3.50.36 \(\int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} (48-16 x-21 x^2-5 x^3+7 x^4-x^5)+(-48-8 x+29 x^2-x^3-5 x^4+x^5) \log (-\frac {10}{-3+x})}{e^{-\frac {4 x}{-4-x+x^2}} (-48-8 x+29 x^2-x^3-5 x^4+x^5)+e^{-\frac {2 x}{-4-x+x^2}} (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5) \log (-\frac {10}{-3+x})+(-48-8 x+29 x^2-x^3-5 x^4+x^5) \log ^2(-\frac {10}{-3+x})} \, dx\)

Optimal. Leaf size=32 \[ \frac {x}{-e^{\frac {2 x}{4+x-x^2}}+\log \left (\frac {10}{3-x}\right )} \]

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Rubi [F]  time = 118.02, 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 {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + (48 - 16*x - 21*x^2 - 5*x^3 + 7*x^4 - x^5)/E^((2*x)/(-4 - x + x^2))
+ (-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)*Log[-10/(-3 + x)])/((-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)/E^((4*
x)/(-4 - x + x^2)) + ((96 + 16*x - 58*x^2 + 2*x^3 + 10*x^4 - 2*x^5)*Log[-10/(-3 + x)])/E^((2*x)/(-4 - x + x^2)
) + (-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)*Log[-10/(-3 + x)]^2),x]

[Out]

Defer[Int][(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^(-2), x] + 3*Defer[Int][1/((-3 + x)*(E^((2*x)/(4 + x
- x^2)) - Log[-10/(-3 + x)])^2), x] - Defer[Int][(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^(-1), x] - (4*D
efer[Int][1/((1 + Sqrt[17] - 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])), x])/Sqrt[17] + (2*(17 + Sqrt
[17])*Defer[Int][1/((-1 - Sqrt[17] + 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])), x])/17 - (4*Defer[In
t][1/((-1 + Sqrt[17] + 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])), x])/Sqrt[17] + (2*(17 - Sqrt[17])*
Defer[Int][1/((-1 + Sqrt[17] + 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])), x])/17 + 8*Defer[Int][1/((
-4 - x + x^2)^2*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])), x] + 18*Defer[Int][x/((-4 - x + x^2)^2*(E^((2*
x)/(4 + x - x^2)) - Log[-10/(-3 + x)])), x] - (4*Defer[Int][Log[-10/(-3 + x)]/((1 + Sqrt[17] - 2*x)*(E^((2*x)/
(4 + x - x^2)) - Log[-10/(-3 + x)])^2), x])/Sqrt[17] + (2*(17 + Sqrt[17])*Defer[Int][Log[-10/(-3 + x)]/((-1 -
Sqrt[17] + 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^2), x])/17 - (4*Defer[Int][Log[-10/(-3 + x)]/((-
1 + Sqrt[17] + 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^2), x])/Sqrt[17] + (2*(17 - Sqrt[17])*Defer[
Int][Log[-10/(-3 + x)]/((-1 + Sqrt[17] + 2*x)*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^2), x])/17 + 8*Def
er[Int][Log[-10/(-3 + x)]/((-4 - x + x^2)^2*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^2), x] + 18*Defer[In
t][(x*Log[-10/(-3 + x)])/((-4 - x + x^2)^2*(E^((2*x)/(4 + x - x^2)) - Log[-10/(-3 + x)])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16 x-8 x^2+7 x^3+2 x^4-x^5-e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )-\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{(3-x) \left (4+x-x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx\\ &=\int \left (-\frac {16-7 x^2-4 x^3+x^4}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}+\frac {x \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{(-3+x) \left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2}\right ) \, dx\\ &=-\int \frac {16-7 x^2-4 x^3+x^4}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx+\int \frac {x \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{(-3+x) \left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx\\ &=-\int \left (\frac {1}{e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )}-\frac {2 (4+9 x)}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}-\frac {2 (1+x)}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}\right ) \, dx+\int \frac {x \left (-\left (4+x-x^2\right )^2-2 \left (-12+4 x-3 x^2+x^3\right ) \log \left (-\frac {10}{-3+x}\right )\right )}{(3-x) \left (4+x-x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx\\ &=2 \int \frac {4+9 x}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx+2 \int \frac {1+x}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx-\int \frac {1}{e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )} \, dx+\int \left (\frac {3 \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{4 (-3+x) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2}-\frac {(4+3 x) \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{2 \left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2}-\frac {3 (2+x) \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{4 \left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {(4+3 x) \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx\right )+\frac {3}{4} \int \frac {16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )}{(-3+x) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx-\frac {3}{4} \int \frac {(2+x) \left (16+8 x-7 x^2-2 x^3+x^4-24 \log \left (-\frac {10}{-3+x}\right )+8 x \log \left (-\frac {10}{-3+x}\right )-6 x^2 \log \left (-\frac {10}{-3+x}\right )+2 x^3 \log \left (-\frac {10}{-3+x}\right )\right )}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx+2 \int \left (\frac {4}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}+\frac {9 x}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}\right ) \, dx+2 \int \left (\frac {1}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}+\frac {x}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )}\right ) \, dx-\int \frac {1}{e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )} \, dx\\ &=-\left (\frac {1}{2} \int \frac {(4+3 x) \left (\left (4+x-x^2\right )^2+2 \left (-12+4 x-3 x^2+x^3\right ) \log \left (-\frac {10}{-3+x}\right )\right )}{\left (4+x-x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx\right )+\frac {3}{4} \int \frac {-\left (4+x-x^2\right )^2-2 \left (-12+4 x-3 x^2+x^3\right ) \log \left (-\frac {10}{-3+x}\right )}{(3-x) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx-\frac {3}{4} \int \frac {(2+x) \left (-\left (4+x-x^2\right )^2-2 \left (-12+4 x-3 x^2+x^3\right ) \log \left (-\frac {10}{-3+x}\right )\right )}{\left (4+x-x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )^2} \, dx+2 \int \frac {1}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx+2 \int \frac {x}{\left (-4-x+x^2\right ) \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx+8 \int \frac {1}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx+18 \int \frac {x}{\left (-4-x+x^2\right )^2 \left (e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )\right )} \, dx-\int \frac {1}{e^{\frac {2 x}{4+x-x^2}}-\log \left (-\frac {10}{-3+x}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 45, normalized size = 1.41 \begin {gather*} \frac {e^{\frac {2 x}{-4-x+x^2}} x}{-1+e^{\frac {2 x}{-4-x+x^2}} \log \left (-\frac {10}{-3+x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + (48 - 16*x - 21*x^2 - 5*x^3 + 7*x^4 - x^5)/E^((2*x)/(-4 - x +
x^2)) + (-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)*Log[-10/(-3 + x)])/((-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)/
E^((4*x)/(-4 - x + x^2)) + ((96 + 16*x - 58*x^2 + 2*x^3 + 10*x^4 - 2*x^5)*Log[-10/(-3 + x)])/E^((2*x)/(-4 - x
+ x^2)) + (-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)*Log[-10/(-3 + x)]^2),x]

[Out]

(E^((2*x)/(-4 - x + x^2))*x)/(-1 + E^((2*x)/(-4 - x + x^2))*Log[-10/(-3 + x)])

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fricas [A]  time = 0.53, size = 30, normalized size = 0.94 \begin {gather*} -\frac {x}{e^{\left (-\frac {2 \, x}{x^{2} - x - 4}\right )} - \log \left (-\frac {10}{x - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(x-3))+(-x^5+7*x^4-5*x^3-21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+
x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(x-3))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+9
6)*exp(-2*x/(x^2-x-4))*log(-10/(x-3))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(-2*x/(x^2-x-4))^2),x, algorithm="frica
s")

[Out]

-x/(e^(-2*x/(x^2 - x - 4)) - log(-10/(x - 3)))

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giac [A]  time = 0.35, size = 43, normalized size = 1.34 \begin {gather*} \frac {x e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )}}{e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )} \log \left (-\frac {10}{x - 3}\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(x-3))+(-x^5+7*x^4-5*x^3-21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+
x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(x-3))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+9
6)*exp(-2*x/(x^2-x-4))*log(-10/(x-3))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(-2*x/(x^2-x-4))^2),x, algorithm="giac"
)

[Out]

x*e^(2*x/(x^2 - x - 4))/(e^(2*x/(x^2 - x - 4))*log(-10/(x - 3)) - 1)

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maple [C]  time = 0.22, size = 73, normalized size = 2.28

method	result	size
risch	\(-\frac {2 i x}{-2 \pi \mathrm {csgn}\left (\frac {i}{x -3}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{x -3}\right )^{3}+2 \pi -2 i \ln \relax (2)-2 i \ln \relax (5)+2 i {\mathrm e}^{-\frac {2 x}{x^{2}-x -4}}+2 i \ln \left (x -3\right )}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5-5*x^4-x^3+29*x^2-8*x-48)*ln(-10/(x-3))+(-x^5+7*x^4-5*x^3-21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+x^5-2*x
^4-7*x^3+8*x^2+16*x)/((x^5-5*x^4-x^3+29*x^2-8*x-48)*ln(-10/(x-3))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+96)*exp(-
2*x/(x^2-x-4))*ln(-10/(x-3))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(-2*x/(x^2-x-4))^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.83, size = 48, normalized size = 1.50 \begin {gather*} \frac {x e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )}}{{\left (\log \relax (5) + \log \relax (2) - \log \left (-x + 3\right )\right )} e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(x-3))+(-x^5+7*x^4-5*x^3-21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+
x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(x-3))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+9
6)*exp(-2*x/(x^2-x-4))*log(-10/(x-3))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(-2*x/(x^2-x-4))^2),x, algorithm="maxim
a")

[Out]

x*e^(2*x/(x^2 - x - 4))/((log(5) + log(2) - log(-x + 3))*e^(2*x/(x^2 - x - 4)) - 1)

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mupad [B]  time = 1.15, size = 293, normalized size = 9.16 \begin {gather*} -\frac {\left (x^5-2\,x^4-7\,x^3+8\,x^2+16\,x\right )\,{\left (-x^5+5\,x^4+x^3-29\,x^2+8\,x+48\right )}^2-\ln \left (-\frac {10}{x-3}\right )\,\left (-2\,x^4+6\,x^3-8\,x^2+24\,x\right )\,{\left (-x^5+5\,x^4+x^3-29\,x^2+8\,x+48\right )}^2}{\left (\ln \left (-\frac {10}{x-3}\right )-{\mathrm {e}}^{\frac {2\,x}{-x^2+x+4}}\right )\,\left (x-3\right )\,{\left (-x^2+x+4\right )}^2\,\left (512\,x-1152\,\ln \left (-\frac {10}{x-3}\right )+472\,x^2\,\ln \left (-\frac {10}{x-3}\right )-208\,x^3\,\ln \left (-\frac {10}{x-3}\right )+78\,x^4\,\ln \left (-\frac {10}{x-3}\right )-36\,x^6\,\ln \left (-\frac {10}{x-3}\right )+16\,x^7\,\ln \left (-\frac {10}{x-3}\right )-2\,x^8\,\ln \left (-\frac {10}{x-3}\right )-736\,x^2-368\,x^3+323\,x^4+83\,x^5-74\,x^6-2\,x^7+7\,x^8-x^9+192\,x\,\ln \left (-\frac {10}{x-3}\right )+768\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-10/(x - 3))*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48) - 16*x + exp((2*x)/(x - x^2 + 4))*(16*x + 21*x^2
+ 5*x^3 - 7*x^4 + x^5 - 48) - 8*x^2 + 7*x^3 + 2*x^4 - x^5)/(log(-10/(x - 3))^2*(8*x - 29*x^2 + x^3 + 5*x^4 -
x^5 + 48) + exp((4*x)/(x - x^2 + 4))*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48) - log(-10/(x - 3))*exp((2*x)/(x -
x^2 + 4))*(16*x - 58*x^2 + 2*x^3 + 10*x^4 - 2*x^5 + 96)),x)

[Out]

-((16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5)*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48)^2 - log(-10/(x - 3))*(24*x - 8*
x^2 + 6*x^3 - 2*x^4)*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48)^2)/((log(-10/(x - 3)) - exp((2*x)/(x - x^2 + 4)))
*(x - 3)*(x - x^2 + 4)^2*(512*x - 1152*log(-10/(x - 3)) + 472*x^2*log(-10/(x - 3)) - 208*x^3*log(-10/(x - 3))
+ 78*x^4*log(-10/(x - 3)) - 36*x^6*log(-10/(x - 3)) + 16*x^7*log(-10/(x - 3)) - 2*x^8*log(-10/(x - 3)) - 736*x
^2 - 368*x^3 + 323*x^4 + 83*x^5 - 74*x^6 - 2*x^7 + 7*x^8 - x^9 + 192*x*log(-10/(x - 3)) + 768))

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sympy [A]  time = 0.93, size = 24, normalized size = 0.75 \begin {gather*} - \frac {x}{- \log {\left (- \frac {10}{x - 3} \right )} + e^{- \frac {2 x}{x^{2} - x - 4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**5-5*x**4-x**3+29*x**2-8*x-48)*ln(-10/(x-3))+(-x**5+7*x**4-5*x**3-21*x**2-16*x+48)*exp(-2*x/(x**
2-x-4))+x**5-2*x**4-7*x**3+8*x**2+16*x)/((x**5-5*x**4-x**3+29*x**2-8*x-48)*ln(-10/(x-3))**2+(-2*x**5+10*x**4+2
*x**3-58*x**2+16*x+96)*exp(-2*x/(x**2-x-4))*ln(-10/(x-3))+(x**5-5*x**4-x**3+29*x**2-8*x-48)*exp(-2*x/(x**2-x-4
))**2),x)

[Out]

-x/(-log(-10/(x - 3)) + exp(-2*x/(x**2 - x - 4)))

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