3.30.11 \(\int \frac {-1+e^{\frac {5+(10+x) \log (3)+\log (3) \log ^2(16)}{\log (3)}}}{(e^{\frac {5+(10+x) \log (3)+\log (3) \log ^2(16)}{\log (3)}}-x) \log (e^{\frac {5+(10+x) \log (3)+\log (3) \log ^2(16)}{\log (3)}}-x)} \, dx\)

Optimal. Leaf size=21 \[ \log \left (\log \left (e^{10+x+\frac {5}{\log (3)}+\log ^2(16)}-x\right )\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 35, normalized size of antiderivative = 1.67, number of steps used = 1, number of rules used = 1, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6684} \begin {gather*} \log \left (\log \left (e^{\frac {5}{\log (3)}} 3^{\frac {x+10}{\log (3)}+\frac {\log ^2(16)}{\log (3)}}-x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + E^((5 + (10 + x)*Log[3] + Log[3]*Log[16]^2)/Log[3]))/((E^((5 + (10 + x)*Log[3] + Log[3]*Log[16]^2)/L
og[3]) - x)*Log[E^((5 + (10 + x)*Log[3] + Log[3]*Log[16]^2)/Log[3]) - x]),x]

[Out]

Log[Log[3^((10 + x)/Log[3] + Log[16]^2/Log[3])*E^(5/Log[3]) - x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (\log \left (3^{\frac {10+x}{\log (3)}+\frac {\log ^2(16)}{\log (3)}} e^{\frac {5}{\log (3)}}-x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 21, normalized size = 1.00 \begin {gather*} \log \left (\log \left (e^{10+x+\frac {5}{\log (3)}+\log ^2(16)}-x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^((5 + (10 + x)*Log[3] + Log[3]*Log[16]^2)/Log[3]))/((E^((5 + (10 + x)*Log[3] + Log[3]*Log[16
]^2)/Log[3]) - x)*Log[E^((5 + (10 + x)*Log[3] + Log[3]*Log[16]^2)/Log[3]) - x]),x]

[Out]

Log[Log[E^(10 + x + 5/Log[3] + Log[16]^2) - x]]

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fricas [A]  time = 0.79, size = 26, normalized size = 1.24 \begin {gather*} \log \left (\log \left (-x + e^{\left (\frac {{\left (16 \, \log \relax (2)^{2} + x + 10\right )} \log \relax (3) + 5}{\log \relax (3)}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3))-1)/(exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3
))-x)/log(exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3))-x),x, algorithm="fricas")

[Out]

log(log(-x + e^(((16*log(2)^2 + x + 10)*log(3) + 5)/log(3))))

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giac [A]  time = 0.27, size = 28, normalized size = 1.33 \begin {gather*} \log \left (\log \left (-x + e^{\left (x + \frac {16 \, \log \relax (3) \log \relax (2)^{2} + 10 \, \log \relax (3) + 5}{\log \relax (3)}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3))-1)/(exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3
))-x)/log(exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3))-x),x, algorithm="giac")

[Out]

log(log(-x + e^(x + (16*log(3)*log(2)^2 + 10*log(3) + 5)/log(3))))

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maple [A]  time = 0.03, size = 29, normalized size = 1.38

method	result	size
derivativedivides	\(\ln \left (\ln \left ({\mathrm e}^{\frac {16 \ln \relax (2)^{2} \ln \relax (3)+\left (x +10\right ) \ln \relax (3)+5}{\ln \relax (3)}}-x \right )\right )\)	\(29\)
default	\(\ln \left (\ln \left ({\mathrm e}^{\frac {16 \ln \relax (2)^{2} \ln \relax (3)+\left (x +10\right ) \ln \relax (3)+5}{\ln \relax (3)}}-x \right )\right )\)	\(29\)
norman	\(\ln \left (\ln \left ({\mathrm e}^{\frac {16 \ln \relax (2)^{2} \ln \relax (3)+\left (x +10\right ) \ln \relax (3)+5}{\ln \relax (3)}}-x \right )\right )\)	\(29\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{\frac {16 \ln \relax (2)^{2} \ln \relax (3)+x \ln \relax (3)+10 \ln \relax (3)+5}{\ln \relax (3)}}-x \right )\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((16*ln(2)^2*ln(3)+(x+10)*ln(3)+5)/ln(3))-1)/(exp((16*ln(2)^2*ln(3)+(x+10)*ln(3)+5)/ln(3))-x)/ln(exp((
16*ln(2)^2*ln(3)+(x+10)*ln(3)+5)/ln(3))-x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp((16*ln(2)^2*ln(3)+(x+10)*ln(3)+5)/ln(3))-x))

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maxima [A]  time = 0.65, size = 22, normalized size = 1.05 \begin {gather*} \log \left (\log \left (-x + e^{\left (16 \, \log \relax (2)^{2} + x + \frac {5}{\log \relax (3)} + 10\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3))-1)/(exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3
))-x)/log(exp((16*log(2)^2*log(3)+(x+10)*log(3)+5)/log(3))-x),x, algorithm="maxima")

[Out]

log(log(-x + e^(16*log(2)^2 + x + 5/log(3) + 10)))

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mupad [B]  time = 2.12, size = 25, normalized size = 1.19 \begin {gather*} \ln \left (\ln \left ({\mathrm {e}}^{10}\,{\mathrm {e}}^{\frac {5}{\ln \relax (3)}}\,{\mathrm {e}}^{16\,{\ln \relax (2)}^2}\,{\mathrm {e}}^x-x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((16*log(2)^2*log(3) + log(3)*(x + 10) + 5)/log(3)) - 1)/(log(exp((16*log(2)^2*log(3) + log(3)*(x + 1
0) + 5)/log(3)) - x)*(x - exp((16*log(2)^2*log(3) + log(3)*(x + 10) + 5)/log(3)))),x)

[Out]

log(log(exp(10)*exp(5/log(3))*exp(16*log(2)^2)*exp(x) - x))

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sympy [A]  time = 0.20, size = 27, normalized size = 1.29 \begin {gather*} \log {\left (\log {\left (- x + e^{\frac {\left (x + 10\right ) \log {\relax (3 )} + 5 + 16 \log {\relax (2 )}^{2} \log {\relax (3 )}}{\log {\relax (3 )}}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp((16*ln(2)**2*ln(3)+(x+10)*ln(3)+5)/ln(3))-1)/(exp((16*ln(2)**2*ln(3)+(x+10)*ln(3)+5)/ln(3))-x)/
ln(exp((16*ln(2)**2*ln(3)+(x+10)*ln(3)+5)/ln(3))-x),x)

[Out]

log(log(-x + exp(((x + 10)*log(3) + 5 + 16*log(2)**2*log(3))/log(3))))

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