3.101.33 \(\int \frac {e^{\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}} (16-4 e^x x+4 e^{2 x} x-32 e^x x \log (6 x)+64 x \log ^2(6 x))}{e^{2 x} x-8 e^x x \log (6 x)+16 x \log ^2(6 x)} \, dx\)

Optimal. Leaf size=18 \[ 8 e^{x+\frac {1}{e^x-4 \log (6 x)}} \]

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Rubi [F]  time = 25.13, 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 {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \left (16-4 e^x x+4 e^{2 x} x-32 e^x x \log (6 x)+64 x \log ^2(6 x)\right )}{e^{2 x} x-8 e^x x \log (6 x)+16 x \log ^2(6 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-1 + E^x*(-x - Log[2]) + (4*x + 4*Log[2])*Log[6*x])/(-E^x + 4*Log[6*x]))*(16 - 4*E^x*x + 4*E^(2*x)*x
- 32*E^x*x*Log[6*x] + 64*x*Log[6*x]^2))/(E^(2*x)*x - 8*E^x*x*Log[6*x] + 16*x*Log[6*x]^2),x]

[Out]

4*Defer[Int][E^((-1 + E^x*(-x - Log[2]) + (4*x + 4*Log[2])*Log[6*x])/(-E^x + 4*Log[6*x])), x] + 16*Defer[Int][
E^((-1 + E^x*(-x - Log[2]) + (4*x + 4*Log[2])*Log[6*x])/(-E^x + 4*Log[6*x]))/(x*(E^x - 4*Log[6*x])^2), x] - 4*
Defer[Int][E^((-1 + E^x*(-x - Log[2]) + (4*x + 4*Log[2])*Log[6*x])/(-E^x + 4*Log[6*x]))/(E^x - 4*Log[6*x]), x]
- 16*Defer[Int][(E^((-1 + E^x*(-x - Log[2]) + (4*x + 4*Log[2])*Log[6*x])/(-E^x + 4*Log[6*x]))*Log[6*x])/(E^x
- 4*Log[6*x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \left (4-e^x x+e^{2 x} x-8 e^x x \log (6 x)+16 x \log ^2(6 x)\right )}{x \left (e^x-4 \log (6 x)\right )^2} \, dx\\ &=4 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \left (4-e^x x+e^{2 x} x-8 e^x x \log (6 x)+16 x \log ^2(6 x)\right )}{x \left (e^x-4 \log (6 x)\right )^2} \, dx\\ &=4 \int \left (\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )-\frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )}{e^x-4 \log (6 x)}-\frac {4 \exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) (-1+x \log (6 x))}{x \left (e^x-4 \log (6 x)\right )^2}\right ) \, dx\\ &=4 \int \exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \, dx-4 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )}{e^x-4 \log (6 x)} \, dx-16 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) (-1+x \log (6 x))}{x \left (e^x-4 \log (6 x)\right )^2} \, dx\\ &=4 \int \exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \, dx-4 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )}{e^x-4 \log (6 x)} \, dx-16 \int \left (-\frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )}{x \left (e^x-4 \log (6 x)\right )^2}+\frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \log (6 x)}{\left (e^x-4 \log (6 x)\right )^2}\right ) \, dx\\ &=4 \int \exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \, dx-4 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )}{e^x-4 \log (6 x)} \, dx+16 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right )}{x \left (e^x-4 \log (6 x)\right )^2} \, dx-16 \int \frac {\exp \left (\frac {-1+e^x (-x-\log (2))+(4 x+4 \log (2)) \log (6 x)}{-e^x+4 \log (6 x)}\right ) \log (6 x)}{\left (e^x-4 \log (6 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 18, normalized size = 1.00 \begin {gather*} 8 e^{x+\frac {1}{e^x-4 \log (6 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 + E^x*(-x - Log[2]) + (4*x + 4*Log[2])*Log[6*x])/(-E^x + 4*Log[6*x]))*(16 - 4*E^x*x + 4*E^(2
*x)*x - 32*E^x*x*Log[6*x] + 64*x*Log[6*x]^2))/(E^(2*x)*x - 8*E^x*x*Log[6*x] + 16*x*Log[6*x]^2),x]

[Out]

8*E^(x + (E^x - 4*Log[6*x])^(-1))

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fricas [A]  time = 1.00, size = 34, normalized size = 1.89 \begin {gather*} 4 \, e^{\left (\frac {{\left (x + \log \relax (2)\right )} e^{x} - 4 \, {\left (x + \log \relax (2)\right )} \log \left (6 \, x\right ) + 1}{e^{x} - 4 \, \log \left (6 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(6*x)^2-32*x*exp(x)*log(6*x)+4*x*exp(x)^2-4*exp(x)*x+16)*exp(((4*log(2)+4*x)*log(6*x)+(-x-l
og(2))*exp(x)-1)/(4*log(6*x)-exp(x)))/(16*x*log(6*x)^2-8*x*exp(x)*log(6*x)+x*exp(x)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 4.28, size = 56, normalized size = 3.11 \begin {gather*} 4 \, e^{\left (\frac {x e^{x} - 4 \, x \log \relax (2) + e^{x} \log \relax (2) - 4 \, \log \relax (2)^{2} - 4 \, x \log \left (3 \, x\right ) - 4 \, \log \relax (2) \log \left (3 \, x\right ) + 1}{e^{x} - 4 \, \log \relax (2) - 4 \, \log \left (3 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(6*x)^2-32*x*exp(x)*log(6*x)+4*x*exp(x)^2-4*exp(x)*x+16)*exp(((4*log(2)+4*x)*log(6*x)+(-x-l
og(2))*exp(x)-1)/(4*log(6*x)-exp(x)))/(16*x*log(6*x)^2-8*x*exp(x)*log(6*x)+x*exp(x)^2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.09, size = 42, normalized size = 2.33

method	result	size
risch	\(4 \,{\mathrm e}^{\frac {{\mathrm e}^{x} \ln \relax (2)+{\mathrm e}^{x} x -4 \ln \left (6 x \right ) \ln \relax (2)-4 x \ln \left (6 x \right )+1}{{\mathrm e}^{x}-4 \ln \left (6 x \right )}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*x*ln(6*x)^2-32*x*exp(x)*ln(6*x)+4*x*exp(x)^2-4*exp(x)*x+16)*exp(((4*ln(2)+4*x)*ln(6*x)+(-x-ln(2))*exp(
x)-1)/(4*ln(6*x)-exp(x)))/(16*x*ln(6*x)^2-8*x*exp(x)*ln(6*x)+x*exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

4*exp((exp(x)*ln(2)+exp(x)*x-4*ln(6*x)*ln(2)-4*x*ln(6*x)+1)/(exp(x)-4*ln(6*x)))

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maxima [B]  time = 0.99, size = 199, normalized size = 11.06 \begin {gather*} 4 \, e^{\left (\frac {x e^{x}}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} - \frac {4 \, x \log \relax (3)}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} - \frac {4 \, x \log \relax (2)}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} + \frac {e^{x} \log \relax (2)}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} - \frac {4 \, \log \relax (3) \log \relax (2)}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} - \frac {4 \, \log \relax (2)^{2}}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} - \frac {4 \, x \log \relax (x)}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} - \frac {4 \, \log \relax (2) \log \relax (x)}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)} + \frac {1}{e^{x} - 4 \, \log \relax (3) - 4 \, \log \relax (2) - 4 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(6*x)^2-32*x*exp(x)*log(6*x)+4*x*exp(x)^2-4*exp(x)*x+16)*exp(((4*log(2)+4*x)*log(6*x)+(-x-l
og(2))*exp(x)-1)/(4*log(6*x)-exp(x)))/(16*x*log(6*x)^2-8*x*exp(x)*log(6*x)+x*exp(x)^2),x, algorithm="maxima")

[Out]

4*e^(x*e^x/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) - 4*x*log(3)/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) - 4*x*lo
g(2)/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) + e^x*log(2)/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) - 4*log(3)*log
(2)/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) - 4*log(2)^2/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) - 4*x*log(x)/(e
^x - 4*log(3) - 4*log(2) - 4*log(x)) - 4*log(2)*log(x)/(e^x - 4*log(3) - 4*log(2) - 4*log(x)) + 1/(e^x - 4*log
(3) - 4*log(2) - 4*log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^x\,\left (x+\ln \relax (2)\right )-\ln \left (6\,x\right )\,\left (4\,x+4\,\ln \relax (2)\right )+1}{4\,\ln \left (6\,x\right )-{\mathrm {e}}^x}}\,\left (64\,x\,{\ln \left (6\,x\right )}^2-32\,x\,{\mathrm {e}}^x\,\ln \left (6\,x\right )+4\,x\,{\mathrm {e}}^{2\,x}-4\,x\,{\mathrm {e}}^x+16\right )}{16\,x\,{\ln \left (6\,x\right )}^2-8\,x\,{\mathrm {e}}^x\,\ln \left (6\,x\right )+x\,{\mathrm {e}}^{2\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(exp(x)*(x + log(2)) - log(6*x)*(4*x + 4*log(2)) + 1)/(4*log(6*x) - exp(x)))*(4*x*exp(2*x) + 64*x*lo
g(6*x)^2 - 4*x*exp(x) - 32*x*log(6*x)*exp(x) + 16))/(x*exp(2*x) + 16*x*log(6*x)^2 - 8*x*log(6*x)*exp(x)),x)

[Out]

int((exp(-(exp(x)*(x + log(2)) - log(6*x)*(4*x + 4*log(2)) + 1)/(4*log(6*x) - exp(x)))*(4*x*exp(2*x) + 64*x*lo
g(6*x)^2 - 4*x*exp(x) - 32*x*log(6*x)*exp(x) + 16))/(x*exp(2*x) + 16*x*log(6*x)^2 - 8*x*log(6*x)*exp(x)), x)

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sympy [B]  time = 0.97, size = 37, normalized size = 2.06 \begin {gather*} 4 e^{\frac {\left (- x - \log {\relax (2 )}\right ) e^{x} + \left (4 x + 4 \log {\relax (2 )}\right ) \log {\left (6 x \right )} - 1}{- e^{x} + 4 \log {\left (6 x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*ln(6*x)**2-32*x*exp(x)*ln(6*x)+4*x*exp(x)**2-4*exp(x)*x+16)*exp(((4*ln(2)+4*x)*ln(6*x)+(-x-ln(
2))*exp(x)-1)/(4*ln(6*x)-exp(x)))/(16*x*ln(6*x)**2-8*x*exp(x)*ln(6*x)+x*exp(x)**2),x)

[Out]

4*exp(((-x - log(2))*exp(x) + (4*x + 4*log(2))*log(6*x) - 1)/(-exp(x) + 4*log(6*x)))

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