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

Optimal. Leaf size=20 \[ \log \left (e^x+\frac {16 x^8 \log ^8(4)}{(1+\log (2))^4}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6684} \begin {gather*} \log \left (16 x^8 \log ^8(4)+e^x (1+\log (2))^4\right ) \end {gather*}

Antiderivative was successfully verified.

[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 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 (e^x (1+\log (2))^4+16 x^8 \log ^8(4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 21, normalized size = 1.05 \begin {gather*} \log \left (e^x (1+\log (2))^4+16 x^8 \log ^8(4)\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.70, size = 36, normalized size = 1.80 \begin {gather*} \log \left (4096 \, x^{8} \log \relax (2)^{8} + {\left (\log \relax (2)^{4} + 4 \, \log \relax (2)^{3} + 6 \, \log \relax (2)^{2} + 4 \, \log \relax (2) + 1\right )} e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.25, size = 42, normalized size = 2.10 \begin {gather*} \log \left (4096 \, x^{8} \log \relax (2)^{8} + e^{x} \log \relax (2)^{4} + 4 \, e^{x} \log \relax (2)^{3} + 6 \, e^{x} \log \relax (2)^{2} + 4 \, e^{x} \log \relax (2) + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 37, normalized size = 1.85

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.36, size = 36, normalized size = 1.80 \begin {gather*} \log \left (4096 \, x^{8} \log \relax (2)^{8} + {\left (\log \relax (2)^{4} + 4 \, \log \relax (2)^{3} + 6 \, \log \relax (2)^{2} + 4 \, \log \relax (2) + 1\right )} e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.89, size = 42, normalized size = 2.10 \begin {gather*} \ln \left ({\mathrm {e}}^x+4096\,x^8\,{\ln \relax (2)}^8+6\,{\mathrm {e}}^x\,{\ln \relax (2)}^2+4\,{\mathrm {e}}^x\,{\ln \relax (2)}^3+{\mathrm {e}}^x\,{\ln \relax (2)}^4+4\,{\mathrm {e}}^x\,\ln \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.24, size = 39, normalized size = 1.95 \begin {gather*} \log {\left (\frac {4096 x^{8} \log {\relax (2 )}^{8}}{\log {\relax (2 )}^{4} + 1 + 4 \log {\relax (2 )}^{3} + 4 \log {\relax (2 )} + 6 \log {\relax (2 )}^{2}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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