3.96.42 \(\int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} (4 x^3-3 x^4)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (-4+e^5-e^x-\frac {4}{e^{10} x^2}+3 x\right ) \]

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Rubi [F]  time = 1.17, 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 {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 - 3*E^10*x^3 + E^(10 + x)*x^3)/(4*x - E^15*x^3 + E^(10 + x)*x^3 + E^10*(4*x^3 - 3*x^4)),x]

[Out]

x - E^10*(7 - E^5)*Defer[Int][x^2/(4 + E^(10 + x)*x^2 + 4*E^10*(1 - E^5/4)*x^2 - 3*E^10*x^3), x] + 3*E^10*Defe
r[Int][x^3/(4 + E^(10 + x)*x^2 + 4*E^10*(1 - E^5/4)*x^2 - 3*E^10*x^3), x] + 4*Defer[Int][(-4 - E^(10 + x)*x^2
- 4*E^10*(1 - E^5/4)*x^2 + 3*E^10*x^3)^(-1), x] + 8*Defer[Int][1/(x*(-4 - E^(10 + x)*x^2 - 4*E^10*(1 - E^5/4)*
x^2 + 3*E^10*x^3)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {-8-4 x-e^{10} \left (7-e^5\right ) x^3+3 e^{10} x^4}{x \left (4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3\right )}\right ) \, dx\\ &=x+\int \frac {-8-4 x-e^{10} \left (7-e^5\right ) x^3+3 e^{10} x^4}{x \left (4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3\right )} \, dx\\ &=x+\int \left (\frac {e^{10} \left (-7+e^5\right ) x^2}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3}+\frac {3 e^{10} x^3}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3}+\frac {4}{-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3}+\frac {8}{x \left (-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3\right )}\right ) \, dx\\ &=x+4 \int \frac {1}{-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3} \, dx+8 \int \frac {1}{x \left (-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3\right )} \, dx+\left (3 e^{10}\right ) \int \frac {x^3}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3} \, dx-\left (e^{10} \left (7-e^5\right )\right ) \int \frac {x^2}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 37, normalized size = 1.68 \begin {gather*} -2 \log (x)+\log \left (4-e^{15} x^2+e^{10+x} x^2+e^{10} (4-3 x) x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 3*E^10*x^3 + E^(10 + x)*x^3)/(4*x - E^15*x^3 + E^(10 + x)*x^3 + E^10*(4*x^3 - 3*x^4)),x]

[Out]

-2*Log[x] + Log[4 - E^15*x^2 + E^(10 + x)*x^2 + E^10*(4 - 3*x)*x^2]

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fricas [A]  time = 0.67, size = 37, normalized size = 1.68 \begin {gather*} \log \left (-\frac {x^{2} e^{15} - x^{2} e^{\left (x + 10\right )} + {\left (3 \, x^{3} - 4 \, x^{2}\right )} e^{10} - 4}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*exp(5)^2*exp(x)-3*x^3*exp(5)^2-8)/(x^3*exp(5)^2*exp(x)-x^3*exp(5)^3+(-3*x^4+4*x^3)*exp(5)^2+4*x
),x, algorithm="fricas")

[Out]

log(-(x^2*e^15 - x^2*e^(x + 10) + (3*x^3 - 4*x^2)*e^10 - 4)/x^2)

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giac [A]  time = 0.19, size = 37, normalized size = 1.68 \begin {gather*} \log \left (-3 \, x^{3} e^{10} - x^{2} e^{15} + 4 \, x^{2} e^{10} + x^{2} e^{\left (x + 10\right )} + 4\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*exp(5)^2*exp(x)-3*x^3*exp(5)^2-8)/(x^3*exp(5)^2*exp(x)-x^3*exp(5)^3+(-3*x^4+4*x^3)*exp(5)^2+4*x
),x, algorithm="giac")

[Out]

log(-3*x^3*e^10 - x^2*e^15 + 4*x^2*e^10 + x^2*e^(x + 10) + 4) - 2*log(x)

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maple [A]  time = 0.18, size = 34, normalized size = 1.55

method	result	size
risch	\(\ln \left ({\mathrm e}^{x}-\frac {\left (x^{2} {\mathrm e}^{15}+3 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-4\right ) {\mathrm e}^{-10}}{x^{2}}\right )\)	\(34\)
norman	\(-2 \ln \relax (x )+\ln \left (x^{2} {\mathrm e}^{15}-x^{2} {\mathrm e}^{10} {\mathrm e}^{x}+3 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-4\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*exp(5)^2*exp(x)-3*x^3*exp(5)^2-8)/(x^3*exp(5)^2*exp(x)-x^3*exp(5)^3+(-3*x^4+4*x^3)*exp(5)^2+4*x),x,me
thod=_RETURNVERBOSE)

[Out]

ln(exp(x)-(x^2*exp(15)+3*x^3*exp(10)-4*x^2*exp(10)-4)/x^2*exp(-10))

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maxima [A]  time = 0.39, size = 37, normalized size = 1.68 \begin {gather*} \log \left (-\frac {{\left (3 \, x^{3} e^{10} + x^{2} {\left (e^{15} - 4 \, e^{10}\right )} - x^{2} e^{\left (x + 10\right )} - 4\right )} e^{\left (-10\right )}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*exp(5)^2*exp(x)-3*x^3*exp(5)^2-8)/(x^3*exp(5)^2*exp(x)-x^3*exp(5)^3+(-3*x^4+4*x^3)*exp(5)^2+4*x
),x, algorithm="maxima")

[Out]

log(-(3*x^3*e^10 + x^2*(e^15 - 4*e^10) - x^2*e^(x + 10) - 4)*e^(-10)/x^2)

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mupad [B]  time = 8.19, size = 33, normalized size = 1.50 \begin {gather*} \ln \left (\frac {x^2\,{\mathrm {e}}^5}{3}-\frac {x^2\,{\mathrm {e}}^x}{3}-\frac {4\,{\mathrm {e}}^{-10}}{3}-\frac {4\,x^2}{3}+x^3\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^3*exp(10) - x^3*exp(10)*exp(x) + 8)/(4*x + exp(10)*(4*x^3 - 3*x^4) - x^3*exp(15) + x^3*exp(10)*exp(x
)),x)

[Out]

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

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sympy [A]  time = 0.24, size = 36, normalized size = 1.64 \begin {gather*} \log {\left (e^{x} + \frac {- 3 x^{3} e^{10} - x^{2} e^{15} + 4 x^{2} e^{10} + 4}{x^{2} e^{10}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3*exp(5)**2*exp(x)-3*x**3*exp(5)**2-8)/(x**3*exp(5)**2*exp(x)-x**3*exp(5)**3+(-3*x**4+4*x**3)*ex
p(5)**2+4*x),x)

[Out]

log(exp(x) + (-3*x**3*exp(10) - x**2*exp(15) + 4*x**2*exp(10) + 4)*exp(-10)/x**2)

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