3.72.30 \(\int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log (\frac {1}{x+4 x^2})+x^2 \log ^2(\frac {1}{x+4 x^2})} (e^{5 x^2} (2+16 x)+e^{10 x^2} (20 x+80 x^2)+(-2 x-16 x^2+e^{5 x^2} (-2-8 x-20 x^2-80 x^3)) \log (\frac {1}{x+4 x^2})+(2 x+8 x^2) \log ^2(\frac {1}{x+4 x^2}))}{1+4 x} \, dx\)

Optimal. Leaf size=31 \[ -1+e^{\left (-e^{5 x^2}+x \log \left (\frac {1}{x+4 x^2}\right )\right )^2}-x \]

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Rubi [F]  time = 41.15, 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 {-1-4 x+\exp \left (e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )\right ) \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 4*x + E^(E^(10*x^2) - 2*E^(5*x^2)*x*Log[(x + 4*x^2)^(-1)] + x^2*Log[(x + 4*x^2)^(-1)]^2)*(E^(5*x^2)*
(2 + 16*x) + E^(10*x^2)*(20*x + 80*x^2) + (-2*x - 16*x^2 + E^(5*x^2)*(-2 - 8*x - 20*x^2 - 80*x^3))*Log[(x + 4*
x^2)^(-1)] + (2*x + 8*x^2)*Log[(x + 4*x^2)^(-1)]^2))/(1 + 4*x),x]

[Out]

-x + 4*Defer[Int][E^(E^(10*x^2) + 5*x^2 + x^2*Log[(x + 4*x^2)^(-1)]^2)/(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x), x] +
20*Defer[Int][(E^(E^(10*x^2) + 10*x^2 + x^2*Log[(x + 4*x^2)^(-1)]^2)*x)/(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x), x]
- 2*Defer[Int][E^(E^(10*x^2) + 5*x^2 + x^2*Log[(x + 4*x^2)^(-1)]^2)/((1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x)*(1 + 4
*x)), x] + Defer[Int][(E^(E^(10*x^2) + x^2*Log[(x + 4*x^2)^(-1)]^2)*Log[1/(x*(1 + 4*x))])/(1/(x*(1 + 4*x)))^(2
*E^(5*x^2)*x), x]/2 - 2*Defer[Int][(E^(E^(10*x^2) + 5*x^2 + x^2*Log[(x + 4*x^2)^(-1)]^2)*Log[1/(x*(1 + 4*x))])
/(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x), x] + Defer[Int][(E^(E^(10*x^2) + x^2*Log[(x + 4*x^2)^(-1)]^2)*Log[1/(x*(1
+ 4*x))])/((-1 - 4*x)*(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x)), x]/2 - 4*Defer[Int][(E^(E^(10*x^2) + x^2*Log[(x + 4*
x^2)^(-1)]^2)*x*Log[1/(x*(1 + 4*x))])/(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x), x] - 20*Defer[Int][(E^(E^(10*x^2) + 5
*x^2 + x^2*Log[(x + 4*x^2)^(-1)]^2)*x^2*Log[1/(x*(1 + 4*x))])/(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x), x] + 2*Defer[
Int][(E^(E^(10*x^2) + x^2*Log[(x + 4*x^2)^(-1)]^2)*x*Log[1/(x*(1 + 4*x))]^2)/(1/(x*(1 + 4*x)))^(2*E^(5*x^2)*x)
, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+\frac {2 e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+8 x+10 e^{5 x^2} x+40 e^{5 x^2} x^2-\log \left (\frac {1}{x+4 x^2}\right )-4 x \log \left (\frac {1}{x+4 x^2}\right )\right ) \left (e^{5 x^2}-x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x}\right ) \, dx\\ &=-x+2 \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+8 x+10 e^{5 x^2} x+40 e^{5 x^2} x^2-\log \left (\frac {1}{x+4 x^2}\right )-4 x \log \left (\frac {1}{x+4 x^2}\right )\right ) \left (e^{5 x^2}-x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx\\ &=-x+2 \int \left (10 e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}+\frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x}-\frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )+10 x^2 \log \left (\frac {1}{x+4 x^2}\right )+40 x^3 \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x}\right ) \, dx\\ &=-x+2 \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (-1-8 x+\log \left (\frac {1}{x+4 x^2}\right )+4 x \log \left (\frac {1}{x+4 x^2}\right )+10 x^2 \log \left (\frac {1}{x+4 x^2}\right )+40 x^3 \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx\\ &=-x+2 \int \left (\frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{1+4 x}+e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right )\right ) \, dx-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (-1-8 x+\left (1+4 x+10 x^2+40 x^3\right ) \log \left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx\\ &=-x+2 \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{1+4 x} \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx-2 \int \left (\frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x}+e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+10 x^2\right ) \log \left (\frac {1}{x (1+4 x)}\right )\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx\\ &=-x-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} (-1-8 x) \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x} \, dx-2 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \left (1+10 x^2\right ) \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx+2 \int \left (\frac {1}{4} e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )+\frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{4 (-1-4 x)}-2 e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx\\ &=-x+\frac {1}{2} \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+\frac {1}{2} \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{-1-4 x} \, dx-2 \int \left (-2 e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}+\frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x}\right ) \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx-2 \int \left (e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )+10 e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x^2 \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )\right ) \, dx-4 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx\\ &=-x+\frac {1}{2} \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+\frac {1}{2} \int \frac {e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right )}{-1-4 x} \, dx-2 \int \frac {e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x}}{1+4 x} \, dx-2 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+2 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log ^2\left (\frac {1}{x (1+4 x)}\right ) \, dx+4 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx-4 \int e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx+20 \int e^{e^{10 x^2}+10 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \, dx-20 \int e^{e^{10 x^2}+5 x^2+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} x^2 \left (\frac {1}{x (1+4 x)}\right )^{-2 e^{5 x^2} x} \log \left (\frac {1}{x (1+4 x)}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 51, normalized size = 1.65 \begin {gather*} -x+e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x+4 x^2}\right )^{-2 e^{5 x^2} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 4*x + E^(E^(10*x^2) - 2*E^(5*x^2)*x*Log[(x + 4*x^2)^(-1)] + x^2*Log[(x + 4*x^2)^(-1)]^2)*(E^(5
*x^2)*(2 + 16*x) + E^(10*x^2)*(20*x + 80*x^2) + (-2*x - 16*x^2 + E^(5*x^2)*(-2 - 8*x - 20*x^2 - 80*x^3))*Log[(
x + 4*x^2)^(-1)] + (2*x + 8*x^2)*Log[(x + 4*x^2)^(-1)]^2))/(1 + 4*x),x]

[Out]

-x + E^(E^(10*x^2) + x^2*Log[(x + 4*x^2)^(-1)]^2)/((x + 4*x^2)^(-1))^(2*E^(5*x^2)*x)

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fricas [A]  time = 0.62, size = 47, normalized size = 1.52 \begin {gather*} -x + e^{\left (x^{2} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} - 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right ) + e^{\left (10 \, x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+2*x)*log(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x^2)-16*x^2-2*x)*log(1/(4*x^2+x))+(80
*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5*x^2))*exp(x^2*log(1/(4*x^2+x))^2-2*x*exp(5*x^2)*log(1/(4*x^2+x))+exp(5*
x^2)^2)-4*x-1)/(4*x+1),x, algorithm="fricas")

[Out]

-x + e^(x^2*log(1/(4*x^2 + x))^2 - 2*x*e^(5*x^2)*log(1/(4*x^2 + x)) + e^(10*x^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (4 \, x^{2} + x\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} + 10 \, {\left (4 \, x^{2} + x\right )} e^{\left (10 \, x^{2}\right )} + {\left (8 \, x + 1\right )} e^{\left (5 \, x^{2}\right )} - {\left (8 \, x^{2} + {\left (40 \, x^{3} + 10 \, x^{2} + 4 \, x + 1\right )} e^{\left (5 \, x^{2}\right )} + x\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right )\right )} e^{\left (x^{2} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} - 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right ) + e^{\left (10 \, x^{2}\right )}\right )} - 4 \, x - 1}{4 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+2*x)*log(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x^2)-16*x^2-2*x)*log(1/(4*x^2+x))+(80
*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5*x^2))*exp(x^2*log(1/(4*x^2+x))^2-2*x*exp(5*x^2)*log(1/(4*x^2+x))+exp(5*
x^2)^2)-4*x-1)/(4*x+1),x, algorithm="giac")

[Out]

integrate((2*((4*x^2 + x)*log(1/(4*x^2 + x))^2 + 10*(4*x^2 + x)*e^(10*x^2) + (8*x + 1)*e^(5*x^2) - (8*x^2 + (4
0*x^3 + 10*x^2 + 4*x + 1)*e^(5*x^2) + x)*log(1/(4*x^2 + x)))*e^(x^2*log(1/(4*x^2 + x))^2 - 2*x*e^(5*x^2)*log(1
/(4*x^2 + x)) + e^(10*x^2)) - 4*x - 1)/(4*x + 1), x)

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maple [C]  time = 0.68, size = 863, normalized size = 27.84

method	result	size
risch	\(-x +x^{4 x^{2} \ln \relax (2)} \left (x +\frac {1}{4}\right )^{4 x^{2} \ln \relax (2)} \left (x +\frac {1}{4}\right )^{2 x^{2} \ln \relax (x )} 2^{-2 i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x}\right )} x^{i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (4 x +1\right )}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{4 x +1}\right )} \left (x +\frac {1}{4}\right )^{-i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{4 x +1}\right )} x^{-i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x}\right )} 2^{2 i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (4 x +1\right )}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{4 x +1}\right )} x^{-i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{4 x +1}\right )} \left (x +\frac {1}{4}\right )^{-i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x}\right )} \left (x +\frac {1}{4}\right )^{i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (4 x +1\right )}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{4 x +1}\right )} 2^{2 i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (4 x +1\right )}\right )} x^{i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (4 x +1\right )}\right )} 16^{x \,{\mathrm e}^{5 x^{2}}} x^{2 x \,{\mathrm e}^{5 x^{2}}} \left (x +\frac {1}{4}\right )^{2 x \,{\mathrm e}^{5 x^{2}}} 2^{-2 i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{4 x +1}\right )} \left (x +\frac {1}{4}\right )^{i x^{2} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (4 x +1\right )}\right )} {\mathrm e}^{x^{2} \ln \relax (x )^{2}+4 x^{2} \ln \relax (2)^{2}+x^{2} \ln \left (x +\frac {1}{4}\right )^{2}+{\mathrm e}^{10 x^{2}}} {\mathrm e}^{\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{5} \mathrm {csgn}\left (\frac {i}{x}\right )}{2}} {\mathrm e}^{\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{5} \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )}{2}} {\mathrm e}^{-\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{4} \mathrm {csgn}\left (\frac {i}{x}\right )^{2}}{4}} {\mathrm e}^{-\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{4} \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )^{2}}{4}} {\mathrm e}^{i x \,{\mathrm e}^{5 x^{2}} \pi \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{3}} {\mathrm e}^{-i x \,{\mathrm e}^{5 x^{2}} \pi \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )} {\mathrm e}^{-i x \,{\mathrm e}^{5 x^{2}} \pi \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )} {\mathrm e}^{\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{3} \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )}{2}} {\mathrm e}^{-\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )^{2}}{4}} {\mathrm e}^{-x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{4} \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )} {\mathrm e}^{\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{3} \mathrm {csgn}\left (\frac {i}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )}{2}} {\mathrm e}^{i x \,{\mathrm e}^{5 x^{2}} \pi \,\mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x +\frac {1}{4}}\right )} {\mathrm e}^{-\frac {x^{2} \pi ^{2} \mathrm {csgn}\left (\frac {i}{x \left (x +\frac {1}{4}\right )}\right )^{6}}{4}}\)	\(863\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2+2*x)*ln(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x^2)-16*x^2-2*x)*ln(1/(4*x^2+x))+(80*x^2+20*
x)*exp(5*x^2)^2+(16*x+2)*exp(5*x^2))*exp(x^2*ln(1/(4*x^2+x))^2-2*x*exp(5*x^2)*ln(1/(4*x^2+x))+exp(5*x^2)^2)-4*
x-1)/(4*x+1),x,method=_RETURNVERBOSE)

[Out]

-x+x^(4*x^2*ln(2))*(x+1/4)^(4*x^2*ln(2))*(x+1/4)^(2*x^2*ln(x))*2^(-2*I*x^2*Pi*csgn(I/x))*x^(I*x^2*Pi*csgn(I/x/
(4*x+1))*csgn(I/x)*csgn(I/(4*x+1)))*(x+1/4)^(-I*x^2*Pi*csgn(I/(4*x+1)))*x^(-I*x^2*Pi*csgn(I/x))*2^(2*I*x^2*Pi*
csgn(I/x/(4*x+1))*csgn(I/x)*csgn(I/(4*x+1)))*x^(-I*x^2*Pi*csgn(I/(4*x+1)))*(x+1/4)^(-I*x^2*Pi*csgn(I/x))*(x+1/
4)^(I*x^2*Pi*csgn(I/x/(4*x+1))*csgn(I/x)*csgn(I/(4*x+1)))*2^(2*I*x^2*Pi*csgn(I/x/(4*x+1)))*x^(I*x^2*Pi*csgn(I/
x/(4*x+1)))*16^(x*exp(5*x^2))*x^(2*x*exp(5*x^2))*(x+1/4)^(2*x*exp(5*x^2))*2^(-2*I*x^2*Pi*csgn(I/(4*x+1)))*(x+1
/4)^(I*x^2*Pi*csgn(I/x/(4*x+1)))*exp(x^2*ln(x)^2+4*x^2*ln(2)^2+x^2*ln(x+1/4)^2+exp(10*x^2))*exp(1/2*x^2*Pi^2*c
sgn(I/x/(x+1/4))^5*csgn(I/x))*exp(1/2*x^2*Pi^2*csgn(I/x/(x+1/4))^5*csgn(I/(x+1/4)))*exp(-1/4*x^2*Pi^2*csgn(I/x
/(x+1/4))^4*csgn(I/x)^2)*exp(-1/4*x^2*Pi^2*csgn(I/x/(x+1/4))^4*csgn(I/(x+1/4))^2)*exp(I*x*exp(5*x^2)*Pi*csgn(I
/x/(x+1/4))^3)*exp(-I*x*exp(5*x^2)*Pi*csgn(I/x/(x+1/4))^2*csgn(I/x))*exp(-I*x*exp(5*x^2)*Pi*csgn(I/x/(x+1/4))^
2*csgn(I/(x+1/4)))*exp(1/2*x^2*Pi^2*csgn(I/x/(x+1/4))^3*csgn(I/(x+1/4))^2*csgn(I/x))*exp(-1/4*x^2*Pi^2*csgn(I/
x/(x+1/4))^2*csgn(I/x)^2*csgn(I/(x+1/4))^2)*exp(-x^2*Pi^2*csgn(I/x/(x+1/4))^4*csgn(I/x)*csgn(I/(x+1/4)))*exp(1
/2*x^2*Pi^2*csgn(I/x/(x+1/4))^3*csgn(I/x)^2*csgn(I/(x+1/4)))*exp(I*x*exp(5*x^2)*Pi*csgn(I/x/(x+1/4))*csgn(I/x)
*csgn(I/(x+1/4)))*exp(-1/4*x^2*Pi^2*csgn(I/x/(x+1/4))^6)

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maxima [B]  time = 0.53, size = 71, normalized size = 2.29 \begin {gather*} -x + e^{\left (x^{2} \log \left (4 \, x + 1\right )^{2} + 2 \, x^{2} \log \left (4 \, x + 1\right ) \log \relax (x) + x^{2} \log \relax (x)^{2} + 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (4 \, x + 1\right ) + 2 \, x e^{\left (5 \, x^{2}\right )} \log \relax (x) + e^{\left (10 \, x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+2*x)*log(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x^2)-16*x^2-2*x)*log(1/(4*x^2+x))+(80
*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5*x^2))*exp(x^2*log(1/(4*x^2+x))^2-2*x*exp(5*x^2)*log(1/(4*x^2+x))+exp(5*
x^2)^2)-4*x-1)/(4*x+1),x, algorithm="maxima")

[Out]

-x + e^(x^2*log(4*x + 1)^2 + 2*x^2*log(4*x + 1)*log(x) + x^2*log(x)^2 + 2*x*e^(5*x^2)*log(4*x + 1) + 2*x*e^(5*
x^2)*log(x) + e^(10*x^2))

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mupad [B]  time = 4.51, size = 50, normalized size = 1.61 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{10\,x^2}}\,{\mathrm {e}}^{x^2\,{\ln \left (\frac {1}{4\,x^2+x}\right )}^2}}{{\left (\frac {1}{4\,x^2+x}\right )}^{2\,x\,{\mathrm {e}}^{5\,x^2}}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - exp(exp(10*x^2) + x^2*log(1/(x + 4*x^2))^2 - 2*x*exp(5*x^2)*log(1/(x + 4*x^2)))*(log(1/(x + 4*x^2)
)^2*(2*x + 8*x^2) + exp(5*x^2)*(16*x + 2) - log(1/(x + 4*x^2))*(2*x + exp(5*x^2)*(8*x + 20*x^2 + 80*x^3 + 2) +
16*x^2) + exp(10*x^2)*(20*x + 80*x^2)) + 1)/(4*x + 1),x)

[Out]

(exp(exp(10*x^2))*exp(x^2*log(1/(x + 4*x^2))^2))/(1/(x + 4*x^2))^(2*x*exp(5*x^2)) - x

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sympy [A]  time = 1.56, size = 44, normalized size = 1.42 \begin {gather*} - x + e^{x^{2} \log {\left (\frac {1}{4 x^{2} + x} \right )}^{2} - 2 x e^{5 x^{2}} \log {\left (\frac {1}{4 x^{2} + x} \right )} + e^{10 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2+2*x)*ln(1/(4*x**2+x))**2+((-80*x**3-20*x**2-8*x-2)*exp(5*x**2)-16*x**2-2*x)*ln(1/(4*x**2+x
))+(80*x**2+20*x)*exp(5*x**2)**2+(16*x+2)*exp(5*x**2))*exp(x**2*ln(1/(4*x**2+x))**2-2*x*exp(5*x**2)*ln(1/(4*x*
*2+x))+exp(5*x**2)**2)-4*x-1)/(4*x+1),x)

[Out]

-x + exp(x**2*log(1/(4*x**2 + x))**2 - 2*x*exp(5*x**2)*log(1/(4*x**2 + x)) + exp(10*x**2))

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