\(\int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+(10 x+2 x^2) \log (3)+(5+x) \log ^2(3)+(16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))) \log (\log (4))} \, dx\) [533]

Optimal result

Integrand size = 94, antiderivative size = 25 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\log \left (4+\frac {e^x}{x+\log (3)}+\frac {5+x}{4 \log (\log (4))}\right ) \]

[Out]

ln(exp(x)/(ln(3)+x)+4+1/4*(5+x)/ln(2*ln(2)))

Rubi [F]

\[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx \]

[In]

Int[(x^2 + 2*x*Log[3] + Log[3]^2 + E^x*(-4 + 4*x + 4*Log[3])*Log[Log[4]])/(5*x^2 + x^3 + (10*x + 2*x^2)*Log[3]
+ (5 + x)*Log[3]^2 + (16*x^2 + 32*x*Log[3] + 16*Log[3]^2 + E^x*(4*x + 4*Log[3]))*Log[Log[4]]),x]

[Out]

x - Log[x + Log[3]] - (5 - Log[81] + 16*(1 - Log[3])*Log[Log[4]])*Defer[Int][(-x^2 - 4*E^x*Log[Log[4]] - 5*Log
[3]*(1 + (16*Log[Log[4]])/5) - 5*x*(1 + (Log[3] + 16*Log[Log[4]])/5))^(-1), x] + Defer[Int][x^2/(-x^2 - 4*E^x*
Log[Log[4]] - 5*Log[3]*(1 + (16*Log[Log[4]])/5) - 5*x*(1 + (Log[3] + 16*Log[Log[4]])/5)), x] - (3 + Log[3] + 1
6*Log[Log[4]])*Defer[Int][x/(x^2 + 4*E^x*Log[Log[4]] + 5*Log[3]*(1 + (16*Log[Log[4]])/5) + 5*x*(1 + (Log[3] +
16*Log[Log[4]])/5)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{(x+\log (3)) \left (x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )\right )} \, dx \\ & = \int \left (\frac {-1+x+\log (3)}{x+\log (3)}+\frac {5-x^2-\log (81)+16 (1-\log (3)) \log (\log (4))-x (3+\log (3)+16 \log (\log (4)))}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}\right ) \, dx \\ & = \int \frac {-1+x+\log (3)}{x+\log (3)} \, dx+\int \frac {5-x^2-\log (81)+16 (1-\log (3)) \log (\log (4))-x (3+\log (3)+16 \log (\log (4)))}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx \\ & = \int \left (1+\frac {1}{-x-\log (3)}\right ) \, dx+\int \left (\frac {x^2}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}+\frac {\log (81) \left (1+\frac {-5+16 (-1+\log (3)) \log (\log (4))}{\log (81)}\right )}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}+\frac {x (-3-\log (3)-16 \log (\log (4)))}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )}\right ) \, dx \\ & = x-\log (x+\log (3))+(-3-\log (3)-16 \log (\log (4))) \int \frac {x}{x^2+4 e^x \log (\log (4))+5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )+5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx+(-5+\log (81)-16 (1-\log (3)) \log (\log (4))) \int \frac {1}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx+\int \frac {x^2}{-x^2-4 e^x \log (\log (4))-5 \log (3) \left (1+\frac {16}{5} \log (\log (4))\right )-5 x \left (1+\frac {1}{5} (\log (3)+16 \log (\log (4)))\right )} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx \]

[In]

Integrate[(x^2 + 2*x*Log[3] + Log[3]^2 + E^x*(-4 + 4*x + 4*Log[3])*Log[Log[4]])/(5*x^2 + x^3 + (10*x + 2*x^2)*
Log[3] + (5 + x)*Log[3]^2 + (16*x^2 + 32*x*Log[3] + 16*Log[3]^2 + E^x*(4*x + 4*Log[3]))*Log[Log[4]]),x]

[Out]

Integrate[(x^2 + 2*x*Log[3] + Log[3]^2 + E^x*(-4 + 4*x + 4*Log[3])*Log[Log[4]])/(5*x^2 + x^3 + (10*x + 2*x^2)*
Log[3] + (5 + x)*Log[3]^2 + (16*x^2 + 32*x*Log[3] + 16*Log[3]^2 + E^x*(4*x + 4*Log[3]))*Log[Log[4]]), x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(50\) vs. \(2(24)=48\).

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04

method	result	size
norman	\(-\ln \left (\ln \left (3\right )+x \right )+\ln \left (16 \ln \left (3\right ) \ln \left (2 \ln \left (2\right )\right )+x \ln \left (3\right )+16 x \ln \left (2 \ln \left (2\right )\right )+4 \,{\mathrm e}^{x} \ln \left (2 \ln \left (2\right )\right )+x^{2}+5 \ln \left (3\right )+5 x \right )\)	\(51\)
parallelrisch	\(-\ln \left (\ln \left (3\right )+x \right )+\ln \left (16 \ln \left (3\right ) \ln \left (2 \ln \left (2\right )\right )+x \ln \left (3\right )+16 x \ln \left (2 \ln \left (2\right )\right )+4 \,{\mathrm e}^{x} \ln \left (2 \ln \left (2\right )\right )+x^{2}+5 \ln \left (3\right )+5 x \right )\)	\(51\)
risch	\(-\ln \left (\ln \left (3\right )+x \right )+\ln \left ({\mathrm e}^{x}+\frac {16 \ln \left (2\right ) \ln \left (3\right )+16 \ln \left (3\right ) \ln \left (\ln \left (2\right )\right )+x \ln \left (3\right )+16 x \ln \left (2\right )+16 x \ln \left (\ln \left (2\right )\right )+x^{2}+5 \ln \left (3\right )+5 x}{4 \ln \left (2\right )+4 \ln \left (\ln \left (2\right )\right )}\right )\)	\(62\)

[In]

int(((4*ln(3)+4*x-4)*exp(x)*ln(2*ln(2))+ln(3)^2+2*x*ln(3)+x^2)/(((4*ln(3)+4*x)*exp(x)+16*ln(3)^2+32*x*ln(3)+16
*x^2)*ln(2*ln(2))+(5+x)*ln(3)^2+(2*x^2+10*x)*ln(3)+x^3+5*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(3)+x)+ln(16*ln(3)*ln(2*ln(2))+x*ln(3)+16*x*ln(2*ln(2))+4*exp(x)*ln(2*ln(2))+x^2+5*ln(3)+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\log \left (x^{2} + {\left (x + 5\right )} \log \left (3\right ) + 4 \, {\left (4 \, x + e^{x} + 4 \, \log \left (3\right )\right )} \log \left (2 \, \log \left (2\right )\right ) + 5 \, x\right ) - \log \left (x + \log \left (3\right )\right ) \]

[In]

integrate(((4*log(3)+4*x-4)*exp(x)*log(2*log(2))+log(3)^2+2*x*log(3)+x^2)/(((4*log(3)+4*x)*exp(x)+16*log(3)^2+
32*x*log(3)+16*x^2)*log(2*log(2))+(5+x)*log(3)^2+(2*x^2+10*x)*log(3)+x^3+5*x^2),x, algorithm="fricas")

[Out]

log(x^2 + (x + 5)*log(3) + 4*(4*x + e^x + 4*log(3))*log(2*log(2)) + 5*x) - log(x + log(3))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (24) = 48\).

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=- \log {\left (x + \log {\left (3 \right )} \right )} + \log {\left (\frac {x^{2} + 16 x \log {\left (\log {\left (2 \right )} \right )} + x \log {\left (3 \right )} + 5 x + 16 x \log {\left (2 \right )} + 16 \log {\left (3 \right )} \log {\left (\log {\left (2 \right )} \right )} + 5 \log {\left (3 \right )} + 16 \log {\left (2 \right )} \log {\left (3 \right )}}{4 \log {\left (\log {\left (2 \right )} \right )} + 4 \log {\left (2 \right )}} + e^{x} \right )} \]

[In]

integrate(((4*ln(3)+4*x-4)*exp(x)*ln(2*ln(2))+ln(3)**2+2*x*ln(3)+x**2)/(((4*ln(3)+4*x)*exp(x)+16*ln(3)**2+32*x
*ln(3)+16*x**2)*ln(2*ln(2))+(5+x)*ln(3)**2+(2*x**2+10*x)*ln(3)+x**3+5*x**2),x)

[Out]

-log(x + log(3)) + log((x**2 + 16*x*log(log(2)) + x*log(3) + 5*x + 16*x*log(2) + 16*log(3)*log(log(2)) + 5*log
(3) + 16*log(2)*log(3))/(4*log(log(2)) + 4*log(2)) + exp(x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (24) = 48\).

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.56 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=-\log \left (x + \log \left (3\right )\right ) + \log \left (\frac {x^{2} + x {\left (\log \left (3\right ) + 16 \, \log \left (2\right ) + 16 \, \log \left (\log \left (2\right )\right ) + 5\right )} + 4 \, {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} e^{x} + {\left (16 \, \log \left (\log \left (2\right )\right ) + 5\right )} \log \left (3\right ) + 16 \, \log \left (3\right ) \log \left (2\right )}{4 \, {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )}}\right ) \]

[In]

integrate(((4*log(3)+4*x-4)*exp(x)*log(2*log(2))+log(3)^2+2*x*log(3)+x^2)/(((4*log(3)+4*x)*exp(x)+16*log(3)^2+
32*x*log(3)+16*x^2)*log(2*log(2))+(5+x)*log(3)^2+(2*x^2+10*x)*log(3)+x^3+5*x^2),x, algorithm="maxima")

[Out]

-log(x + log(3)) + log(1/4*(x^2 + x*(log(3) + 16*log(2) + 16*log(log(2)) + 5) + 4*(log(2) + log(log(2)))*e^x +
(16*log(log(2)) + 5)*log(3) + 16*log(3)*log(2))/(log(2) + log(log(2))))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).

Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.44 \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\log \left (x^{2} + x \log \left (3\right ) + 16 \, x \log \left (2\right ) + 4 \, e^{x} \log \left (2\right ) + 16 \, \log \left (3\right ) \log \left (2\right ) + 16 \, x \log \left (\log \left (2\right )\right ) + 4 \, e^{x} \log \left (\log \left (2\right )\right ) + 16 \, \log \left (3\right ) \log \left (\log \left (2\right )\right ) + 5 \, x + 5 \, \log \left (3\right )\right ) - \log \left (x + \log \left (3\right )\right ) \]

[In]

integrate(((4*log(3)+4*x-4)*exp(x)*log(2*log(2))+log(3)^2+2*x*log(3)+x^2)/(((4*log(3)+4*x)*exp(x)+16*log(3)^2+
32*x*log(3)+16*x^2)*log(2*log(2))+(5+x)*log(3)^2+(2*x^2+10*x)*log(3)+x^3+5*x^2),x, algorithm="giac")

[Out]

log(x^2 + x*log(3) + 16*x*log(2) + 4*e^x*log(2) + 16*log(3)*log(2) + 16*x*log(log(2)) + 4*e^x*log(log(2)) + 16
*log(3)*log(log(2)) + 5*x + 5*log(3)) - log(x + log(3))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2+2 x \log (3)+\log ^2(3)+e^x (-4+4 x+4 \log (3)) \log (\log (4))}{5 x^2+x^3+\left (10 x+2 x^2\right ) \log (3)+(5+x) \log ^2(3)+\left (16 x^2+32 x \log (3)+16 \log ^2(3)+e^x (4 x+4 \log (3))\right ) \log (\log (4))} \, dx=\int \frac {2\,x\,\ln \left (3\right )+{\ln \left (3\right )}^2+x^2+\ln \left (2\,\ln \left (2\right )\right )\,{\mathrm {e}}^x\,\left (4\,x+4\,\ln \left (3\right )-4\right )}{{\ln \left (3\right )}^2\,\left (x+5\right )+\ln \left (3\right )\,\left (2\,x^2+10\,x\right )+5\,x^2+x^3+\ln \left (2\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^x\,\left (4\,x+4\,\ln \left (3\right )\right )+32\,x\,\ln \left (3\right )+16\,{\ln \left (3\right )}^2+16\,x^2\right )} \,d x \]

[In]

int((2*x*log(3) + log(3)^2 + x^2 + log(2*log(2))*exp(x)*(4*x + 4*log(3) - 4))/(log(3)^2*(x + 5) + log(3)*(10*x
+ 2*x^2) + 5*x^2 + x^3 + log(2*log(2))*(exp(x)*(4*x + 4*log(3)) + 32*x*log(3) + 16*log(3)^2 + 16*x^2)),x)

[Out]

int((2*x*log(3) + log(3)^2 + x^2 + log(2*log(2))*exp(x)*(4*x + 4*log(3) - 4))/(log(3)^2*(x + 5) + log(3)*(10*x
+ 2*x^2) + 5*x^2 + x^3 + log(2*log(2))*(exp(x)*(4*x + 4*log(3)) + 32*x*log(3) + 16*log(3)^2 + 16*x^2)), x)