3.93.36 \(\int \frac {e^{3+2 x+\frac {e^{3+2 x} (3-4 x+x^2)}{x \log (2)}} (-3+6 x-7 x^2+2 x^3)+5 x^2 \log (2)}{x^2 \log (2)} \, dx\)

Optimal. Leaf size=28 \[ e^{\frac {e^{3+2 x} (3+(-4+x) x)}{x \log (2)}}+5 x \]

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Rubi [F]  time = 2.03, 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 {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2]))*(-3 + 6*x - 7*x^2 + 2*x^3) + 5*x^2*Log[2])/(x^2*Lo
g[2]),x]

[Out]

(x*Log[32])/Log[2] - (7*Defer[Int][E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2])), x])/Log[2] - (3*Def
er[Int][E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2]))/x^2, x])/Log[2] + (6*Defer[Int][E^(3 + 2*x + (E
^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2]))/x, x])/Log[2] + (2*Defer[Int][E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2
))/(x*Log[2]))*x, x])/Log[2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2} \, dx}{\log (2)}\\ &=\frac {\int \left (\frac {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) \left (-3+6 x-7 x^2+2 x^3\right )}{x^2}+\log (32)\right ) \, dx}{\log (2)}\\ &=\frac {x \log (32)}{\log (2)}+\frac {\int \frac {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) \left (-3+6 x-7 x^2+2 x^3\right )}{x^2} \, dx}{\log (2)}\\ &=\frac {x \log (32)}{\log (2)}+\frac {\int \left (-7 \exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right )-\frac {3 \exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right )}{x^2}+\frac {6 \exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right )}{x}+2 \exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) x\right ) \, dx}{\log (2)}\\ &=\frac {x \log (32)}{\log (2)}+\frac {2 \int \exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) x \, dx}{\log (2)}-\frac {3 \int \frac {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right )}{x^2} \, dx}{\log (2)}+\frac {6 \int \frac {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right )}{x} \, dx}{\log (2)}-\frac {7 \int \exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) \, dx}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.72, size = 29, normalized size = 1.04 \begin {gather*} e^{\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}}+5 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2]))*(-3 + 6*x - 7*x^2 + 2*x^3) + 5*x^2*Log[2])/(
x^2*Log[2]),x]

[Out]

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

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fricas [B]  time = 0.64, size = 54, normalized size = 1.93 \begin {gather*} {\left (5 \, x e^{\left (2 \, x + 3\right )} + e^{\left (\frac {{\left (x^{2} - 4 \, x + 3\right )} e^{\left (2 \, x + 3\right )} + {\left (2 \, x^{2} + 3 \, x\right )} \log \relax (2)}{x \log \relax (2)}\right )}\right )} e^{\left (-2 \, x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-7*x^2+6*x-3)*exp(2*x+3)*exp((x^2-4*x+3)*exp(2*x+3)/x/log(2))+5*x^2*log(2))/x^2/log(2),x, alg
orithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 \, x^{2} \log \relax (2) + {\left (2 \, x^{3} - 7 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x + \frac {{\left (x^{2} - 4 \, x + 3\right )} e^{\left (2 \, x + 3\right )}}{x \log \relax (2)} + 3\right )}}{x^{2} \log \relax (2)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-7*x^2+6*x-3)*exp(2*x+3)*exp((x^2-4*x+3)*exp(2*x+3)/x/log(2))+5*x^2*log(2))/x^2/log(2),x, alg
orithm="giac")

[Out]

integrate((5*x^2*log(2) + (2*x^3 - 7*x^2 + 6*x - 3)*e^(2*x + (x^2 - 4*x + 3)*e^(2*x + 3)/(x*log(2)) + 3))/(x^2
*log(2)), x)

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maple [A]  time = 0.12, size = 26, normalized size = 0.93

method	result	size
risch	\(5 x +{\mathrm e}^{\frac {\left (x -1\right ) \left (x -3\right ) {\mathrm e}^{2 x +3}}{x \ln \relax (2)}}\)	\(26\)
norman	\(\frac {x \,{\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{2 x +3}}{x \ln \relax (2)}}+5 x^{2}}{x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-7*x^2+6*x-3)*exp(2*x+3)*exp((x^2-4*x+3)*exp(2*x+3)/x/ln(2))+5*x^2*ln(2))/x^2/ln(2),x,method=_RETUR
NVERBOSE)

[Out]

5*x+exp((x-1)*(x-3)*exp(2*x+3)/x/ln(2))

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maxima [B]  time = 0.61, size = 55, normalized size = 1.96 \begin {gather*} \frac {5 \, x \log \relax (2) + e^{\left (\frac {x e^{\left (2 \, x + 3\right )}}{\log \relax (2)} - \frac {4 \, e^{\left (2 \, x + 3\right )}}{\log \relax (2)} + \frac {3 \, e^{\left (2 \, x + 3\right )}}{x \log \relax (2)}\right )} \log \relax (2)}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-7*x^2+6*x-3)*exp(2*x+3)*exp((x^2-4*x+3)*exp(2*x+3)/x/log(2))+5*x^2*log(2))/x^2/log(2),x, alg
orithm="maxima")

[Out]

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

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mupad [B]  time = 7.82, size = 47, normalized size = 1.68 \begin {gather*} 5\,x+{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{x\,\ln \relax (2)}}\,{\mathrm {e}}^{-\frac {4\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{\ln \relax (2)}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{\ln \relax (2)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2*log(2) + exp((exp(2*x + 3)*(x^2 - 4*x + 3))/(x*log(2)))*exp(2*x + 3)*(6*x - 7*x^2 + 2*x^3 - 3))/(x^
2*log(2)),x)

[Out]

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

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sympy [A]  time = 0.35, size = 24, normalized size = 0.86 \begin {gather*} 5 x + e^{\frac {\left (x^{2} - 4 x + 3\right ) e^{2 x + 3}}{x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-7*x**2+6*x-3)*exp(2*x+3)*exp((x**2-4*x+3)*exp(2*x+3)/x/ln(2))+5*x**2*ln(2))/x**2/ln(2),x)

[Out]

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

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