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

Optimal. Leaf size=25 \[ \log \left (4+\frac {e^x}{x+\log (3)}+\frac {5+x}{4 \log (\log (4))}\right ) \]

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Rubi [F]  time = 1.90, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [F]  time = 2.13, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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]

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fricas [A]  time = 0.60, size = 39, normalized size = 1.56 \begin {gather*} \log \left (x^{2} + {\left (x + 5\right )} \log \relax (3) + 4 \, {\left (4 \, x + e^{x} + 4 \, \log \relax (3)\right )} \log \left (2 \, \log \relax (2)\right ) + 5 \, x\right ) - \log \left (x + \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 0.39, size = 61, normalized size = 2.44 \begin {gather*} \log \left (x^{2} + x \log \relax (3) + 16 \, x \log \relax (2) + 4 \, e^{x} \log \relax (2) + 16 \, \log \relax (3) \log \relax (2) + 16 \, x \log \left (\log \relax (2)\right ) + 4 \, e^{x} \log \left (\log \relax (2)\right ) + 16 \, \log \relax (3) \log \left (\log \relax (2)\right ) + 5 \, x + 5 \, \log \relax (3)\right ) - \log \left (x + \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [B]  time = 0.36, size = 51, normalized size = 2.04

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 0.50, size = 64, normalized size = 2.56 \begin {gather*} -\log \left (x + \log \relax (3)\right ) + \log \left (\frac {x^{2} + x {\left (\log \relax (3) + 16 \, \log \relax (2) + 16 \, \log \left (\log \relax (2)\right ) + 5\right )} + 4 \, {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} e^{x} + {\left (16 \, \log \left (\log \relax (2)\right ) + 5\right )} \log \relax (3) + 16 \, \log \relax (3) \log \relax (2)}{4 \, {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,x\,\ln \relax (3)+{\ln \relax (3)}^2+x^2+\ln \left (2\,\ln \relax (2)\right )\,{\mathrm {e}}^x\,\left (4\,x+4\,\ln \relax (3)-4\right )}{{\ln \relax (3)}^2\,\left (x+5\right )+\ln \relax (3)\,\left (2\,x^2+10\,x\right )+5\,x^2+x^3+\ln \left (2\,\ln \relax (2)\right )\,\left ({\mathrm {e}}^x\,\left (4\,x+4\,\ln \relax (3)\right )+32\,x\,\ln \relax (3)+16\,{\ln \relax (3)}^2+16\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 0.60, size = 73, normalized size = 2.92 \begin {gather*} - \log {\left (x + \log {\relax (3 )} \right )} + \log {\left (\frac {x^{2} + 16 x \log {\left (\log {\relax (2 )} \right )} + x \log {\relax (3 )} + 5 x + 16 x \log {\relax (2 )} + 16 \log {\relax (3 )} \log {\left (\log {\relax (2 )} \right )} + 5 \log {\relax (3 )} + 16 \log {\relax (2 )} \log {\relax (3 )}}{4 \log {\left (\log {\relax (2 )} \right )} + 4 \log {\relax (2 )}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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