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

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

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Rubi [A]  time = 0.21, antiderivative size = 36, normalized size of antiderivative = 1.38, number of steps used = 6, number of rules used = 5, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 12, 6688, 2230, 2209} \begin {gather*} -\exp \left (\frac {2}{5 \left (e^5-\log (4)\right )^2}-\frac {2}{(x+5) \left (e^5-\log (4)\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

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 {2 \exp \left (\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}\right )}{e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \left (e^{10}+\log ^2(4)\right )} \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}\right )}{e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \left (e^{10}+\log ^2(4)\right )} \, dx\right )\\ &=-\left (2 \int \frac {e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}}}{(5+x)^2 \left (e^5-\log (4)\right )^2} \, dx\right )\\ &=-\frac {2 \int \frac {e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}}}{(5+x)^2} \, dx}{\left (e^5-\log (4)\right )^2}\\ &=-\frac {2 \int \frac {\exp \left (\frac {2}{5 \left (e^5-\log (4)\right )^2}-\frac {2}{(5+x) \left (e^5-\log (4)\right )^2}\right )}{(5+x)^2} \, dx}{\left (e^5-\log (4)\right )^2}\\ &=-\exp \left (\frac {2}{5 \left (e^5-\log (4)\right )^2}-\frac {2}{(5+x) \left (e^5-\log (4)\right )^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.92 \begin {gather*} -e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 34, normalized size = 1.31 \begin {gather*} -e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, {\left (x + 5\right )} e^{5} \log \relax (2) - 4 \, {\left (x + 5\right )} \log \relax (2)^{2} - {\left (x + 5\right )} e^{10}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, {\left (x + 5\right )} e^{5} \log \relax (2) - 4 \, {\left (x + 5\right )} \log \relax (2)^{2} - {\left (x + 5\right )} e^{10}\right )}}\right )}}{4 \, {\left (x^{2} + 10 \, x + 25\right )} e^{5} \log \relax (2) - 4 \, {\left (x^{2} + 10 \, x + 25\right )} \log \relax (2)^{2} - {\left (x^{2} + 10 \, x + 25\right )} e^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(2*e^(-2/5*x/(4*(x + 5)*e^5*log(2) - 4*(x + 5)*log(2)^2 - (x + 5)*e^10))/(4*(x^2 + 10*x + 25)*e^5*log
(2) - 4*(x^2 + 10*x + 25)*log(2)^2 - (x^2 + 10*x + 25)*e^10), x)

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maple [A]  time = 0.29, size = 29, normalized size = 1.12

method	result	size
risch	\(-{\mathrm e}^{\frac {2 x}{5 \left (5+x \right ) \left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right )}}\)	\(29\)
gosper	\(-{\mathrm e}^{\frac {2 x}{5 \left (x \,{\mathrm e}^{10}-4 x \,{\mathrm e}^{5} \ln \relax (2)+4 x \ln \relax (2)^{2}+5 \,{\mathrm e}^{10}-20 \,{\mathrm e}^{5} \ln \relax (2)+20 \ln \relax (2)^{2}\right )}}\)	\(48\)
derivativedivides	\(\frac {25 \left (4 \,{\mathrm e}^{5} \ln \relax (2)-4 \ln \relax (2)^{2}-{\mathrm e}^{10}\right ) \left (\frac {{\mathrm e}^{10}}{25}-\frac {4 \,{\mathrm e}^{5} \ln \relax (2)}{25}+\frac {4 \ln \relax (2)^{2}}{25}\right ) {\mathrm e}^{-\frac {2}{\left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right ) \left (5+x \right )}+\frac {2}{5 \left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right )}}}{\left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right )^{2}}\)	\(109\)
default	\(\frac {25 \left (4 \,{\mathrm e}^{5} \ln \relax (2)-4 \ln \relax (2)^{2}-{\mathrm e}^{10}\right ) \left (\frac {{\mathrm e}^{10}}{25}-\frac {4 \,{\mathrm e}^{5} \ln \relax (2)}{25}+\frac {4 \ln \relax (2)^{2}}{25}\right ) {\mathrm e}^{-\frac {2}{\left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right ) \left (5+x \right )}+\frac {2}{5 \left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right )}}}{\left (-4 \,{\mathrm e}^{5} \ln \relax (2)+4 \ln \relax (2)^{2}+{\mathrm e}^{10}\right )^{2}}\)	\(109\)
norman	\(\frac {\left (-5 \,{\mathrm e}^{5}+10 \ln \relax (2)\right ) {\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \relax (2)^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \relax (2)+\left (25+5 x \right ) {\mathrm e}^{10}}}+\left (-{\mathrm e}^{5}+2 \ln \relax (2)\right ) x \,{\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \relax (2)^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \relax (2)+\left (25+5 x \right ) {\mathrm e}^{10}}}}{\left (5+x \right ) \left ({\mathrm e}^{5}-2 \ln \relax (2)\right )}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(2/5*x/(5+x)/(-4*exp(5)*ln(2)+4*ln(2)^2+exp(10)))

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maxima [B]  time = 0.48, size = 65, normalized size = 2.50 \begin {gather*} -e^{\left (\frac {2}{{\left (4 \, e^{5} \log \relax (2) - 4 \, \log \relax (2)^{2} - e^{10}\right )} x + 20 \, e^{5} \log \relax (2) - 20 \, \log \relax (2)^{2} - 5 \, e^{10}} - \frac {2}{5 \, {\left (4 \, e^{5} \log \relax (2) - 4 \, \log \relax (2)^{2} - e^{10}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(2/((4*e^5*log(2) - 4*log(2)^2 - e^10)*x + 20*e^5*log(2) - 20*log(2)^2 - 5*e^10) - 2/5/(4*e^5*log(2) - 4*lo
g(2)^2 - e^10))

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mupad [B]  time = 0.85, size = 44, normalized size = 1.69 \begin {gather*} -{\mathrm {e}}^{\frac {2\,x}{25\,{\mathrm {e}}^{10}-100\,{\mathrm {e}}^5\,\ln \relax (2)+5\,x\,{\mathrm {e}}^{10}+20\,x\,{\ln \relax (2)}^2+100\,{\ln \relax (2)}^2-20\,x\,{\mathrm {e}}^5\,\ln \relax (2)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp((2*x)/(25*exp(10) - 100*exp(5)*log(2) + 5*x*exp(10) + 20*x*log(2)^2 + 100*log(2)^2 - 20*x*exp(5)*log(2)))

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sympy [A]  time = 0.42, size = 37, normalized size = 1.42 \begin {gather*} - e^{\frac {2 x}{\left (- 20 x - 100\right ) e^{5} \log {\relax (2 )} + \left (5 x + 25\right ) e^{10} + \left (20 x + 100\right ) \log {\relax (2 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(2*x/((-20*x - 100)*exp(5)*log(2) + (5*x + 25)*exp(10) + (20*x + 100)*log(2)**2))

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