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

Optimal. Leaf size=27 \[ x+\frac {1}{-2 x+\log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \]

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Rubi [F]  time = 4.50, 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-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x + 2*Defer[Int][(2*x - Log[-x + (16*(1 + Log[5])^2)/E^(2*x)])^(-2), x] - Defer[Int][1/(x*(2*x - Log[-x + (16*
(1 + Log[5])^2)/E^(2*x)])^2), x] + 32*(1 + Log[5])^2*Defer[Int][1/((-(E^(2*x)*x) + 16*(1 + Log[5]*(2 + Log[5])
))*(2*x - Log[-x + (16*(1 + Log[5])^2)/E^(2*x)])^2), x] + 16*(1 + Log[5])^2*Defer[Int][1/(x*(-(E^(2*x)*x) + 16
*(1 + Log[5]*(2 + Log[5])))*(2*x - Log[-x + (16*(1 + Log[5])^2)/E^(2*x)])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-1+2 x+4 x^3\right )-64 \left (1+x^2\right ) (1+\log (5))^2-4 x \left (e^{2 x} x-16 (1+\log (5))^2\right ) \log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )+\left (e^{2 x} x-16 (1+\log (5))^2\right ) \log ^2\left (-x+16 e^{-2 x} (1+\log (5))^2\right )}{\left (e^{2 x} x-16 (1+\log (5))^2\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx\\ &=\int \left (\frac {16 (-1-2 x) (1+\log (5))^2}{x \left (e^{2 x} x-16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}+\frac {-1+2 x+4 x^3-4 x^2 \log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )+x \log ^2\left (-x+16 e^{-2 x} (1+\log (5))^2\right )}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx\\ &=\left (16 (1+\log (5))^2\right ) \int \frac {-1-2 x}{x \left (e^{2 x} x-16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\int \frac {-1+2 x+4 x^3-4 x^2 \log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )+x \log ^2\left (-x+16 e^{-2 x} (1+\log (5))^2\right )}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx\\ &=\left (16 (1+\log (5))^2\right ) \int \left (\frac {2}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}+\frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx+\int \left (1+\frac {-1+2 x}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx\\ &=x+\left (16 (1+\log (5))^2\right ) \int \frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (32 (1+\log (5))^2\right ) \int \frac {1}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\int \frac {-1+2 x}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx\\ &=x+\left (16 (1+\log (5))^2\right ) \int \frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (32 (1+\log (5))^2\right ) \int \frac {1}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\int \left (\frac {2}{\left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}-\frac {1}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx\\ &=x+2 \int \frac {1}{\left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (16 (1+\log (5))^2\right ) \int \frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (32 (1+\log (5))^2\right ) \int \frac {1}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx-\int \frac {1}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 26, normalized size = 0.96 \begin {gather*} x+\frac {1}{-2 x+\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x - 4*x^3 + ((4 + 4*x^2)*(4 + 4*Log[5])^2)/E^(2*x) + (4*x^2 - (4*x*(4 + 4*Log[5])^2)/E^(2*x))
*Log[-x + (4 + 4*Log[5])^2/E^(2*x)] + (-x + (4 + 4*Log[5])^2/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)]^2)/(-
4*x^3 + (4*x^2*(4 + 4*Log[5])^2)/E^(2*x) + (4*x^2 - (4*x*(4 + 4*Log[5])^2)/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/
E^(2*x)] + (-x + (4 + 4*Log[5])^2/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)]^2),x]

[Out]

x + (-2*x + Log[-x + (16*(1 + Log[5])^2)/E^(2*x)])^(-1)

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fricas [A]  time = 0.97, size = 57, normalized size = 2.11 \begin {gather*} \frac {2 \, x^{2} - x \log \left (-x + e^{\left (-2 \, x + 2 \, \log \left (4 \, \log \relax (5) + 4\right )\right )}\right ) - 1}{2 \, x - \log \left (-x + e^{\left (-2 \, x + 2 \, \log \left (4 \, \log \relax (5) + 4\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.80, size = 77, normalized size = 2.85 \begin {gather*} \frac {2 \, x^{2} - x \log \left (16 \, e^{\left (-2 \, x\right )} \log \relax (5)^{2} + 32 \, e^{\left (-2 \, x\right )} \log \relax (5) - x + 16 \, e^{\left (-2 \, x\right )}\right ) - 1}{2 \, x - \log \left (16 \, e^{\left (-2 \, x\right )} \log \relax (5)^{2} + 32 \, e^{\left (-2 \, x\right )} \log \relax (5) - x + 16 \, e^{\left (-2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^2 - x*log(16*e^(-2*x)*log(5)^2 + 32*e^(-2*x)*log(5) - x + 16*e^(-2*x)) - 1)/(2*x - log(16*e^(-2*x)*log(5)
^2 + 32*e^(-2*x)*log(5) - x + 16*e^(-2*x)))

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maple [A]  time = 0.06, size = 31, normalized size = 1.15

method	result	size
risch	\(x -\frac {1}{2 x -\ln \left (\left (4 \ln \relax (5)+4\right )^{2} {\mathrm e}^{-2 x}-x \right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.82, size = 59, normalized size = 2.19 \begin {gather*} \frac {4 \, x^{2} - x \log \left (-x e^{\left (2 \, x\right )} + 16 \, \log \relax (5)^{2} + 32 \, \log \relax (5) + 16\right ) - 1}{4 \, x - \log \left (-x e^{\left (2 \, x\right )} + 16 \, \log \relax (5)^{2} + 32 \, \log \relax (5) + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x^2 - x*log(-x*e^(2*x) + 16*log(5)^2 + 32*log(5) + 16) - 1)/(4*x - log(-x*e^(2*x) + 16*log(5)^2 + 32*log(5)
+ 16))

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mupad [B]  time = 3.73, size = 41, normalized size = 1.52 \begin {gather*} x-\frac {1}{2\,x-\ln \left (16\,{\mathrm {e}}^{-2\,x}-x+32\,{\mathrm {e}}^{-2\,x}\,\ln \relax (5)+16\,{\mathrm {e}}^{-2\,x}\,{\ln \relax (5)}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - exp(2*log(4*log(5) + 4) - 2*x)*(4*x^2 + 4) + log(exp(2*log(4*log(5) + 4) - 2*x) - x)*(4*x*exp(2*log
(4*log(5) + 4) - 2*x) - 4*x^2) + log(exp(2*log(4*log(5) + 4) - 2*x) - x)^2*(x - exp(2*log(4*log(5) + 4) - 2*x)
) + 4*x^3 - 1)/(log(exp(2*log(4*log(5) + 4) - 2*x) - x)*(4*x*exp(2*log(4*log(5) + 4) - 2*x) - 4*x^2) + log(exp
(2*log(4*log(5) + 4) - 2*x) - x)^2*(x - exp(2*log(4*log(5) + 4) - 2*x)) - 4*x^2*exp(2*log(4*log(5) + 4) - 2*x)
+ 4*x^3),x)

[Out]

x - 1/(2*x - log(16*exp(-2*x) - x + 32*exp(-2*x)*log(5) + 16*exp(-2*x)*log(5)^2))

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sympy [A]  time = 0.69, size = 22, normalized size = 0.81 \begin {gather*} x + \frac {1}{- 2 x + \log {\left (- x + \left (4 + 4 \log {\relax (5 )}\right )^{2} e^{- 2 x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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