3.49.31 \(\int \frac {e^{\frac {2+x}{x}} (-32-16 x)+(16 e^{\frac {2+x}{x}} x+16 x^2) \log (e^{\frac {2+x}{x}} x+x^2)}{4 e^{\frac {2+x}{x}} x^3+4 x^4+(-4 e^{\frac {2+x}{x}} x^3-4 x^4) \log (e^{\frac {2+x}{x}} x+x^2)+(e^{\frac {2+x}{x}} x^3+x^4) \log ^2(e^{\frac {2+x}{x}} x+x^2)} \, dx\)

Optimal. Leaf size=27 \[ \frac {16}{x \left (2-\log \left (e^{\frac {2+x}{x}} x+x^2\right )\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.79, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 147, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 12, 6687} \begin {gather*} \frac {16}{x \left (2-\log \left (x \left (x+e^{\frac {2}{x}+1}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2 + x)/x)*(-32 - 16*x) + (16*E^((2 + x)/x)*x + 16*x^2)*Log[E^((2 + x)/x)*x + x^2])/(4*E^((2 + x)/x)*x
^3 + 4*x^4 + (-4*E^((2 + x)/x)*x^3 - 4*x^4)*Log[E^((2 + x)/x)*x + x^2] + (E^((2 + x)/x)*x^3 + x^4)*Log[E^((2 +
x)/x)*x + x^2]^2),x]

[Out]

16/(x*(2 - Log[x*(E^(1 + 2/x) + x)]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 \left (-e^{1+\frac {2}{x}} (2+x)+x \left (e^{1+\frac {2}{x}}+x\right ) \log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )\right )}{x^3 \left (e^{1+\frac {2}{x}}+x\right ) \left (2-\log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )\right )^2} \, dx\\ &=16 \int \frac {-e^{1+\frac {2}{x}} (2+x)+x \left (e^{1+\frac {2}{x}}+x\right ) \log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )}{x^3 \left (e^{1+\frac {2}{x}}+x\right ) \left (2-\log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )\right )^2} \, dx\\ &=\frac {16}{x \left (2-\log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.48, size = 23, normalized size = 0.85 \begin {gather*} -\frac {16}{x \left (-2+\log \left (x \left (e^{1+\frac {2}{x}}+x\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2 + x)/x)*(-32 - 16*x) + (16*E^((2 + x)/x)*x + 16*x^2)*Log[E^((2 + x)/x)*x + x^2])/(4*E^((2 + x
)/x)*x^3 + 4*x^4 + (-4*E^((2 + x)/x)*x^3 - 4*x^4)*Log[E^((2 + x)/x)*x + x^2] + (E^((2 + x)/x)*x^3 + x^4)*Log[E
^((2 + x)/x)*x + x^2]^2),x]

[Out]

-16/(x*(-2 + Log[x*(E^(1 + 2/x) + x)]))

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 25, normalized size = 0.93 \begin {gather*} -\frac {16}{x \log \left (x^{2} + x e^{\left (\frac {x + 2}{x}\right )}\right ) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*log(x*exp(1/2*(2+x)/x)^2+x^2)+(-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3
*exp(1/2*(2+x)/x)^2+x^4)*log(x*exp(1/2*(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*log(x*exp(1/2*(2+x)
/x)^2+x^2)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x, algorithm="fricas")

[Out]

-16/(x*log(x^2 + x*e^((x + 2)/x)) - 2*x)

________________________________________________________________________________________

giac [A]  time = 0.48, size = 25, normalized size = 0.93 \begin {gather*} -\frac {16}{x \log \left (x^{2} + x e^{\left (\frac {x + 2}{x}\right )}\right ) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*log(x*exp(1/2*(2+x)/x)^2+x^2)+(-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3
*exp(1/2*(2+x)/x)^2+x^4)*log(x*exp(1/2*(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*log(x*exp(1/2*(2+x)
/x)^2+x^2)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x, algorithm="giac")

[Out]

-16/(x*log(x^2 + x*e^((x + 2)/x)) - 2*x)

________________________________________________________________________________________

maple [C]  time = 0.14, size = 145, normalized size = 5.37

method	result	size
risch	\(-\frac {32 i}{x \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )^{2}+\pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )\right )^{3}+2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{\frac {2+x}{x}}+x \right )-4 i\right )}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*ln(x*exp(1/2*(2+x)/x)^2+x^2)+(-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3*exp(1/
2*(2+x)/x)^2+x^4)*ln(x*exp(1/2*(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*ln(x*exp(1/2*(2+x)/x)^2+x^2
)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x,method=_RETURNVERBOSE)

[Out]

-32*I/x/(Pi*csgn(I*x)*csgn(I*(exp((2+x)/x)+x))*csgn(I*x*(exp((2+x)/x)+x))-Pi*csgn(I*x)*csgn(I*x*(exp((2+x)/x)+
x))^2-Pi*csgn(I*(exp((2+x)/x)+x))*csgn(I*x*(exp((2+x)/x)+x))^2+Pi*csgn(I*x*(exp((2+x)/x)+x))^3+2*I*ln(x)+2*I*l
n(exp((2+x)/x)+x)-4*I)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 25, normalized size = 0.93 \begin {gather*} -\frac {16}{x \log \left (x + e^{\left (\frac {2}{x} + 1\right )}\right ) + x \log \relax (x) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(1/2*(2+x)/x)^2+16*x^2)*log(x*exp(1/2*(2+x)/x)^2+x^2)+(-16*x-32)*exp(1/2*(2+x)/x)^2)/((x^3
*exp(1/2*(2+x)/x)^2+x^4)*log(x*exp(1/2*(2+x)/x)^2+x^2)^2+(-4*x^3*exp(1/2*(2+x)/x)^2-4*x^4)*log(x*exp(1/2*(2+x)
/x)^2+x^2)+4*x^3*exp(1/2*(2+x)/x)^2+4*x^4),x, algorithm="maxima")

[Out]

-16/(x*log(x + e^(2/x + 1)) + x*log(x) - 2*x)

________________________________________________________________________________________

mupad [B]  time = 4.46, size = 24, normalized size = 0.89 \begin {gather*} -\frac {16}{x\,\left (\ln \left (x^2+x\,\mathrm {e}\,{\mathrm {e}}^{2/x}\right )-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(x/2 + 1))/x)*(16*x + 32) - log(x*exp((2*(x/2 + 1))/x) + x^2)*(16*x*exp((2*(x/2 + 1))/x) + 16*x^2
))/(4*x^3*exp((2*(x/2 + 1))/x) - log(x*exp((2*(x/2 + 1))/x) + x^2)*(4*x^3*exp((2*(x/2 + 1))/x) + 4*x^4) + log(
x*exp((2*(x/2 + 1))/x) + x^2)^2*(x^3*exp((2*(x/2 + 1))/x) + x^4) + 4*x^4),x)

[Out]

-16/(x*(log(x^2 + x*exp(1)*exp(2/x)) - 2))

________________________________________________________________________________________

sympy [A]  time = 0.38, size = 20, normalized size = 0.74 \begin {gather*} - \frac {16}{x \log {\left (x^{2} + x e^{\frac {2 \left (\frac {x}{2} + 1\right )}{x}} \right )} - 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*exp(1/2*(2+x)/x)**2+16*x**2)*ln(x*exp(1/2*(2+x)/x)**2+x**2)+(-16*x-32)*exp(1/2*(2+x)/x)**2)/(
(x**3*exp(1/2*(2+x)/x)**2+x**4)*ln(x*exp(1/2*(2+x)/x)**2+x**2)**2+(-4*x**3*exp(1/2*(2+x)/x)**2-4*x**4)*ln(x*ex
p(1/2*(2+x)/x)**2+x**2)+4*x**3*exp(1/2*(2+x)/x)**2+4*x**4),x)

[Out]

-16/(x*log(x**2 + x*exp(2*(x/2 + 1)/x)) - 2*x)

________________________________________________________________________________________