\(\int \frac {1}{5} e^{\frac {1}{5} (-60+15 e x+15 x^2-(3 e+3 x) \log (x))} (20 x-6 x^2+60 x^3+e (-6 x+30 x^2)-6 x^2 \log (x)) \, dx\) [4055]

Optimal result

Integrand size = 64, antiderivative size = 24 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 e^{-3 \left (4-(e+x) \left (x-\frac {\log (x)}{5}\right )\right )} x^2 \]

[Out]

2*x^2/exp(12-3*(x+exp(1))*(x-1/5*ln(x)))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(71\) vs. \(2(24)=48\).

Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 2326} \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=-\frac {2 e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3}{5} (x+e)} \left (-10 x^3+x^2+e \left (x-5 x^2\right )+x^2 \log (x)\right )}{10 x-\frac {x+e}{x}-\log (x)+5 e} \]

[In]

Int[(E^((-60 + 15*E*x + 15*x^2 - (3*E + 3*x)*Log[x])/5)*(20*x - 6*x^2 + 60*x^3 + E*(-6*x + 30*x^2) - 6*x^2*Log
[x]))/5,x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \exp \left (\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )\right ) \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx \\ & = -\frac {2 e^{-3 \left (4-e x-x^2\right )} x^{-\frac {3}{5} (e+x)} \left (x^2-10 x^3+e \left (x-5 x^2\right )+x^2 \log (x)\right )}{5 e+10 x-\frac {e+x}{x}-\log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 e^{3 \left (-4+e x+x^2\right )} x^{2-\frac {3 e}{5}-\frac {3 x}{5}} \]

[In]

Integrate[(E^((-60 + 15*E*x + 15*x^2 - (3*E + 3*x)*Log[x])/5)*(20*x - 6*x^2 + 60*x^3 + E*(-6*x + 30*x^2) - 6*x
^2*Log[x]))/5,x]

[Out]

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

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29

method	result	size
risch	\(2 x^{2} x^{-\frac {3 x}{5}-\frac {3 \,{\mathrm e}}{5}} {\mathrm e}^{-12+3 x \,{\mathrm e}+3 x^{2}}\)	\(31\)
norman	\(2 x^{2} {\mathrm e}^{\left (-\frac {3 x}{5}-\frac {3 \,{\mathrm e}}{5}\right ) \ln \left (x \right )+3 x \,{\mathrm e}+3 x^{2}-12}\)	\(33\)
parallelrisch	\(2 x^{2} {\mathrm e}^{\left (-\frac {3 x}{5}-\frac {3 \,{\mathrm e}}{5}\right ) \ln \left (x \right )+3 x \,{\mathrm e}+3 x^{2}-12}\)	\(33\)

[In]

int(1/5*(-6*x^2*ln(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/5*(3*exp(1)+3*x)*ln(x)-3*x*exp(1)-3*x^2+12)
,x,method=_RETURNVERBOSE)

[Out]

2*x^2/(x^(3/5*x+3/5*exp(1)))*exp(-12+3*x*exp(1)+3*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 \, x^{2} e^{\left (3 \, x^{2} + 3 \, x e - \frac {3}{5} \, {\left (x + e\right )} \log \left (x\right ) - 12\right )} \]

[In]

integrate(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/5*(3*exp(1)+3*x)*log(x)-3*x*exp(1)-3
*x^2+12),x, algorithm="fricas")

[Out]

2*x^2*e^(3*x^2 + 3*x*e - 3/5*(x + e)*log(x) - 12)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 x^{2} e^{3 x^{2} + 3 e x - \left (\frac {3 x}{5} + \frac {3 e}{5}\right ) \log {\left (x \right )} - 12} \]

[In]

integrate(1/5*(-6*x**2*ln(x)+(30*x**2-6*x)*exp(1)+60*x**3-6*x**2+20*x)/exp(1/5*(3*exp(1)+3*x)*ln(x)-3*x*exp(1)
-3*x**2+12),x)

[Out]

2*x**2*exp(3*x**2 + 3*E*x - (3*x/5 + 3*E/5)*log(x) - 12)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 \, x^{2} e^{\left (3 \, x^{2} + 3 \, x e - \frac {3}{5} \, x \log \left (x\right ) - \frac {3}{5} \, e \log \left (x\right ) - 12\right )} \]

[In]

integrate(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/5*(3*exp(1)+3*x)*log(x)-3*x*exp(1)-3
*x^2+12),x, algorithm="maxima")

[Out]

2*x^2*e^(3*x^2 + 3*x*e - 3/5*x*log(x) - 3/5*e*log(x) - 12)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 \, x^{2} e^{\left (3 \, x^{2} + 3 \, x e - \frac {3}{5} \, x \log \left (x\right ) - \frac {3}{5} \, e \log \left (x\right ) - 12\right )} \]

[In]

integrate(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/5*(3*exp(1)+3*x)*log(x)-3*x*exp(1)-3
*x^2+12),x, algorithm="giac")

[Out]

2*x^2*e^(3*x^2 + 3*x*e - 3/5*x*log(x) - 3/5*e*log(x) - 12)

Mupad [B] (verification not implemented)

Time = 10.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2\,x^{2-\frac {3\,\mathrm {e}}{5}-\frac {3\,x}{5}}\,{\mathrm {e}}^{-12}\,{\mathrm {e}}^{3\,x^2}\,{\mathrm {e}}^{3\,x\,\mathrm {e}} \]

[In]

int(-exp(3*x*exp(1) - (log(x)*(3*x + 3*exp(1)))/5 + 3*x^2 - 12)*((6*x^2*log(x))/5 - 4*x + (exp(1)*(6*x - 30*x^
2))/5 + (6*x^2)/5 - 12*x^3),x)

[Out]

2*x^(2 - (3*exp(1))/5 - (3*x)/5)*exp(-12)*exp(3*x^2)*exp(3*x*exp(1))