3.2.96 \(\int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x (\frac {1}{12 x+4 x^2})^{e^{-3+x}}} (\frac {1}{12 x+4 x^2})^{e^{-3+x}} (e^{2 x} (-6-4 x)+e^{3+x} (6 x+2 x^2)+e^{2 x} (6 x+2 x^2) \log (\frac {5}{12 x+4 x^2}))}{3 x+x^2} \, dx\)

Optimal. Leaf size=41 \[ e^{-8+2^{1-2 e^{-3+x}} 5^{e^{-3+x}} e^x \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}}} \]

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Rubi [F]  time = 25.28, 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 {5^{e^{-3+x}} \exp \left (-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5^E^(-3 + x)*E^(-11 + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*((12*x + 4*x^2)^(-1))^E^(-3 +
x)*(E^(2*x)*(-6 - 4*x) + E^(3 + x)*(6*x + 2*x^2) + E^(2*x)*(6*x + 2*x^2)*Log[5/(12*x + 4*x^2)]))/(3*x + x^2),x
]

[Out]

2*Defer[Int][5^E^(-3 + x)*E^(-8 + x + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*(1/(x*(12 + 4*x)))^
E^(-3 + x), x] + 2*Log[5/(4*x*(3 + x))]*Defer[Int][5^E^(-3 + x)*E^(-11 + 2*x + 2*5^E^(-3 + x)*E^x*((12*x + 4*x
^2)^(-1))^E^(-3 + x))*(1/(x*(12 + 4*x)))^E^(-3 + x), x] + 2*Defer[Int][(5^E^(-3 + x)*E^(-11 + 2*x + 2*5^E^(-3
+ x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*(1/(x*(12 + 4*x)))^E^(-3 + x))/(-3 - x), x] - 2*Defer[Int][(5^E^(-3
+ x)*E^(-11 + 2*x + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*(1/(x*(12 + 4*x)))^E^(-3 + x))/x, x]
- 2*Defer[Int][Defer[Int][(5/4)^E^(-3 + x)*E^(-11 + 2*x + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x)
)*(1/(x*(3 + x)))^E^(-3 + x), x]/(-3 - x), x] + 2*Defer[Int][Defer[Int][(5/4)^E^(-3 + x)*E^(-11 + 2*x + 2*5^E^
(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*(1/(x*(3 + x)))^E^(-3 + x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5^{e^{-3+x}} \exp \left (-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{x (3+x)} \, dx\\ &=\int \frac {5^{e^{-3+x}} \exp \left (-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{x (3+x)} \, dx\\ &=\int \left (2\ 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}+\frac {2\ 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \left (-3-2 x+3 x \log \left (\frac {5}{4 x (3+x)}\right )+x^2 \log \left (\frac {5}{4 x (3+x)}\right )\right )}{x (3+x)}\right ) \, dx\\ &=2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \left (-3-2 x+3 x \log \left (\frac {5}{4 x (3+x)}\right )+x^2 \log \left (\frac {5}{4 x (3+x)}\right )\right )}{x (3+x)} \, dx\\ &=2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \left (\frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) (-3-2 x) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x (3+x)}+5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \log \left (\frac {5}{4 x (3+x)}\right )\right ) \, dx\\ &=2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) (-3-2 x) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x (3+x)} \, dx+2 \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \log \left (\frac {5}{4 x (3+x)}\right ) \, dx\\ &=2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \left (\frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{-3-x}-\frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x}\right ) \, dx-2 \int \frac {(-3-2 x) \int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{x (3+x)} \, dx+\left (2 \log \left (\frac {5}{4 x (3+x)}\right )\right ) \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx\\ &=2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{-3-x} \, dx-2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x} \, dx-2 \int \left (\frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{-3-x}-\frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{x}\right ) \, dx+\left (2 \log \left (\frac {5}{4 x (3+x)}\right )\right ) \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx\\ &=2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{-3-x} \, dx-2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x} \, dx-2 \int \frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{-3-x} \, dx+2 \int \frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{x} \, dx+\left (2 \log \left (\frac {5}{4 x (3+x)}\right )\right ) \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.69, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(5^E^(-3 + x)*E^(-11 + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*((12*x + 4*x^2)^(-1))^E^
(-3 + x)*(E^(2*x)*(-6 - 4*x) + E^(3 + x)*(6*x + 2*x^2) + E^(2*x)*(6*x + 2*x^2)*Log[5/(12*x + 4*x^2)]))/(3*x +
x^2),x]

[Out]

Integrate[(5^E^(-3 + x)*E^(-11 + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 + x))*((12*x + 4*x^2)^(-1))^E^
(-3 + x)*(E^(2*x)*(-6 - 4*x) + E^(3 + x)*(6*x + 2*x^2) + E^(2*x)*(6*x + 2*x^2)*Log[5/(12*x + 4*x^2)]))/(3*x +
x^2), x]

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fricas [A]  time = 0.71, size = 33, normalized size = 0.80 \begin {gather*} e^{\left ({\left (2 \, \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{\left (x + 3\right )} - 11 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+6*x)*exp(x)^2*log(5/(4*x^2+12*x))+(-4*x-6)*exp(x)^2+(2*x^2+6*x)*exp(3)*exp(x))*exp(exp(x)*lo
g(5/(4*x^2+12*x))/exp(3))*exp(exp(x)*exp(exp(x)*log(5/(4*x^2+12*x))/exp(3))-4)^2/(x^2+3*x)/exp(3),x, algorithm
="fricas")

[Out]

e^((2*(5/4/(x^2 + 3*x))^e^(x - 3)*e^(x + 3) - 11*e^3)*e^(-3) + 3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )} \log \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right ) - {\left (2 \, x + 3\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 3 \, x\right )} e^{\left (x + 3\right )}\right )} \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{\left (2 \, \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{x} - 11\right )}}{x^{2} + 3 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+6*x)*exp(x)^2*log(5/(4*x^2+12*x))+(-4*x-6)*exp(x)^2+(2*x^2+6*x)*exp(3)*exp(x))*exp(exp(x)*lo
g(5/(4*x^2+12*x))/exp(3))*exp(exp(x)*exp(exp(x)*log(5/(4*x^2+12*x))/exp(3))-4)^2/(x^2+3*x)/exp(3),x, algorithm
="giac")

[Out]

integrate(2*((x^2 + 3*x)*e^(2*x)*log(5/4/(x^2 + 3*x)) - (2*x + 3)*e^(2*x) + (x^2 + 3*x)*e^(x + 3))*(5/4/(x^2 +
3*x))^e^(x - 3)*e^(2*(5/4/(x^2 + 3*x))^e^(x - 3)*e^x - 11)/(x^2 + 3*x), x)

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maple [C]  time = 1.02, size = 164, normalized size = 4.00

method	result	size
risch	\({\mathrm e}^{2 \,{\mathrm e}^{x -\ln \relax (x ) {\mathrm e}^{x -3}-2 \ln \relax (2) {\mathrm e}^{x -3}-\ln \left (3+x \right ) {\mathrm e}^{x -3}+\ln \relax (5) {\mathrm e}^{x -3}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (\frac {i}{x \left (3+x \right )}\right )^{3} {\mathrm e}^{x -3}}{2}} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (\frac {i}{x \left (3+x \right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x -3}}{2}} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (\frac {i}{x \left (3+x \right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{3+x}\right ) {\mathrm e}^{x -3}}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x \left (3+x \right )}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{3+x}\right ) {\mathrm e}^{x -3}}{2}}-8}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+6*x)*exp(x)^2*ln(5/(4*x^2+12*x))+(-4*x-6)*exp(x)^2+(2*x^2+6*x)*exp(3)*exp(x))*exp(exp(x)*ln(5/(4*x
^2+12*x))/exp(3))*exp(exp(x)*exp(exp(x)*ln(5/(4*x^2+12*x))/exp(3))-4)^2/(x^2+3*x)/exp(3),x,method=_RETURNVERBO
SE)

[Out]

exp(2*exp(x-ln(x)*exp(x-3)-2*ln(2)*exp(x-3)-ln(3+x)*exp(x-3)+ln(5)*exp(x-3))*exp(-1/2*I*Pi*csgn(I/x/(3+x))^3*e
xp(x-3))*exp(1/2*I*Pi*csgn(I/x/(3+x))^2*csgn(I/x)*exp(x-3))*exp(1/2*I*Pi*csgn(I/x/(3+x))^2*csgn(I/(3+x))*exp(x
-3))*exp(-1/2*I*Pi*csgn(I/x/(3+x))*csgn(I/x)*csgn(I/(3+x))*exp(x-3))-8)

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maxima [A]  time = 0.94, size = 41, normalized size = 1.00 \begin {gather*} e^{\left (2 \, e^{\left (e^{\left (x - 3\right )} \log \relax (5) - 2 \, e^{\left (x - 3\right )} \log \relax (2) - e^{\left (x - 3\right )} \log \left (x + 3\right ) - e^{\left (x - 3\right )} \log \relax (x) + x\right )} - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+6*x)*exp(x)^2*log(5/(4*x^2+12*x))+(-4*x-6)*exp(x)^2+(2*x^2+6*x)*exp(3)*exp(x))*exp(exp(x)*lo
g(5/(4*x^2+12*x))/exp(3))*exp(exp(x)*exp(exp(x)*log(5/(4*x^2+12*x))/exp(3))-4)^2/(x^2+3*x)/exp(3),x, algorithm
="maxima")

[Out]

e^(2*e^(e^(x - 3)*log(5) - 2*e^(x - 3)*log(2) - e^(x - 3)*log(x + 3) - e^(x - 3)*log(x) + x) - 8)

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mupad [B]  time = 0.78, size = 27, normalized size = 0.66 \begin {gather*} {\mathrm {e}}^{-8}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x\,{\left (\frac {5}{4\,x^2+12\,x}\right )}^{{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*exp(exp(-3)*exp(x)*log(5/(12*x + 4*x^2)))*exp(2*exp(exp(-3)*exp(x)*log(5/(12*x + 4*x^2)))*exp(x)
- 8)*(exp(3)*exp(x)*(6*x + 2*x^2) - exp(2*x)*(4*x + 6) + exp(2*x)*log(5/(12*x + 4*x^2))*(6*x + 2*x^2)))/(3*x +
x^2),x)

[Out]

exp(-8)*exp(2*exp(x)*(5/(12*x + 4*x^2))^(exp(-3)*exp(x)))

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sympy [A]  time = 38.67, size = 27, normalized size = 0.66 \begin {gather*} e^{2 e^{x} e^{\frac {e^{x} \log {\left (\frac {5}{4 x^{2} + 12 x} \right )}}{e^{3}}} - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+6*x)*exp(x)**2*ln(5/(4*x**2+12*x))+(-4*x-6)*exp(x)**2+(2*x**2+6*x)*exp(3)*exp(x))*exp(exp(x
)*ln(5/(4*x**2+12*x))/exp(3))*exp(exp(x)*exp(exp(x)*ln(5/(4*x**2+12*x))/exp(3))-4)**2/(x**2+3*x)/exp(3),x)

[Out]

exp(2*exp(x)*exp(exp(-3)*exp(x)*log(5/(4*x**2 + 12*x))) - 8)

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