3.23.25 \(\int \frac {-19-4 e^3+e^x (48+24 x+3 x^2)}{204+83 x+8 x^2+e^3 (-4 x-x^2)+e^x (48+24 x+3 x^2)} \, dx\)

Optimal. Leaf size=30 \[ -4+\log \left (-3 \left (4+e^x\right )+\frac {\left (3+e^3+\frac {-3+x}{x}\right ) x}{4+x}\right ) \]

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Rubi [F]  time = 1.27, 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 {-19-4 e^3+e^x \left (48+24 x+3 x^2\right )}{204+83 x+8 x^2+e^3 \left (-4 x-x^2\right )+e^x \left (48+24 x+3 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-19 - 4*E^3 + E^x*(48 + 24*x + 3*x^2))/(204 + 83*x + 8*x^2 + E^3*(-4*x - x^2) + E^x*(48 + 24*x + 3*x^2)),
x]

[Out]

x + 51*Defer[Int][(-51 - 12*E^x - 3*E^x*x - 8*(1 - E^3/8)*x)^(-1), x] - (8 - E^3)*Defer[Int][x/(51 + 12*E^x +
3*E^x*x + 8*(1 - E^3/8)*x), x] - (19 + 4*E^3)*Defer[Int][1/((4 + x)*(51 + 12*E^x + 3*E^x*x + 8*(1 - E^3/8)*x))
, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-19 \left (1+\frac {4 e^3}{19}\right )+e^x \left (48+24 x+3 x^2\right )}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )} \, dx\\ &=\int \left (1+\frac {-223-4 e^3-\left (83-4 e^3\right ) x-\left (8-e^3\right ) x^2}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )}\right ) \, dx\\ &=x+\int \frac {-223-4 e^3-\left (83-4 e^3\right ) x-\left (8-e^3\right ) x^2}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )} \, dx\\ &=x+\int \left (\frac {51}{-51-12 e^x-3 e^x x-8 \left (1-\frac {e^3}{8}\right ) x}+\frac {\left (-8+e^3\right ) x}{51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x}+\frac {-19-4 e^3}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )}\right ) \, dx\\ &=x+51 \int \frac {1}{-51-12 e^x-3 e^x x-8 \left (1-\frac {e^3}{8}\right ) x} \, dx+\left (-19-4 e^3\right ) \int \frac {1}{(4+x) \left (51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x\right )} \, dx+\left (-8+e^3\right ) \int \frac {x}{51+12 e^x+3 e^x x+8 \left (1-\frac {e^3}{8}\right ) x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 30, normalized size = 1.00 \begin {gather*} -\log (4+x)+\log \left (51+12 e^x+8 x-e^3 x+3 e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-19 - 4*E^3 + E^x*(48 + 24*x + 3*x^2))/(204 + 83*x + 8*x^2 + E^3*(-4*x - x^2) + E^x*(48 + 24*x + 3*
x^2)),x]

[Out]

-Log[4 + x] + Log[51 + 12*E^x + 8*x - E^3*x + 3*E^x*x]

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fricas [A]  time = 1.03, size = 24, normalized size = 0.80 \begin {gather*} \log \left (-\frac {x e^{3} - 3 \, {\left (x + 4\right )} e^{x} - 8 \, x - 51}{x + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+24*x+48)*exp(x)-4*exp(3)-19)/((3*x^2+24*x+48)*exp(x)+(-x^2-4*x)*exp(3)+8*x^2+83*x+204),x, al
gorithm="fricas")

[Out]

log(-(x*e^3 - 3*(x + 4)*e^x - 8*x - 51)/(x + 4))

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giac [A]  time = 0.22, size = 26, normalized size = 0.87 \begin {gather*} \log \left (x e^{3} - 3 \, x e^{x} - 8 \, x - 12 \, e^{x} - 51\right ) - \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+24*x+48)*exp(x)-4*exp(3)-19)/((3*x^2+24*x+48)*exp(x)+(-x^2-4*x)*exp(3)+8*x^2+83*x+204),x, al
gorithm="giac")

[Out]

log(x*e^3 - 3*x*e^x - 8*x - 12*e^x - 51) - log(x + 4)

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maple [A]  time = 0.10, size = 21, normalized size = 0.70

method	result	size
risch	\(\ln \left ({\mathrm e}^{x}-\frac {x \,{\mathrm e}^{3}-8 x -51}{3 \left (4+x \right )}\right )\)	\(21\)
norman	\(-\ln \left (4+x \right )+\ln \left (x \,{\mathrm e}^{3}-3 \,{\mathrm e}^{x} x -8 x -12 \,{\mathrm e}^{x}-51\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2+24*x+48)*exp(x)-4*exp(3)-19)/((3*x^2+24*x+48)*exp(x)+(-x^2-4*x)*exp(3)+8*x^2+83*x+204),x,method=_R
ETURNVERBOSE)

[Out]

ln(exp(x)-1/3*(x*exp(3)-8*x-51)/(4+x))

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maxima [A]  time = 0.61, size = 23, normalized size = 0.77 \begin {gather*} \log \left (-\frac {x {\left (e^{3} - 8\right )} - 3 \, {\left (x + 4\right )} e^{x} - 51}{3 \, {\left (x + 4\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+24*x+48)*exp(x)-4*exp(3)-19)/((3*x^2+24*x+48)*exp(x)+(-x^2-4*x)*exp(3)+8*x^2+83*x+204),x, al
gorithm="maxima")

[Out]

log(-1/3*(x*(e^3 - 8) - 3*(x + 4)*e^x - 51)/(x + 4))

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mupad [B]  time = 1.44, size = 27, normalized size = 0.90 \begin {gather*} \ln \left (8\,x+12\,{\mathrm {e}}^x-x\,{\mathrm {e}}^3+3\,x\,{\mathrm {e}}^x+51\right )-\ln \left (x+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(3) - exp(x)*(24*x + 3*x^2 + 48) + 19)/(83*x + exp(x)*(24*x + 3*x^2 + 48) - exp(3)*(4*x + x^2) + 8*
x^2 + 204),x)

[Out]

log(8*x + 12*exp(x) - x*exp(3) + 3*x*exp(x) + 51) - log(x + 4)

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sympy [A]  time = 0.37, size = 19, normalized size = 0.63 \begin {gather*} \log {\left (e^{x} + \frac {- x e^{3} + 8 x + 51}{3 x + 12} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2+24*x+48)*exp(x)-4*exp(3)-19)/((3*x**2+24*x+48)*exp(x)+(-x**2-4*x)*exp(3)+8*x**2+83*x+204),x
)

[Out]

log(exp(x) + (-x*exp(3) + 8*x + 51)/(3*x + 12))

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