3.96.8 \(\int \frac {(4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} (-20-5\ 2^{5+8 x} x \log (2))) \log (\frac {1}{25} (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)))}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx\) [9508]

Optimal. Leaf size=24 \[ \log ^2\left (4+\left (-e^{2^{8 x}}+\frac {x}{5}\right )^2+\log (2)\right ) \]

[Out]

ln((1/5*x-exp(exp(8*x*ln(2))))^2+4+ln(2))^2

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Rubi [A]
time = 0.53, antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 1, number of rules used = 2, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6816, 6818} \begin {gather*} \log ^2\left (\frac {1}{25} \left (x^2-10 e^{2^{8 x}} x+25 e^{2^{8 x+1}}+25 (4+\log (2))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 25*E^2^
(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])
,x]

[Out]

Log[(25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*(4 + Log[2]))/25]^2

Rule 6816

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

Rule 6818

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2\left (\frac {1}{25} \left (25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 (4+\log (2))\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]
time = 6.90, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 2
5*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*L
og[2]),x]

[Out]

Integrate[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 2
5*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*L
og[2]), x]

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Maple [A]
time = 1.38, size = 26, normalized size = 1.08

method	result	size
risch	\(\ln \left ({\mathrm e}^{2 \,256^{x}}-\frac {2 x \,{\mathrm e}^{256^{x}}}{5}+\ln \left (2\right )+\frac {x^{2}}{25}+4\right )^{2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((800*ln(2)*exp(8*x*ln(2))*exp(exp(8*x*ln(2)))^2+(-160*x*ln(2)*exp(8*x*ln(2))-20)*exp(exp(8*x*ln(2)))+4*x)*
ln(exp(exp(8*x*ln(2)))^2-2/5*x*exp(exp(8*x*ln(2)))+ln(2)+1/25*x^2+4)/(25*exp(exp(8*x*ln(2)))^2-10*x*exp(exp(8*
x*ln(2)))+25*ln(2)+x^2+100),x,method=_RETURNVERBOSE)

[Out]

ln(exp(2*256^x)-2/5*x*exp(256^x)+ln(2)+1/25*x^2+4)^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (21) = 42\).
time = 0.56, size = 92, normalized size = 3.83 \begin {gather*} -\log \left (x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \left (2\right ) + 100\right )^{2} + 2 \, \log \left (x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \left (2\right ) + 100\right ) \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2^{8 \, x + 1}\right )} + \log \left (2\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp(8*x*log(2))-20)*exp(exp(8*x*lo
g(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^
2-10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="maxima")

[Out]

-log(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100)^2 + 2*log(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2
^(8*x)) + 25*log(2) + 100)*log(1/25*x^2 - 2/5*x*e^(2^(8*x)) + e^(2^(8*x + 1)) + log(2) + 4)

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Fricas [A]
time = 0.42, size = 29, normalized size = 1.21 \begin {gather*} \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2 \cdot 2^{8 \, x}\right )} + \log \left (2\right ) + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp(8*x*log(2))-20)*exp(exp(8*x*lo
g(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^
2-10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="fricas")

[Out]

log(1/25*x^2 - 2/5*x*e^(2^(8*x)) + e^(2*2^(8*x)) + log(2) + 4)^2

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Sympy [A]
time = 1.14, size = 39, normalized size = 1.62 \begin {gather*} \log {\left (\frac {x^{2}}{25} - \frac {2 x e^{e^{8 x \log {\left (2 \right )}}}}{5} + e^{2 e^{8 x \log {\left (2 \right )}}} + \log {\left (2 \right )} + 4 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*ln(2)*exp(8*x*ln(2))*exp(exp(8*x*ln(2)))**2+(-160*x*ln(2)*exp(8*x*ln(2))-20)*exp(exp(8*x*ln(2))
)+4*x)*ln(exp(exp(8*x*ln(2)))**2-2/5*x*exp(exp(8*x*ln(2)))+ln(2)+1/25*x**2+4)/(25*exp(exp(8*x*ln(2)))**2-10*x*
exp(exp(8*x*ln(2)))+25*ln(2)+x**2+100),x)

[Out]

log(x**2/25 - 2*x*exp(exp(8*x*log(2)))/5 + exp(2*exp(8*x*log(2))) + log(2) + 4)**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp(8*x*log(2))-20)*exp(exp(8*x*lo
g(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^
2-10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="giac")

[Out]

integrate(4*(200*2^(8*x)*e^(2*2^(8*x))*log(2) - 5*(8*2^(8*x)*x*log(2) + 1)*e^(2^(8*x)) + x)*log(1/25*x^2 - 2/5
*x*e^(2^(8*x)) + e^(2*2^(8*x)) + log(2) + 4)/(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100), x)

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Mupad [B]
time = 8.25, size = 29, normalized size = 1.21 \begin {gather*} {\ln \left (\ln \left (2\right )+{\mathrm {e}}^{2\,2^{8\,x}}-\frac {2\,x\,{\mathrm {e}}^{2^{8\,x}}}{5}+\frac {x^2}{25}+4\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(2) + exp(2*exp(8*x*log(2))) + x^2/25 - (2*x*exp(exp(8*x*log(2))))/5 + 4)*(4*x - exp(exp(8*x*log(2
)))*(160*x*exp(8*x*log(2))*log(2) + 20) + 800*exp(2*exp(8*x*log(2)))*exp(8*x*log(2))*log(2)))/(25*log(2) + 25*
exp(2*exp(8*x*log(2))) + x^2 - 10*x*exp(exp(8*x*log(2))) + 100),x)

[Out]

log(log(2) + exp(2*2^(8*x)) - (2*x*exp(2^(8*x)))/5 + x^2/25 + 4)^2

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