3.56.4 \(\int \frac {e^9 (-10-100 x^3)+e^{10} (-25-50 x-250 x^3-125 x^4)}{4 e^8 x^2+e^9 (20 x^2+20 x^3)+e^{10} (25 x^2+50 x^3+25 x^4)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1-5 x^3}{x \left (1+\frac {2}{5 e}+x\right )} \]

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Rubi [A]  time = 0.18, antiderivative size = 38, normalized size of antiderivative = 1.65, number of steps used = 5, number of rules used = 5, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {1680, 12, 1814, 21, 8} \begin {gather*} \frac {5 e^2-(2+5 e)^2 x}{e x (5 e x+5 e+2)}-5 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^9*(-10 - 100*x^3) + E^10*(-25 - 50*x - 250*x^3 - 125*x^4))/(4*E^8*x^2 + E^9*(20*x^2 + 20*x^3) + E^10*(2
5*x^2 + 50*x^3 + 25*x^4)),x]

[Out]

-5*x + (5*E^2 - (2 + 5*E)^2*x)/(E*x*(2 + 5*E + 5*E*x))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {5 \left (3 (2+5 e)^4-80 e \left (8+60 e+150 e^2+175 e^3\right ) x+600 e^2 (2+5 e)^2 x^2-10000 e^4 x^4\right )}{\left ((2+5 e)^2-100 e^2 x^2\right )^2} \, dx,x,\frac {20 e^9+50 e^{10}}{100 e^{10}}+x\right )\\ &=5 \operatorname {Subst}\left (\int \frac {3 (2+5 e)^4-80 e \left (8+60 e+150 e^2+175 e^3\right ) x+600 e^2 (2+5 e)^2 x^2-10000 e^4 x^4}{\left ((2+5 e)^2-100 e^2 x^2\right )^2} \, dx,x,\frac {20 e^9+50 e^{10}}{100 e^{10}}+x\right )\\ &=\frac {5 e^2-(2+5 e)^2 x}{e x (2+5 e+5 e x)}-\frac {5 \operatorname {Subst}\left (\int \frac {2 (2+5 e)^4-200 e^2 (2+5 e)^2 x^2}{(2+5 e)^2-100 e^2 x^2} \, dx,x,\frac {20 e^9+50 e^{10}}{100 e^{10}}+x\right )}{2 (2+5 e)^2}\\ &=\frac {5 e^2-(2+5 e)^2 x}{e x (2+5 e+5 e x)}-5 \operatorname {Subst}\left (\int 1 \, dx,x,\frac {20 e^9+50 e^{10}}{100 e^{10}}+x\right )\\ &=-5 x+\frac {5 e^2-(2+5 e)^2 x}{e x (2+5 e+5 e x)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 49, normalized size = 2.13 \begin {gather*} -\frac {4 x+10 e x (2+x)+5 e^2 \left (-1+5 x+5 x^2+5 x^3\right )}{e x (2+5 e (1+x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^9*(-10 - 100*x^3) + E^10*(-25 - 50*x - 250*x^3 - 125*x^4))/(4*E^8*x^2 + E^9*(20*x^2 + 20*x^3) + E
^10*(25*x^2 + 50*x^3 + 25*x^4)),x]

[Out]

-((4*x + 10*E*x*(2 + x) + 5*E^2*(-1 + 5*x + 5*x^2 + 5*x^3))/(E*x*(2 + 5*E*(1 + x))))

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fricas [B]  time = 1.71, size = 53, normalized size = 2.30 \begin {gather*} -\frac {5 \, {\left (5 \, x^{3} + 5 \, x^{2} + 5 \, x - 1\right )} e^{2} + 10 \, {\left (x^{2} + 2 \, x\right )} e + 4 \, x}{5 \, {\left (x^{2} + x\right )} e^{2} + 2 \, x e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x^4-250*x^3-50*x-25)*exp(5)^2+(-100*x^3-10)*exp(4)*exp(5))/((25*x^4+50*x^3+25*x^2)*exp(5)^2+(
20*x^3+20*x^2)*exp(4)*exp(5)+4*x^2*exp(4)^2),x, algorithm="fricas")

[Out]

-(5*(5*x^3 + 5*x^2 + 5*x - 1)*e^2 + 10*(x^2 + 2*x)*e + 4*x)/(5*(x^2 + x)*e^2 + 2*x*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x^4-250*x^3-50*x-25)*exp(5)^2+(-100*x^3-10)*exp(4)*exp(5))/((25*x^4+50*x^3+25*x^2)*exp(5)^2+(
20*x^3+20*x^2)*exp(4)*exp(5)+4*x^2*exp(4)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -5*(25*sageVARx*exp(10)*1/25/exp(10)+(-2
0*exp(8)*exp(10)-8*exp(8)*exp(9)-125*exp(10)^2-150*exp(10)*exp(9)-40*exp(9)^2)/(4*exp(8)+25*exp(10)+20*exp(9))
^2/sageVARx+(-20*exp(

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maple [A]  time = 0.19, size = 31, normalized size = 1.35

method	result	size
gosper	\(-\frac {5 \left (5 x^{3}-1\right ) {\mathrm e}^{5}}{x \left (5 x \,{\mathrm e}^{5}+5 \,{\mathrm e}^{5}+2 \,{\mathrm e}^{4}\right )}\)	\(31\)
norman	\(\frac {-25 x^{3} {\mathrm e}^{5}+5 \,{\mathrm e}^{5}}{x \left (5 x \,{\mathrm e}^{5}+5 \,{\mathrm e}^{5}+2 \,{\mathrm e}^{4}\right )}\)	\(33\)
risch	\(-5 x +\frac {-\left (4+25 \,{\mathrm e}^{2}+20 \,{\mathrm e}\right ) {\mathrm e}^{-1} x +5 \,{\mathrm e}}{\left (5 x \,{\mathrm e}+5 \,{\mathrm e}+2\right ) x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-125*x^4-250*x^3-50*x-25)*exp(5)^2+(-100*x^3-10)*exp(4)*exp(5))/((25*x^4+50*x^3+25*x^2)*exp(5)^2+(20*x^3
+20*x^2)*exp(4)*exp(5)+4*x^2*exp(4)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 44, normalized size = 1.91 \begin {gather*} -5 \, x - \frac {x {\left (25 \, e^{2} + 20 \, e + 4\right )} - 5 \, e^{2}}{5 \, x^{2} e^{2} + x {\left (5 \, e^{2} + 2 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x^4-250*x^3-50*x-25)*exp(5)^2+(-100*x^3-10)*exp(4)*exp(5))/((25*x^4+50*x^3+25*x^2)*exp(5)^2+(
20*x^3+20*x^2)*exp(4)*exp(5)+4*x^2*exp(4)^2),x, algorithm="maxima")

[Out]

-5*x - (x*(25*e^2 + 20*e + 4) - 5*e^2)/(5*x^2*e^2 + x*(5*e^2 + 2*e))

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mupad [B]  time = 3.69, size = 39, normalized size = 1.70 \begin {gather*} \frac {5\,\mathrm {e}-x\,\left (4\,{\mathrm {e}}^{-1}+25\,\mathrm {e}+20\right )}{x\,\left (5\,\mathrm {e}+5\,x\,\mathrm {e}+2\right )}-5\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(9)*(100*x^3 + 10) + exp(10)*(50*x + 250*x^3 + 125*x^4 + 25))/(exp(9)*(20*x^2 + 20*x^3) + 4*x^2*exp(8
) + exp(10)*(25*x^2 + 50*x^3 + 25*x^4)),x)

[Out]

(5*exp(1) - x*(4*exp(-1) + 25*exp(1) + 20))/(x*(5*exp(1) + 5*x*exp(1) + 2)) - 5*x

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sympy [B]  time = 0.89, size = 42, normalized size = 1.83 \begin {gather*} - 5 x - \frac {x \left (4 + 20 e + 25 e^{2}\right ) - 5 e^{2}}{5 x^{2} e^{2} + x \left (2 e + 5 e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x**4-250*x**3-50*x-25)*exp(5)**2+(-100*x**3-10)*exp(4)*exp(5))/((25*x**4+50*x**3+25*x**2)*exp
(5)**2+(20*x**3+20*x**2)*exp(4)*exp(5)+4*x**2*exp(4)**2),x)

[Out]

-5*x - (x*(4 + 20*E + 25*exp(2)) - 5*exp(2))/(5*x**2*exp(2) + x*(2*E + 5*exp(2)))

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