\(\int \frac {e^4 (44-28 x+111 x^2-90 x^3+e^6 (4-12 x+9 x^2))}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} (4 x^2-12 x^3+9 x^4)+e^6 (88 x^2-160 x^3+102 x^4-90 x^5)} \, dx\) [5980]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 106, antiderivative size = 30 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {e^4}{x \left (1-e^6+5 x+\frac {16}{-\frac {4}{3}+2 x}\right )} \]

[Out]

exp(4)/(5*x-exp(3)^2+16/(2*x-4/3)+1)/x

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2099, 652, 632, 212} \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {e^4 (15 x+26)}{\left (11+e^6\right ) \left (15 x^2-\left (7+3 e^6\right ) x+2 \left (11+e^6\right )\right )}-\frac {e^4}{\left (11+e^6\right ) x} \]

[In]

Int[(E^4*(44 - 28*x + 111*x^2 - 90*x^3 + E^6*(4 - 12*x + 9*x^2)))/(484*x^2 - 308*x^3 + 709*x^4 - 210*x^5 + 225
*x^6 + E^12*(4*x^2 - 12*x^3 + 9*x^4) + E^6*(88*x^2 - 160*x^3 + 102*x^4 - 90*x^5)),x]

[Out]

-(E^4/((11 + E^6)*x)) + (E^4*(26 + 15*x))/((11 + E^6)*(2*(11 + E^6) - (7 + 3*E^6)*x + 15*x^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = e^4 \int \frac {44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx \\ & = e^4 \int \left (\frac {1}{\left (11+e^6\right ) x^2}+\frac {2 \left (421+69 e^6\right )-15 \left (59+3 e^6\right ) x}{\left (11+e^6\right ) \left (2 \left (11+e^6\right )-\left (7+3 e^6\right ) x+15 x^2\right )^2}+\frac {15}{\left (-11-e^6\right ) \left (2 \left (11+e^6\right )-\left (7+3 e^6\right ) x+15 x^2\right )}\right ) \, dx \\ & = -\frac {e^4}{\left (11+e^6\right ) x}+\frac {e^4 \int \frac {2 \left (421+69 e^6\right )-15 \left (59+3 e^6\right ) x}{\left (2 \left (11+e^6\right )+\left (-7-3 e^6\right ) x+15 x^2\right )^2} \, dx}{11+e^6}-\frac {\left (15 e^4\right ) \int \frac {1}{2 \left (11+e^6\right )-\left (7+3 e^6\right ) x+15 x^2} \, dx}{11+e^6} \\ & = -\frac {e^4}{\left (11+e^6\right ) x}+\frac {e^4 (26+15 x)}{\left (11+e^6\right ) \left (2 \left (11+e^6\right )-\left (7+3 e^6\right ) x+15 x^2\right )}+\frac {\left (15 e^4\right ) \int \frac {1}{2 \left (11+e^6\right )+\left (-7-3 e^6\right ) x+15 x^2} \, dx}{11+e^6}+\frac {\left (30 e^4\right ) \text {Subst}\left (\int \frac {1}{-1271-78 e^6+9 e^{12}-x^2} \, dx,x,-7-3 e^6+30 x\right )}{11+e^6} \\ & = -\frac {e^4}{\left (11+e^6\right ) x}+\frac {e^4 (26+15 x)}{\left (11+e^6\right ) \left (2 \left (11+e^6\right )-\left (7+3 e^6\right ) x+15 x^2\right )}-\frac {30 e^4 \tanh ^{-1}\left (\frac {7+3 e^6-30 x}{\sqrt {-1271-78 e^6+9 e^{12}}}\right )}{\left (11+e^6\right ) \sqrt {-1271-78 e^6+9 e^{12}}}-\frac {\left (30 e^4\right ) \text {Subst}\left (\int \frac {1}{-1271-78 e^6+9 e^{12}-x^2} \, dx,x,-7-3 e^6+30 x\right )}{11+e^6} \\ & = -\frac {e^4}{\left (11+e^6\right ) x}+\frac {e^4 (26+15 x)}{\left (11+e^6\right ) \left (2 \left (11+e^6\right )-\left (7+3 e^6\right ) x+15 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {e^4 (-2+3 x)}{x \left (22+e^6 (2-3 x)-7 x+15 x^2\right )} \]

[In]

Integrate[(E^4*(44 - 28*x + 111*x^2 - 90*x^3 + E^6*(4 - 12*x + 9*x^2)))/(484*x^2 - 308*x^3 + 709*x^4 - 210*x^5
 + 225*x^6 + E^12*(4*x^2 - 12*x^3 + 9*x^4) + E^6*(88*x^2 - 160*x^3 + 102*x^4 - 90*x^5)),x]

[Out]

(E^4*(-2 + 3*x))/(x*(22 + E^6*(2 - 3*x) - 7*x + 15*x^2))

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13

method result size
risch \(\frac {3 \,{\mathrm e}^{4} \left (\frac {2}{3}-x \right )}{\left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right ) x}\) \(34\)
gosper \(-\frac {\left (-2+3 x \right ) {\mathrm e}^{4}}{x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\) \(38\)
parallelrisch \(-\frac {{\mathrm e}^{4} \left (-30+45 x \right )}{15 x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\) \(38\)
norman \(\frac {-3 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}}{x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\) \(40\)
default \({\mathrm e}^{4} \left (\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (225 \textit {\_Z}^{4}+\left (-90 \,{\mathrm e}^{6}-210\right ) \textit {\_Z}^{3}+\left (9 \,{\mathrm e}^{12}+102 \,{\mathrm e}^{6}+709\right ) \textit {\_Z}^{2}+\left (-12 \,{\mathrm e}^{12}-160 \,{\mathrm e}^{6}-308\right ) \textit {\_Z} +4 \,{\mathrm e}^{12}+88 \,{\mathrm e}^{6}+484\right )}{\sum }\frac {\left (681472+225 \left (-1331-33 \,{\mathrm e}^{12}-363 \,{\mathrm e}^{6}-{\mathrm e}^{18}\right ) \textit {\_R}^{2}+780 \left (-1331-33 \,{\mathrm e}^{12}-363 \,{\mathrm e}^{6}-{\mathrm e}^{18}\right ) \textit {\_R} +56100 \,{\mathrm e}^{12}+329604 \,{\mathrm e}^{6}+4076 \,{\mathrm e}^{18}+108 \,{\mathrm e}^{24}\right ) \ln \left (x -\textit {\_R} \right )}{-154+9 \textit {\_R} \,{\mathrm e}^{12}-6 \,{\mathrm e}^{12}-135 \textit {\_R}^{2} {\mathrm e}^{6}+102 \textit {\_R} \,{\mathrm e}^{6}+450 \textit {\_R}^{3}-80 \,{\mathrm e}^{6}-315 \textit {\_R}^{2}+709 \textit {\_R}}}{2 \left ({\mathrm e}^{12}+22 \,{\mathrm e}^{6}+121\right )^{2}}-\frac {{\mathrm e}^{18}+33 \,{\mathrm e}^{12}+363 \,{\mathrm e}^{6}+1331}{\left ({\mathrm e}^{12}+22 \,{\mathrm e}^{6}+121\right )^{2} x}\right )\) \(202\)

[In]

int(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4-12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*x^4-16
0*x^3+88*x^2)*exp(3)^2+225*x^6-210*x^5+709*x^4-308*x^3+484*x^2),x,method=_RETURNVERBOSE)

[Out]

3*exp(4)*(2/3-x)/(3*x*exp(6)-2*exp(6)-15*x^2+7*x-22)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {{\left (3 \, x - 2\right )} e^{4}}{15 \, x^{3} - 7 \, x^{2} - {\left (3 \, x^{2} - 2 \, x\right )} e^{6} + 22 \, x} \]

[In]

integrate(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4-12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*
x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-210*x^5+709*x^4-308*x^3+484*x^2),x, algorithm="fricas")

[Out]

(3*x - 2)*e^4/(15*x^3 - 7*x^2 - (3*x^2 - 2*x)*e^6 + 22*x)

Sympy [A] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=- \frac {- 3 x e^{4} + 2 e^{4}}{15 x^{3} + x^{2} \left (- 3 e^{6} - 7\right ) + x \left (22 + 2 e^{6}\right )} \]

[In]

integrate(((9*x**2-12*x+4)*exp(3)**2-90*x**3+111*x**2-28*x+44)*exp(4)/((9*x**4-12*x**3+4*x**2)*exp(3)**4+(-90*
x**5+102*x**4-160*x**3+88*x**2)*exp(3)**2+225*x**6-210*x**5+709*x**4-308*x**3+484*x**2),x)

[Out]

-(-3*x*exp(4) + 2*exp(4))/(15*x**3 + x**2*(-3*exp(6) - 7) + x*(22 + 2*exp(6)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {{\left (3 \, x - 2\right )} e^{4}}{15 \, x^{3} - x^{2} {\left (3 \, e^{6} + 7\right )} + 2 \, x {\left (e^{6} + 11\right )}} \]

[In]

integrate(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4-12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*
x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-210*x^5+709*x^4-308*x^3+484*x^2),x, algorithm="maxima")

[Out]

(3*x - 2)*e^4/(15*x^3 - x^2*(3*e^6 + 7) + 2*x*(e^6 + 11))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {{\left (3 \, x - 2\right )} e^{4}}{15 \, x^{3} - 3 \, x^{2} e^{6} - 7 \, x^{2} + 2 \, x e^{6} + 22 \, x} \]

[In]

integrate(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4-12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*
x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-210*x^5+709*x^4-308*x^3+484*x^2),x, algorithm="giac")

[Out]

(3*x - 2)*e^4/(15*x^3 - 3*x^2*e^6 - 7*x^2 + 2*x*e^6 + 22*x)

Mupad [B] (verification not implemented)

Time = 10.96 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=-\frac {2\,{\mathrm {e}}^4-3\,x\,{\mathrm {e}}^4}{15\,x^3+\left (-3\,{\mathrm {e}}^6-7\right )\,x^2+\left (2\,{\mathrm {e}}^6+22\right )\,x} \]

[In]

int((exp(4)*(exp(6)*(9*x^2 - 12*x + 4) - 28*x + 111*x^2 - 90*x^3 + 44))/(exp(12)*(4*x^2 - 12*x^3 + 9*x^4) + 48
4*x^2 - 308*x^3 + 709*x^4 - 210*x^5 + 225*x^6 + exp(6)*(88*x^2 - 160*x^3 + 102*x^4 - 90*x^5)),x)

[Out]

-(2*exp(4) - 3*x*exp(4))/(15*x^3 - x^2*(3*exp(6) + 7) + x*(2*exp(6) + 22))