3.91.100 \(\int \frac {-20 x+5 e^4 x+e^{\frac {2+x}{x}} (10+5 x)}{16 x-8 e^4 x+e^8 x+e^{\frac {2 (2+x)}{x}} x+e^{\frac {2+x}{x}} (-8 x+2 e^4 x)} \, dx\)

Optimal. Leaf size=19 \[ \frac {5 x}{-4+e^4+e^{1+\frac {2}{x}}} \]

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Rubi [A]  time = 0.24, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 4, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6, 6688, 12, 6687} \begin {gather*} -\frac {5 x}{-e^{\frac {2}{x}+1}+4-e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x + 5*E^4*x + E^((2 + x)/x)*(10 + 5*x))/(16*x - 8*E^4*x + E^8*x + E^((2*(2 + x))/x)*x + E^((2 + x)/x)
*(-8*x + 2*E^4*x)),x]

[Out]

(-5*x)/(4 - E^4 - E^(1 + 2/x))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 {\left (-20+5 e^4\right ) x+e^{\frac {2+x}{x}} (10+5 x)}{16 x-8 e^4 x+e^8 x+e^{\frac {2 (2+x)}{x}} x+e^{\frac {2+x}{x}} \left (-8 x+2 e^4 x\right )} \, dx\\ &=\int \frac {\left (-20+5 e^4\right ) x+e^{\frac {2+x}{x}} (10+5 x)}{e^8 x+e^{\frac {2 (2+x)}{x}} x+\left (16-8 e^4\right ) x+e^{\frac {2+x}{x}} \left (-8 x+2 e^4 x\right )} \, dx\\ &=\int \frac {\left (-20+5 e^4\right ) x+e^{\frac {2+x}{x}} (10+5 x)}{e^{\frac {2 (2+x)}{x}} x+\left (16-8 e^4+e^8\right ) x+e^{\frac {2+x}{x}} \left (-8 x+2 e^4 x\right )} \, dx\\ &=\int \frac {5 \left (-4 \left (1-\frac {e^4}{4}\right ) x+e^{1+\frac {2}{x}} (2+x)\right )}{\left (e^{1+\frac {2}{x}}-4 \left (1-\frac {e^4}{4}\right )\right )^2 x} \, dx\\ &=5 \int \frac {-4 \left (1-\frac {e^4}{4}\right ) x+e^{1+\frac {2}{x}} (2+x)}{\left (e^{1+\frac {2}{x}}-4 \left (1-\frac {e^4}{4}\right )\right )^2 x} \, dx\\ &=-\frac {5 x}{4-e^4-e^{1+\frac {2}{x}}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 19, normalized size = 1.00 \begin {gather*} \frac {5 x}{-4+e^4+e^{1+\frac {2}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x + 5*E^4*x + E^((2 + x)/x)*(10 + 5*x))/(16*x - 8*E^4*x + E^8*x + E^((2*(2 + x))/x)*x + E^((2 +
x)/x)*(-8*x + 2*E^4*x)),x]

[Out]

(5*x)/(-4 + E^4 + E^(1 + 2/x))

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fricas [A]  time = 1.47, size = 17, normalized size = 0.89 \begin {gather*} \frac {5 \, x}{e^{4} + e^{\left (\frac {x + 2}{x}\right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x+10)*exp((2+x)/x)+5*x*exp(4)-20*x)/(x*exp((2+x)/x)^2+(2*x*exp(4)-8*x)*exp((2+x)/x)+x*exp(4)^2-8
*x*exp(4)+16*x),x, algorithm="fricas")

[Out]

5*x/(e^4 + e^((x + 2)/x) - 4)

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giac [A]  time = 0.19, size = 28, normalized size = 1.47 \begin {gather*} \frac {5}{\frac {e^{4}}{x} + \frac {e^{\left (\frac {2}{x} + 1\right )}}{x} - \frac {4}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x+10)*exp((2+x)/x)+5*x*exp(4)-20*x)/(x*exp((2+x)/x)^2+(2*x*exp(4)-8*x)*exp((2+x)/x)+x*exp(4)^2-8
*x*exp(4)+16*x),x, algorithm="giac")

[Out]

5/(e^4/x + e^(2/x + 1)/x - 4/x)

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maple [A]  time = 0.84, size = 18, normalized size = 0.95

method	result	size
norman	\(\frac {5 x}{-4+{\mathrm e}^{4}+{\mathrm e}^{\frac {2+x}{x}}}\)	\(18\)
risch	\(\frac {5 x}{-4+{\mathrm e}^{4}+{\mathrm e}^{\frac {2+x}{x}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x+10)*exp((2+x)/x)+5*x*exp(4)-20*x)/(x*exp((2+x)/x)^2+(2*x*exp(4)-8*x)*exp((2+x)/x)+x*exp(4)^2-8*x*exp
(4)+16*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 17, normalized size = 0.89 \begin {gather*} \frac {5 \, x}{e^{4} + e^{\left (\frac {2}{x} + 1\right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x+10)*exp((2+x)/x)+5*x*exp(4)-20*x)/(x*exp((2+x)/x)^2+(2*x*exp(4)-8*x)*exp((2+x)/x)+x*exp(4)^2-8
*x*exp(4)+16*x),x, algorithm="maxima")

[Out]

5*x/(e^4 + e^(2/x + 1) - 4)

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mupad [B]  time = 7.91, size = 17, normalized size = 0.89 \begin {gather*} \frac {5\,x}{{\mathrm {e}}^4+{\mathrm {e}}^{\frac {2}{x}+1}-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x + 2)/x)*(5*x + 10) - 20*x + 5*x*exp(4))/(16*x - 8*x*exp(4) + x*exp(8) - exp((x + 2)/x)*(8*x - 2*x*
exp(4)) + x*exp((2*(x + 2))/x)),x)

[Out]

(5*x)/(exp(4) + exp(2/x + 1) - 4)

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sympy [A]  time = 0.13, size = 14, normalized size = 0.74 \begin {gather*} \frac {5 x}{e^{\frac {x + 2}{x}} - 4 + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x+10)*exp((2+x)/x)+5*x*exp(4)-20*x)/(x*exp((2+x)/x)**2+(2*x*exp(4)-8*x)*exp((2+x)/x)+x*exp(4)**2
-8*x*exp(4)+16*x),x)

[Out]

5*x/(exp((x + 2)/x) - 4 + exp(4))

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