3.32.2 \(\int \frac {e^{4+4 x} (4 x^5-16 x^6+e^x (4 x^4-16 x^5))+e^{4+4 x} (8 x^2+8 e^x x^2) \log (e^x+x)+e^{4+4 x} (-8 x^2-16 x^3+e^x (-8 x-16 x^2)) \log ^2(e^x+x)}{25 e^x x^6+25 x^7+e^{8+8 x} (e^x x^4+x^5)+e^{4+4 x} (-10 e^x x^5-10 x^6)+(50 e^x x^3+50 x^4+e^{4+4 x} (-10 e^x x^2-10 x^3)) \log ^2(e^x+x)+(25 e^x+25 x) \log ^4(e^x+x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {4}{-5+\frac {e^{4 (1+x)}}{x+\frac {\log ^2\left (e^x+x\right )}{x^2}}} \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(4 + 4*x)*(4*x^5 - 16*x^6 + E^x*(4*x^4 - 16*x^5)) + E^(4 + 4*x)*(8*x^2 + 8*E^x*x^2)*Log[E^x + x] + E^(4
+ 4*x)*(-8*x^2 - 16*x^3 + E^x*(-8*x - 16*x^2))*Log[E^x + x]^2)/(25*E^x*x^6 + 25*x^7 + E^(8 + 8*x)*(E^x*x^4 +
x^5) + E^(4 + 4*x)*(-10*E^x*x^5 - 10*x^6) + (50*E^x*x^3 + 50*x^4 + E^(4 + 4*x)*(-10*E^x*x^2 - 10*x^3))*Log[E^x
+ x]^2 + (25*E^x + 25*x)*Log[E^x + x]^4),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.17, size = 44, normalized size = 1.47 \begin {gather*} -\frac {4 e^{4+4 x} x^2}{5 \left (-e^{4+4 x} x^2+5 x^3+5 \log ^2\left (e^x+x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 + 4*x)*(4*x^5 - 16*x^6 + E^x*(4*x^4 - 16*x^5)) + E^(4 + 4*x)*(8*x^2 + 8*E^x*x^2)*Log[E^x + x]
+ E^(4 + 4*x)*(-8*x^2 - 16*x^3 + E^x*(-8*x - 16*x^2))*Log[E^x + x]^2)/(25*E^x*x^6 + 25*x^7 + E^(8 + 8*x)*(E^x*
x^4 + x^5) + E^(4 + 4*x)*(-10*E^x*x^5 - 10*x^6) + (50*E^x*x^3 + 50*x^4 + E^(4 + 4*x)*(-10*E^x*x^2 - 10*x^3))*L
og[E^x + x]^2 + (25*E^x + 25*x)*Log[E^x + x]^4),x]

[Out]

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

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fricas [A]  time = 0.56, size = 39, normalized size = 1.30 \begin {gather*} -\frac {4 \, x^{2} e^{\left (4 \, x + 4\right )}}{5 \, {\left (5 \, x^{3} - x^{2} e^{\left (4 \, x + 4\right )} + 5 \, \log \left (x + e^{x}\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^2-8*x)*exp(x)-16*x^3-8*x^2)*exp(4*x+4)*log(exp(x)+x)^2+(8*exp(x)*x^2+8*x^2)*exp(4*x+4)*log(
exp(x)+x)+((-16*x^5+4*x^4)*exp(x)-16*x^6+4*x^5)*exp(4*x+4))/((25*exp(x)+25*x)*log(exp(x)+x)^4+((-10*exp(x)*x^2
-10*x^3)*exp(4*x+4)+50*exp(x)*x^3+50*x^4)*log(exp(x)+x)^2+(exp(x)*x^4+x^5)*exp(4*x+4)^2+(-10*x^5*exp(x)-10*x^6
)*exp(4*x+4)+25*x^6*exp(x)+25*x^7),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^2-8*x)*exp(x)-16*x^3-8*x^2)*exp(4*x+4)*log(exp(x)+x)^2+(8*exp(x)*x^2+8*x^2)*exp(4*x+4)*log(
exp(x)+x)+((-16*x^5+4*x^4)*exp(x)-16*x^6+4*x^5)*exp(4*x+4))/((25*exp(x)+25*x)*log(exp(x)+x)^4+((-10*exp(x)*x^2
-10*x^3)*exp(4*x+4)+50*exp(x)*x^3+50*x^4)*log(exp(x)+x)^2+(exp(x)*x^4+x^5)*exp(4*x+4)^2+(-10*x^5*exp(x)-10*x^6
)*exp(4*x+4)+25*x^6*exp(x)+25*x^7),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.06, size = 39, normalized size = 1.30

method	result	size
risch	\(\frac {4 x^{2} {\mathrm e}^{4 x +4}}{5 \left (x^{2} {\mathrm e}^{4 x +4}-5 x^{3}-5 \ln \left ({\mathrm e}^{x}+x \right )^{2}\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x^2-8*x)*exp(x)-16*x^3-8*x^2)*exp(4*x+4)*ln(exp(x)+x)^2+(8*exp(x)*x^2+8*x^2)*exp(4*x+4)*ln(exp(x)+x
)+((-16*x^5+4*x^4)*exp(x)-16*x^6+4*x^5)*exp(4*x+4))/((25*exp(x)+25*x)*ln(exp(x)+x)^4+((-10*exp(x)*x^2-10*x^3)*
exp(4*x+4)+50*exp(x)*x^3+50*x^4)*ln(exp(x)+x)^2+(exp(x)*x^4+x^5)*exp(4*x+4)^2+(-10*x^5*exp(x)-10*x^6)*exp(4*x+
4)+25*x^6*exp(x)+25*x^7),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.78, size = 39, normalized size = 1.30 \begin {gather*} -\frac {4 \, x^{2} e^{\left (4 \, x + 4\right )}}{5 \, {\left (5 \, x^{3} - x^{2} e^{\left (4 \, x + 4\right )} + 5 \, \log \left (x + e^{x}\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x^2-8*x)*exp(x)-16*x^3-8*x^2)*exp(4*x+4)*log(exp(x)+x)^2+(8*exp(x)*x^2+8*x^2)*exp(4*x+4)*log(
exp(x)+x)+((-16*x^5+4*x^4)*exp(x)-16*x^6+4*x^5)*exp(4*x+4))/((25*exp(x)+25*x)*log(exp(x)+x)^4+((-10*exp(x)*x^2
-10*x^3)*exp(4*x+4)+50*exp(x)*x^3+50*x^4)*log(exp(x)+x)^2+(exp(x)*x^4+x^5)*exp(4*x+4)^2+(-10*x^5*exp(x)-10*x^6
)*exp(4*x+4)+25*x^6*exp(x)+25*x^7),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {-{\mathrm {e}}^{4\,x+4}\,\left ({\mathrm {e}}^x\,\left (16\,x^2+8\,x\right )+8\,x^2+16\,x^3\right )\,{\ln \left (x+{\mathrm {e}}^x\right )}^2+{\mathrm {e}}^{4\,x+4}\,\left (8\,x^2\,{\mathrm {e}}^x+8\,x^2\right )\,\ln \left (x+{\mathrm {e}}^x\right )+{\mathrm {e}}^{4\,x+4}\,\left ({\mathrm {e}}^x\,\left (4\,x^4-16\,x^5\right )+4\,x^5-16\,x^6\right )}{25\,x^6\,{\mathrm {e}}^x+{\ln \left (x+{\mathrm {e}}^x\right )}^2\,\left (50\,x^3\,{\mathrm {e}}^x-{\mathrm {e}}^{4\,x+4}\,\left (10\,x^2\,{\mathrm {e}}^x+10\,x^3\right )+50\,x^4\right )+{\mathrm {e}}^{8\,x+8}\,\left (x^4\,{\mathrm {e}}^x+x^5\right )+{\ln \left (x+{\mathrm {e}}^x\right )}^4\,\left (25\,x+25\,{\mathrm {e}}^x\right )-{\mathrm {e}}^{4\,x+4}\,\left (10\,x^5\,{\mathrm {e}}^x+10\,x^6\right )+25\,x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x + 4)*(exp(x)*(4*x^4 - 16*x^5) + 4*x^5 - 16*x^6) + log(x + exp(x))*exp(4*x + 4)*(8*x^2*exp(x) + 8*
x^2) - log(x + exp(x))^2*exp(4*x + 4)*(exp(x)*(8*x + 16*x^2) + 8*x^2 + 16*x^3))/(25*x^6*exp(x) + log(x + exp(x
))^2*(50*x^3*exp(x) - exp(4*x + 4)*(10*x^2*exp(x) + 10*x^3) + 50*x^4) + exp(8*x + 8)*(x^4*exp(x) + x^5) + log(
x + exp(x))^4*(25*x + 25*exp(x)) - exp(4*x + 4)*(10*x^5*exp(x) + 10*x^6) + 25*x^7),x)

[Out]

int((exp(4*x + 4)*(exp(x)*(4*x^4 - 16*x^5) + 4*x^5 - 16*x^6) + log(x + exp(x))*exp(4*x + 4)*(8*x^2*exp(x) + 8*
x^2) - log(x + exp(x))^2*exp(4*x + 4)*(exp(x)*(8*x + 16*x^2) + 8*x^2 + 16*x^3))/(25*x^6*exp(x) + log(x + exp(x
))^2*(50*x^3*exp(x) - exp(4*x + 4)*(10*x^2*exp(x) + 10*x^3) + 50*x^4) + exp(8*x + 8)*(x^4*exp(x) + x^5) + log(
x + exp(x))^4*(25*x + 25*exp(x)) - exp(4*x + 4)*(10*x^5*exp(x) + 10*x^6) + 25*x^7), x)

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sympy [A]  time = 0.45, size = 42, normalized size = 1.40 \begin {gather*} - \frac {4 x^{2} e^{4} e^{4 x}}{25 x^{3} - 5 x^{2} e^{4} e^{4 x} + 25 \log {\left (x + e^{x} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x**2-8*x)*exp(x)-16*x**3-8*x**2)*exp(4*x+4)*ln(exp(x)+x)**2+(8*exp(x)*x**2+8*x**2)*exp(4*x+4)
*ln(exp(x)+x)+((-16*x**5+4*x**4)*exp(x)-16*x**6+4*x**5)*exp(4*x+4))/((25*exp(x)+25*x)*ln(exp(x)+x)**4+((-10*ex
p(x)*x**2-10*x**3)*exp(4*x+4)+50*exp(x)*x**3+50*x**4)*ln(exp(x)+x)**2+(exp(x)*x**4+x**5)*exp(4*x+4)**2+(-10*x*
*5*exp(x)-10*x**6)*exp(4*x+4)+25*x**6*exp(x)+25*x**7),x)

[Out]

-4*x**2*exp(4)*exp(4*x)/(25*x**3 - 5*x**2*exp(4)*exp(4*x) + 25*log(x + exp(x))**2)

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