3.71.80 \(\int \frac {-16 e^{\frac {1+x}{x}} x+e^{x+\frac {1+x}{x}} (4 x+4 x^2)+(16 e^{\frac {1+x}{x}}-4 e^{x+\frac {1+x}{x}}) \log (-4 x+e^x x)}{-4 x^2+e^x x^2} \, dx\)

Optimal. Leaf size=19 \[ 4 e^{\frac {1+x}{x}} \log \left (\left (-4+e^x\right ) x\right ) \]

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Rubi [A]  time = 0.37, antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6688, 12, 2288} \begin {gather*} 4 e^{\frac {1}{x}+1} \log \left (-\left (\left (4-e^x\right ) x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*E^((1 + x)/x)*x + E^(x + (1 + x)/x)*(4*x + 4*x^2) + (16*E^((1 + x)/x) - 4*E^(x + (1 + x)/x))*Log[-4*x
+ E^x*x])/(-4*x^2 + E^x*x^2),x]

[Out]

4*E^(1 + x^(-1))*Log[-((4 - E^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 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 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 {4 e^{1+\frac {1}{x}} \left (-x \left (-4+e^x (1+x)\right )+\left (-4+e^x\right ) \log \left (\left (-4+e^x\right ) x\right )\right )}{\left (4-e^x\right ) x^2} \, dx\\ &=4 \int \frac {e^{1+\frac {1}{x}} \left (-x \left (-4+e^x (1+x)\right )+\left (-4+e^x\right ) \log \left (\left (-4+e^x\right ) x\right )\right )}{\left (4-e^x\right ) x^2} \, dx\\ &=4 e^{1+\frac {1}{x}} \log \left (-\left (\left (4-e^x\right ) x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 17, normalized size = 0.89 \begin {gather*} 4 e^{1+\frac {1}{x}} \log \left (\left (-4+e^x\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^((1 + x)/x)*x + E^(x + (1 + x)/x)*(4*x + 4*x^2) + (16*E^((1 + x)/x) - 4*E^(x + (1 + x)/x))*Lo
g[-4*x + E^x*x])/(-4*x^2 + E^x*x^2),x]

[Out]

4*E^(1 + x^(-1))*Log[(-4 + E^x)*x]

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fricas [A]  time = 1.25, size = 26, normalized size = 1.37 \begin {gather*} 4 \, e^{\left (-x + \frac {x^{2} + x + 1}{x}\right )} \log \left (x e^{x} - 4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*e^(-x + (x^2 + x + 1)/x)*log(x*e^x - 4*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left ({\left (x^{2} + x\right )} e^{\left (x + \frac {x + 1}{x}\right )} - 4 \, x e^{\left (\frac {x + 1}{x}\right )} - {\left (e^{\left (x + \frac {x + 1}{x}\right )} - 4 \, e^{\left (\frac {x + 1}{x}\right )}\right )} \log \left (x e^{x} - 4 \, x\right )\right )}}{x^{2} e^{x} - 4 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(4*((x^2 + x)*e^(x + (x + 1)/x) - 4*x*e^((x + 1)/x) - (e^(x + (x + 1)/x) - 4*e^((x + 1)/x))*log(x*e^x
- 4*x))/(x^2*e^x - 4*x^2), x)

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maple [C]  time = 0.07, size = 145, normalized size = 7.63

method	result	size
risch	\(4 \,{\mathrm e}^{\frac {x +1}{x}} \ln \left ({\mathrm e}^{x}-4\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right ) {\mathrm e}^{\frac {x +1}{x}}+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right )^{2} {\mathrm e}^{\frac {x +1}{x}}+2 i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right )^{2} {\mathrm e}^{\frac {x +1}{x}}-2 i \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}-4\right )\right )^{3} {\mathrm e}^{\frac {x +1}{x}}+4 \,{\mathrm e}^{\frac {x +1}{x}} \ln \relax (x )\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*exp((x+1)/x)*ln(exp(x)-4)-2*I*Pi*csgn(I*x)*csgn(I*(exp(x)-4))*csgn(I*x*(exp(x)-4))*exp((x+1)/x)+2*I*Pi*csgn(
I*x)*csgn(I*x*(exp(x)-4))^2*exp((x+1)/x)+2*I*Pi*csgn(I*(exp(x)-4))*csgn(I*x*(exp(x)-4))^2*exp((x+1)/x)-2*I*Pi*
csgn(I*x*(exp(x)-4))^3*exp((x+1)/x)+4*exp((x+1)/x)*ln(x)

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maxima [A]  time = 0.42, size = 24, normalized size = 1.26 \begin {gather*} 4 \, e^{\left (\frac {1}{x} + 1\right )} \log \relax (x) + 4 \, e^{\left (\frac {1}{x} + 1\right )} \log \left (e^{x} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*e^(1/x + 1)*log(x) + 4*e^(1/x + 1)*log(e^x - 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {\ln \left (x\,{\mathrm {e}}^x-4\,x\right )\,\left (16\,{\mathrm {e}}^{\frac {x+1}{x}}-4\,{\mathrm {e}}^{\frac {x+1}{x}}\,{\mathrm {e}}^x\right )-16\,x\,{\mathrm {e}}^{\frac {x+1}{x}}+{\mathrm {e}}^{\frac {x+1}{x}}\,{\mathrm {e}}^x\,\left (4\,x^2+4\,x\right )}{x^2\,{\mathrm {e}}^x-4\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(((-4*exp((x+1)/x)*exp(x)+16*exp((x+1)/x))*ln(exp(x)*x-4*x)+(4*x**2+4*x)*exp((x+1)/x)*exp(x)-16*x*exp
((x+1)/x))/(exp(x)*x**2-4*x**2),x)

[Out]

Timed out

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