3.8.87 \(\int \frac {e^{\frac {2 (4-x^3 \log (-e^x-x+\log (-15+5 x)))}{\log (-e^x-x+\log (-15+5 x))}} (-32+8 x+e^x (-24+8 x)+(-18 x^3+6 x^4+e^x (-18 x^2+6 x^3)+(18 x^2-6 x^3) \log (-15+5 x)) \log ^2(-e^x-x+\log (-15+5 x)))}{(e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)) \log ^2(-e^x-x+\log (-15+5 x))} \, dx\)

Optimal. Leaf size=30 \[ 2+e^{-2 x^3+\frac {8}{\log \left (-e^x-x+\log (5 (-3+x))\right )}} \]

________________________________________________________________________________________

Rubi [A]  time = 2.76, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 171, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6688, 12, 6706} \begin {gather*} \exp \left (\frac {8}{\log \left (-x-e^x+\log (-5 (3-x))\right )}-2 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*(4 - x^3*Log[-E^x - x + Log[-15 + 5*x]]))/Log[-E^x - x + Log[-15 + 5*x]])*(-32 + 8*x + E^x*(-24 + 8
*x) + (-18*x^3 + 6*x^4 + E^x*(-18*x^2 + 6*x^3) + (18*x^2 - 6*x^3)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15 + 5*x
]]^2))/((E^x*(3 - x) + 3*x - x^2 + (-3 + x)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15 + 5*x]]^2),x]

[Out]

E^(-2*x^3 + 8/Log[-E^x - x + Log[-5*(3 - 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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 2 \exp \left (-2 x^3+\frac {8}{\log \left (-e^x-x+\log (5 (-3+x))\right )}\right ) \left (-3 x^2-\frac {4 \left (-4+e^x (-3+x)+x\right )}{(-3+x) \left (e^x+x-\log (5 (-3+x))\right ) \log ^2\left (-e^x-x+\log (5 (-3+x))\right )}\right ) \, dx\\ &=2 \int \exp \left (-2 x^3+\frac {8}{\log \left (-e^x-x+\log (5 (-3+x))\right )}\right ) \left (-3 x^2-\frac {4 \left (-4+e^x (-3+x)+x\right )}{(-3+x) \left (e^x+x-\log (5 (-3+x))\right ) \log ^2\left (-e^x-x+\log (5 (-3+x))\right )}\right ) \, dx\\ &=\exp \left (-2 x^3+\frac {8}{\log \left (-e^x-x+\log (-5 (3-x))\right )}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 28, normalized size = 0.93 \begin {gather*} e^{-2 x^3+\frac {8}{\log \left (-e^x-x+\log (5 (-3+x))\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(4 - x^3*Log[-E^x - x + Log[-15 + 5*x]]))/Log[-E^x - x + Log[-15 + 5*x]])*(-32 + 8*x + E^x*(-
24 + 8*x) + (-18*x^3 + 6*x^4 + E^x*(-18*x^2 + 6*x^3) + (18*x^2 - 6*x^3)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15
+ 5*x]]^2))/((E^x*(3 - x) + 3*x - x^2 + (-3 + x)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15 + 5*x]]^2),x]

[Out]

E^(-2*x^3 + 8/Log[-E^x - x + Log[5*(-3 + x)]])

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 41, normalized size = 1.37 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{3} \log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right ) - 4\right )}}{\log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^3+18*x^2)*log(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3)*log(log(5*x-15)-exp(x)-x)^2+(8*x-2
4)*exp(x)+8*x-32)*exp((-x^3*log(log(5*x-15)-exp(x)-x)+4)/log(log(5*x-15)-exp(x)-x))^2/((x-3)*log(5*x-15)+(3-x)
*exp(x)-x^2+3*x)/log(log(5*x-15)-exp(x)-x)^2,x, algorithm="fricas")

[Out]

e^(-2*(x^3*log(-x - e^x + log(5*x - 15)) - 4)/log(-x - e^x + log(5*x - 15)))

________________________________________________________________________________________

giac [A]  time = 20.02, size = 26, normalized size = 0.87 \begin {gather*} e^{\left (-2 \, x^{3} + \frac {8}{\log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^3+18*x^2)*log(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3)*log(log(5*x-15)-exp(x)-x)^2+(8*x-2
4)*exp(x)+8*x-32)*exp((-x^3*log(log(5*x-15)-exp(x)-x)+4)/log(log(5*x-15)-exp(x)-x))^2/((x-3)*log(5*x-15)+(3-x)
*exp(x)-x^2+3*x)/log(log(5*x-15)-exp(x)-x)^2,x, algorithm="giac")

[Out]

e^(-2*x^3 + 8/log(-x - e^x + log(5*x - 15)))

________________________________________________________________________________________

maple [A]  time = 0.06, size = 42, normalized size = 1.40

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (x^{3} \ln \left (\ln \left (5 x -15\right )-{\mathrm e}^{x}-x \right )-4\right )}{\ln \left (\ln \left (5 x -15\right )-{\mathrm e}^{x}-x \right )}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x^3+18*x^2)*ln(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3)*ln(ln(5*x-15)-exp(x)-x)^2+(8*x-24)*exp(x)
+8*x-32)*exp((-x^3*ln(ln(5*x-15)-exp(x)-x)+4)/ln(ln(5*x-15)-exp(x)-x))^2/((x-3)*ln(5*x-15)+(3-x)*exp(x)-x^2+3*
x)/ln(ln(5*x-15)-exp(x)-x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(-2*(x^3*ln(ln(5*x-15)-exp(x)-x)-4)/ln(ln(5*x-15)-exp(x)-x))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {x e^{\left (-2 \, x^{3} + x + \frac {8}{\log \left (-x - e^{x} + \log \relax (5) + \log \left (x - 3\right )\right )}\right )}}{{\left (x - 3\right )} e^{x} + x - 4} + \frac {x e^{\left (-2 \, x^{3} + \frac {8}{\log \left (-x - e^{x} + \log \relax (5) + \log \left (x - 3\right )\right )}\right )}}{{\left (x - 3\right )} e^{x} + x - 4} - \frac {3 \, e^{\left (-2 \, x^{3} + x + \frac {8}{\log \left (-x - e^{x} + \log \relax (5) + \log \left (x - 3\right )\right )}\right )}}{{\left (x - 3\right )} e^{x} + x - 4} - \frac {4 \, e^{\left (-2 \, x^{3} + \frac {8}{\log \left (-x - e^{x} + \log \relax (5) + \log \left (x - 3\right )\right )}\right )}}{{\left (x - 3\right )} e^{x} + x - 4} - 6 \, \int x^{2} e^{\left (-2 \, x^{3} + \frac {8}{\log \left (-x - e^{x} + \log \relax (5) + \log \left (x - 3\right )\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^3+18*x^2)*log(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3)*log(log(5*x-15)-exp(x)-x)^2+(8*x-2
4)*exp(x)+8*x-32)*exp((-x^3*log(log(5*x-15)-exp(x)-x)+4)/log(log(5*x-15)-exp(x)-x))^2/((x-3)*log(5*x-15)+(3-x)
*exp(x)-x^2+3*x)/log(log(5*x-15)-exp(x)-x)^2,x, algorithm="maxima")

[Out]

x*e^(-2*x^3 + x + 8/log(-x - e^x + log(5) + log(x - 3)))/((x - 3)*e^x + x - 4) + x*e^(-2*x^3 + 8/log(-x - e^x
+ log(5) + log(x - 3)))/((x - 3)*e^x + x - 4) - 3*e^(-2*x^3 + x + 8/log(-x - e^x + log(5) + log(x - 3)))/((x -
3)*e^x + x - 4) - 4*e^(-2*x^3 + 8/log(-x - e^x + log(5) + log(x - 3)))/((x - 3)*e^x + x - 4) - 6*integrate(x^
2*e^(-2*x^3 + 8/log(-x - e^x + log(5) + log(x - 3))), x)

________________________________________________________________________________________

mupad [B]  time = 1.09, size = 27, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{\frac {8}{\ln \left (\ln \left (5\,x-15\right )-x-{\mathrm {e}}^x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*(x^3*log(log(5*x - 15) - x - exp(x)) - 4))/log(log(5*x - 15) - x - exp(x)))*(8*x + exp(x)*(8*x -
24) - log(log(5*x - 15) - x - exp(x))^2*(exp(x)*(18*x^2 - 6*x^3) - log(5*x - 15)*(18*x^2 - 6*x^3) + 18*x^3 - 6
*x^4) - 32))/(log(log(5*x - 15) - x - exp(x))^2*(3*x - exp(x)*(x - 3) - x^2 + log(5*x - 15)*(x - 3))),x)

[Out]

exp(-2*x^3)*exp(8/log(log(5*x - 15) - x - exp(x)))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x**3+18*x**2)*ln(5*x-15)+(6*x**3-18*x**2)*exp(x)+6*x**4-18*x**3)*ln(ln(5*x-15)-exp(x)-x)**2+(8
*x-24)*exp(x)+8*x-32)*exp((-x**3*ln(ln(5*x-15)-exp(x)-x)+4)/ln(ln(5*x-15)-exp(x)-x))**2/((x-3)*ln(5*x-15)+(3-x
)*exp(x)-x**2+3*x)/ln(ln(5*x-15)-exp(x)-x)**2,x)

[Out]

Timed out

________________________________________________________________________________________