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

Optimal. Leaf size=30 \[ \frac {e^4}{x \left (1-e^6+5 x+\frac {16}{-\frac {4}{3}+2 x}\right )} \]

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Rubi [A]  time = 0.29, antiderivative size = 57, normalized size of antiderivative = 1.90, number of steps used = 8, number of rules used = 5, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2074, 638, 618, 206} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 206

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

Rule 618

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 638

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)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), 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 2074

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 {gather*} \begin {aligned} \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 ) \operatorname {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 ) \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 33, normalized size = 1.10 \begin {gather*} \frac {e^4 (-2+3 x)}{x \left (22+e^6 (2-3 x)-7 x+15 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.80, size = 37, normalized size = 1.23 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [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(((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]

Timed out

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maple [A]  time = 0.25, size = 34, normalized size = 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	\(-\frac {\left (3 x -2\right ) {\mathrm e}^{4}}{x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\)	\(38\)
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	\({\mathrm e}^{4} \left (-\frac {\munderset {\textit {\_R} =\RootOf \left (225 \textit {\_Z}^{4}+\left (-90 \,{\mathrm e}^{6}-210\right ) \textit {\_Z}^{3}+\left (102 \,{\mathrm e}^{6}+9 \,{\mathrm e}^{12}+709\right ) \textit {\_Z}^{2}+\left (-160 \,{\mathrm e}^{6}-12 \,{\mathrm e}^{12}-308\right ) \textit {\_Z} +88 \,{\mathrm e}^{6}+4 \,{\mathrm e}^{12}+484\right )}{\sum }\frac {\left (681472+225 \left (-1331-363 \,{\mathrm e}^{6}-33 \,{\mathrm e}^{12}-{\mathrm e}^{18}\right ) \textit {\_R}^{2}+780 \left (-1331-363 \,{\mathrm e}^{6}-33 \,{\mathrm e}^{12}-{\mathrm e}^{18}\right ) \textit {\_R} +329604 \,{\mathrm e}^{6}+56100 \,{\mathrm e}^{12}+4076 \,{\mathrm e}^{18}+108 \,{\mathrm e}^{24}\right ) \ln \left (x -\textit {\_R} \right )}{154+135 \textit {\_R}^{2} {\mathrm e}^{6}-450 \textit {\_R}^{3}-102 \textit {\_R} \,{\mathrm e}^{6}-9 \textit {\_R} \,{\mathrm e}^{12}+315 \textit {\_R}^{2}+80 \,{\mathrm e}^{6}+6 \,{\mathrm e}^{12}-709 \textit {\_R}}}{2 \left (22 \,{\mathrm e}^{6}+{\mathrm e}^{12}+121\right )^{2}}-\frac {1331+363 \,{\mathrm e}^{6}+33 \,{\mathrm e}^{12}+{\mathrm e}^{18}}{\left (22 \,{\mathrm e}^{6}+{\mathrm e}^{12}+121\right )^{2} x}\right )\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.35, size = 34, normalized size = 1.13 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.39, size = 39, normalized size = 1.30 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 1.58, size = 37, normalized size = 1.23 \begin {gather*} - \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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