3.22.52 \(\int \frac {5030+2000 x+2096 x^2+e^4 (10240+25600 x+16000 x^2)}{10000+21800 x+7881 x^2-4360 x^3+400 x^4+e^4 (80000+174400 x+63048 x^2-34880 x^3+3200 x^4)} \, dx\)

Optimal. Leaf size=32 \[ \frac {6}{\left (4+32 e^4\right ) \left (5+\frac {4}{x}\right )}+\frac {5}{\frac {25}{4}-x} \]

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Rubi [A]  time = 0.12, antiderivative size = 45, normalized size of antiderivative = 1.41, number of steps used = 4, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1680, 12, 1814, 8} \begin {gather*} \frac {2 \left (2 \left (131+1000 e^4\right ) x+25 \left (5+64 e^4\right )\right )}{5 \left (1+8 e^4\right ) \left (-20 x^2+109 x+100\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5030 + 2000*x + 2096*x^2 + E^4*(10240 + 25600*x + 16000*x^2))/(10000 + 21800*x + 7881*x^2 - 4360*x^3 + 40
0*x^4 + E^4*(80000 + 174400*x + 63048*x^2 - 34880*x^3 + 3200*x^4)),x]

[Out]

(2*(25*(5 + 64*E^4) + 2*(131 + 1000*E^4)*x))/(5*(1 + 8*E^4)*(100 + 109*x - 20*x^2))

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 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 {64 \left (19881 \left (131+1000 e^4\right )+11280 \left (119+1000 e^4\right ) x+1600 \left (131+1000 e^4\right ) x^2\right )}{\left (1+8 e^4\right ) \left (19881-1600 x^2\right )^2} \, dx,x,\frac {-4360-34880 e^4}{4 \left (400+3200 e^4\right )}+x\right )\\ &=\frac {64 \operatorname {Subst}\left (\int \frac {19881 \left (131+1000 e^4\right )+11280 \left (119+1000 e^4\right ) x+1600 \left (131+1000 e^4\right ) x^2}{\left (19881-1600 x^2\right )^2} \, dx,x,\frac {-4360-34880 e^4}{4 \left (400+3200 e^4\right )}+x\right )}{1+8 e^4}\\ &=\frac {2 \left (25 \left (5+64 e^4\right )+2 \left (131+1000 e^4\right ) x\right )}{5 \left (1+8 e^4\right ) \left (100+109 x-20 x^2\right )}-\frac {32 \operatorname {Subst}\left (\int 0 \, dx,x,\frac {-4360-34880 e^4}{4 \left (400+3200 e^4\right )}+x\right )}{19881 \left (1+8 e^4\right )}\\ &=\frac {2 \left (25 \left (5+64 e^4\right )+2 \left (131+1000 e^4\right ) x\right )}{5 \left (1+8 e^4\right ) \left (100+109 x-20 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 38, normalized size = 1.19 \begin {gather*} \frac {2 \left (\frac {10+80 e^4}{25-4 x}-\frac {3}{5 (4+5 x)}\right )}{1+8 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5030 + 2000*x + 2096*x^2 + E^4*(10240 + 25600*x + 16000*x^2))/(10000 + 21800*x + 7881*x^2 - 4360*x^
3 + 400*x^4 + E^4*(80000 + 174400*x + 63048*x^2 - 34880*x^3 + 3200*x^4)),x]

[Out]

(2*((10 + 80*E^4)/(25 - 4*x) - 3/(5*(4 + 5*x))))/(1 + 8*E^4)

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fricas [A]  time = 0.59, size = 42, normalized size = 1.31 \begin {gather*} -\frac {2 \, {\left (400 \, {\left (5 \, x + 4\right )} e^{4} + 262 \, x + 125\right )}}{5 \, {\left (20 \, x^{2} + 8 \, {\left (20 \, x^{2} - 109 \, x - 100\right )} e^{4} - 109 \, x - 100\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16000*x^2+25600*x+10240)*exp(4)+2096*x^2+2000*x+5030)/((3200*x^4-34880*x^3+63048*x^2+174400*x+8000
0)*exp(4)+400*x^4-4360*x^3+7881*x^2+21800*x+10000),x, algorithm="fricas")

[Out]

-2/5*(400*(5*x + 4)*e^4 + 262*x + 125)/(20*x^2 + 8*(20*x^2 - 109*x - 100)*e^4 - 109*x - 100)

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giac [A]  time = 0.22, size = 36, normalized size = 1.12 \begin {gather*} -\frac {2 \, {\left (2000 \, x e^{4} + 262 \, x + 1600 \, e^{4} + 125\right )}}{5 \, {\left (20 \, x^{2} - 109 \, x - 100\right )} {\left (8 \, e^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16000*x^2+25600*x+10240)*exp(4)+2096*x^2+2000*x+5030)/((3200*x^4-34880*x^3+63048*x^2+174400*x+8000
0)*exp(4)+400*x^4-4360*x^3+7881*x^2+21800*x+10000),x, algorithm="giac")

[Out]

-2/5*(2000*x*e^4 + 262*x + 1600*e^4 + 125)/((20*x^2 - 109*x - 100)*(8*e^4 + 1))

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maple [A]  time = 0.08, size = 36, normalized size = 1.12

method	result	size
default	\(\frac {-\frac {2 \left (80 \,{\mathrm e}^{4}+10\right )}{4 x -25}-\frac {6}{5 \left (4+5 x \right )}}{8 \,{\mathrm e}^{4}+1}\)	\(36\)
risch	\(\frac {\left (-5 \,{\mathrm e}^{4}-\frac {131}{200}\right ) x -4 \,{\mathrm e}^{4}-\frac {5}{16}}{x^{2} {\mathrm e}^{4}-\frac {109 x \,{\mathrm e}^{4}}{20}+\frac {x^{2}}{8}-5 \,{\mathrm e}^{4}-\frac {109 x}{160}-\frac {5}{8}}\)	\(43\)
gosper	\(-\frac {2 \left (2000 x \,{\mathrm e}^{4}+1600 \,{\mathrm e}^{4}+262 x +125\right )}{5 \left (160 x^{2} {\mathrm e}^{4}-872 x \,{\mathrm e}^{4}+20 x^{2}-800 \,{\mathrm e}^{4}-109 x -100\right )}\)	\(45\)
norman	\(\frac {-\frac {5 \left (10+128 \,{\mathrm e}^{4}\right )}{8 \,{\mathrm e}^{4}+1}-\frac {\left (2096+16000 \,{\mathrm e}^{4}\right ) x}{20 \left (8 \,{\mathrm e}^{4}+1\right )}}{20 x^{2}-109 x -100}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16000*x^2+25600*x+10240)*exp(4)+2096*x^2+2000*x+5030)/((3200*x^4-34880*x^3+63048*x^2+174400*x+80000)*exp
(4)+400*x^4-4360*x^3+7881*x^2+21800*x+10000),x,method=_RETURNVERBOSE)

[Out]

2/(8*exp(4)+1)*(-(80*exp(4)+10)/(4*x-25)-3/5/(4+5*x))

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maxima [A]  time = 0.34, size = 45, normalized size = 1.41 \begin {gather*} -\frac {2 \, {\left (2 \, x {\left (1000 \, e^{4} + 131\right )} + 1600 \, e^{4} + 125\right )}}{5 \, {\left (20 \, x^{2} {\left (8 \, e^{4} + 1\right )} - 109 \, x {\left (8 \, e^{4} + 1\right )} - 800 \, e^{4} - 100\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16000*x^2+25600*x+10240)*exp(4)+2096*x^2+2000*x+5030)/((3200*x^4-34880*x^3+63048*x^2+174400*x+8000
0)*exp(4)+400*x^4-4360*x^3+7881*x^2+21800*x+10000),x, algorithm="maxima")

[Out]

-2/5*(2*x*(1000*e^4 + 131) + 1600*e^4 + 125)/(20*x^2*(8*e^4 + 1) - 109*x*(8*e^4 + 1) - 800*e^4 - 100)

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mupad [B]  time = 0.19, size = 27, normalized size = 0.84 \begin {gather*} -\frac {6}{5\,\left (8\,{\mathrm {e}}^4+1\right )\,\left (5\,x+4\right )}-\frac {20}{4\,x-25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2000*x + exp(4)*(25600*x + 16000*x^2 + 10240) + 2096*x^2 + 5030)/(21800*x + exp(4)*(174400*x + 63048*x^2
- 34880*x^3 + 3200*x^4 + 80000) + 7881*x^2 - 4360*x^3 + 400*x^4 + 10000),x)

[Out]

- 6/(5*(8*exp(4) + 1)*(5*x + 4)) - 20/(4*x - 25)

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sympy [B]  time = 0.67, size = 42, normalized size = 1.31 \begin {gather*} \frac {x \left (- 4000 e^{4} - 524\right ) - 3200 e^{4} - 250}{x^{2} \left (100 + 800 e^{4}\right ) + x \left (- 4360 e^{4} - 545\right ) - 4000 e^{4} - 500} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16000*x**2+25600*x+10240)*exp(4)+2096*x**2+2000*x+5030)/((3200*x**4-34880*x**3+63048*x**2+174400*x
+80000)*exp(4)+400*x**4-4360*x**3+7881*x**2+21800*x+10000),x)

[Out]

(x*(-4000*exp(4) - 524) - 3200*exp(4) - 250)/(x**2*(100 + 800*exp(4)) + x*(-4360*exp(4) - 545) - 4000*exp(4) -
500)

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