3.13.40 \(\int \frac {e^{2 e^5-2 x} (-3 x^2-2 x^3+(-1-2 x) \log (4))}{40 x^6+80 x^4 \log (4)+40 x^2 \log ^2(4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{2 e^5-2 x}}{40 x \left (x^2+\log (4)\right )} \]

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Rubi [A]  time = 1.03, antiderivative size = 34, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {1594, 28, 6741, 2288} \begin {gather*} \frac {e^{2 e^5-2 x} \left (x^3+x \log (4)\right )}{40 x^2 \left (x^2+\log (4)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*E^5 - 2*x)*(-3*x^2 - 2*x^3 + (-1 - 2*x)*Log[4]))/(40*x^6 + 80*x^4*Log[4] + 40*x^2*Log[4]^2),x]

[Out]

(E^(2*E^5 - 2*x)*(x^3 + x*Log[4]))/(40*x^2*(x^2 + Log[4])^2)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 e^5-2 x} \left (-3 x^2-2 x^3+(-1-2 x) \log (4)\right )}{x^2 \left (40 x^4+80 x^2 \log (4)+40 \log ^2(4)\right )} \, dx\\ &=40 \int \frac {e^{2 e^5-2 x} \left (-3 x^2-2 x^3+(-1-2 x) \log (4)\right )}{x^2 \left (40 x^2+40 \log (4)\right )^2} \, dx\\ &=40 \int \frac {e^{2 e^5-2 x} \left (-3 x^2-2 x^3-\log (4)-2 x \log (4)\right )}{x^2 \left (40 x^2+40 \log (4)\right )^2} \, dx\\ &=\frac {e^{2 e^5-2 x} \left (x^3+x \log (4)\right )}{40 x^2 \left (x^2+\log (4)\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 36, normalized size = 1.38 \begin {gather*} \frac {e^{2 e^5-2 x} \left (2 x^3+x \log (16)\right )}{80 x^2 \left (x^2+\log (4)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^5 - 2*x)*(-3*x^2 - 2*x^3 + (-1 - 2*x)*Log[4]))/(40*x^6 + 80*x^4*Log[4] + 40*x^2*Log[4]^2),x]

[Out]

(E^(2*E^5 - 2*x)*(2*x^3 + x*Log[16]))/(80*x^2*(x^2 + Log[4])^2)

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fricas [A]  time = 0.65, size = 22, normalized size = 0.85 \begin {gather*} \frac {e^{\left (-2 \, x + 2 \, e^{5}\right )}}{40 \, {\left (x^{3} + 2 \, x \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-2*x-1)*log(2)-2*x^3-3*x^2)/(160*x^2*log(2)^2+160*x^4*log(2)+40*x^6)/exp(-exp(5)+x)^2,x, algorit
hm="fricas")

[Out]

1/40*e^(-2*x + 2*e^5)/(x^3 + 2*x*log(2))

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giac [A]  time = 0.56, size = 22, normalized size = 0.85 \begin {gather*} \frac {e^{\left (-2 \, x + 2 \, e^{5}\right )}}{40 \, {\left (x^{3} + 2 \, x \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-2*x-1)*log(2)-2*x^3-3*x^2)/(160*x^2*log(2)^2+160*x^4*log(2)+40*x^6)/exp(-exp(5)+x)^2,x, algorit
hm="giac")

[Out]

1/40*e^(-2*x + 2*e^5)/(x^3 + 2*x*log(2))

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maple [A]  time = 0.16, size = 25, normalized size = 0.96

method	result	size
gosper	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{5}-2 x}}{40 \left (x^{2}+2 \ln \relax (2)\right ) x}\)	\(25\)
norman	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{5}-2 x}}{40 \left (x^{2}+2 \ln \relax (2)\right ) x}\)	\(25\)
risch	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{5}-2 x}}{40 \left (x^{2}+2 \ln \relax (2)\right ) x}\)	\(25\)
derivativedivides	\(\text {Expression too large to display}\)	\(2581\)
default	\(\text {Expression too large to display}\)	\(2581\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-2*x-1)*ln(2)-2*x^3-3*x^2)/(160*x^2*ln(2)^2+160*x^4*ln(2)+40*x^6)/exp(-exp(5)+x)^2,x,method=_RETURNVER
BOSE)

[Out]

1/40/(x^2+2*ln(2))/x/exp(-exp(5)+x)^2

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maxima [A]  time = 0.62, size = 22, normalized size = 0.85 \begin {gather*} \frac {e^{\left (-2 \, x + 2 \, e^{5}\right )}}{40 \, {\left (x^{3} + 2 \, x \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-2*x-1)*log(2)-2*x^3-3*x^2)/(160*x^2*log(2)^2+160*x^4*log(2)+40*x^6)/exp(-exp(5)+x)^2,x, algorit
hm="maxima")

[Out]

1/40*e^(-2*x + 2*e^5)/(x^3 + 2*x*log(2))

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mupad [B]  time = 1.08, size = 23, normalized size = 0.88 \begin {gather*} \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^5}\,{\mathrm {e}}^{-2\,x}}{40\,\left (x^3+2\,\ln \relax (2)\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(5) - 2*x)*(2*log(2)*(2*x + 1) + 3*x^2 + 2*x^3))/(160*x^2*log(2)^2 + 160*x^4*log(2) + 40*x^6),x
)

[Out]

(exp(2*exp(5))*exp(-2*x))/(40*(2*x*log(2) + x^3))

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sympy [A]  time = 0.15, size = 20, normalized size = 0.77 \begin {gather*} \frac {e^{- 2 x + 2 e^{5}}}{40 x^{3} + 80 x \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-2*x-1)*ln(2)-2*x**3-3*x**2)/(160*x**2*ln(2)**2+160*x**4*ln(2)+40*x**6)/exp(-exp(5)+x)**2,x)

[Out]

exp(-2*x + 2*exp(5))/(40*x**3 + 80*x*log(2))

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