\(\int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} (e^4 (10-5 x)+x^2)}{-2 x^2+x^3} \, dx\) [7939]

Optimal result

Integrand size = 45, antiderivative size = 18 \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=-3+e^{\frac {5 e^4}{x}} (6-3 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1607, 6820, 12, 2326} \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=3 e^{\frac {5 e^4}{x}} (2-x) \]

[In]

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

[Out]

3*E^((5*E^4)/x)*(2 - x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{(-2+x) x^2} \, dx \\ & = \int \frac {3 e^{\frac {5 e^4}{x}} \left (-10 e^4+5 e^4 x-x^2\right )}{x^2} \, dx \\ & = 3 \int \frac {e^{\frac {5 e^4}{x}} \left (-10 e^4+5 e^4 x-x^2\right )}{x^2} \, dx \\ & = 3 e^{\frac {5 e^4}{x}} (2-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=-3 e^{\frac {5 e^4}{x}} (-2+x) \]

[In]

Integrate[(E^((5*E^4 + x*Log[6 - 3*x])/x)*(E^4*(10 - 5*x) + x^2))/(-2*x^2 + x^3),x]

[Out]

-3*E^((5*E^4)/x)*(-2 + x)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
risch	\(\left (-3 x +6\right ) {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{x}}\)	\(15\)
gosper	\({\mathrm e}^{\frac {x \ln \left (-3 x +6\right )+5 \,{\mathrm e}^{4}}{x}}\)	\(19\)
norman	\({\mathrm e}^{\frac {x \ln \left (-3 x +6\right )+5 \,{\mathrm e}^{4}}{x}}\)	\(19\)
parallelrisch	\({\mathrm e}^{\frac {x \ln \left (-3 x +6\right )+5 \,{\mathrm e}^{4}}{x}}\)	\(19\)
default	\(-\frac {3 \,{\mathrm e}^{4} \left (-125 \,{\mathrm e}^{12} \left (-\frac {2 \left ({\mathrm e}^{-8}\right )^{2} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}\right )}{25}+\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} \left (-\frac {x \,{\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{x}}}{5}-\operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}\right )\right )}{5}+\frac {2 \left ({\mathrm e}^{-8}\right )^{2} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{25}\right )+50 \,{\mathrm e}^{8} \left (-\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}\right )}{5}+\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{5}\right )+125 \,{\mathrm e}^{12} \left (-\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}\right )}{5}+\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{5}\right )-50 \,{\mathrm e}^{8} {\mathrm e}^{-8} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )+20 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{-8} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{x}}}{2}+\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \operatorname {Ei}_{1}\left (-\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{4}\right )\right )}{5}\)	\(261\)

[In]

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

[Out]

(-3*x+6)*exp(5*exp(4)/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=e^{\left (\frac {x \log \left (-3 \, x + 6\right ) + 5 \, e^{4}}{x}\right )} \]

[In]

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

[Out]

e^((x*log(-3*x + 6) + 5*e^4)/x)

Sympy [A] (verification not implemented)

Time = 11.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=e^{\frac {x \log {\left (6 - 3 x \right )} + 5 e^{4}}{x}} \]

[In]

integrate(((-5*x+10)*exp(4)+x**2)*exp((x*ln(-3*x+6)+5*exp(4))/x)/(x**3-2*x**2),x)

[Out]

exp((x*log(6 - 3*x) + 5*exp(4))/x)

Maxima [F]

\[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=\int { \frac {{\left (x^{2} - 5 \, {\left (x - 2\right )} e^{4}\right )} e^{\left (\frac {x \log \left (-3 \, x + 6\right ) + 5 \, e^{4}}{x}\right )}}{x^{3} - 2 \, x^{2}} \,d x } \]

[In]

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

[Out]

-3*x*e^(5*e^4/x) - 30*integrate(e^(5*e^4/x + 4)/x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=3 \, x {\left (\frac {2 \, e^{\left (\frac {5 \, e^{4}}{x} + 8\right )}}{x} - e^{\left (\frac {5 \, e^{4}}{x} + 8\right )}\right )} e^{\left (-8\right )} \]

[In]

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

[Out]

3*x*(2*e^(5*e^4/x + 8)/x - e^(5*e^4/x + 8))*e^(-8)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {5 e^4+x \log (6-3 x)}{x}} \left (e^4 (10-5 x)+x^2\right )}{-2 x^2+x^3} \, dx=\int -\frac {{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^4+x\,\ln \left (6-3\,x\right )}{x}}\,\left (x^2-{\mathrm {e}}^4\,\left (5\,x-10\right )\right )}{2\,x^2-x^3} \,d x \]

[In]

int(-(exp((5*exp(4) + x*log(6 - 3*x))/x)*(x^2 - exp(4)*(5*x - 10)))/(2*x^2 - x^3),x)

[Out]

int(-(exp((5*exp(4) + x*log(6 - 3*x))/x)*(x^2 - exp(4)*(5*x - 10)))/(2*x^2 - x^3), x)