3.100.66 \(\int \frac {(-4 e^x x+e^{x+x^2} (4 x+8 x^2)) \log ^3(e^x-e^{x+x^2})+(4 e^x-4 e^{x+x^2}) \log ^4(e^x-e^{x+x^2})}{-e^x x^5+e^{x+x^2} x^5} \, dx\)

Optimal. Leaf size=20 \[ \frac {\log ^4\left (e^x-e^{x+x^2}\right )}{x^4} \]

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Rubi [F]  time = 1.07, 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 {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 21, normalized size = 1.05 \begin {gather*} -1+\frac {\log ^4\left (-e^x \left (-1+e^{x^2}\right )\right )}{x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4*E^x*x + E^(x + x^2)*(4*x + 8*x^2))*Log[E^x - E^(x + x^2)]^3 + (4*E^x - 4*E^(x + x^2))*Log[E^x -
E^(x + x^2)]^4)/(-(E^x*x^5) + E^(x + x^2)*x^5),x]

[Out]

-1 + Log[-(E^x*(-1 + E^x^2))]^4/x^4

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fricas [A]  time = 0.86, size = 18, normalized size = 0.90 \begin {gather*} \frac {\log \left (-e^{\left (x^{2} + x\right )} + e^{x}\right )^{4}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*
x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3)/(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x, algorithm="fricas")

[Out]

log(-e^(x^2 + x) + e^x)^4/x^4

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giac [A]  time = 2.01, size = 18, normalized size = 0.90 \begin {gather*} \frac {\log \left (-e^{\left (x^{2} + x\right )} + e^{x}\right )^{4}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*
x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3)/(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x, algorithm="giac")

[Out]

log(-e^(x^2 + x) + e^x)^4/x^4

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maple [A]  time = 0.04, size = 19, normalized size = 0.95

method	result	size
risch	\(\frac {\ln \left (-{\mathrm e}^{\left (x +1\right ) x}+{\mathrm e}^{x}\right )^{4}}{x^{4}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*ln(-exp(x^2+x)+exp(1/2*x)^2)^4+((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*x)^2)*l
n(-exp(x^2+x)+exp(1/2*x)^2)^3)/(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x,method=_RETURNVERBOSE)

[Out]

ln(-exp((x+1)*x)+exp(x))^4/x^4

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maxima [B]  time = 0.40, size = 60, normalized size = 3.00 \begin {gather*} \frac {4 \, x^{3} \log \left (-e^{\left (x^{2}\right )} + 1\right ) + 6 \, x^{2} \log \left (-e^{\left (x^{2}\right )} + 1\right )^{2} + 4 \, x \log \left (-e^{\left (x^{2}\right )} + 1\right )^{3} + \log \left (-e^{\left (x^{2}\right )} + 1\right )^{4}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*
x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3)/(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x, algorithm="maxima")

[Out]

(4*x^3*log(-e^(x^2) + 1) + 6*x^2*log(-e^(x^2) + 1)^2 + 4*x*log(-e^(x^2) + 1)^3 + log(-e^(x^2) + 1)^4)/x^4

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mupad [B]  time = 8.46, size = 18, normalized size = 0.90 \begin {gather*} \frac {{\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x\right )}^4}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(x) - exp(x + x^2))^3*(4*x*exp(x) - exp(x + x^2)*(4*x + 8*x^2)) + log(exp(x) - exp(x + x^2))^4*(4*
exp(x + x^2) - 4*exp(x)))/(x^5*exp(x) - x^5*exp(x + x^2)),x)

[Out]

log(exp(x) - exp(x^2)*exp(x))^4/x^4

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sympy [A]  time = 0.49, size = 15, normalized size = 0.75 \begin {gather*} \frac {\log {\left (e^{x} - e^{x^{2} + x} \right )}^{4}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x**2+x)+4*exp(1/2*x)**2)*ln(-exp(x**2+x)+exp(1/2*x)**2)**4+((8*x**2+4*x)*exp(x**2+x)-4*x*ex
p(1/2*x)**2)*ln(-exp(x**2+x)+exp(1/2*x)**2)**3)/(x**5*exp(x**2+x)-x**5*exp(1/2*x)**2),x)

[Out]

log(exp(x) - exp(x**2 + x))**4/x**4

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