3.47.98 \(\int \frac {(4 x+8 x^2+4 x^3) \log (x)+(e^x (-x^2-3 x^3-3 x^4-x^5+(-2 x-6 x^2-6 x^3-2 x^4) \log (2))+(8 x+8 x^2+(16+16 x) \log (2)) \log (x)+(-8 x^2-16 x \log (2)) \log ^2(x)) \log (\frac {1}{2} (x+2 \log (2)))+(2 x+6 x^2+6 x^3+2 x^4+(4 x+8 x^2+4 x^3+(8+16 x+8 x^2) \log (2)+(-4 x^2-4 x^3+(-8 x-8 x^2) \log (2)) \log (x)) \log (\frac {1}{2} (x+2 \log (2)))) \log (\log (\frac {1}{2} (x+2 \log (2))))}{(x^2+3 x^3+3 x^4+x^5+(2 x+6 x^2+6 x^3+2 x^4) \log (2)) \log (\frac {1}{2} (x+2 \log (2)))} \, dx\)

Optimal. Leaf size=28 \[ -e^x+\left (\frac {2 \log (x)}{1+x}+\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )\right )^2 \]

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Rubi [F]  time = 5.68, 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 {\left (4 x+8 x^2+4 x^3\right ) \log (x)+\left (e^x \left (-x^2-3 x^3-3 x^4-x^5+\left (-2 x-6 x^2-6 x^3-2 x^4\right ) \log (2)\right )+\left (8 x+8 x^2+(16+16 x) \log (2)\right ) \log (x)+\left (-8 x^2-16 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {1}{2} (x+2 \log (2))\right )+\left (2 x+6 x^2+6 x^3+2 x^4+\left (4 x+8 x^2+4 x^3+\left (8+16 x+8 x^2\right ) \log (2)+\left (-4 x^2-4 x^3+\left (-8 x-8 x^2\right ) \log (2)\right ) \log (x)\right ) \log \left (\frac {1}{2} (x+2 \log (2))\right )\right ) \log \left (\log \left (\frac {1}{2} (x+2 \log (2))\right )\right )}{\left (x^2+3 x^3+3 x^4+x^5+\left (2 x+6 x^2+6 x^3+2 x^4\right ) \log (2)\right ) \log \left (\frac {1}{2} (x+2 \log (2))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((4*x + 8*x^2 + 4*x^3)*Log[x] + (E^x*(-x^2 - 3*x^3 - 3*x^4 - x^5 + (-2*x - 6*x^2 - 6*x^3 - 2*x^4)*Log[2])
+ (8*x + 8*x^2 + (16 + 16*x)*Log[2])*Log[x] + (-8*x^2 - 16*x*Log[2])*Log[x]^2)*Log[(x + 2*Log[2])/2] + (2*x +
6*x^2 + 6*x^3 + 2*x^4 + (4*x + 8*x^2 + 4*x^3 + (8 + 16*x + 8*x^2)*Log[2] + (-4*x^2 - 4*x^3 + (-8*x - 8*x^2)*Lo
g[2])*Log[x])*Log[(x + 2*Log[2])/2])*Log[Log[(x + 2*Log[2])/2]])/((x^2 + 3*x^3 + 3*x^4 + x^5 + (2*x + 6*x^2 +
6*x^3 + 2*x^4)*Log[2])*Log[(x + 2*Log[2])/2]),x]

[Out]

-E^x + (4*Log[x]^2)/(1 + x)^2 + Log[Log[x/2 + Log[2]]]^2 - (4*Defer[Int][Log[x]/((1 + x)*Log[x/2 + Log[2]]), x
])/(1 - Log[4]) + 4*Defer[Int][Log[Log[x/2 + Log[2]]]/x, x] - 4*Defer[Int][Log[Log[x/2 + Log[2]]]/(1 + x), x]
- 4*Defer[Int][(Log[x]*Log[Log[x/2 + Log[2]]])/(1 + x)^2, x] + (8*Log[2]*Defer[Subst][Defer[Int][Log[2*x - 2*L
og[2]]/(x*Log[x]), x], x, x/2 + Log[2]])/((1 - Log[4])*Log[4])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x (1+x)^2 \log (x)-(x+\log (4)) \left (e^x x (1+x)^3-8 (1+x) \log (x)+8 x \log ^2(x)\right ) \log \left (\frac {x}{2}+\log (2)\right )+2 \left (x+3 x^2+3 x^3+x^4-2 (1+x) (x+\log (4)) (-1-x+x \log (x)) \log \left (\frac {x}{2}+\log (2)\right )\right ) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x (1+x)^3 (x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )} \, dx\\ &=\int \left (-e^x+\frac {8 \log (x)}{x (1+x)^2}-\frac {8 \log ^2(x)}{(1+x)^3}+\frac {4 \log (x)}{(1+x) (x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )}+\frac {2 \left (x+2 x^2+x^3+2 x^2 \log \left (\frac {x}{2}+\log (2)\right )+2 \log (4) \log \left (\frac {x}{2}+\log (2)\right )+2 x (1+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )-2 x^2 \log (x) \log \left (\frac {x}{2}+\log (2)\right )-2 x \log (4) \log (x) \log \left (\frac {x}{2}+\log (2)\right )\right ) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x (1+x)^2 (x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )}\right ) \, dx\\ &=2 \int \frac {\left (x+2 x^2+x^3+2 x^2 \log \left (\frac {x}{2}+\log (2)\right )+2 \log (4) \log \left (\frac {x}{2}+\log (2)\right )+2 x (1+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )-2 x^2 \log (x) \log \left (\frac {x}{2}+\log (2)\right )-2 x \log (4) \log (x) \log \left (\frac {x}{2}+\log (2)\right )\right ) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x (1+x)^2 (x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )} \, dx+4 \int \frac {\log (x)}{(1+x) (x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )} \, dx+8 \int \frac {\log (x)}{x (1+x)^2} \, dx-8 \int \frac {\log ^2(x)}{(1+x)^3} \, dx-\int e^x \, dx\\ &=-e^x+\frac {4 \log ^2(x)}{(1+x)^2}+2 \int \left (\frac {2}{x+x^2}-\frac {2 \log (x)}{(1+x)^2}+\frac {1}{(x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )}\right ) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right ) \, dx+4 \int \left (\frac {\log (x)}{(1+x) (-1+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )}-\frac {\log (x)}{(-1+\log (4)) (x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )}\right ) \, dx-8 \int \frac {\log (x)}{(1+x)^2} \, dx-8 \int \frac {\log (x)}{x (1+x)^2} \, dx+8 \int \frac {\log (x)}{x (1+x)} \, dx\\ &=-e^x-\frac {8 x \log (x)}{1+x}+\frac {4 \log ^2(x)}{(1+x)^2}+2 \int \left (\frac {2 \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x (1+x)}-\frac {2 \log (x) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{(1+x)^2}+\frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{(x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )}\right ) \, dx+8 \int \frac {1}{1+x} \, dx+8 \int \frac {\log (x)}{x} \, dx+8 \int \frac {\log (x)}{(1+x)^2} \, dx-8 \int \frac {\log (x)}{1+x} \, dx-8 \int \frac {\log (x)}{x (1+x)} \, dx-\frac {4 \int \frac {\log (x)}{(1+x) \log \left (\frac {x}{2}+\log (2)\right )} \, dx}{1-\log (4)}+\frac {4 \int \frac {\log (x)}{(x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )} \, dx}{1-\log (4)}\\ &=-e^x+4 \log ^2(x)+\frac {4 \log ^2(x)}{(1+x)^2}+8 \log (1+x)-8 \log (x) \log (1+x)+2 \int \frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{(x+\log (4)) \log \left (\frac {x}{2}+\log (2)\right )} \, dx+4 \int \frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x (1+x)} \, dx-4 \int \frac {\log (x) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{(1+x)^2} \, dx-8 \int \frac {1}{1+x} \, dx-8 \int \frac {\log (x)}{x} \, dx+8 \int \frac {\log (x)}{1+x} \, dx+8 \int \frac {\log (1+x)}{x} \, dx-\frac {4 \int \frac {\log (x)}{(1+x) \log \left (\frac {x}{2}+\log (2)\right )} \, dx}{1-\log (4)}+\frac {8 \operatorname {Subst}\left (\int \frac {\log (2) \log (2 x-2 \log (2))}{x \log (4) \log (x)} \, dx,x,\frac {x}{2}+\log (2)\right )}{1-\log (4)}\\ &=-e^x+\frac {4 \log ^2(x)}{(1+x)^2}+\log ^2\left (\log \left (\frac {x}{2}+\log (2)\right )\right )-8 \text {Li}_2(-x)-4 \int \frac {\log (x) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{(1+x)^2} \, dx+4 \int \left (\frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x}-\frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{1+x}\right ) \, dx-8 \int \frac {\log (1+x)}{x} \, dx-\frac {4 \int \frac {\log (x)}{(1+x) \log \left (\frac {x}{2}+\log (2)\right )} \, dx}{1-\log (4)}+\frac {(8 \log (2)) \operatorname {Subst}\left (\int \frac {\log (2 x-2 \log (2))}{x \log (x)} \, dx,x,\frac {x}{2}+\log (2)\right )}{(1-\log (4)) \log (4)}\\ &=-e^x+\frac {4 \log ^2(x)}{(1+x)^2}+\log ^2\left (\log \left (\frac {x}{2}+\log (2)\right )\right )+4 \int \frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{x} \, dx-4 \int \frac {\log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{1+x} \, dx-4 \int \frac {\log (x) \log \left (\log \left (\frac {x}{2}+\log (2)\right )\right )}{(1+x)^2} \, dx-\frac {4 \int \frac {\log (x)}{(1+x) \log \left (\frac {x}{2}+\log (2)\right )} \, dx}{1-\log (4)}+\frac {(8 \log (2)) \operatorname {Subst}\left (\int \frac {\log (2 x-2 \log (2))}{x \log (x)} \, dx,x,\frac {x}{2}+\log (2)\right )}{(1-\log (4)) \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (4 x+8 x^2+4 x^3\right ) \log (x)+\left (e^x \left (-x^2-3 x^3-3 x^4-x^5+\left (-2 x-6 x^2-6 x^3-2 x^4\right ) \log (2)\right )+\left (8 x+8 x^2+(16+16 x) \log (2)\right ) \log (x)+\left (-8 x^2-16 x \log (2)\right ) \log ^2(x)\right ) \log \left (\frac {1}{2} (x+2 \log (2))\right )+\left (2 x+6 x^2+6 x^3+2 x^4+\left (4 x+8 x^2+4 x^3+\left (8+16 x+8 x^2\right ) \log (2)+\left (-4 x^2-4 x^3+\left (-8 x-8 x^2\right ) \log (2)\right ) \log (x)\right ) \log \left (\frac {1}{2} (x+2 \log (2))\right )\right ) \log \left (\log \left (\frac {1}{2} (x+2 \log (2))\right )\right )}{\left (x^2+3 x^3+3 x^4+x^5+\left (2 x+6 x^2+6 x^3+2 x^4\right ) \log (2)\right ) \log \left (\frac {1}{2} (x+2 \log (2))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((4*x + 8*x^2 + 4*x^3)*Log[x] + (E^x*(-x^2 - 3*x^3 - 3*x^4 - x^5 + (-2*x - 6*x^2 - 6*x^3 - 2*x^4)*Lo
g[2]) + (8*x + 8*x^2 + (16 + 16*x)*Log[2])*Log[x] + (-8*x^2 - 16*x*Log[2])*Log[x]^2)*Log[(x + 2*Log[2])/2] + (
2*x + 6*x^2 + 6*x^3 + 2*x^4 + (4*x + 8*x^2 + 4*x^3 + (8 + 16*x + 8*x^2)*Log[2] + (-4*x^2 - 4*x^3 + (-8*x - 8*x
^2)*Log[2])*Log[x])*Log[(x + 2*Log[2])/2])*Log[Log[(x + 2*Log[2])/2]])/((x^2 + 3*x^3 + 3*x^4 + x^5 + (2*x + 6*
x^2 + 6*x^3 + 2*x^4)*Log[2])*Log[(x + 2*Log[2])/2]),x]

[Out]

Integrate[((4*x + 8*x^2 + 4*x^3)*Log[x] + (E^x*(-x^2 - 3*x^3 - 3*x^4 - x^5 + (-2*x - 6*x^2 - 6*x^3 - 2*x^4)*Lo
g[2]) + (8*x + 8*x^2 + (16 + 16*x)*Log[2])*Log[x] + (-8*x^2 - 16*x*Log[2])*Log[x]^2)*Log[(x + 2*Log[2])/2] + (
2*x + 6*x^2 + 6*x^3 + 2*x^4 + (4*x + 8*x^2 + 4*x^3 + (8 + 16*x + 8*x^2)*Log[2] + (-4*x^2 - 4*x^3 + (-8*x - 8*x
^2)*Log[2])*Log[x])*Log[(x + 2*Log[2])/2])*Log[Log[(x + 2*Log[2])/2]])/((x^2 + 3*x^3 + 3*x^4 + x^5 + (2*x + 6*
x^2 + 6*x^3 + 2*x^4)*Log[2])*Log[(x + 2*Log[2])/2]), x]

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fricas [B]  time = 0.82, size = 64, normalized size = 2.29 \begin {gather*} \frac {4 \, {\left (x + 1\right )} \log \relax (x) \log \left (\log \left (\frac {1}{2} \, x + \log \relax (2)\right )\right ) + {\left (x^{2} + 2 \, x + 1\right )} \log \left (\log \left (\frac {1}{2} \, x + \log \relax (2)\right )\right )^{2} - {\left (x^{2} + 2 \, x + 1\right )} e^{x} + 4 \, \log \relax (x)^{2}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((-8*x^2-8*x)*log(2)-4*x^3-4*x^2)*log(x)+(8*x^2+16*x+8)*log(2)+4*x^3+8*x^2+4*x)*log(log(2)+1/2*x)
+2*x^4+6*x^3+6*x^2+2*x)*log(log(log(2)+1/2*x))+((-16*x*log(2)-8*x^2)*log(x)^2+((16*x+16)*log(2)+8*x^2+8*x)*log
(x)+((-2*x^4-6*x^3-6*x^2-2*x)*log(2)-x^5-3*x^4-3*x^3-x^2)*exp(x))*log(log(2)+1/2*x)+(4*x^3+8*x^2+4*x)*log(x))/
((2*x^4+6*x^3+6*x^2+2*x)*log(2)+x^5+3*x^4+3*x^3+x^2)/log(log(2)+1/2*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.43, size = 121, normalized size = 4.32 \begin {gather*} \frac {x^{2} \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right )^{2} - x^{2} e^{x} + 4 \, x \log \relax (x) \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right ) + 2 \, x \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right )^{2} - 2 \, x e^{x} + 4 \, \log \relax (x)^{2} + 4 \, \log \relax (x) \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right ) + \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right )^{2} - e^{x}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((-8*x^2-8*x)*log(2)-4*x^3-4*x^2)*log(x)+(8*x^2+16*x+8)*log(2)+4*x^3+8*x^2+4*x)*log(log(2)+1/2*x)
+2*x^4+6*x^3+6*x^2+2*x)*log(log(log(2)+1/2*x))+((-16*x*log(2)-8*x^2)*log(x)^2+((16*x+16)*log(2)+8*x^2+8*x)*log
(x)+((-2*x^4-6*x^3-6*x^2-2*x)*log(2)-x^5-3*x^4-3*x^3-x^2)*exp(x))*log(log(2)+1/2*x)+(4*x^3+8*x^2+4*x)*log(x))/
((2*x^4+6*x^3+6*x^2+2*x)*log(2)+x^5+3*x^4+3*x^3+x^2)/log(log(2)+1/2*x),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.12, size = 56, normalized size = 2.00

method	result	size
risch	\(\ln \left (\ln \left (\ln \relax (2)+\frac {x}{2}\right )\right )^{2}+\frac {4 \ln \relax (x ) \ln \left (\ln \left (\ln \relax (2)+\frac {x}{2}\right )\right )}{x +1}-\frac {{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{x} x -4 \ln \relax (x )^{2}+{\mathrm e}^{x}}{\left (x +1\right )^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((((-8*x^2-8*x)*ln(2)-4*x^3-4*x^2)*ln(x)+(8*x^2+16*x+8)*ln(2)+4*x^3+8*x^2+4*x)*ln(ln(2)+1/2*x)+2*x^4+6*x^
3+6*x^2+2*x)*ln(ln(ln(2)+1/2*x))+((-16*x*ln(2)-8*x^2)*ln(x)^2+((16*x+16)*ln(2)+8*x^2+8*x)*ln(x)+((-2*x^4-6*x^3
-6*x^2-2*x)*ln(2)-x^5-3*x^4-3*x^3-x^2)*exp(x))*ln(ln(2)+1/2*x)+(4*x^3+8*x^2+4*x)*ln(x))/((2*x^4+6*x^3+6*x^2+2*
x)*ln(2)+x^5+3*x^4+3*x^3+x^2)/ln(ln(2)+1/2*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.61, size = 74, normalized size = 2.64 \begin {gather*} \frac {4 \, {\left (x + 1\right )} \log \relax (x) \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right ) + {\left (x^{2} + 2 \, x + 1\right )} \log \left (-\log \relax (2) + \log \left (x + 2 \, \log \relax (2)\right )\right )^{2} - {\left (x^{2} + 2 \, x + 1\right )} e^{x} + 4 \, \log \relax (x)^{2}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((-8*x^2-8*x)*log(2)-4*x^3-4*x^2)*log(x)+(8*x^2+16*x+8)*log(2)+4*x^3+8*x^2+4*x)*log(log(2)+1/2*x)
+2*x^4+6*x^3+6*x^2+2*x)*log(log(log(2)+1/2*x))+((-16*x*log(2)-8*x^2)*log(x)^2+((16*x+16)*log(2)+8*x^2+8*x)*log
(x)+((-2*x^4-6*x^3-6*x^2-2*x)*log(2)-x^5-3*x^4-3*x^3-x^2)*exp(x))*log(log(2)+1/2*x)+(4*x^3+8*x^2+4*x)*log(x))/
((2*x^4+6*x^3+6*x^2+2*x)*log(2)+x^5+3*x^4+3*x^3+x^2)/log(log(2)+1/2*x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (\ln \left (\frac {x}{2}+\ln \relax (2)\right )\right )\,\left (2\,x+\ln \left (\frac {x}{2}+\ln \relax (2)\right )\,\left (4\,x+\ln \relax (2)\,\left (8\,x^2+16\,x+8\right )-\ln \relax (x)\,\left (\ln \relax (2)\,\left (8\,x^2+8\,x\right )+4\,x^2+4\,x^3\right )+8\,x^2+4\,x^3\right )+6\,x^2+6\,x^3+2\,x^4\right )-\ln \left (\frac {x}{2}+\ln \relax (2)\right )\,\left (\left (8\,x^2+16\,\ln \relax (2)\,x\right )\,{\ln \relax (x)}^2+\left (-8\,x-\ln \relax (2)\,\left (16\,x+16\right )-8\,x^2\right )\,\ln \relax (x)+{\mathrm {e}}^x\,\left (\ln \relax (2)\,\left (2\,x^4+6\,x^3+6\,x^2+2\,x\right )+x^2+3\,x^3+3\,x^4+x^5\right )\right )+\ln \relax (x)\,\left (4\,x^3+8\,x^2+4\,x\right )}{\ln \left (\frac {x}{2}+\ln \relax (2)\right )\,\left (\ln \relax (2)\,\left (2\,x^4+6\,x^3+6\,x^2+2\,x\right )+x^2+3\,x^3+3\,x^4+x^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x/2 + log(2)))*(2*x + log(x/2 + log(2))*(4*x + log(2)*(16*x + 8*x^2 + 8) - log(x)*(log(2)*(8*x +
8*x^2) + 4*x^2 + 4*x^3) + 8*x^2 + 4*x^3) + 6*x^2 + 6*x^3 + 2*x^4) - log(x/2 + log(2))*(exp(x)*(log(2)*(2*x + 6
*x^2 + 6*x^3 + 2*x^4) + x^2 + 3*x^3 + 3*x^4 + x^5) + log(x)^2*(16*x*log(2) + 8*x^2) - log(x)*(8*x + log(2)*(16
*x + 16) + 8*x^2)) + log(x)*(4*x + 8*x^2 + 4*x^3))/(log(x/2 + log(2))*(log(2)*(2*x + 6*x^2 + 6*x^3 + 2*x^4) +
x^2 + 3*x^3 + 3*x^4 + x^5)),x)

[Out]

int((log(log(x/2 + log(2)))*(2*x + log(x/2 + log(2))*(4*x + log(2)*(16*x + 8*x^2 + 8) - log(x)*(log(2)*(8*x +
8*x^2) + 4*x^2 + 4*x^3) + 8*x^2 + 4*x^3) + 6*x^2 + 6*x^3 + 2*x^4) - log(x/2 + log(2))*(exp(x)*(log(2)*(2*x + 6
*x^2 + 6*x^3 + 2*x^4) + x^2 + 3*x^3 + 3*x^4 + x^5) + log(x)^2*(16*x*log(2) + 8*x^2) - log(x)*(8*x + log(2)*(16
*x + 16) + 8*x^2)) + log(x)*(4*x + 8*x^2 + 4*x^3))/(log(x/2 + log(2))*(log(2)*(2*x + 6*x^2 + 6*x^3 + 2*x^4) +
x^2 + 3*x^3 + 3*x^4 + x^5)), x)

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sympy [B]  time = 3.26, size = 48, normalized size = 1.71 \begin {gather*} - e^{x} + \log {\left (\log {\left (\frac {x}{2} + \log {\relax (2 )} \right )} \right )}^{2} + \frac {4 \log {\relax (x )}^{2}}{x^{2} + 2 x + 1} + \frac {4 \log {\relax (x )} \log {\left (\log {\left (\frac {x}{2} + \log {\relax (2 )} \right )} \right )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((-8*x**2-8*x)*ln(2)-4*x**3-4*x**2)*ln(x)+(8*x**2+16*x+8)*ln(2)+4*x**3+8*x**2+4*x)*ln(ln(2)+1/2*x
)+2*x**4+6*x**3+6*x**2+2*x)*ln(ln(ln(2)+1/2*x))+((-16*x*ln(2)-8*x**2)*ln(x)**2+((16*x+16)*ln(2)+8*x**2+8*x)*ln
(x)+((-2*x**4-6*x**3-6*x**2-2*x)*ln(2)-x**5-3*x**4-3*x**3-x**2)*exp(x))*ln(ln(2)+1/2*x)+(4*x**3+8*x**2+4*x)*ln
(x))/((2*x**4+6*x**3+6*x**2+2*x)*ln(2)+x**5+3*x**4+3*x**3+x**2)/ln(ln(2)+1/2*x),x)

[Out]

-exp(x) + log(log(x/2 + log(2)))**2 + 4*log(x)**2/(x**2 + 2*x + 1) + 4*log(x)*log(log(x/2 + log(2)))/(x + 1)

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