\(\int \frac {1}{8} e^{\frac {1}{2} (-2 x \log (3)-e^x (1-2 x^2 \log (3)))} (8+(10-8 x) \log (3)+e^x (5-4 x+(-20 x+6 x^2+8 x^3) \log (3))) \, dx\) [6296]

Optimal result

Integrand size = 66, antiderivative size = 32 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {1}{4} e^{-\frac {e^x}{2}-x \left (1-e^x x\right ) \log (3)} (-5+4 x) \]

[Out]

1/4*(-5+4*x)/exp(ln(3)*x*(-exp(x)*x+1)+1/2*exp(x))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(92\) vs. \(2(32)=64\).

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {12, 2326} \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {3^{e^x x^2-x} e^{-\frac {e^x}{2}} \left (e^x \left (-2 \left (-4 x^3-3 x^2+10 x\right ) \log (3)-4 x+5\right )+2 (5-4 x) \log (3)\right )}{4 \left (-e^x \left (1-2 x^2 \log (3)\right )+4 e^x x \log (3)-\log (9)\right )} \]

[In]

Int[(E^((-2*x*Log[3] - E^x*(1 - 2*x^2*Log[3]))/2)*(8 + (10 - 8*x)*Log[3] + E^x*(5 - 4*x + (-20*x + 6*x^2 + 8*x
^3)*Log[3])))/8,x]

[Out]

(3^(-x + E^x*x^2)*(2*(5 - 4*x)*Log[3] + E^x*(5 - 4*x - 2*(10*x - 3*x^2 - 4*x^3)*Log[3])))/(4*E^(E^x/2)*(4*E^x*
x*Log[3] - E^x*(1 - 2*x^2*Log[3]) - Log[9]))

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}{8} \int \exp \left (\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )\right ) \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx \\ & = \frac {3^{-x+e^x x^2} e^{-\frac {e^x}{2}} \left (2 (5-4 x) \log (3)+e^x \left (5-4 x-2 \left (10 x-3 x^2-4 x^3\right ) \log (3)\right )\right )}{4 \left (4 e^x x \log (3)-e^x \left (1-2 x^2 \log (3)\right )-\log (9)\right )} \\ \end{align*}

Mathematica [F]

\[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx \]

[In]

Integrate[(E^((-2*x*Log[3] - E^x*(1 - 2*x^2*Log[3]))/2)*(8 + (10 - 8*x)*Log[3] + E^x*(5 - 4*x + (-20*x + 6*x^2
+ 8*x^3)*Log[3])))/8,x]

[Out]

Integrate[E^((-2*x*Log[3] - E^x*(1 - 2*x^2*Log[3]))/2)*(8 + (10 - 8*x)*Log[3] + E^x*(5 - 4*x + (-20*x + 6*x^2
+ 8*x^3)*Log[3])), x]/8

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81

method	result	size
norman	\(\left (x -\frac {5}{4}\right ) {\mathrm e}^{-\frac {\left (-2 x^{2} \ln \left (3\right )+1\right ) {\mathrm e}^{x}}{2}-x \ln \left (3\right )}\)	\(26\)
risch	\(\frac {\left (8 x -10\right ) 3^{{\mathrm e}^{x} x^{2}} 3^{-x} {\mathrm e}^{-\frac {{\mathrm e}^{x}}{2}}}{8}\)	\(28\)
parallelrisch	\(\frac {\left (16 x -20\right ) {\mathrm e}^{x^{2} \ln \left (3\right ) {\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{2}-x \ln \left (3\right )}}{16}\)	\(29\)

[In]

int(1/8*(((8*x^3+6*x^2-20*x)*ln(3)-4*x+5)*exp(x)+(-8*x+10)*ln(3)+8)/exp(1/2*(-2*x^2*ln(3)+1)*exp(x)+x*ln(3)),x
,method=_RETURNVERBOSE)

[Out]

(x-5/4)/exp(1/2*(-2*x^2*ln(3)+1)*exp(x)+x*ln(3))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {1}{4} \, {\left (4 \, x - 5\right )} e^{\left (\frac {1}{2} \, {\left (2 \, x^{2} \log \left (3\right ) - 1\right )} e^{x} - x \log \left (3\right )\right )} \]

[In]

integrate(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8)/exp(1/2*(-2*x^2*log(3)+1)*exp(x)+x
*log(3)),x, algorithm="fricas")

[Out]

1/4*(4*x - 5)*e^(1/2*(2*x^2*log(3) - 1)*e^x - x*log(3))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {\left (4 x - 5\right ) e^{- x \log {\left (3 \right )} - \left (- x^{2} \log {\left (3 \right )} + \frac {1}{2}\right ) e^{x}}}{4} \]

[In]

integrate(1/8*(((8*x**3+6*x**2-20*x)*ln(3)-4*x+5)*exp(x)+(-8*x+10)*ln(3)+8)/exp(1/2*(-2*x**2*ln(3)+1)*exp(x)+x
*ln(3)),x)

[Out]

(4*x - 5)*exp(-x*log(3) - (-x**2*log(3) + 1/2)*exp(x))/4

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {1}{4} \, {\left (4 \, x - 5\right )} e^{\left (x^{2} e^{x} \log \left (3\right ) - x \log \left (3\right ) - \frac {1}{2} \, e^{x}\right )} \]

[In]

integrate(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8)/exp(1/2*(-2*x^2*log(3)+1)*exp(x)+x
*log(3)),x, algorithm="maxima")

[Out]

1/4*(4*x - 5)*e^(x^2*e^x*log(3) - x*log(3) - 1/2*e^x)

Giac [F]

\[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\int { \frac {1}{8} \, {\left ({\left (2 \, {\left (4 \, x^{3} + 3 \, x^{2} - 10 \, x\right )} \log \left (3\right ) - 4 \, x + 5\right )} e^{x} - 2 \, {\left (4 \, x - 5\right )} \log \left (3\right ) + 8\right )} e^{\left (\frac {1}{2} \, {\left (2 \, x^{2} \log \left (3\right ) - 1\right )} e^{x} - x \log \left (3\right )\right )} \,d x } \]

[In]

integrate(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8)/exp(1/2*(-2*x^2*log(3)+1)*exp(x)+x
*log(3)),x, algorithm="giac")

[Out]

integrate(1/8*((2*(4*x^3 + 3*x^2 - 10*x)*log(3) - 4*x + 5)*e^x - 2*(4*x - 5)*log(3) + 8)*e^(1/2*(2*x^2*log(3)
- 1)*e^x - x*log(3)), x)

Mupad [B] (verification not implemented)

Time = 11.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {3^{x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{2}}\,\left (4\,x-5\right )}{4\,3^x} \]

[In]

int(exp((exp(x)*(2*x^2*log(3) - 1))/2 - x*log(3))*((exp(x)*(log(3)*(6*x^2 - 20*x + 8*x^3) - 4*x + 5))/8 - (log
(3)*(8*x - 10))/8 + 1),x)

[Out]

(3^(x^2*exp(x))*exp(-exp(x)/2)*(4*x - 5))/(4*3^x)