3.44.93 \(\int \frac {4 x+4 x^2+e^{5+e} (4 x+4 x^2)}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx\)

Optimal. Leaf size=21 \[ \frac {2 \left (1+e^{5+e}\right )}{4+\frac {1}{x^2}+\frac {2}{x}} \]

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Rubi [A]  time = 0.06, antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {1680, 12, 1814, 8} \begin {gather*} -\frac {2 \left (1+e^{5+e}\right ) (2 x+1)}{16 \left (x+\frac {1}{4}\right )^2+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + 4*x^2 + E^(5 + E)*(4*x + 4*x^2))/(1 + 4*x + 12*x^2 + 16*x^3 + 16*x^4),x]

[Out]

(-2*(1 + E^(5 + E))*(1 + 2*x))/(3 + 16*(1/4 + 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 {4 \left (1+e^{5+e}\right ) \left (-3+8 x+16 x^2\right )}{\left (3+16 x^2\right )^2} \, dx,x,\frac {1}{4}+x\right )\\ &=\left (4 \left (1+e^{5+e}\right )\right ) \operatorname {Subst}\left (\int \frac {-3+8 x+16 x^2}{\left (3+16 x^2\right )^2} \, dx,x,\frac {1}{4}+x\right )\\ &=-\frac {2 \left (1+e^{5+e}\right ) (1+2 x)}{3+(1+4 x)^2}-\frac {1}{3} \left (2 \left (1+e^{5+e}\right )\right ) \operatorname {Subst}\left (\int 0 \, dx,x,\frac {1}{4}+x\right )\\ &=-\frac {2 \left (1+e^{5+e}\right ) (1+2 x)}{3+(1+4 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.33 \begin {gather*} \frac {\left (1+e^{5+e}\right ) (-1-2 x)}{2 \left (1+2 x+4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + 4*x^2 + E^(5 + E)*(4*x + 4*x^2))/(1 + 4*x + 12*x^2 + 16*x^3 + 16*x^4),x]

[Out]

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

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fricas [A]  time = 0.51, size = 30, normalized size = 1.43 \begin {gather*} -\frac {{\left (2 \, x + 1\right )} e^{\left (e + 5\right )} + 2 \, x + 1}{2 \, {\left (4 \, x^{2} + 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+4*x)*exp(exp(1)+5)+4*x^2+4*x)/(16*x^4+16*x^3+12*x^2+4*x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 32, normalized size = 1.52 \begin {gather*} -\frac {2 \, x e^{\left (e + 5\right )} + 2 \, x + e^{\left (e + 5\right )} + 1}{2 \, {\left (4 \, x^{2} + 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+4*x)*exp(exp(1)+5)+4*x^2+4*x)/(16*x^4+16*x^3+12*x^2+4*x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 26, normalized size = 1.24

method	result	size
default	\(\frac {\left (4 \,{\mathrm e}^{{\mathrm e}+5}+4\right ) \left (-\frac {x}{16}-\frac {1}{32}\right )}{x^{2}+\frac {1}{2} x +\frac {1}{4}}\)	\(26\)
gosper	\(-\frac {\left (2 x +1\right ) \left ({\mathrm e}^{{\mathrm e}+5}+1\right )}{2 \left (4 x^{2}+2 x +1\right )}\)	\(27\)
risch	\(\frac {\left (-\frac {{\mathrm e}^{{\mathrm e}+5}}{4}-\frac {1}{4}\right ) x -\frac {{\mathrm e}^{{\mathrm e}+5}}{8}-\frac {1}{8}}{x^{2}+\frac {1}{2} x +\frac {1}{4}}\)	\(32\)
norman	\(\frac {\left (-{\mathrm e}^{{\mathrm e}} {\mathrm e}^{5}-1\right ) x -\frac {{\mathrm e}^{{\mathrm e}} {\mathrm e}^{5}}{2}-\frac {1}{2}}{4 x^{2}+2 x +1}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+4*x)*exp(exp(1)+5)+4*x^2+4*x)/(16*x^4+16*x^3+12*x^2+4*x+1),x,method=_RETURNVERBOSE)

[Out]

(4*exp(exp(1)+5)+4)*(-1/16*x-1/32)/(x^2+1/2*x+1/4)

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maxima [A]  time = 0.36, size = 31, normalized size = 1.48 \begin {gather*} -\frac {2 \, x {\left (e^{\left (e + 5\right )} + 1\right )} + e^{\left (e + 5\right )} + 1}{2 \, {\left (4 \, x^{2} + 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+4*x)*exp(exp(1)+5)+4*x^2+4*x)/(16*x^4+16*x^3+12*x^2+4*x+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.09, size = 26, normalized size = 1.24 \begin {gather*} -\frac {\left (2\,x+1\right )\,\left ({\mathrm {e}}^{\mathrm {e}+5}+1\right )}{2\,\left (4\,x^2+2\,x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + 4*x^2 + exp(exp(1) + 5)*(4*x + 4*x^2))/(4*x + 12*x^2 + 16*x^3 + 16*x^4 + 1),x)

[Out]

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

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sympy [A]  time = 0.34, size = 34, normalized size = 1.62 \begin {gather*} \frac {x \left (- 2 e^{5} e^{e} - 2\right ) - e^{5} e^{e} - 1}{8 x^{2} + 4 x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+4*x)*exp(exp(1)+5)+4*x**2+4*x)/(16*x**4+16*x**3+12*x**2+4*x+1),x)

[Out]

(x*(-2*exp(5)*exp(E) - 2) - exp(5)*exp(E) - 1)/(8*x**2 + 4*x + 2)

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