3.89.21 \(\int \frac {e^{\frac {e^x+x^2}{x^3 \log (-2+e^{3/2}) \log (4+2 x)}} (-e^x x-x^3+(-2 x^2-x^3+e^x (-6-x+x^2)) \log (4+2 x))}{(2 x^4+x^5) \log (-2+e^{3/2}) \log ^2(4+2 x)} \, dx\)

Optimal. Leaf size=33 \[ e^{\frac {\frac {e^x}{x}+x}{x^2 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \]

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Rubi [F]  time = 32.70, 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 {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))*(-(E^x*x) - x^3 + (-2*x^2 - x^3 + E^x*(-6 - x + x^2)
)*Log[4 + 2*x]))/((2*x^4 + x^5)*Log[-2 + E^(3/2)]*Log[4 + 2*x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )}\\ &=\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{x^4 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )}\\ &=\frac {\int \left (\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{x^2 (2+x) \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^4 (2+x) \log ^2(4+2 x)}\right ) \, dx}{\log \left (-2+e^{3/2}\right )}\\ &=\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{x^2 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^4 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )}\\ &=\frac {\int \left (\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{4 x^3 \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{16 x \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{2 x^4 \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{8 x^2 \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{16 (2+x) \log ^2(4+2 x)}\right ) \, dx}{\log \left (-2+e^{3/2}\right )}+\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-(2+x) \log (2 (2+x)))}{x^2 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )}\\ &=\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{x \log ^2(4+2 x)} \, dx}{16 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{(2+x) \log ^2(4+2 x)} \, dx}{16 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^2 \log ^2(4+2 x)} \, dx}{8 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{x^3 \log ^2(4+2 x)} \, dx}{4 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^4 \log ^2(4+2 x)} \, dx}{2 \log \left (-2+e^{3/2}\right )}+\frac {\int \left (\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{2 x^2 \log ^2(4+2 x)}+\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{4 (2+x) \log ^2(4+2 x)}+\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (x+2 \log (2 (2+x))+x \log (2 (2+x)))}{4 x \log ^2(4+2 x)}\right ) \, dx}{\log \left (-2+e^{3/2}\right )}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 31, normalized size = 0.94 \begin {gather*} e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (2 (2+x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))*(-(E^x*x) - x^3 + (-2*x^2 - x^3 + E^x*(-6 - x
+ x^2))*Log[4 + 2*x]))/((2*x^4 + x^5)*Log[-2 + E^(3/2)]*Log[4 + 2*x]^2),x]

[Out]

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

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fricas [A]  time = 0.47, size = 26, normalized size = 0.79 \begin {gather*} e^{\left (\frac {x^{2} + e^{x}}{x^{3} \log \left (2 \, x + 4\right ) \log \left (e^{\frac {3}{2}} - 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-x-6)*exp(x)-x^3-2*x^2)*log(2*x+4)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/log(2*x+4)/log(exp(3/2)-
2))/(x^5+2*x^4)/log(2*x+4)^2/log(exp(3/2)-2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.68, size = 42, normalized size = 1.27 \begin {gather*} e^{\left (\frac {1}{x \log \left (2 \, x + 4\right ) \log \left (e^{\frac {3}{2}} - 2\right )} + \frac {e^{x}}{x^{3} \log \left (2 \, x + 4\right ) \log \left (e^{\frac {3}{2}} - 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-x-6)*exp(x)-x^3-2*x^2)*log(2*x+4)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/log(2*x+4)/log(exp(3/2)-
2))/(x^5+2*x^4)/log(2*x+4)^2/log(exp(3/2)-2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.43, size = 27, normalized size = 0.82

method	result	size
risch	\({\mathrm e}^{\frac {x^{2}+{\mathrm e}^{x}}{x^{3} \ln \left (2 x +4\right ) \ln \left ({\mathrm e}^{\frac {3}{2}}-2\right )}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-x-6)*exp(x)-x^3-2*x^2)*ln(2*x+4)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/ln(2*x+4)/ln(exp(3/2)-2))/(x^5+
2*x^4)/ln(2*x+4)^2/ln(exp(3/2)-2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-x-6)*exp(x)-x^3-2*x^2)*log(2*x+4)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/log(2*x+4)/log(exp(3/2)-
2))/(x^5+2*x^4)/log(2*x+4)^2/log(exp(3/2)-2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 5.84, size = 26, normalized size = 0.79 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^x+x^2}{x^3\,\ln \left ({\mathrm {e}}^{3/2}-2\right )\,\ln \left (2\,x+4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(x) + x^2)/(x^3*log(exp(3/2) - 2)*log(2*x + 4)))*(x*exp(x) + log(2*x + 4)*(exp(x)*(x - x^2 + 6)
+ 2*x^2 + x^3) + x^3))/(log(exp(3/2) - 2)*log(2*x + 4)^2*(2*x^4 + x^5)),x)

[Out]

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

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sympy [A]  time = 0.92, size = 26, normalized size = 0.79 \begin {gather*} e^{\frac {x^{2} + e^{x}}{x^{3} \log {\left (-2 + e^{\frac {3}{2}} \right )} \log {\left (2 x + 4 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-x-6)*exp(x)-x**3-2*x**2)*ln(2*x+4)-exp(x)*x-x**3)*exp((x**2+exp(x))/x**3/ln(2*x+4)/ln(exp(3/
2)-2))/(x**5+2*x**4)/ln(2*x+4)**2/ln(exp(3/2)-2),x)

[Out]

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

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