3.33.94 $\int \frac{x^2+e^{-e^{-4+4x}+x^2}(4+16e^{-4+4x}x-8x^2)}{x^2}dx$

Optimal. Leaf size=22 $$-\frac{4e^{-e^{-4+4x}+x^2}}{x}+x$$

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Rubi [B]  time = 0.17, antiderivative size = 55, normalized size of antiderivative = 2.50, number of steps used = 3, number of rules used = 2, integrand size = 41, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.049, Rules used = {14, 2288} $$\begin{array}{c}x-\frac{4e^{x^2-e^{4x-4}-4}(2e^{4x}x-e^4{x}^2)}{(2e^{4x-4}-x)x^2}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

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

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Mathematica [A]  time = 0.12, size = 22, normalized size = 1.00 $$\begin{array}{c}-\frac{4e^{-e^{-4+4x}+x^2}}{x}+x\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*E^(-E^(-4 + 4*x) + x^2))/x + x

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fricas [A]  time = 0.60, size = 23, normalized size = 1.05 $$\begin{array}{c}\frac{x^2-4e^{(x^2-e^{(4x-4)})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(2*x-2)^2-8*x^2+4)*exp(-exp(2*x-2)^2+x^2)+x^2)/x^2,x, algorithm="fricas")

[Out]

(x^2 - 4*e^(x^2 - e^(4*x - 4)))/x

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giac [A]  time = 0.24, size = 23, normalized size = 1.05 $$\begin{array}{c}\frac{x^2-4e^{(x^2-e^{(4x-4)})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(2*x-2)^2-8*x^2+4)*exp(-exp(2*x-2)^2+x^2)+x^2)/x^2,x, algorithm="giac")

[Out]

(x^2 - 4*e^(x^2 - e^(4*x - 4)))/x

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maple [A]  time = 0.06, size = 21, normalized size = 0.95

method	result	size
risch	$x-\frac{4{\mathrm{e}}^{-{\mathrm{e}}^{4x-4}+x^2}}{x}$	$21$
norman	$\frac{x^2-4{\mathrm{e}}^{-{\mathrm{e}}^{4x-4}+x^2}}{x}$	$26$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x*exp(2*x-2)^2-8*x^2+4)*exp(-exp(2*x-2)^2+x^2)+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 20, normalized size = 0.91 $$\begin{array}{c}x-\frac{4e^{(x^2-e^{(4x-4)})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(2*x-2)^2-8*x^2+4)*exp(-exp(2*x-2)^2+x^2)+x^2)/x^2,x, algorithm="maxima")

[Out]

x - 4*e^(x^2 - e^(4*x - 4))/x

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mupad [B]  time = 1.98, size = 20, normalized size = 0.91 $$\begin{array}{c}x-\frac{4{\mathrm{e}}^{x^2}{\mathrm{e}}^{-{\mathrm{e}}^{4x}{\mathrm{e}}^{-4}}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2 - exp(4*x - 4))*(16*x*exp(4*x - 4) - 8*x^2 + 4) + x^2)/x^2,x)

[Out]

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

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sympy [A]  time = 0.39, size = 15, normalized size = 0.68 $$\begin{array}{c}x-\frac{4e^{x^2-e^{4x-4}}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(2*x-2)**2-8*x**2+4)*exp(-exp(2*x-2)**2+x**2)+x**2)/x**2,x)

[Out]

x - 4*exp(x**2 - exp(4*x - 4))/x

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