\(\int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} (8+5 x-2 x^2) (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6))}{-8-5 x+2 x^2} \, dx\) [1818]

Optimal result

Integrand size = 103, antiderivative size = 32 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=e^{e^{(3+x)^4} x} (-x+(4-x) (2+2 x))-\log (5) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(32)=64\).

Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 4.12, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {1600, 2326} \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=\frac {\left (-8 x^6-52 x^5-4 x^4+612 x^3+1402 x^2+869 x+8\right ) \exp \left (x^4+12 x^3+54 x^2+e^{x^4+12 x^3+54 x^2+108 x+81} x+108 x+81\right )}{4 e^{x^4+12 x^3+54 x^2+108 x+81} x \left (x^3+9 x^2+27 x+27\right )+e^{x^4+12 x^3+54 x^2+108 x+81}} \]

[In]

Int[(E^(E^(81 + 108*x + 54*x^2 + 12*x^3 + x^4)*x)*(8 + 5*x - 2*x^2)*(-5 + 4*x + E^(81 + 108*x + 54*x^2 + 12*x^
3 + x^4)*(-8 - 869*x - 1402*x^2 - 612*x^3 + 4*x^4 + 52*x^5 + 8*x^6)))/(-8 - 5*x + 2*x^2),x]

[Out]

(E^(81 + 108*x + E^(81 + 108*x + 54*x^2 + 12*x^3 + x^4)*x + 54*x^2 + 12*x^3 + x^4)*(8 + 869*x + 1402*x^2 + 612
*x^3 - 4*x^4 - 52*x^5 - 8*x^6))/(E^(81 + 108*x + 54*x^2 + 12*x^3 + x^4) + 4*E^(81 + 108*x + 54*x^2 + 12*x^3 +
x^4)*x*(27 + 27*x + 9*x^2 + x^3))

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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}& = -\int e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right ) \, dx \\ & = \frac {\exp \left (81+108 x+e^{81+108 x+54 x^2+12 x^3+x^4} x+54 x^2+12 x^3+x^4\right ) \left (8+869 x+1402 x^2+612 x^3-4 x^4-52 x^5-8 x^6\right )}{e^{81+108 x+54 x^2+12 x^3+x^4}+4 e^{81+108 x+54 x^2+12 x^3+x^4} x \left (27+27 x+9 x^2+x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=e^{e^{(3+x)^4} x} \left (8+5 x-2 x^2\right ) \]

[In]

Integrate[(E^(E^(81 + 108*x + 54*x^2 + 12*x^3 + x^4)*x)*(8 + 5*x - 2*x^2)*(-5 + 4*x + E^(81 + 108*x + 54*x^2 +
12*x^3 + x^4)*(-8 - 869*x - 1402*x^2 - 612*x^3 + 4*x^4 + 52*x^5 + 8*x^6)))/(-8 - 5*x + 2*x^2),x]

[Out]

E^(E^(3 + x)^4*x)*(8 + 5*x - 2*x^2)

Maple [A] (verified)

Time = 4.96 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66

method	result	size
risch	\(\left (-2 x^{2}+5 x +8\right ) {\mathrm e}^{{\mathrm e}^{\left (3+x \right )^{4}} x}\)	\(21\)
parallelrisch	\({\mathrm e}^{\ln \left (-2 x^{2}+5 x +8\right )+x \,{\mathrm e}^{x^{4}+12 x^{3}+54 x^{2}+108 x +81}}\)	\(35\)

[In]

int(((8*x^6+52*x^5+4*x^4-612*x^3-1402*x^2-869*x-8)*exp(x^4+12*x^3+54*x^2+108*x+81)+4*x-5)*exp(ln(-2*x^2+5*x+8)
+x*exp(x^4+12*x^3+54*x^2+108*x+81))/(2*x^2-5*x-8),x,method=_RETURNVERBOSE)

[Out]

(-2*x^2+5*x+8)*exp(exp((3+x)^4)*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=e^{\left (x e^{\left (x^{4} + 12 \, x^{3} + 54 \, x^{2} + 108 \, x + 81\right )} + \log \left (-2 \, x^{2} + 5 \, x + 8\right )\right )} \]

[In]

integrate(((8*x^6+52*x^5+4*x^4-612*x^3-1402*x^2-869*x-8)*exp(x^4+12*x^3+54*x^2+108*x+81)+4*x-5)*exp(log(-2*x^2
+5*x+8)+x*exp(x^4+12*x^3+54*x^2+108*x+81))/(2*x^2-5*x-8),x, algorithm="fricas")

[Out]

e^(x*e^(x^4 + 12*x^3 + 54*x^2 + 108*x + 81) + log(-2*x^2 + 5*x + 8))

Sympy [A] (verification not implemented)

Time = 3.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=\left (- 2 x^{2} + 5 x + 8\right ) e^{x e^{x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81}} \]

[In]

integrate(((8*x**6+52*x**5+4*x**4-612*x**3-1402*x**2-869*x-8)*exp(x**4+12*x**3+54*x**2+108*x+81)+4*x-5)*exp(ln
(-2*x**2+5*x+8)+x*exp(x**4+12*x**3+54*x**2+108*x+81))/(2*x**2-5*x-8),x)

[Out]

(-2*x**2 + 5*x + 8)*exp(x*exp(x**4 + 12*x**3 + 54*x**2 + 108*x + 81))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=-{\left (2 \, x^{2} - 5 \, x - 8\right )} e^{\left (x e^{\left (x^{4} + 12 \, x^{3} + 54 \, x^{2} + 108 \, x + 81\right )}\right )} \]

[In]

integrate(((8*x^6+52*x^5+4*x^4-612*x^3-1402*x^2-869*x-8)*exp(x^4+12*x^3+54*x^2+108*x+81)+4*x-5)*exp(log(-2*x^2
+5*x+8)+x*exp(x^4+12*x^3+54*x^2+108*x+81))/(2*x^2-5*x-8),x, algorithm="maxima")

[Out]

-(2*x^2 - 5*x - 8)*e^(x*e^(x^4 + 12*x^3 + 54*x^2 + 108*x + 81))

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx=e^{\left (x e^{\left (x^{4} + 12 \, x^{3} + 54 \, x^{2} + 108 \, x + 81\right )} + \log \left (-2 \, x^{2} + 5 \, x + 8\right )\right )} \]

[In]

integrate(((8*x^6+52*x^5+4*x^4-612*x^3-1402*x^2-869*x-8)*exp(x^4+12*x^3+54*x^2+108*x+81)+4*x-5)*exp(log(-2*x^2
+5*x+8)+x*exp(x^4+12*x^3+54*x^2+108*x+81))/(2*x^2-5*x-8),x, algorithm="giac")

[Out]

e^(x*e^(x^4 + 12*x^3 + 54*x^2 + 108*x + 81) + log(-2*x^2 + 5*x + 8))

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{e^{81+108 x+54 x^2+12 x^3+x^4} x} \left (8+5 x-2 x^2\right ) \left (-5+4 x+e^{81+108 x+54 x^2+12 x^3+x^4} \left (-8-869 x-1402 x^2-612 x^3+4 x^4+52 x^5+8 x^6\right )\right )}{-8-5 x+2 x^2} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^{108\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{81}\,{\mathrm {e}}^{12\,x^3}\,{\mathrm {e}}^{54\,x^2}}\,\left (-2\,x^2+5\,x+8\right ) \]

[In]

int((exp(log(5*x - 2*x^2 + 8) + x*exp(108*x + 54*x^2 + 12*x^3 + x^4 + 81))*(exp(108*x + 54*x^2 + 12*x^3 + x^4
+ 81)*(869*x + 1402*x^2 + 612*x^3 - 4*x^4 - 52*x^5 - 8*x^6 + 8) - 4*x + 5))/(5*x - 2*x^2 + 8),x)

[Out]

exp(x*exp(108*x)*exp(x^4)*exp(81)*exp(12*x^3)*exp(54*x^2))*(5*x - 2*x^2 + 8)