3.55.26 \(\int \frac {-5 x+x^2+e^{x^2} (10 x^3-2 x^4+e^3 (-10 x^2+2 x^3))+(-20 e^3+e^{x^2} (5 e^3-5 x)+20 x+(-5 e^3+5 x) \log (e^3-x)) \log (\frac {5}{4-e^{x^2}+\log (e^3-x)})}{(4 e^3 x^2-4 x^3+e^{x^2} (-e^3 x^2+x^3)+(e^3 x^2-x^3) \log (e^3-x)) \log ^2(\frac {5}{4-e^{x^2}+\log (e^3-x)})} \, dx\)

Optimal. Leaf size=35 \[ 3+\frac {5-x}{x \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \]

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Rubi [F]  time = 20.49, 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 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

10*Defer[Int][Log[5/(4 - E^x^2 + Log[E^3 - x])]^(-2), x] - 2*Defer[Int][x/Log[5/(4 - E^x^2 + Log[E^3 - x])]^2,
x] + 40*Defer[Int][1/((-4 + E^x^2 - Log[E^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x] + ((5 - E^3)*Defer
[Int][1/((E^3 - x)*(-4 + E^x^2 - Log[E^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x])/E^3 + (5*Defer[Int][1
/(x*(-4 + E^x^2 - Log[E^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x])/E^3 - (40*Defer[Int][x/((-4 + E^x^2
- Log[E^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x])/E^3 + (8*(5 - E^3)*Defer[Int][x/((-4 + E^x^2 - Log[E
^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x])/E^3 + 10*Defer[Int][Log[E^3 - x]/((-4 + E^x^2 - Log[E^3 - x
])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x] - (10*Defer[Int][(x*Log[E^3 - x])/((-4 + E^x^2 - Log[E^3 - x])*Log
[5/(4 - E^x^2 + Log[E^3 - x])]^2), x])/E^3 + (2*(5 - E^3)*Defer[Int][(x*Log[E^3 - x])/((-4 + E^x^2 - Log[E^3 -
x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2), x])/E^3 - 5*Defer[Int][1/(x^2*Log[5/(4 - E^x^2 + Log[E^3 - x])]), x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {(-5+x) x \left (-1-2 e^{3+x^2} x+2 e^{x^2} x^2\right )}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right )}-5 \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{x^2 \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx\\ &=\int \left (-\frac {(-5+x) \left (-1-8 e^3 x+8 x^2-2 e^3 x \log \left (e^3-x\right )+2 x^2 \log \left (e^3-x\right )\right )}{x \left (-e^3+x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}+\frac {10 x^2-2 x^3-5 \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{x^2 \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}\right ) \, dx\\ &=-\int \frac {(-5+x) \left (-1-8 e^3 x+8 x^2-2 e^3 x \log \left (e^3-x\right )+2 x^2 \log \left (e^3-x\right )\right )}{x \left (-e^3+x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx+\int \frac {10 x^2-2 x^3-5 \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{x^2 \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx\\ &=-\int \left (\frac {\left (-5+e^3\right ) \left (1+8 e^3 x-8 x^2+2 e^3 x \log \left (e^3-x\right )-2 x^2 \log \left (e^3-x\right )\right )}{e^3 \left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}-\frac {5 \left (1+8 e^3 x-8 x^2+2 e^3 x \log \left (e^3-x\right )-2 x^2 \log \left (e^3-x\right )\right )}{e^3 x \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}\right ) \, dx+\int \left (-\frac {2 (-5+x)}{\log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}-\frac {5}{x^2 \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {-5+x}{\log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx\right )-5 \int \frac {1}{x^2 \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx+\frac {5 \int \frac {1+8 e^3 x-8 x^2+2 e^3 x \log \left (e^3-x\right )-2 x^2 \log \left (e^3-x\right )}{x \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx}{e^3}-\frac {\left (-5+e^3\right ) \int \frac {1+8 e^3 x-8 x^2+2 e^3 x \log \left (e^3-x\right )-2 x^2 \log \left (e^3-x\right )}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx}{e^3}\\ &=-\left (2 \int \left (-\frac {5}{\log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}+\frac {x}{\log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}\right ) \, dx\right )-5 \int \frac {1}{x^2 \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx+\frac {5 \int \left (\frac {8 e^3}{\left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}+\frac {1}{x \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}-\frac {8 x}{\left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}+\frac {2 e^3 \log \left (e^3-x\right )}{\left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}-\frac {2 x \log \left (e^3-x\right )}{\left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}\right ) \, dx}{e^3}-\frac {\left (-5+e^3\right ) \int \left (\frac {1}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}+\frac {8 e^3 x}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}-\frac {8 x^2}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}+\frac {2 e^3 x \log \left (e^3-x\right )}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}-\frac {2 x^2 \log \left (e^3-x\right )}{\left (e^3-x\right ) \left (-4+e^{x^2}-\log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}\right ) \, dx}{e^3}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 33, normalized size = 0.94 \begin {gather*} \frac {5-x}{x \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 30, normalized size = 0.86 \begin {gather*} -\frac {x - 5}{x \log \left (-\frac {5}{e^{\left (x^{2}\right )} - \log \left (-x + e^{3}\right ) - 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp(3)+20*x)*log(5/(log(-x+exp(3))+4-exp
(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+10*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x
^3)*exp(x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.98, size = 569, normalized size = 16.26 \begin {gather*} \frac {2 \, {\left (2 \, x \log \relax (5) - x \log \left (\frac {1}{2} \, \pi ^{2} \mathrm {sgn}\left (x - e^{3}\right ) + \frac {1}{2} \, \pi ^{2} - 2 \, e^{\left (x^{2}\right )} \log \left ({\left | x - e^{3} \right |}\right ) + \log \left ({\left | x - e^{3} \right |}\right )^{2} + e^{\left (2 \, x^{2}\right )} - 8 \, e^{\left (x^{2}\right )} + 8 \, \log \left ({\left | x - e^{3} \right |}\right ) + 16\right ) - 10 \, \log \relax (5) + 5 \, \log \left (\frac {1}{2} \, \pi ^{2} \mathrm {sgn}\left (x - e^{3}\right ) + \frac {1}{2} \, \pi ^{2} - 2 \, e^{\left (x^{2}\right )} \log \left ({\left | x - e^{3} \right |}\right ) + \log \left ({\left | x - e^{3} \right |}\right )^{2} + e^{\left (2 \, x^{2}\right )} - 8 \, e^{\left (x^{2}\right )} + 8 \, \log \left ({\left | x - e^{3} \right |}\right ) + 16\right )\right )}}{4 \, \pi ^{2} x \mathrm {sgn}\left (\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )\right ) \mathrm {sgn}\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right ) + 4 \, \pi x \arctan \left (\frac {\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )}{2 \, {\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right )}}\right ) \mathrm {sgn}\left (\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )\right ) \mathrm {sgn}\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right ) - 4 \, \pi ^{2} x \mathrm {sgn}\left (\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )\right ) - 4 \, \pi x \arctan \left (\frac {\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )}{2 \, {\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right )}}\right ) \mathrm {sgn}\left (\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )\right ) + 2 \, \pi ^{2} x \mathrm {sgn}\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right ) - 6 \, \pi ^{2} x - 8 \, \pi x \arctan \left (\frac {\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )}{2 \, {\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right )}}\right ) - 4 \, x \arctan \left (\frac {\pi + \pi \mathrm {sgn}\left (x - e^{3}\right )}{2 \, {\left (e^{\left (x^{2}\right )} - \log \left ({\left | x - e^{3} \right |}\right ) - 4\right )}}\right )^{2} - 4 \, x \log \relax (5)^{2} + 4 \, x \log \relax (5) \log \left (\frac {1}{2} \, \pi ^{2} \mathrm {sgn}\left (x - e^{3}\right ) + \frac {1}{2} \, \pi ^{2} - 2 \, e^{\left (x^{2}\right )} \log \left ({\left | x - e^{3} \right |}\right ) + \log \left ({\left | x - e^{3} \right |}\right )^{2} + e^{\left (2 \, x^{2}\right )} - 8 \, e^{\left (x^{2}\right )} + 8 \, \log \left ({\left | x - e^{3} \right |}\right ) + 16\right ) - x \log \left (\frac {1}{2} \, \pi ^{2} \mathrm {sgn}\left (x - e^{3}\right ) + \frac {1}{2} \, \pi ^{2} - 2 \, e^{\left (x^{2}\right )} \log \left ({\left | x - e^{3} \right |}\right ) + \log \left ({\left | x - e^{3} \right |}\right )^{2} + e^{\left (2 \, x^{2}\right )} - 8 \, e^{\left (x^{2}\right )} + 8 \, \log \left ({\left | x - e^{3} \right |}\right ) + 16\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp(3)+20*x)*log(5/(log(-x+exp(3))+4-exp
(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+10*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x
^3)*exp(x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x, algorithm="giac")

[Out]

2*(2*x*log(5) - x*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x^2)*log(abs(x - e^3)) + log(abs(x - e^3))^2 + e
^(2*x^2) - 8*e^(x^2) + 8*log(abs(x - e^3)) + 16) - 10*log(5) + 5*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x
^2)*log(abs(x - e^3)) + log(abs(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log(abs(x - e^3)) + 16))/(4*pi^2*x*sgn
(pi + pi*sgn(x - e^3))*sgn(e^(x^2) - log(abs(x - e^3)) - 4) + 4*pi*x*arctan(1/2*(pi + pi*sgn(x - e^3))/(e^(x^2
) - log(abs(x - e^3)) - 4))*sgn(pi + pi*sgn(x - e^3))*sgn(e^(x^2) - log(abs(x - e^3)) - 4) - 4*pi^2*x*sgn(pi +
pi*sgn(x - e^3)) - 4*pi*x*arctan(1/2*(pi + pi*sgn(x - e^3))/(e^(x^2) - log(abs(x - e^3)) - 4))*sgn(pi + pi*sg
n(x - e^3)) + 2*pi^2*x*sgn(e^(x^2) - log(abs(x - e^3)) - 4) - 6*pi^2*x - 8*pi*x*arctan(1/2*(pi + pi*sgn(x - e^
3))/(e^(x^2) - log(abs(x - e^3)) - 4)) - 4*x*arctan(1/2*(pi + pi*sgn(x - e^3))/(e^(x^2) - log(abs(x - e^3)) -
4))^2 - 4*x*log(5)^2 + 4*x*log(5)*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x^2)*log(abs(x - e^3)) + log(abs
(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log(abs(x - e^3)) + 16) - x*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*
e^(x^2)*log(abs(x - e^3)) + log(abs(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log(abs(x - e^3)) + 16)^2)

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maple [C]  time = 0.24, size = 92, normalized size = 2.63

method	result	size
risch	\(\frac {2 i \left (x -5\right )}{x \left (-2 \pi \mathrm {csgn}\left (\frac {i}{-\ln \left (-x +{\mathrm e}^{3}\right )-4+{\mathrm e}^{x^{2}}}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{-\ln \left (-x +{\mathrm e}^{3}\right )-4+{\mathrm e}^{x^{2}}}\right )^{3}+2 \pi -2 i \ln \relax (5)+2 i \ln \left (-\ln \left (-x +{\mathrm e}^{3}\right )-4+{\mathrm e}^{x^{2}}\right )\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*exp(3)+5*x)*ln(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp(3)+20*x)*ln(5/(ln(-x+exp(3))+4-exp(x^2)))+(
(2*x^3-10*x^2)*exp(3)-2*x^4+10*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*ln(-x+exp(3))+(-x^2*exp(3)+x^3)*exp(x^
2)+4*x^2*exp(3)-4*x^3)/ln(5/(ln(-x+exp(3))+4-exp(x^2)))^2,x,method=_RETURNVERBOSE)

[Out]

2*I*(x-5)/x/(-2*Pi*csgn(I/(-ln(-x+exp(3))-4+exp(x^2)))^2+2*Pi*csgn(I/(-ln(-x+exp(3))-4+exp(x^2)))^3+2*Pi-2*I*l
n(5)+2*I*ln(-ln(-x+exp(3))-4+exp(x^2)))

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maxima [A]  time = 0.85, size = 31, normalized size = 0.89 \begin {gather*} -\frac {x - 5}{x \log \relax (5) - x \log \left (-e^{\left (x^{2}\right )} + \log \left (-x + e^{3}\right ) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp(3)+20*x)*log(5/(log(-x+exp(3))+4-exp
(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+10*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x
^3)*exp(x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x, algorithm="maxima")

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [A]  time = 2.53, size = 20, normalized size = 0.57 \begin {gather*} \frac {5 - x}{x \log {\left (\frac {5}{- e^{x^{2}} + \log {\left (- x + e^{3} \right )} + 4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*exp(3)+5*x)*ln(-x+exp(3))+(5*exp(3)-5*x)*exp(x**2)-20*exp(3)+20*x)*ln(5/(ln(-x+exp(3))+4-exp(x
**2)))+((2*x**3-10*x**2)*exp(3)-2*x**4+10*x**3)*exp(x**2)+x**2-5*x)/((x**2*exp(3)-x**3)*ln(-x+exp(3))+(-x**2*e
xp(3)+x**3)*exp(x**2)+4*x**2*exp(3)-4*x**3)/ln(5/(ln(-x+exp(3))+4-exp(x**2)))**2,x)

[Out]

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

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