3.77.43 \(\int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} (240-60 x-15 x^2)+(120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} (-60 x+15 x^2)) \log (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x)}{(8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} (-4 x+x^2)) \log ^2(-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {15 x}{\log \left (-2+e^{-\frac {2 (2+x)}{x}} (-4+x)-x\right )} \]

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Rubi [F]  time = 4.87, 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 {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60*x^2 + 15*x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(240 - 60*x - 15*x^2) + (120*x + 30*x^2 - 15*x^3 + E^
((-4 - 2*x + x*Log[-4 + x])/x)*(-60*x + 15*x^2))*Log[-2 + E^((-4 - 2*x + x*Log[-4 + x])/x) - x])/((8*x + 2*x^2
- x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(-4*x + x^2))*Log[-2 + E^((-4 - 2*x + x*Log[-4 + x])/x) - x]^2),x]

[Out]

-15*Defer[Int][Log[-2 + E^(-2 - 4/x)*(-4 + x) - x]^(-2), x] + 30*Defer[Int][1/((2 + x)*Log[-2 + E^(-2 - 4/x)*(
-4 + x) - x]^2), x] + 150*Defer[Int][1/((4 - x + E^(2 + 4/x)*(2 + x))*Log[-2 + E^(-2 - 4/x)*(-4 + x) - x]^2),
x] - 240*Defer[Int][1/(x*(4 - x + E^(2 + 4/x)*(2 + x))*Log[-2 + E^(-2 - 4/x)*(-4 + x) - x]^2), x] - 180*Defer[
Int][1/((2 + x)*(4 - x + E^(2 + 4/x)*(2 + x))*Log[-2 + E^(-2 - 4/x)*(-4 + x) - x]^2), x] + 15*Defer[Int][Log[-
2 + E^(-2 - 4/x)*(-4 + x) - x]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 \left (-16+4 x-\left (-1+e^{2+\frac {4}{x}}\right ) x^2+x \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )\right )}{x \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ &=15 \int \frac {-16+4 x-\left (-1+e^{2+\frac {4}{x}}\right ) x^2+x \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}{x \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ &=15 \int \left (\frac {2 \left (-16-4 x+5 x^2\right )}{x (2+x) \left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}+\frac {-x+2 \log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )+x \log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}{(2+x) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}\right ) \, dx\\ &=15 \int \frac {-x+2 \log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )+x \log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}{(2+x) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx+30 \int \frac {-16-4 x+5 x^2}{x (2+x) \left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ &=15 \int \frac {-\frac {x}{2+x}+\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}{\log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx+30 \int \frac {-16-4 x+5 x^2}{x (2+x) \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ &=15 \int \left (-\frac {x}{(2+x) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}+\frac {1}{\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}\right ) \, dx+30 \int \left (\frac {5}{\left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}-\frac {8}{x \left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}-\frac {6}{(2+x) \left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}\right ) \, dx\\ &=-\left (15 \int \frac {x}{(2+x) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\right )+15 \int \frac {1}{\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx+150 \int \frac {1}{\left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx-180 \int \frac {1}{(2+x) \left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx-240 \int \frac {1}{x \left (4+2 e^{2+\frac {4}{x}}-x+e^{2+\frac {4}{x}} x\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ &=-\left (15 \int \left (\frac {1}{\log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}-\frac {2}{(2+x) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )}\right ) \, dx\right )+15 \int \frac {1}{\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx+150 \int \frac {1}{\left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx-180 \int \frac {1}{(2+x) \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx-240 \int \frac {1}{x \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ &=-\left (15 \int \frac {1}{\log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\right )+15 \int \frac {1}{\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx+30 \int \frac {1}{(2+x) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx+150 \int \frac {1}{\left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx-180 \int \frac {1}{(2+x) \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx-240 \int \frac {1}{x \left (4-x+e^{2+\frac {4}{x}} (2+x)\right ) \log ^2\left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 24, normalized size = 0.96 \begin {gather*} \frac {15 x}{\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60*x^2 + 15*x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(240 - 60*x - 15*x^2) + (120*x + 30*x^2 - 15*x^
3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(-60*x + 15*x^2))*Log[-2 + E^((-4 - 2*x + x*Log[-4 + x])/x) - x])/((8*x +
2*x^2 - x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(-4*x + x^2))*Log[-2 + E^((-4 - 2*x + x*Log[-4 + x])/x) - x]^2
),x]

[Out]

(15*x)/Log[-2 + E^(-2 - 4/x)*(-4 + x) - x]

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fricas [A]  time = 1.29, size = 27, normalized size = 1.08 \begin {gather*} \frac {15 \, x}{\log \left (-x + e^{\left (\frac {x \log \left (x - 4\right ) - 2 \, x - 4}{x}\right )} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x^2-60*x)*exp((x*log(x-4)-2*x-4)/x)-15*x^3+30*x^2+120*x)*log(exp((x*log(x-4)-2*x-4)/x)-x-2)+(-
15*x^2-60*x+240)*exp((x*log(x-4)-2*x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(x-4)-2*x-4)/x)-x^3+2*x^2+8*x)/
log(exp((x*log(x-4)-2*x-4)/x)-x-2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 41, normalized size = 1.64 \begin {gather*} \frac {15 \, x^{2}}{x \log \left (-x e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} + x - 2 \, e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} - 4\right ) - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x^2-60*x)*exp((x*log(x-4)-2*x-4)/x)-15*x^3+30*x^2+120*x)*log(exp((x*log(x-4)-2*x-4)/x)-x-2)+(-
15*x^2-60*x+240)*exp((x*log(x-4)-2*x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(x-4)-2*x-4)/x)-x^3+2*x^2+8*x)/
log(exp((x*log(x-4)-2*x-4)/x)-x-2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 1.00

method	result	size
risch	\(\frac {15 x}{\ln \left (\left (x -4\right ) {\mathrm e}^{-\frac {2 \left (2+x \right )}{x}}-x -2\right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((15*x^2-60*x)*exp((x*ln(x-4)-2*x-4)/x)-15*x^3+30*x^2+120*x)*ln(exp((x*ln(x-4)-2*x-4)/x)-x-2)+(-15*x^2-60
*x+240)*exp((x*ln(x-4)-2*x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*ln(x-4)-2*x-4)/x)-x^3+2*x^2+8*x)/ln(exp((x*l
n(x-4)-2*x-4)/x)-x-2)^2,x,method=_RETURNVERBOSE)

[Out]

15*x/ln((x-4)*exp(-2*(2+x)/x)-x-2)

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maxima [A]  time = 0.45, size = 35, normalized size = 1.40 \begin {gather*} \frac {15 \, x^{2}}{x \log \left (-{\left (x e^{2} + 2 \, e^{2}\right )} e^{\frac {4}{x}} + x - 4\right ) - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x^2-60*x)*exp((x*log(x-4)-2*x-4)/x)-15*x^3+30*x^2+120*x)*log(exp((x*log(x-4)-2*x-4)/x)-x-2)+(-
15*x^2-60*x+240)*exp((x*log(x-4)-2*x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(x-4)-2*x-4)/x)-x^3+2*x^2+8*x)/
log(exp((x*log(x-4)-2*x-4)/x)-x-2)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int -\frac {\ln \left ({\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}-x-2\right )\,\left (120\,x-{\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}\,\left (60\,x-15\,x^2\right )+30\,x^2-15\,x^3\right )-{\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}\,\left (15\,x^2+60\,x-240\right )-60\,x^2+15\,x^3}{{\ln \left ({\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}-x-2\right )}^2\,\left (8\,x-{\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}\,\left (4\,x-x^2\right )+2\,x^2-x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(-(2*x - x*log(x - 4) + 4)/x) - x - 2)*(120*x - exp(-(2*x - x*log(x - 4) + 4)/x)*(60*x - 15*x^2) +
30*x^2 - 15*x^3) - exp(-(2*x - x*log(x - 4) + 4)/x)*(60*x + 15*x^2 - 240) - 60*x^2 + 15*x^3)/(log(exp(-(2*x -
x*log(x - 4) + 4)/x) - x - 2)^2*(8*x - exp(-(2*x - x*log(x - 4) + 4)/x)*(4*x - x^2) + 2*x^2 - x^3)),x)

[Out]

-int(-(log(exp(-(2*x - x*log(x - 4) + 4)/x) - x - 2)*(120*x - exp(-(2*x - x*log(x - 4) + 4)/x)*(60*x - 15*x^2)
+ 30*x^2 - 15*x^3) - exp(-(2*x - x*log(x - 4) + 4)/x)*(60*x + 15*x^2 - 240) - 60*x^2 + 15*x^3)/(log(exp(-(2*x
- x*log(x - 4) + 4)/x) - x - 2)^2*(8*x - exp(-(2*x - x*log(x - 4) + 4)/x)*(4*x - x^2) + 2*x^2 - x^3)), x)

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sympy [A]  time = 1.48, size = 22, normalized size = 0.88 \begin {gather*} \frac {15 x}{\log {\left (- x + e^{\frac {x \log {\left (x - 4 \right )} - 2 x - 4}{x}} - 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x**2-60*x)*exp((x*ln(x-4)-2*x-4)/x)-15*x**3+30*x**2+120*x)*ln(exp((x*ln(x-4)-2*x-4)/x)-x-2)+(-
15*x**2-60*x+240)*exp((x*ln(x-4)-2*x-4)/x)+15*x**3-60*x**2)/((x**2-4*x)*exp((x*ln(x-4)-2*x-4)/x)-x**3+2*x**2+8
*x)/ln(exp((x*ln(x-4)-2*x-4)/x)-x-2)**2,x)

[Out]

15*x/log(-x + exp((x*log(x - 4) - 2*x - 4)/x) - 2)

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