\(\int \frac {e^{16-4 x^2} (4 x^3-8 x^5-8 x^6)}{(1+x)^5} \, dx\) [3142]

Optimal result

Integrand size = 31, antiderivative size = 18 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {e^{16-4 x^2} x^4}{(1+x)^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1608, 2326} \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {e^{16-4 x^2} x^2 \left (x^3+x^2\right )}{(x+1)^5} \]

[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 1608

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 2326

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{align*} \text {integral}& = \int \frac {e^{16-4 x^2} x^3 \left (4-8 x^2-8 x^3\right )}{(1+x)^5} \, dx \\ & = \frac {e^{16-4 x^2} x^2 \left (x^2+x^3\right )}{(1+x)^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {e^{16-4 x^2} x^4}{(1+x)^4} \]

[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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {x^{4} {\mathrm e}^{-4 \left (-2+x \right ) \left (2+x \right )}}{\left (1+x \right )^{4}}\)	\(19\)
gosper	\(\frac {x^{4} {\mathrm e}^{-4 x^{2}+16}}{\left (1+x \right )^{4}}\)	\(21\)
default	\(\frac {x^{4} {\mathrm e}^{-4 x^{2}+16}}{\left (1+x \right )^{4}}\)	\(21\)
parallelrisch	\(\frac {x^{4} {\mathrm e}^{-4 x^{2}+16}}{\left (1+x \right )^{4}}\)	\(21\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=x^{4} e^{\left (-4 \, x^{2} - 4 \, \log \left (x + 1\right ) + 16\right )} \]

[In]

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

[Out]

x^4*e^(-4*x^2 - 4*log(x + 1) + 16)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^{4} e^{16 - 4 x^{2}}}{x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^{4} e^{\left (-4 \, x^{2} + 16\right )}}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^{4} e^{\left (-4 \, x^{2} + 16\right )}}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{16-4 x^2} \left (4 x^3-8 x^5-8 x^6\right )}{(1+x)^5} \, dx=\frac {x^4\,{\mathrm {e}}^{16}\,{\mathrm {e}}^{-4\,x^2}}{{\left (x+1\right )}^4} \]

[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