\(\int \frac {e^{3+2 x+\frac {e^{3+2 x} (3-4 x+x^2)}{x \log (2)}} (-3+6 x-7 x^2+2 x^3)+5 x^2 \log (2)}{x^2 \log (2)} \, dx\) [9236]

Optimal result

Integrand size = 62, antiderivative size = 28 \[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=e^{\frac {e^{3+2 x} (3+(-4+x) x)}{x \log (2)}}+5 x \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=\int \frac {\exp \left (3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}\right ) \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=e^{\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}}+5 x \]

[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

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
risch	\(5 x +{\mathrm e}^{\frac {\left (-1+x \right ) \left (-3+x \right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}\)	\(26\)
parts	\(5 x +{\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}\)	\(28\)
norman	\(\frac {x \,{\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}+5 x^{2}}{x}\)	\(36\)
parallelrisch	\(\frac {5 x \ln \left (2\right )+\ln \left (2\right ) {\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}}{\ln \left (2\right )}\)	\(38\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93 \[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx={\left (5 \, x e^{\left (2 \, x + 3\right )} + e^{\left (\frac {{\left (x^{2} - 4 \, x + 3\right )} e^{\left (2 \, x + 3\right )} + {\left (2 \, x^{2} + 3 \, x\right )} \log \left (2\right )}{x \log \left (2\right )}\right )}\right )} e^{\left (-2 \, x - 3\right )} \]

[In]

integrate(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/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)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=5 x + e^{\frac {\left (x^{2} - 4 x + 3\right ) e^{2 x + 3}}{x \log {\left (2 \right )}}} \]

[In]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).

Time = 0.44 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=\frac {5 \, x \log \left (2\right ) + e^{\left (\frac {x e^{\left (2 \, x + 3\right )}}{\log \left (2\right )} - \frac {4 \, e^{\left (2 \, x + 3\right )}}{\log \left (2\right )} + \frac {3 \, e^{\left (2 \, x + 3\right )}}{x \log \left (2\right )}\right )} \log \left (2\right )}{\log \left (2\right )} \]

[In]

integrate(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/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)

Giac [F]

\[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=\int { \frac {5 \, x^{2} \log \left (2\right ) + {\left (2 \, x^{3} - 7 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x + \frac {{\left (x^{2} - 4 \, x + 3\right )} e^{\left (2 \, x + 3\right )}}{x \log \left (2\right )} + 3\right )}}{x^{2} \log \left (2\right )} \,d x } \]

[In]

integrate(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/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)

Mupad [B] (verification not implemented)

Time = 14.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=5\,x+{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{x\,\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {4\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{\ln \left (2\right )}} \]

[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))