∫ 12x2+3e2x2−10x3+4x4+e3(8x+2e2x−5x2)__ 16+e4−40x+57x2−40x3+16x4+e2(8−10x+8x2) dx

3.70.53 \(\int \frac {12 x^2+3 e^2 x^2-10 x^3+4 x^4+e^3 (8 x+2 e^2 x-5 x^2)}{16+e^4-40 x+57 x^2-40 x^3+16 x^4+e^2 (8-10 x+8 x^2)} \, dx\)

Optimal. Leaf size=24 \[ \frac {x^2 \left (e^3+x\right )}{4+e^2-5 x+4 x^2} \]

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Rubi [B] time = 0.17, antiderivative size = 64, normalized size of antiderivative = 2.67, number of steps used = 6, number of rules used = 6, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6, 1680, 12, 1814, 21, 8} \begin {gather*} \frac {x}{4}-\frac {-8 \left (9-4 e^2+20 e^3\right ) \left (x-\frac {5}{8}\right )+32 e^5+28 e^3+60 e^2+115}{8 \left (64 \left (x-\frac {5}{8}\right )^2+16 e^2+39\right )} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(12*x^2 + 3*E^2*x^2 - 10*x^3 + 4*x^4 + E^3*(8*x + 2*E^2*x - 5*x^2))/(16 + E^4 - 40*x + 57*x^2 - 40*x^3 + 1

6*x^4 + E^2*(8 - 10*x + 8*x^2)),x]

[Out]

-1/8*(115 + 60*E^2 + 28*E^3 + 32*E^5 - 8*(9 - 4*E^2 + 20*E^3)*(-5/8 + x))/(39 + 16*E^2 + 64*(-5/8 + x)^2) + x/

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x

] && !FreeQ[v, 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} &=\int \frac {\left (12+3 e^2\right ) x^2-10 x^3+4 x^4+e^3 \left (8 x+2 e^2 x-5 x^2\right )}{16+e^4-40 x+57 x^2-40 x^3+16 x^4+e^2 \left (8-10 x+8 x^2\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {5 \left (39+16 e^2\right ) \left (15+16 e^3\right )+64 \left (115+60 e^2+28 e^3+32 e^5\right ) x+128 \left (21+24 e^2-40 e^3\right ) x^2+4096 x^4}{4 \left (39+16 e^2+64 x^2\right )^2} \, dx,x,-\frac {5}{8}+x\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {5 \left (39+16 e^2\right ) \left (15+16 e^3\right )+64 \left (115+60 e^2+28 e^3+32 e^5\right ) x+128 \left (21+24 e^2-40 e^3\right ) x^2+4096 x^4}{\left (39+16 e^2+64 x^2\right )^2} \, dx,x,-\frac {5}{8}+x\right )\\ &=-\frac {115+60 e^2+28 e^3+32 e^5+\left (9-4 e^2+20 e^3\right ) (5-8 x)}{8 \left (39+16 e^2+(-5+8 x)^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 \left (39+16 e^2\right )^2-128 \left (39+16 e^2\right ) x^2}{39+16 e^2+64 x^2} \, dx,x,-\frac {5}{8}+x\right )}{8 \left (39+16 e^2\right )}\\ &=-\frac {115+60 e^2+28 e^3+32 e^5+\left (9-4 e^2+20 e^3\right ) (5-8 x)}{8 \left (39+16 e^2+(-5+8 x)^2\right )}+\frac {1}{4} \operatorname {Subst}\left (\int 1 \, dx,x,-\frac {5}{8}+x\right )\\ &=\frac {x}{4}-\frac {115+60 e^2+28 e^3+32 e^5+\left (9-4 e^2+20 e^3\right ) (5-8 x)}{8 \left (39+16 e^2+(-5+8 x)^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B] time = 0.03, size = 54, normalized size = 2.25 \begin {gather*} \frac {-20-5 e^2-4 e^5+25 x-20 x^2+16 x^3+4 e^3 (-4+5 x)}{16 \left (4+e^2-5 x+4 x^2\right )} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(12*x^2 + 3*E^2*x^2 - 10*x^3 + 4*x^4 + E^3*(8*x + 2*E^2*x - 5*x^2))/(16 + E^4 - 40*x + 57*x^2 - 40*x

^3 + 16*x^4 + E^2*(8 - 10*x + 8*x^2)),x]

[Out]

(-20 - 5*E^2 - 4*E^5 + 25*x - 20*x^2 + 16*x^3 + 4*E^3*(-4 + 5*x))/(16*(4 + E^2 - 5*x + 4*x^2))

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fricas [B] time = 0.76, size = 48, normalized size = 2.00 \begin {gather*} \frac {16 \, x^{3} - 20 \, x^{2} + 4 \, {\left (5 \, x - 4\right )} e^{3} + 25 \, x - 4 \, e^{5} - 5 \, e^{2} - 20}{16 \, {\left (4 \, x^{2} - 5 \, x + e^{2} + 4\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-5*x^2+8*x)*exp(3)+3*x^2*exp(2)+4*x^4-10*x^3+12*x^2)/(exp(2)^2+(8*x^2-10*x+8)*exp(2)+16*

x^4-40*x^3+57*x^2-40*x+16),x, algorithm="fricas")

[Out]

1/16*(16*x^3 - 20*x^2 + 4*(5*x - 4)*e^3 + 25*x - 4*e^5 - 5*e^2 - 20)/(4*x^2 - 5*x + e^2 + 4)

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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, x^{4} - 10 \, x^{3} + 3 \, x^{2} e^{2} + 12 \, x^{2} - {\left (5 \, x^{2} - 2 \, x e^{2} - 8 \, x\right )} e^{3}}{16 \, x^{4} - 40 \, x^{3} + 57 \, x^{2} + 2 \, {\left (4 \, x^{2} - 5 \, x + 4\right )} e^{2} - 40 \, x + e^{4} + 16}\,{d x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-5*x^2+8*x)*exp(3)+3*x^2*exp(2)+4*x^4-10*x^3+12*x^2)/(exp(2)^2+(8*x^2-10*x+8)*exp(2)+16*

x^4-40*x^3+57*x^2-40*x+16),x, algorithm="giac")

[Out]

integrate((4*x^4 - 10*x^3 + 3*x^2*e^2 + 12*x^2 - (5*x^2 - 2*x*e^2 - 8*x)*e^3)/(16*x^4 - 40*x^3 + 57*x^2 + 2*(4

*x^2 - 5*x + 4)*e^2 - 40*x + e^4 + 16), x)

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maple [A] time = 0.08, size = 35, normalized size = 1.46


method	result	size

norman	\(\frac {x^{3}+\frac {5 x \,{\mathrm e}^{3}}{4}-\frac {{\mathrm e}^{2} {\mathrm e}^{3}}{4}-{\mathrm e}^{3}}{4+4 x^{2}-5 x +{\mathrm e}^{2}}\)	\(35\)
gosper	\(-\frac {-4 x^{3}+{\mathrm e}^{2} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3}+4 \,{\mathrm e}^{3}}{4 \left (4+4 x^{2}-5 x +{\mathrm e}^{2}\right )}\)	\(37\)
risch	\(\frac {x}{4}+\frac {\frac {\left (5 \,{\mathrm e}^{3}-{\mathrm e}^{2}+\frac {9}{4}\right ) x}{4}-\frac {{\mathrm e}^{5}}{4}-{\mathrm e}^{3}-\frac {5 \,{\mathrm e}^{2}}{16}-\frac {5}{4}}{4+4 x^{2}-5 x +{\mathrm e}^{2}}\)	\(47\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(2)*x-5*x^2+8*x)*exp(3)+3*x^2*exp(2)+4*x^4-10*x^3+12*x^2)/(exp(2)^2+(8*x^2-10*x+8)*exp(2)+16*x^4-40

*x^3+57*x^2-40*x+16),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B] time = 0.39, size = 46, normalized size = 1.92 \begin {gather*} \frac {1}{4} \, x + \frac {x {\left (20 \, e^{3} - 4 \, e^{2} + 9\right )} - 4 \, e^{5} - 16 \, e^{3} - 5 \, e^{2} - 20}{16 \, {\left (4 \, x^{2} - 5 \, x + e^{2} + 4\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-5*x^2+8*x)*exp(3)+3*x^2*exp(2)+4*x^4-10*x^3+12*x^2)/(exp(2)^2+(8*x^2-10*x+8)*exp(2)+16*

x^4-40*x^3+57*x^2-40*x+16),x, algorithm="maxima")

[Out]

1/4*x + 1/16*(x*(20*e^3 - 4*e^2 + 9) - 4*e^5 - 16*e^3 - 5*e^2 - 20)/(4*x^2 - 5*x + e^2 + 4)

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mupad [B] time = 4.28, size = 47, normalized size = 1.96 \begin {gather*} \frac {x}{4}-\frac {\frac {5\,{\mathrm {e}}^2}{4}+4\,{\mathrm {e}}^3+{\mathrm {e}}^5-x\,\left (5\,{\mathrm {e}}^3-{\mathrm {e}}^2+\frac {9}{4}\right )+5}{16\,x^2-20\,x+4\,{\mathrm {e}}^2+16} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(8*x + 2*x*exp(2) - 5*x^2) + 3*x^2*exp(2) + 12*x^2 - 10*x^3 + 4*x^4)/(exp(4) - 40*x + exp(2)*(8*x^

2 - 10*x + 8) + 57*x^2 - 40*x^3 + 16*x^4 + 16),x)

[Out]

x/4 - ((5*exp(2))/4 + 4*exp(3) + exp(5) - x*(5*exp(3) - exp(2) + 9/4) + 5)/(4*exp(2) - 20*x + 16*x^2 + 16)

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sympy [B] time = 1.43, size = 48, normalized size = 2.00 \begin {gather*} \frac {x}{4} + \frac {x \left (- 4 e^{2} + 9 + 20 e^{3}\right ) - 4 e^{5} - 16 e^{3} - 5 e^{2} - 20}{64 x^{2} - 80 x + 64 + 16 e^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-5*x**2+8*x)*exp(3)+3*x**2*exp(2)+4*x**4-10*x**3+12*x**2)/(exp(2)**2+(8*x**2-10*x+8)*exp

(2)+16*x**4-40*x**3+57*x**2-40*x+16),x)

[Out]

x/4 + (x*(-4*exp(2) + 9 + 20*exp(3)) - 4*exp(5) - 16*exp(3) - 5*exp(2) - 20)/(64*x**2 - 80*x + 64 + 16*exp(2))

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