3.36.90 \(\int \frac {-e^x \log (5)+e^{5+e^5 (2+x)} \log (5)-\log (5) \log (16)}{1+e^{2 x}+e^{2 e^5 (2+x)}-2 x \log (16)+x^2 \log ^2(16)+e^{e^5 (2+x)} (2-2 e^x-2 x \log (16))+e^x (-2+2 x \log (16))} \, dx\)

Optimal. Leaf size=25 \[ \frac {\log (5)}{-1+e^x-e^{e^5 (2+x)}+x \log (16)} \]

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Rubi [A]  time = 0.40, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {\log (5)}{-e^x+e^{e^5 (x+2)}+x (-\log (16))+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^x*Log[5]) + E^(5 + E^5*(2 + x))*Log[5] - Log[5]*Log[16])/(1 + E^(2*x) + E^(2*E^5*(2 + x)) - 2*x*Log[1
6] + x^2*Log[16]^2 + E^(E^5*(2 + x))*(2 - 2*E^x - 2*x*Log[16]) + E^x*(-2 + 2*x*Log[16])),x]

[Out]

-(Log[5]/(1 - E^x + E^(E^5*(2 + x)) - x*Log[16]))

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, 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 {\log (5) \left (-e^x+e^{5+e^5 (2+x)}-\log (16)\right )}{\left (1-e^x+e^{e^5 (2+x)}-x \log (16)\right )^2} \, dx\\ &=\log (5) \int \frac {-e^x+e^{5+e^5 (2+x)}-\log (16)}{\left (1-e^x+e^{e^5 (2+x)}-x \log (16)\right )^2} \, dx\\ &=-\frac {\log (5)}{1-e^x+e^{e^5 (2+x)}-x \log (16)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.00 \begin {gather*} \frac {\log (5)}{-1+e^x-e^{e^5 (2+x)}+x \log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^x*Log[5]) + E^(5 + E^5*(2 + x))*Log[5] - Log[5]*Log[16])/(1 + E^(2*x) + E^(2*E^5*(2 + x)) - 2*x
*Log[16] + x^2*Log[16]^2 + E^(E^5*(2 + x))*(2 - 2*E^x - 2*x*Log[16]) + E^x*(-2 + 2*x*Log[16])),x]

[Out]

Log[5]/(-1 + E^x - E^(E^5*(2 + x)) + x*Log[16])

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fricas [A]  time = 0.68, size = 34, normalized size = 1.36 \begin {gather*} \frac {e^{5} \log \relax (5)}{4 \, x e^{5} \log \relax (2) - e^{5} - e^{\left ({\left (x + 2\right )} e^{5} + 5\right )} + e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log(5)*exp((2+x)*exp(5))-exp(x)*log(5)-4*log(2)*log(5))/(exp((2+x)*exp(5))^2+(-2*exp(x)-8*x*
log(2)+2)*exp((2+x)*exp(5))+exp(x)^2+(8*x*log(2)-2)*exp(x)+16*x^2*log(2)^2-8*x*log(2)+1),x, algorithm="fricas"
)

[Out]

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

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giac [A]  time = 0.29, size = 26, normalized size = 1.04 \begin {gather*} \frac {\log \relax (5)}{4 \, x \log \relax (2) - e^{\left (x e^{5} + 2 \, e^{5}\right )} + e^{x} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log(5)*exp((2+x)*exp(5))-exp(x)*log(5)-4*log(2)*log(5))/(exp((2+x)*exp(5))^2+(-2*exp(x)-8*x*
log(2)+2)*exp((2+x)*exp(5))+exp(x)^2+(8*x*log(2)-2)*exp(x)+16*x^2*log(2)^2-8*x*log(2)+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.69, size = 24, normalized size = 0.96

method	result	size
norman	\(\frac {\ln \relax (5)}{{\mathrm e}^{x}-1-{\mathrm e}^{\left (2+x \right ) {\mathrm e}^{5}}+4 x \ln \relax (2)}\)	\(24\)
risch	\(\frac {\ln \relax (5)}{{\mathrm e}^{x}-1-{\mathrm e}^{\left (2+x \right ) {\mathrm e}^{5}}+4 x \ln \relax (2)}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*ln(5)*exp((2+x)*exp(5))-exp(x)*ln(5)-4*ln(2)*ln(5))/(exp((2+x)*exp(5))^2+(-2*exp(x)-8*x*ln(2)+2)*e
xp((2+x)*exp(5))+exp(x)^2+(8*x*ln(2)-2)*exp(x)+16*x^2*ln(2)^2-8*x*ln(2)+1),x,method=_RETURNVERBOSE)

[Out]

ln(5)/(exp(x)-1-exp((2+x)*exp(5))+4*x*ln(2))

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maxima [A]  time = 0.94, size = 26, normalized size = 1.04 \begin {gather*} \frac {\log \relax (5)}{4 \, x \log \relax (2) - e^{\left (x e^{5} + 2 \, e^{5}\right )} + e^{x} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log(5)*exp((2+x)*exp(5))-exp(x)*log(5)-4*log(2)*log(5))/(exp((2+x)*exp(5))^2+(-2*exp(x)-8*x*
log(2)+2)*exp((2+x)*exp(5))+exp(x)^2+(8*x*log(2)-2)*exp(x)+16*x^2*log(2)^2-8*x*log(2)+1),x, algorithm="maxima"
)

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {4\,\ln \relax (2)\,\ln \relax (5)+{\mathrm {e}}^x\,\ln \relax (5)-{\mathrm {e}}^{{\mathrm {e}}^5\,\left (x+2\right )}\,{\mathrm {e}}^5\,\ln \relax (5)}{{\mathrm {e}}^{2\,{\mathrm {e}}^5\,\left (x+2\right )}+{\mathrm {e}}^{2\,x}+16\,x^2\,{\ln \relax (2)}^2-{\mathrm {e}}^{{\mathrm {e}}^5\,\left (x+2\right )}\,\left (2\,{\mathrm {e}}^x+8\,x\,\ln \relax (2)-2\right )-8\,x\,\ln \relax (2)+{\mathrm {e}}^x\,\left (8\,x\,\ln \relax (2)-2\right )+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(2)*log(5) + exp(x)*log(5) - exp(exp(5)*(x + 2))*exp(5)*log(5))/(exp(2*exp(5)*(x + 2)) + exp(2*x) +
16*x^2*log(2)^2 - exp(exp(5)*(x + 2))*(2*exp(x) + 8*x*log(2) - 2) - 8*x*log(2) + exp(x)*(8*x*log(2) - 2) + 1)
,x)

[Out]

int(-(4*log(2)*log(5) + exp(x)*log(5) - exp(exp(5)*(x + 2))*exp(5)*log(5))/(exp(2*exp(5)*(x + 2)) + exp(2*x) +
16*x^2*log(2)^2 - exp(exp(5)*(x + 2))*(2*exp(x) + 8*x*log(2) - 2) - 8*x*log(2) + exp(x)*(8*x*log(2) - 2) + 1)
, 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((exp(5)*ln(5)*exp((2+x)*exp(5))-exp(x)*ln(5)-4*ln(2)*ln(5))/(exp((2+x)*exp(5))**2+(-2*exp(x)-8*x*ln(
2)+2)*exp((2+x)*exp(5))+exp(x)**2+(8*x*ln(2)-2)*exp(x)+16*x**2*ln(2)**2-8*x*ln(2)+1),x)

[Out]

Timed out

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