3.47.2 \(\int \frac {e^{2 e^e+2 (-5+x) \log ^4(2+x)} (e^{2 x^2} (8 x+4 x^2)+e^{2 x^2} (-40+8 x) \log ^3(2+x)+e^{2 x^2} (4+2 x) \log ^4(2+x))}{2+x} \, dx\)

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

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Rubi [B]  time = 0.42, antiderivative size = 95, normalized size of antiderivative = 3.65, number of steps used = 1, number of rules used = 1, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2288} \begin {gather*} \frac {e^{2 e^e-2 (5-x) \log ^4(x+2)} \left (4 e^{2 x^2} (5-x) \log ^3(x+2)-e^{2 x^2} (x+2) \log ^4(x+2)\right )}{(x+2) \left (\frac {4 (5-x) \log ^3(x+2)}{x+2}-\log ^4(x+2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*E^E + 2*(-5 + x)*Log[2 + x]^4)*(E^(2*x^2)*(8*x + 4*x^2) + E^(2*x^2)*(-40 + 8*x)*Log[2 + x]^3 + E^(2*
x^2)*(4 + 2*x)*Log[2 + x]^4))/(2 + x),x]

[Out]

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

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

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Mathematica [A]  time = 0.75, size = 21, normalized size = 0.81 \begin {gather*} e^{2 \left (e^e+x^2+(-5+x) \log ^4(2+x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^E + 2*(-5 + x)*Log[2 + x]^4)*(E^(2*x^2)*(8*x + 4*x^2) + E^(2*x^2)*(-40 + 8*x)*Log[2 + x]^3 +
E^(2*x^2)*(4 + 2*x)*Log[2 + x]^4))/(2 + x),x]

[Out]

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

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fricas [A]  time = 0.67, size = 23, normalized size = 0.88 \begin {gather*} e^{\left (2 \, {\left (x - 5\right )} \log \left (x + 2\right )^{4} + 2 \, x^{2} + 2 \, e^{e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+4)*exp(x^2)^2*log(2+x)^4+(8*x-40)*exp(x^2)^2*log(2+x)^3+(4*x^2+8*x)*exp(x^2)^2)*exp((x-5)*log(
2+x)^4+exp(exp(1)))^2/(2+x),x, algorithm="fricas")

[Out]

e^(2*(x - 5)*log(x + 2)^4 + 2*x^2 + 2*e^e)

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giac [A]  time = 0.27, size = 29, normalized size = 1.12 \begin {gather*} e^{\left (2 \, x \log \left (x + 2\right )^{4} - 10 \, \log \left (x + 2\right )^{4} + 2 \, x^{2} + 2 \, e^{e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+4)*exp(x^2)^2*log(2+x)^4+(8*x-40)*exp(x^2)^2*log(2+x)^3+(4*x^2+8*x)*exp(x^2)^2)*exp((x-5)*log(
2+x)^4+exp(exp(1)))^2/(2+x),x, algorithm="giac")

[Out]

e^(2*x*log(x + 2)^4 - 10*log(x + 2)^4 + 2*x^2 + 2*e^e)

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maple [A]  time = 0.08, size = 30, normalized size = 1.15

method	result	size
risch	\({\mathrm e}^{2 x^{2}+2 \ln \left (2+x \right )^{4} x -10 \ln \left (2+x \right )^{4}+2 \,{\mathrm e}^{{\mathrm e}}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+4)*exp(x^2)^2*ln(2+x)^4+(8*x-40)*exp(x^2)^2*ln(2+x)^3+(4*x^2+8*x)*exp(x^2)^2)*exp((x-5)*ln(2+x)^4+ex
p(exp(1)))^2/(2+x),x,method=_RETURNVERBOSE)

[Out]

exp(2*x^2+2*ln(2+x)^4*x-10*ln(2+x)^4+2*exp(exp(1)))

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maxima [A]  time = 0.52, size = 29, normalized size = 1.12 \begin {gather*} e^{\left (2 \, x \log \left (x + 2\right )^{4} - 10 \, \log \left (x + 2\right )^{4} + 2 \, x^{2} + 2 \, e^{e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+4)*exp(x^2)^2*log(2+x)^4+(8*x-40)*exp(x^2)^2*log(2+x)^3+(4*x^2+8*x)*exp(x^2)^2)*exp((x-5)*log(
2+x)^4+exp(exp(1)))^2/(2+x),x, algorithm="maxima")

[Out]

e^(2*x*log(x + 2)^4 - 10*log(x + 2)^4 + 2*x^2 + 2*e^e)

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mupad [B]  time = 3.44, size = 32, normalized size = 1.23 \begin {gather*} {\mathrm {e}}^{-10\,{\ln \left (x+2\right )}^4}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{\mathrm {e}}}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{2\,x\,{\ln \left (x+2\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(exp(1)) + 2*log(x + 2)^4*(x - 5))*(exp(2*x^2)*(8*x + 4*x^2) + log(x + 2)^4*exp(2*x^2)*(2*x + 4)
+ log(x + 2)^3*exp(2*x^2)*(8*x - 40)))/(x + 2),x)

[Out]

exp(-10*log(x + 2)^4)*exp(2*exp(exp(1)))*exp(2*x^2)*exp(2*x*log(x + 2)^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+4)*exp(x**2)**2*ln(2+x)**4+(8*x-40)*exp(x**2)**2*ln(2+x)**3+(4*x**2+8*x)*exp(x**2)**2)*exp((x-
5)*ln(2+x)**4+exp(exp(1)))**2/(2+x),x)

[Out]

Timed out

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