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

Optimal. Leaf size=18 \[ -3+e^{\frac {5 e^4}{x}} (6-3 x) \]

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Rubi [A]  time = 0.33, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1593, 6688, 12, 2288} \begin {gather*} 3 e^{\frac {5 e^4}{x}} (2-x) \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 2288

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.83 \begin {gather*} -3 e^{\frac {5 e^4}{x}} (-2+x) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 1.06, size = 18, normalized size = 1.00 \begin {gather*} e^{\left (\frac {x \log \left (-3 \, x + 6\right ) + 5 \, e^{4}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.26, size = 33, normalized size = 1.83 \begin {gather*} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.25, size = 15, normalized size = 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	\({\mathrm e}^{\frac {x \ln \left (-3 x +6\right )+5 \,{\mathrm e}^{4}}{x}}\)	\(19\)
default	\(-\frac {3 \,{\mathrm e}^{4} \left (-125 \,{\mathrm e}^{12} \left (\frac {2 \left ({\mathrm e}^{-8}\right )^{2} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{25}-\frac {2 \left ({\mathrm e}^{-8}\right )^{2} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}\right )}{25}+\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} \left (-\frac {x \,{\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{x}}}{5}-\expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}\right )\right )}{5}\right )+50 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{5}-\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}\right )}{5}\right )+125 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{5}-\frac {{\mathrm e}^{-8} {\mathrm e}^{-4} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}\right )}{5}\right )-50 \,{\mathrm e}^{8} {\mathrm e}^{-8} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )+20 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{-8} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{x}}}{2}+\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{\frac {5 \,{\mathrm e}^{4}}{2}} \expIntegralEi \left (1, -\frac {5 \,{\mathrm e}^{4}}{x}+\frac {5 \,{\mathrm e}^{4}}{2}\right )}{4}\right )\right )}{5}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -3 \, x e^{\left (\frac {5 \, e^{4}}{x}\right )} - 30 \, \int \frac {e^{\left (\frac {5 \, e^{4}}{x} + 4\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 23.67, size = 15, normalized size = 0.83 \begin {gather*} e^{\frac {x \log {\left (6 - 3 x \right )} + 5 e^{4}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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