time = 0.13, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, integrand size = 31, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.065, Rules used = {1594, 2288} $$\begin{array}{c}\frac{e^{16-4x^2}{x}^2(x^3+x^2)}{(x+1{)}^5}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(16 - 4*x^2)*(4*x^3 - 8*x^5 - 8*x^6))/(1 + x)^5,x]

[Out]

(E^(16 - 4*x^2)*x^2*(x^2 + x^3))/(1 + x)^5

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{16-4x^2}{x}^3(4-8x^2-8x^3)}{(1+x{)}^5}dx\\ & =\frac{e^{16-4x^2}{x}^2(x^2+x^3)}{(1+x{)}^5}\end{array}\end{array}$$

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.00 $$\begin{array}{c}\frac{e^{16-4x^2}{x}^4}{(1+x{)}^4}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(16 - 4*x^2)*(4*x^3 - 8*x^5 - 8*x^6))/(1 + x)^5,x]

[Out]

(E^(16 - 4*x^2)*x^4)/(1 + x)^4

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fricas [A]  time = 0.54, size = 18, normalized size = 1.00 $$\begin{array}{c}{x}^4{e}^{(-4x^2-4\log (x+1)+16)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^6-8*x^5+4*x^3)*exp(-log(x+1)-x^2+4)^4/(x+1),x, algorithm="fricas")

[Out]

x^4*e^(-4*x^2 - 4*log(x + 1) + 16)

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giac [A]  time = 0.20, size = 32, normalized size = 1.78 $$\begin{array}{c}\frac{x^4{e}^{(-4x^2+16)}}{x^4+4x^3+6x^2+4x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^6-8*x^5+4*x^3)*exp(-log(x+1)-x^2+4)^4/(x+1),x, algorithm="giac")

[Out]

x^4*e^(-4*x^2 + 16)/(x^4 + 4*x^3 + 6*x^2 + 4*x + 1)

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maple [A]  time = 0.53, size = 19, normalized size = 1.06

method	result	size
risch	$\frac{x^4{\mathrm{e}}^{-4(x-2)(2+x)}}{{(x+1)}^4}$	$19$
gosper	$\frac{x^4{\mathrm{e}}^{-4x^2+16}}{{(x+1)}^4}$	$21$
meijerg	$-\frac{8{\mathrm{e}}^{16}{(x+1)}^{-4+4{\mathrm{e}}^{-4x^2}}\mathrm{hypergeom}([7,1+4{\mathrm{e}}^{-4x^2}],[8],-x)x^7{\mathrm{e}}^{-4x^2}}{7}-\frac{4{\mathrm{e}}^{16}{(x+1)}^{-4+4{\mathrm{e}}^{-4x^2}}\mathrm{hypergeom}([6,1+4{\mathrm{e}}^{-4x^2}],[7],-x)x^6{\mathrm{e}}^{-4x^2}}{3}+{\mathrm{e}}^{16}{(x+1)}^{-4+4{\mathrm{e}}^{-4x^2}}\mathrm{hypergeom}([4,1+4{\mathrm{e}}^{-4x^2}],[5],-x)x^4{\mathrm{e}}^{-4x^2}$	$136$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x^6-8*x^5+4*x^3)*exp(-ln(x+1)-x^2+4)^4/(x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.73, size = 32, normalized size = 1.78 $$\begin{array}{c}\frac{x^4{e}^{(-4x^2+16)}}{x^4+4x^3+6x^2+4x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^6-8*x^5+4*x^3)*exp(-log(x+1)-x^2+4)^4/(x+1),x, algorithm="maxima")

[Out]

x^4*e^(-4*x^2 + 16)/(x^4 + 4*x^3 + 6*x^2 + 4*x + 1)

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mupad [B]  time = 0.15, size = 17, normalized size = 0.94 $$\begin{array}{c}\frac{x^4{\mathrm{e}}^{16}{\mathrm{e}}^{-4x^2}}{{(x+1)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(16 - 4*x^2 - 4*log(x + 1))*(8*x^5 - 4*x^3 + 8*x^6))/(x + 1),x)

[Out]

(x^4*exp(16)*exp(-4*x^2))/(x + 1)^4

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sympy [A]  time = 0.12, size = 29, normalized size = 1.61 $$\begin{array}{c}\frac{x^4{e}^{16-4x^2}}{x^4+4x^3+6x^2+4x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x**6-8*x**5+4*x**3)*exp(-ln(x+1)-x**2+4)**4/(x+1),x)

[Out]

x**4*exp(16 - 4*x**2)/(x**4 + 4*x**3 + 6*x**2 + 4*x + 1)

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