3.62.23 \(\int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} (-6 x-12 x^2) \log ^2(1+2 x) \log (\log (x)))}{(9+18 x) \log (x)+(12 x+24 x^2) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} (x^2+2 x^3) \log (x) \log ^2(1+2 x)+(4 x^2+8 x^3) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} ((-6 x-12 x^2) \log (x) \log (1+2 x)+(-4 x^2-8 x^3) \log (x) \log ^2(1+2 x))} \, dx\)

Optimal. Leaf size=28 \[ e^{\frac {3}{-2+e^{\log ^2(\log (x))}-\frac {3}{x \log (1+2 x)}}} \]

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Rubi [F]  time = 11.57, 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 {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((1 + 2*x)^((3*x)/(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x]))*(-18*x*Log[x] + (-9 - 18*x)*Lo
g[x]*Log[1 + 2*x] + E^Log[Log[x]]^2*(-6*x - 12*x^2)*Log[1 + 2*x]^2*Log[Log[x]]))/((9 + 18*x)*Log[x] + (12*x +
24*x^2)*Log[x]*Log[1 + 2*x] + E^(2*Log[Log[x]]^2)*(x^2 + 2*x^3)*Log[x]*Log[1 + 2*x]^2 + (4*x^2 + 8*x^3)*Log[x]
*Log[1 + 2*x]^2 + E^Log[Log[x]]^2*((-6*x - 12*x^2)*Log[x]*Log[1 + 2*x] + (-4*x^2 - 8*x^3)*Log[x]*Log[1 + 2*x]^
2)),x]

[Out]

-18*Defer[Int][(x*(1 + 2*x)^(-1 + (3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 2*x])))/(-3 - 2*x*Log[1 + 2*x]
+ E^Log[Log[x]]^2*x*Log[1 + 2*x])^2, x] - 9*Defer[Int][((1 + 2*x)^(-1 + (3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*L
og[1 + 2*x]))*Log[1 + 2*x])/(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x])^2, x] - 18*Defer[Int][(x*
(1 + 2*x)^(-1 + (3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 2*x]))*Log[1 + 2*x])/(-3 - 2*x*Log[1 + 2*x] + E^L
og[Log[x]]^2*x*Log[1 + 2*x])^2, x] - 18*Defer[Int][((1 + 2*x)^(-1 + (3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1
+ 2*x]))*Log[1 + 2*x]*Log[Log[x]])/(Log[x]*(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x])^2), x] -
36*Defer[Int][(x*(1 + 2*x)^(-1 + (3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 2*x]))*Log[1 + 2*x]*Log[Log[x]])
/(Log[x]*(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x])^2), x] - 12*Defer[Int][(x*(1 + 2*x)^(-1 + (3
*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 2*x]))*Log[1 + 2*x]^2*Log[Log[x]])/(Log[x]*(-3 - 2*x*Log[1 + 2*x] +
E^Log[Log[x]]^2*x*Log[1 + 2*x])^2), x] - 24*Defer[Int][(x^2*(1 + 2*x)^(-1 + (3*x)/(-3 + (-2 + E^Log[Log[x]]^2
)*x*Log[1 + 2*x]))*Log[1 + 2*x]^2*Log[Log[x]])/(Log[x]*(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x]
)^2), x] - 6*Defer[Int][((1 + 2*x)^((3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 2*x]))*Log[1 + 2*x]*Log[Log[x
]])/(Log[x]*(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \left (-9 \log (x) (2 x+(1+2 x) \log (1+2 x))-6 e^{\log ^2(\log (x))} x (1+2 x) \log ^2(1+2 x) \log (\log (x))\right )}{\log (x) \left (3-\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)\right )^2} \, dx\\ &=\int \left (-\frac {6 (1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )}-\frac {3 (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \left (6 x \log (x)+3 \log (x) \log (1+2 x)+6 x \log (x) \log (1+2 x)+6 \log (1+2 x) \log (\log (x))+12 x \log (1+2 x) \log (\log (x))+4 x \log ^2(1+2 x) \log (\log (x))+8 x^2 \log ^2(1+2 x) \log (\log (x))\right )}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}\right ) \, dx\\ &=-\left (3 \int \frac {(1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \left (6 x \log (x)+3 \log (x) \log (1+2 x)+6 x \log (x) \log (1+2 x)+6 \log (1+2 x) \log (\log (x))+12 x \log (1+2 x) \log (\log (x))+4 x \log ^2(1+2 x) \log (\log (x))+8 x^2 \log ^2(1+2 x) \log (\log (x))\right )}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx\right )-6 \int \frac {(1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )} \, dx\\ &=-\left (3 \int \frac {(1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} (3 \log (x) (2 x+(1+2 x) \log (1+2 x))+2 (1+2 x) \log (1+2 x) (3+2 x \log (1+2 x)) \log (\log (x)))}{\log (x) \left (3-\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)\right )^2} \, dx\right )-6 \int \frac {(1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )} \, dx\\ &=-\left (3 \int \left (\frac {6 x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}}}{\left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}+\frac {3 (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x)}{\left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}+\frac {6 x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x)}{\left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}+\frac {6 (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}+\frac {12 x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}+\frac {4 x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log ^2(1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}+\frac {8 x^2 (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log ^2(1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2}\right ) \, dx\right )-6 \int \frac {(1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )} \, dx\\ &=-\left (6 \int \frac {(1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )} \, dx\right )-9 \int \frac {(1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x)}{\left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx-12 \int \frac {x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log ^2(1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx-18 \int \frac {x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}}}{\left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx-18 \int \frac {x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x)}{\left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx-18 \int \frac {(1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx-24 \int \frac {x^2 (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log ^2(1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx-36 \int \frac {x (1+2 x)^{-1+\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \log (1+2 x) \log (\log (x))}{\log (x) \left (-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 30, normalized size = 1.07 \begin {gather*} (1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + 2*x)^((3*x)/(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2*x]))*(-18*x*Log[x] + (-9 - 18
*x)*Log[x]*Log[1 + 2*x] + E^Log[Log[x]]^2*(-6*x - 12*x^2)*Log[1 + 2*x]^2*Log[Log[x]]))/((9 + 18*x)*Log[x] + (1
2*x + 24*x^2)*Log[x]*Log[1 + 2*x] + E^(2*Log[Log[x]]^2)*(x^2 + 2*x^3)*Log[x]*Log[1 + 2*x]^2 + (4*x^2 + 8*x^3)*
Log[x]*Log[1 + 2*x]^2 + E^Log[Log[x]]^2*((-6*x - 12*x^2)*Log[x]*Log[1 + 2*x] + (-4*x^2 - 8*x^3)*Log[x]*Log[1 +
2*x]^2)),x]

[Out]

(1 + 2*x)^((3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 2*x]))

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fricas [A]  time = 0.60, size = 36, normalized size = 1.29 \begin {gather*} {\left (2 \, x + 1\right )}^{\frac {3 \, x}{x e^{\left (\log \left (\log \relax (x)\right )^{2}\right )} \log \left (2 \, x + 1\right ) - 2 \, x \log \left (2 \, x + 1\right ) - 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^2-6*x)*log(2*x+1)^2*log(log(x))*exp(log(log(x))^2)+(-18*x-9)*log(x)*log(2*x+1)-18*x*log(x))*
exp(3*x*log(2*x+1)/(x*log(2*x+1)*exp(log(log(x))^2)-2*x*log(2*x+1)-3))/((2*x^3+x^2)*log(x)*log(2*x+1)^2*exp(lo
g(log(x))^2)^2+((-8*x^3-4*x^2)*log(x)*log(2*x+1)^2+(-12*x^2-6*x)*log(x)*log(2*x+1))*exp(log(log(x))^2)+(8*x^3+
4*x^2)*log(x)*log(2*x+1)^2+(24*x^2+12*x)*log(x)*log(2*x+1)+(18*x+9)*log(x)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {3 \, {\left (2 \, {\left (2 \, x^{2} + x\right )} e^{\left (\log \left (\log \relax (x)\right )^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (\log \relax (x)\right ) + 3 \, {\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) \log \relax (x) + 6 \, x \log \relax (x)\right )} {\left (2 \, x + 1\right )}^{\frac {3 \, x}{x e^{\left (\log \left (\log \relax (x)\right )^{2}\right )} \log \left (2 \, x + 1\right ) - 2 \, x \log \left (2 \, x + 1\right ) - 3}}}{{\left (2 \, x^{3} + x^{2}\right )} e^{\left (2 \, \log \left (\log \relax (x)\right )^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \relax (x) + 4 \, {\left (2 \, x^{3} + x^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \relax (x) + 12 \, {\left (2 \, x^{2} + x\right )} \log \left (2 \, x + 1\right ) \log \relax (x) - 2 \, {\left (2 \, {\left (2 \, x^{3} + x^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \relax (x) + 3 \, {\left (2 \, x^{2} + x\right )} \log \left (2 \, x + 1\right ) \log \relax (x)\right )} e^{\left (\log \left (\log \relax (x)\right )^{2}\right )} + 9 \, {\left (2 \, x + 1\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^2-6*x)*log(2*x+1)^2*log(log(x))*exp(log(log(x))^2)+(-18*x-9)*log(x)*log(2*x+1)-18*x*log(x))*
exp(3*x*log(2*x+1)/(x*log(2*x+1)*exp(log(log(x))^2)-2*x*log(2*x+1)-3))/((2*x^3+x^2)*log(x)*log(2*x+1)^2*exp(lo
g(log(x))^2)^2+((-8*x^3-4*x^2)*log(x)*log(2*x+1)^2+(-12*x^2-6*x)*log(x)*log(2*x+1))*exp(log(log(x))^2)+(8*x^3+
4*x^2)*log(x)*log(2*x+1)^2+(24*x^2+12*x)*log(x)*log(2*x+1)+(18*x+9)*log(x)),x, algorithm="giac")

[Out]

integrate(-3*(2*(2*x^2 + x)*e^(log(log(x))^2)*log(2*x + 1)^2*log(log(x)) + 3*(2*x + 1)*log(2*x + 1)*log(x) + 6
*x*log(x))*(2*x + 1)^(3*x/(x*e^(log(log(x))^2)*log(2*x + 1) - 2*x*log(2*x + 1) - 3))/((2*x^3 + x^2)*e^(2*log(l
og(x))^2)*log(2*x + 1)^2*log(x) + 4*(2*x^3 + x^2)*log(2*x + 1)^2*log(x) + 12*(2*x^2 + x)*log(2*x + 1)*log(x) -
2*(2*(2*x^3 + x^2)*log(2*x + 1)^2*log(x) + 3*(2*x^2 + x)*log(2*x + 1)*log(x))*e^(log(log(x))^2) + 9*(2*x + 1)
*log(x)), x)

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maple [A]  time = 0.07, size = 37, normalized size = 1.32

method	result	size
risch	\(\left (2 x +1\right )^{\frac {3 x}{x \ln \left (2 x +1\right ) {\mathrm e}^{\ln \left (\ln \relax (x )\right )^{2}}-2 x \ln \left (2 x +1\right )-3}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*x^2-6*x)*ln(2*x+1)^2*ln(ln(x))*exp(ln(ln(x))^2)+(-18*x-9)*ln(x)*ln(2*x+1)-18*x*ln(x))*exp(3*x*ln(2*x
+1)/(x*ln(2*x+1)*exp(ln(ln(x))^2)-2*x*ln(2*x+1)-3))/((2*x^3+x^2)*ln(x)*ln(2*x+1)^2*exp(ln(ln(x))^2)^2+((-8*x^3
-4*x^2)*ln(x)*ln(2*x+1)^2+(-12*x^2-6*x)*ln(x)*ln(2*x+1))*exp(ln(ln(x))^2)+(8*x^3+4*x^2)*ln(x)*ln(2*x+1)^2+(24*
x^2+12*x)*ln(x)*ln(2*x+1)+(18*x+9)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

(2*x+1)^(3*x/(x*ln(2*x+1)*exp(ln(ln(x))^2)-2*x*ln(2*x+1)-3))

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maxima [B]  time = 0.77, size = 64, normalized size = 2.29 \begin {gather*} e^{\left (\frac {9}{x e^{\left (2 \, \log \left (\log \relax (x)\right )^{2}\right )} \log \left (2 \, x + 1\right ) - {\left (4 \, x \log \left (2 \, x + 1\right ) + 3\right )} e^{\left (\log \left (\log \relax (x)\right )^{2}\right )} + 4 \, x \log \left (2 \, x + 1\right ) + 6} + \frac {3}{e^{\left (\log \left (\log \relax (x)\right )^{2}\right )} - 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^2-6*x)*log(2*x+1)^2*log(log(x))*exp(log(log(x))^2)+(-18*x-9)*log(x)*log(2*x+1)-18*x*log(x))*
exp(3*x*log(2*x+1)/(x*log(2*x+1)*exp(log(log(x))^2)-2*x*log(2*x+1)-3))/((2*x^3+x^2)*log(x)*log(2*x+1)^2*exp(lo
g(log(x))^2)^2+((-8*x^3-4*x^2)*log(x)*log(2*x+1)^2+(-12*x^2-6*x)*log(x)*log(2*x+1))*exp(log(log(x))^2)+(8*x^3+
4*x^2)*log(x)*log(2*x+1)^2+(24*x^2+12*x)*log(x)*log(2*x+1)+(18*x+9)*log(x)),x, algorithm="maxima")

[Out]

e^(9/(x*e^(2*log(log(x))^2)*log(2*x + 1) - (4*x*log(2*x + 1) + 3)*e^(log(log(x))^2) + 4*x*log(2*x + 1) + 6) +
3/(e^(log(log(x))^2) - 2))

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mupad [B]  time = 5.04, size = 38, normalized size = 1.36 \begin {gather*} {\mathrm {e}}^{-\frac {3\,x\,\ln \left (2\,x+1\right )}{2\,x\,\ln \left (2\,x+1\right )-x\,{\mathrm {e}}^{{\ln \left (\ln \relax (x)\right )}^2}\,\ln \left (2\,x+1\right )+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(3*x*log(2*x + 1))/(2*x*log(2*x + 1) - x*exp(log(log(x))^2)*log(2*x + 1) + 3))*(18*x*log(x) + log(2
*x + 1)*log(x)*(18*x + 9) + log(log(x))*exp(log(log(x))^2)*log(2*x + 1)^2*(6*x + 12*x^2)))/(log(x)*(18*x + 9)
- exp(log(log(x))^2)*(log(2*x + 1)^2*log(x)*(4*x^2 + 8*x^3) + log(2*x + 1)*log(x)*(6*x + 12*x^2)) + log(2*x +
1)^2*log(x)*(4*x^2 + 8*x^3) + log(2*x + 1)*log(x)*(12*x + 24*x^2) + exp(2*log(log(x))^2)*log(2*x + 1)^2*log(x)
*(x^2 + 2*x^3)),x)

[Out]

exp(-(3*x*log(2*x + 1))/(2*x*log(2*x + 1) - x*exp(log(log(x))^2)*log(2*x + 1) + 3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x**2-6*x)*ln(2*x+1)**2*ln(ln(x))*exp(ln(ln(x))**2)+(-18*x-9)*ln(x)*ln(2*x+1)-18*x*ln(x))*exp(3
*x*ln(2*x+1)/(x*ln(2*x+1)*exp(ln(ln(x))**2)-2*x*ln(2*x+1)-3))/((2*x**3+x**2)*ln(x)*ln(2*x+1)**2*exp(ln(ln(x))*
*2)**2+((-8*x**3-4*x**2)*ln(x)*ln(2*x+1)**2+(-12*x**2-6*x)*ln(x)*ln(2*x+1))*exp(ln(ln(x))**2)+(8*x**3+4*x**2)*
ln(x)*ln(2*x+1)**2+(24*x**2+12*x)*ln(x)*ln(2*x+1)+(18*x+9)*ln(x)),x)

[Out]

Timed out

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