\(\int \frac {1}{16} e^{\frac {1}{4} (4 x-65 e^x x)} (12+12 x+e^x (-195 x-195 x^2)) \, dx\) [8410]

Optimal result

Integrand size = 38, antiderivative size = 17 \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3}{4} e^{x-\frac {65 e^x x}{4}} x \]

[Out]

3/4/exp(65/4*exp(x)*x-x)*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(17)=34\).

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 2326} \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3 e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (4 x-65 e^x \left (x^2+x\right )\right )}{4 \left (-65 e^x x-65 e^x+4\right )} \]

[In]

Int[(E^((4*x - 65*E^x*x)/4)*(12 + 12*x + E^x*(-195*x - 195*x^2)))/16,x]

[Out]

(3*E^((4*x - 65*E^x*x)/4)*(4*x - 65*E^x*(x + x^2)))/(4*(4 - 65*E^x - 65*E^x*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}{16} \int e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx \\ & = \frac {3 e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (4 x-65 e^x \left (x+x^2\right )\right )}{4 \left (4-65 e^x-65 e^x x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3}{4} e^{x-\frac {65 e^x x}{4}} x \]

[In]

Integrate[(E^((4*x - 65*E^x*x)/4)*(12 + 12*x + E^x*(-195*x - 195*x^2)))/16,x]

[Out]

(3*E^(x - (65*E^x*x)/4)*x)/4

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{-\frac {x \left (65 \,{\mathrm e}^{x}-4\right )}{4}} x}{4}\)	\(14\)
norman	\(\frac {3 \,{\mathrm e}^{-\frac {65 \,{\mathrm e}^{x} x}{4}+x} x}{4}\)	\(16\)
parallelrisch	\(\frac {3 \,{\mathrm e}^{-\frac {x \left (65 \,{\mathrm e}^{x}-4\right )}{4}} x}{4}\)	\(16\)

[In]

int(1/16*((-195*x^2-195*x)*exp(x)+12*x+12)/exp(65/4*exp(x)*x-x),x,method=_RETURNVERBOSE)

[Out]

3/4*exp(-1/4*x*(65*exp(x)-4))*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3}{4} \, x e^{\left (-\frac {65}{4} \, x e^{x} + x\right )} \]

[In]

integrate(1/16*((-195*x^2-195*x)*exp(x)+12*x+12)/exp(65/4*exp(x)*x-x),x, algorithm="fricas")

[Out]

3/4*x*e^(-65/4*x*e^x + x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3 x e^{- \frac {65 x e^{x}}{4} + x}}{4} \]

[In]

integrate(1/16*((-195*x**2-195*x)*exp(x)+12*x+12)/exp(65/4*exp(x)*x-x),x)

[Out]

3*x*exp(-65*x*exp(x)/4 + x)/4

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3}{4} \, x e^{\left (-\frac {65}{4} \, x e^{x} + x\right )} \]

[In]

integrate(1/16*((-195*x^2-195*x)*exp(x)+12*x+12)/exp(65/4*exp(x)*x-x),x, algorithm="maxima")

[Out]

3/4*x*e^(-65/4*x*e^x + x)

Giac [F]

\[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\int { -\frac {3}{16} \, {\left (65 \, {\left (x^{2} + x\right )} e^{x} - 4 \, x - 4\right )} e^{\left (-\frac {65}{4} \, x e^{x} + x\right )} \,d x } \]

[In]

integrate(1/16*((-195*x^2-195*x)*exp(x)+12*x+12)/exp(65/4*exp(x)*x-x),x, algorithm="giac")

[Out]

integrate(-3/16*(65*(x^2 + x)*e^x - 4*x - 4)*e^(-65/4*x*e^x + x), x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{16} e^{\frac {1}{4} \left (4 x-65 e^x x\right )} \left (12+12 x+e^x \left (-195 x-195 x^2\right )\right ) \, dx=\frac {3\,x\,{\mathrm {e}}^{-\frac {65\,x\,{\mathrm {e}}^x}{4}}\,{\mathrm {e}}^x}{4} \]

[In]

int(exp(x - (65*x*exp(x))/4)*((3*x)/4 - (exp(x)*(195*x + 195*x^2))/16 + 3/4),x)

[Out]

(3*x*exp(-(65*x*exp(x))/4)*exp(x))/4