3.98.18 \(\int \frac {e^{\frac {5}{\log (-4+e^{e^{2 x}+x}+\log (5+x))}} (-5+e^{e^{2 x}+x} (-25+e^{2 x} (-50-10 x)-5 x))}{(-20-4 x+e^{e^{2 x}+x} (5+x)+(5+x) \log (5+x)) \log ^2(-4+e^{e^{2 x}+x}+\log (5+x))} \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {5}{\log \left (-4+e^{e^{2 x}+x}+\log (5+x)\right )}} \]

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Rubi [A]  time = 1.27, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6706} \begin {gather*} e^{\frac {5}{\log \left (e^{x+e^{2 x}}+\log (x+5)-4\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5/Log[-4 + E^(E^(2*x) + x) + Log[5 + x]])*(-5 + E^(E^(2*x) + x)*(-25 + E^(2*x)*(-50 - 10*x) - 5*x)))/(
(-20 - 4*x + E^(E^(2*x) + x)*(5 + x) + (5 + x)*Log[5 + x])*Log[-4 + E^(E^(2*x) + x) + Log[5 + x]]^2),x]

[Out]

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 22, normalized size = 1.00 \begin {gather*} e^{\frac {5}{\log \left (-4+e^{e^{2 x}+x}+\log (5+x)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5/Log[-4 + E^(E^(2*x) + x) + Log[5 + x]])*(-5 + E^(E^(2*x) + x)*(-25 + E^(2*x)*(-50 - 10*x) - 5*
x)))/((-20 - 4*x + E^(E^(2*x) + x)*(5 + x) + (5 + x)*Log[5 + x])*Log[-4 + E^(E^(2*x) + x) + Log[5 + x]]^2),x]

[Out]

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

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fricas [A]  time = 0.59, size = 19, normalized size = 0.86 \begin {gather*} e^{\left (\frac {5}{\log \left (e^{\left (x + e^{\left (2 \, x\right )}\right )} + \log \left (x + 5\right ) - 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-50)*exp(2*x)-5*x-25)*exp(exp(2*x)+x)-5)*exp(5/log(exp(exp(2*x)+x)+log(5+x)-4))/((5+x)*exp(e
xp(2*x)+x)+(5+x)*log(5+x)-4*x-20)/log(exp(exp(2*x)+x)+log(5+x)-4)^2,x, algorithm="fricas")

[Out]

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

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giac [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((((-10*x-50)*exp(2*x)-5*x-25)*exp(exp(2*x)+x)-5)*exp(5/log(exp(exp(2*x)+x)+log(5+x)-4))/((5+x)*exp(e
xp(2*x)+x)+(5+x)*log(5+x)-4*x-20)/log(exp(exp(2*x)+x)+log(5+x)-4)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.06, size = 20, normalized size = 0.91

method	result	size
risch	\({\mathrm e}^{\frac {5}{\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}+x}+\ln \left (5+x \right )-4\right )}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x-50)*exp(2*x)-5*x-25)*exp(exp(2*x)+x)-5)*exp(5/ln(exp(exp(2*x)+x)+ln(5+x)-4))/((5+x)*exp(exp(2*x)+
x)+(5+x)*ln(5+x)-4*x-20)/ln(exp(exp(2*x)+x)+ln(5+x)-4)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.57, size = 264, normalized size = 12.00 \begin {gather*} \frac {2 \, x e^{\left (3 \, x + \frac {5}{\log \left (e^{\left (x + e^{\left (2 \, x\right )}\right )} + \log \left (x + 5\right ) - 4\right )} + e^{\left (2 \, x\right )}\right )}}{{\left (2 \, {\left (x + 5\right )} e^{\left (3 \, x\right )} + {\left (x + 5\right )} e^{x}\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 1} + \frac {x e^{\left (x + \frac {5}{\log \left (e^{\left (x + e^{\left (2 \, x\right )}\right )} + \log \left (x + 5\right ) - 4\right )} + e^{\left (2 \, x\right )}\right )}}{{\left (2 \, {\left (x + 5\right )} e^{\left (3 \, x\right )} + {\left (x + 5\right )} e^{x}\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 1} + \frac {10 \, e^{\left (3 \, x + \frac {5}{\log \left (e^{\left (x + e^{\left (2 \, x\right )}\right )} + \log \left (x + 5\right ) - 4\right )} + e^{\left (2 \, x\right )}\right )}}{{\left (2 \, {\left (x + 5\right )} e^{\left (3 \, x\right )} + {\left (x + 5\right )} e^{x}\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 1} + \frac {5 \, e^{\left (x + \frac {5}{\log \left (e^{\left (x + e^{\left (2 \, x\right )}\right )} + \log \left (x + 5\right ) - 4\right )} + e^{\left (2 \, x\right )}\right )}}{{\left (2 \, {\left (x + 5\right )} e^{\left (3 \, x\right )} + {\left (x + 5\right )} e^{x}\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 1} + \frac {e^{\left (\frac {5}{\log \left (e^{\left (x + e^{\left (2 \, x\right )}\right )} + \log \left (x + 5\right ) - 4\right )}\right )}}{{\left (2 \, {\left (x + 5\right )} e^{\left (3 \, x\right )} + {\left (x + 5\right )} e^{x}\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-50)*exp(2*x)-5*x-25)*exp(exp(2*x)+x)-5)*exp(5/log(exp(exp(2*x)+x)+log(5+x)-4))/((5+x)*exp(e
xp(2*x)+x)+(5+x)*log(5+x)-4*x-20)/log(exp(exp(2*x)+x)+log(5+x)-4)^2,x, algorithm="maxima")

[Out]

2*x*e^(3*x + 5/log(e^(x + e^(2*x)) + log(x + 5) - 4) + e^(2*x))/((2*(x + 5)*e^(3*x) + (x + 5)*e^x)*e^(e^(2*x))
+ 1) + x*e^(x + 5/log(e^(x + e^(2*x)) + log(x + 5) - 4) + e^(2*x))/((2*(x + 5)*e^(3*x) + (x + 5)*e^x)*e^(e^(2
*x)) + 1) + 10*e^(3*x + 5/log(e^(x + e^(2*x)) + log(x + 5) - 4) + e^(2*x))/((2*(x + 5)*e^(3*x) + (x + 5)*e^x)*
e^(e^(2*x)) + 1) + 5*e^(x + 5/log(e^(x + e^(2*x)) + log(x + 5) - 4) + e^(2*x))/((2*(x + 5)*e^(3*x) + (x + 5)*e
^x)*e^(e^(2*x)) + 1) + e^(5/log(e^(x + e^(2*x)) + log(x + 5) - 4))/((2*(x + 5)*e^(3*x) + (x + 5)*e^x)*e^(e^(2*
x)) + 1)

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mupad [B]  time = 6.37, size = 20, normalized size = 0.91 \begin {gather*} {\mathrm {e}}^{\frac {5}{\ln \left (\ln \left (x+5\right )+{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x-4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5/log(log(x + 5) + exp(x + exp(2*x)) - 4))*(exp(x + exp(2*x))*(5*x + exp(2*x)*(10*x + 50) + 25) + 5))
/(log(log(x + 5) + exp(x + exp(2*x)) - 4)^2*(4*x - log(x + 5)*(x + 5) - exp(x + exp(2*x))*(x + 5) + 20)),x)

[Out]

exp(5/log(log(x + 5) + exp(exp(2*x))*exp(x) - 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((((-10*x-50)*exp(2*x)-5*x-25)*exp(exp(2*x)+x)-5)*exp(5/ln(exp(exp(2*x)+x)+ln(5+x)-4))/((5+x)*exp(exp
(2*x)+x)+(5+x)*ln(5+x)-4*x-20)/ln(exp(exp(2*x)+x)+ln(5+x)-4)**2,x)

[Out]

Timed out

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