3.55.46 \(\int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+(-2 x-2 e x-2 x^2) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx\)

Optimal. Leaf size=24 \[ \log \left (e^{25+x}-(-1-e-x+x \log (\log (2)))^2\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 46, normalized size of antiderivative = 1.92, number of steps used = 3, number of rules used = 2, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6, 6684} \begin {gather*} \log \left (x^2 \left (1+\log ^2(\log (2))\right )-2 \left (x^2+e x+x\right ) \log (\log (2))+2 x-e^{x+25}+2 e (x+1)+e^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + E^2 - E^(25 + x) + 2*x + 2*E*(1 + x) - 2*(x + E*x + x^2)*Log[Log[2]] + x^2*(1 + Log[Log[2]]^2)]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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} &=\int \frac {2+2 e-e^{25+x}+(-2-2 e-4 x) \log (\log (2))+x \left (2+2 \log ^2(\log (2))\right )}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx\\ &=\int \frac {2+2 e-e^{25+x}+(-2-2 e-4 x) \log (\log (2))+x \left (2+2 \log ^2(\log (2))\right )}{1+e^2-e^{25+x}+2 x+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \left (1+\log ^2(\log (2))\right )} \, dx\\ &=\log \left (1+e^2-e^{25+x}+2 x+2 e (1+x)-2 \left (x+e x+x^2\right ) \log (\log (2))+x^2 \left (1+\log ^2(\log (2))\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.45, size = 37, normalized size = 1.54 \begin {gather*} \log \left (-e^2+e^{25+x}+2 e (-1+x (-1+\log (\log (2))))-(-1+x (-1+\log (\log (2))))^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[-E^2 + E^(25 + x) + 2*E*(-1 + x*(-1 + Log[Log[2]])) - (-1 + x*(-1 + Log[Log[2]]))^2]

________________________________________________________________________________________

fricas [B]  time = 1.42, size = 50, normalized size = 2.08 \begin {gather*} \log \left (-x^{2} \log \left (\log \relax (2)\right )^{2} - x^{2} - 2 \, {\left (x + 1\right )} e + 2 \, {\left (x^{2} + x e + x\right )} \log \left (\log \relax (2)\right ) - 2 \, x - e^{2} + e^{\left (x + 25\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp(1)+2*x+2)/(x^2*log(log(2))^2+(-2*x*
exp(1)-2*x^2-2*x)*log(log(2))-exp(x+25)+exp(1)^2+(2*x+2)*exp(1)+x^2+2*x+1),x, algorithm="fricas")

[Out]

log(-x^2*log(log(2))^2 - x^2 - 2*(x + 1)*e + 2*(x^2 + x*e + x)*log(log(2)) - 2*x - e^2 + e^(x + 25) - 1)

________________________________________________________________________________________

giac [B]  time = 0.16, size = 60, normalized size = 2.50 \begin {gather*} \log \left (-x^{2} \log \left (\log \relax (2)\right )^{2} + 2 \, x^{2} \log \left (\log \relax (2)\right ) + 2 \, x e \log \left (\log \relax (2)\right ) - x^{2} - 2 \, x e + 2 \, x \log \left (\log \relax (2)\right ) - 2 \, x - e^{2} - 2 \, e + e^{\left (x + 25\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp(1)+2*x+2)/(x^2*log(log(2))^2+(-2*x*
exp(1)-2*x^2-2*x)*log(log(2))-exp(x+25)+exp(1)^2+(2*x+2)*exp(1)+x^2+2*x+1),x, algorithm="giac")

[Out]

log(-x^2*log(log(2))^2 + 2*x^2*log(log(2)) + 2*x*e*log(log(2)) - x^2 - 2*x*e + 2*x*log(log(2)) - 2*x - e^2 - 2
*e + e^(x + 25) - 1)

________________________________________________________________________________________

maple [B]  time = 0.09, size = 55, normalized size = 2.29

method	result	size
derivativedivides	\(\ln \left (x^{2} \ln \left (\ln \relax (2)\right )^{2}+\left (-2 x \,{\mathrm e}-2 x^{2}-2 x \right ) \ln \left (\ln \relax (2)\right )-{\mathrm e}^{x +25}+{\mathrm e}^{2}+\left (2 x +2\right ) {\mathrm e}+x^{2}+2 x +1\right )\)	\(55\)
default	\(\ln \left (x^{2} \ln \left (\ln \relax (2)\right )^{2}+\left (-2 x \,{\mathrm e}-2 x^{2}-2 x \right ) \ln \left (\ln \relax (2)\right )-{\mathrm e}^{x +25}+{\mathrm e}^{2}+\left (2 x +2\right ) {\mathrm e}+x^{2}+2 x +1\right )\)	\(55\)
norman	\(\ln \left (x^{2} \ln \left (\ln \relax (2)\right )^{2}-2 \,{\mathrm e} \ln \left (\ln \relax (2)\right ) x -2 x^{2} \ln \left (\ln \relax (2)\right )+{\mathrm e}^{2}+2 x \,{\mathrm e}-2 x \ln \left (\ln \relax (2)\right )+x^{2}+2 \,{\mathrm e}-{\mathrm e}^{x +25}+2 x +1\right )\)	\(60\)
risch	\(-25+\ln \left (-x^{2} \ln \left (\ln \relax (2)\right )^{2}+2 \,{\mathrm e} \ln \left (\ln \relax (2)\right ) x +2 x^{2} \ln \left (\ln \relax (2)\right )-{\mathrm e}^{2}-2 x \,{\mathrm e}+2 x \ln \left (\ln \relax (2)\right )-x^{2}-2 \,{\mathrm e}-2 x +{\mathrm e}^{x +25}-1\right )\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(ln(2))^2+(-2*exp(1)-4*x-2)*ln(ln(2))-exp(x+25)+2*exp(1)+2*x+2)/(x^2*ln(ln(2))^2+(-2*x*exp(1)-2*x^2
-2*x)*ln(ln(2))-exp(x+25)+exp(1)^2+(2*x+2)*exp(1)+x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

ln(x^2*ln(ln(2))^2+(-2*x*exp(1)-2*x^2-2*x)*ln(ln(2))-exp(x+25)+exp(1)^2+(2*x+2)*exp(1)+x^2+2*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 47, normalized size = 1.96 \begin {gather*} \log \left (x^{2} \log \left (\log \relax (2)\right )^{2} + x^{2} + 2 \, {\left (x + 1\right )} e - 2 \, {\left (x^{2} + x e + x\right )} \log \left (\log \relax (2)\right ) + 2 \, x + e^{2} - e^{\left (x + 25\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp(1)+2*x+2)/(x^2*log(log(2))^2+(-2*x*
exp(1)-2*x^2-2*x)*log(log(2))-exp(x+25)+exp(1)^2+(2*x+2)*exp(1)+x^2+2*x+1),x, algorithm="maxima")

[Out]

log(x^2*log(log(2))^2 + x^2 + 2*(x + 1)*e - 2*(x^2 + x*e + x)*log(log(2)) + 2*x + e^2 - e^(x + 25) + 1)

________________________________________________________________________________________

mupad [B]  time = 4.28, size = 53, normalized size = 2.21 \begin {gather*} \ln \left (2\,x-{\mathrm {e}}^{x+25}+2\,\mathrm {e}+{\mathrm {e}}^2-2\,x^2\,\ln \left (\ln \relax (2)\right )+2\,x\,\mathrm {e}+x^2\,{\ln \left (\ln \relax (2)\right )}^2+x^2-2\,x\,\ln \left (\ln \relax (2)\right )\,\left (\mathrm {e}+1\right )+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - exp(x + 25) + 2*exp(1) + 2*x*log(log(2))^2 - log(log(2))*(4*x + 2*exp(1) + 2) + 2)/(2*x - exp(x + 2
5) + exp(2) + x^2*log(log(2))^2 + x^2 - log(log(2))*(2*x + 2*x*exp(1) + 2*x^2) + exp(1)*(2*x + 2) + 1),x)

[Out]

log(2*x - exp(x + 25) + 2*exp(1) + exp(2) - 2*x^2*log(log(2)) + 2*x*exp(1) + x^2*log(log(2))^2 + x^2 - 2*x*log
(log(2))*(exp(1) + 1) + 1)

________________________________________________________________________________________

sympy [B]  time = 0.21, size = 70, normalized size = 2.92 \begin {gather*} \log {\left (- x^{2} + 2 x^{2} \log {\left (\log {\relax (2 )} \right )} - x^{2} \log {\left (\log {\relax (2 )} \right )}^{2} - 2 e x - 2 x + 2 e x \log {\left (\log {\relax (2 )} \right )} + 2 x \log {\left (\log {\relax (2 )} \right )} + e^{x + 25} - e^{2} - 2 e - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(ln(2))**2+(-2*exp(1)-4*x-2)*ln(ln(2))-exp(x+25)+2*exp(1)+2*x+2)/(x**2*ln(ln(2))**2+(-2*x*exp
(1)-2*x**2-2*x)*ln(ln(2))-exp(x+25)+exp(1)**2+(2*x+2)*exp(1)+x**2+2*x+1),x)

[Out]

log(-x**2 + 2*x**2*log(log(2)) - x**2*log(log(2))**2 - 2*E*x - 2*x + 2*E*x*log(log(2)) + 2*x*log(log(2)) + exp
(x + 25) - exp(2) - 2*E - 1)

________________________________________________________________________________________