\(\int \frac {e^x (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2))+128 x^7 \log ^8(4)}{e^x (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2))+16 x^8 \log ^8(4)} \, dx\) [56]

Optimal result

Integrand size = 75, antiderivative size = 20 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log \left (e^x+\frac {16 x^8 \log ^8(4)}{(1+\log (2))^4}\right ) \]

[Out]

ln(4096*ln(2)^8*x^8/(1+ln(2))^4+exp(x))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6816} \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log \left (16 x^8 \log ^8(4)+e^x (1+\log (2))^4\right ) \]

[In]

Int[(E^x*(1 + 4*Log[2] + 6*Log[2]^2 + 4*Log[2]^3 + Log[2]^4) + 128*x^7*Log[4]^8)/(E^x*(1 + 4*Log[2] + 6*Log[2]
^2 + 4*Log[2]^3 + Log[2]^4) + 16*x^8*Log[4]^8),x]

[Out]

Log[E^x*(1 + Log[2])^4 + 16*x^8*Log[4]^8]

Rule 6816

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

Rubi steps \begin{align*} \text {integral}& = \log \left (e^x (1+\log (2))^4+16 x^8 \log ^8(4)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log \left (e^x (1+\log (2))^4+16 x^8 \log ^8(4)\right ) \]

[In]

Integrate[(E^x*(1 + 4*Log[2] + 6*Log[2]^2 + 4*Log[2]^3 + Log[2]^4) + 128*x^7*Log[4]^8)/(E^x*(1 + 4*Log[2] + 6*
Log[2]^2 + 4*Log[2]^3 + Log[2]^4) + 16*x^8*Log[4]^8),x]

[Out]

Log[E^x*(1 + Log[2])^4 + 16*x^8*Log[4]^8]

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85

method	result	size
derivativedivides	\(\ln \left (\left (\ln \left (2\right )^{4}+4 \ln \left (2\right )^{3}+6 \ln \left (2\right )^{2}+4 \ln \left (2\right )+1\right ) {\mathrm e}^{x}+4096 x^{8} \ln \left (2\right )^{8}\right )\)	\(37\)
default	\(\ln \left (\left (\ln \left (2\right )^{4}+4 \ln \left (2\right )^{3}+6 \ln \left (2\right )^{2}+4 \ln \left (2\right )+1\right ) {\mathrm e}^{x}+4096 x^{8} \ln \left (2\right )^{8}\right )\)	\(37\)
risch	\(\ln \left ({\mathrm e}^{x}+\frac {4096 x^{8} \ln \left (2\right )^{8}}{\ln \left (2\right )^{4}+4 \ln \left (2\right )^{3}+6 \ln \left (2\right )^{2}+4 \ln \left (2\right )+1}\right )\)	\(38\)
norman	\(\ln \left (4096 x^{8} \ln \left (2\right )^{8}+\ln \left (2\right )^{4} {\mathrm e}^{x}+4 \ln \left (2\right )^{3} {\mathrm e}^{x}+6 \ln \left (2\right )^{2} {\mathrm e}^{x}+4 \,{\mathrm e}^{x} \ln \left (2\right )+{\mathrm e}^{x}\right )\)	\(43\)
parallelrisch	\(\ln \left (\frac {4096 x^{8} \ln \left (2\right )^{8}+\ln \left (2\right )^{4} {\mathrm e}^{x}+4 \ln \left (2\right )^{3} {\mathrm e}^{x}+6 \ln \left (2\right )^{2} {\mathrm e}^{x}+4 \,{\mathrm e}^{x} \ln \left (2\right )+{\mathrm e}^{x}}{4096 \ln \left (2\right )^{8}}\right )\)	\(49\)

[In]

int(((ln(2)^4+4*ln(2)^3+6*ln(2)^2+4*ln(2)+1)*exp(x)+32768*x^7*ln(2)^8)/((ln(2)^4+4*ln(2)^3+6*ln(2)^2+4*ln(2)+1
)*exp(x)+4096*x^8*ln(2)^8),x,method=_RETURNVERBOSE)

[Out]

ln((ln(2)^4+4*ln(2)^3+6*ln(2)^2+4*ln(2)+1)*exp(x)+4096*x^8*ln(2)^8)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log \left (4096 \, x^{8} \log \left (2\right )^{8} + {\left (\log \left (2\right )^{4} + 4 \, \log \left (2\right )^{3} + 6 \, \log \left (2\right )^{2} + 4 \, \log \left (2\right ) + 1\right )} e^{x}\right ) \]

[In]

integrate(((log(2)^4+4*log(2)^3+6*log(2)^2+4*log(2)+1)*exp(x)+32768*x^7*log(2)^8)/((log(2)^4+4*log(2)^3+6*log(
2)^2+4*log(2)+1)*exp(x)+4096*x^8*log(2)^8),x, algorithm="fricas")

[Out]

log(4096*x^8*log(2)^8 + (log(2)^4 + 4*log(2)^3 + 6*log(2)^2 + 4*log(2) + 1)*e^x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log {\left (\frac {4096 x^{8} \log {\left (2 \right )}^{8}}{\log {\left (2 \right )}^{4} + 1 + 4 \log {\left (2 \right )}^{3} + 4 \log {\left (2 \right )} + 6 \log {\left (2 \right )}^{2}} + e^{x} \right )} \]

[In]

integrate(((ln(2)**4+4*ln(2)**3+6*ln(2)**2+4*ln(2)+1)*exp(x)+32768*x**7*ln(2)**8)/((ln(2)**4+4*ln(2)**3+6*ln(2
)**2+4*ln(2)+1)*exp(x)+4096*x**8*ln(2)**8),x)

[Out]

log(4096*x**8*log(2)**8/(log(2)**4 + 1 + 4*log(2)**3 + 4*log(2) + 6*log(2)**2) + exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log \left (4096 \, x^{8} \log \left (2\right )^{8} + {\left (\log \left (2\right )^{4} + 4 \, \log \left (2\right )^{3} + 6 \, \log \left (2\right )^{2} + 4 \, \log \left (2\right ) + 1\right )} e^{x}\right ) \]

[In]

integrate(((log(2)^4+4*log(2)^3+6*log(2)^2+4*log(2)+1)*exp(x)+32768*x^7*log(2)^8)/((log(2)^4+4*log(2)^3+6*log(
2)^2+4*log(2)+1)*exp(x)+4096*x^8*log(2)^8),x, algorithm="maxima")

[Out]

log(4096*x^8*log(2)^8 + (log(2)^4 + 4*log(2)^3 + 6*log(2)^2 + 4*log(2) + 1)*e^x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.10 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\log \left (4096 \, x^{8} \log \left (2\right )^{8} + e^{x} \log \left (2\right )^{4} + 4 \, e^{x} \log \left (2\right )^{3} + 6 \, e^{x} \log \left (2\right )^{2} + 4 \, e^{x} \log \left (2\right ) + e^{x}\right ) \]

[In]

integrate(((log(2)^4+4*log(2)^3+6*log(2)^2+4*log(2)+1)*exp(x)+32768*x^7*log(2)^8)/((log(2)^4+4*log(2)^3+6*log(
2)^2+4*log(2)+1)*exp(x)+4096*x^8*log(2)^8),x, algorithm="giac")

[Out]

log(4096*x^8*log(2)^8 + e^x*log(2)^4 + 4*e^x*log(2)^3 + 6*e^x*log(2)^2 + 4*e^x*log(2) + e^x)

Mupad [B] (verification not implemented)

Time = 8.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.10 \[ \int \frac {e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+128 x^7 \log ^8(4)}{e^x \left (1+4 \log (2)+6 \log ^2(2)+4 \log ^3(2)+\log ^4(2)\right )+16 x^8 \log ^8(4)} \, dx=\ln \left ({\mathrm {e}}^x+4096\,x^8\,{\ln \left (2\right )}^8+6\,{\mathrm {e}}^x\,{\ln \left (2\right )}^2+4\,{\mathrm {e}}^x\,{\ln \left (2\right )}^3+{\mathrm {e}}^x\,{\ln \left (2\right )}^4+4\,{\mathrm {e}}^x\,\ln \left (2\right )\right ) \]

[In]

int((32768*x^7*log(2)^8 + exp(x)*(4*log(2) + 6*log(2)^2 + 4*log(2)^3 + log(2)^4 + 1))/(4096*x^8*log(2)^8 + exp
(x)*(4*log(2) + 6*log(2)^2 + 4*log(2)^3 + log(2)^4 + 1)),x)

[Out]

log(exp(x) + 4096*x^8*log(2)^8 + 6*exp(x)*log(2)^2 + 4*exp(x)*log(2)^3 + exp(x)*log(2)^4 + 4*exp(x)*log(2))