3.80.52 \(\int \frac {2 x+4 e x+2 e^2 x-15 x^2-25 x^4-12 x^6}{1+2 x^2+x^4} \, dx\)

Optimal. Leaf size=25 \[ x^2 \left (-4 x+\frac {(1+e)^2-x}{1+x^2}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 5, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6, 28, 1811, 1804, 1149} \begin {gather*} -4 x^3+\frac {\left ((1+e)^2 x+1\right ) x}{x^2+1}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x + 4*E*x + 2*E^2*x - 15*x^2 - 25*x^4 - 12*x^6)/(1 + 2*x^2 + x^4),x]

[Out]

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

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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1149

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p +
q)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2
, 0] && IntegerQ[p]

Rule 1804

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

Rule 1811

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient[Pq, x, x]*(a + b*x^2)^p, x] /; Fre
eQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[Coeff[Pq, x, 0], 0] &&  !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^2 x+(2+4 e) x-15 x^2-25 x^4-12 x^6}{1+2 x^2+x^4} \, dx\\ &=\int \frac {\left (2+4 e+2 e^2\right ) x-15 x^2-25 x^4-12 x^6}{1+2 x^2+x^4} \, dx\\ &=\int \frac {\left (2+4 e+2 e^2\right ) x-15 x^2-25 x^4-12 x^6}{\left (1+x^2\right )^2} \, dx\\ &=\int \frac {x \left (2+4 e+2 e^2-15 x-25 x^3-12 x^5\right )}{\left (1+x^2\right )^2} \, dx\\ &=\frac {x \left (1+(1+e)^2 x\right )}{1+x^2}-\frac {1}{2} \int \frac {2+26 x^2+24 x^4}{1+x^2} \, dx\\ &=\frac {x \left (1+(1+e)^2 x\right )}{1+x^2}-\frac {1}{2} \int \left (2+24 x^2\right ) \, dx\\ &=-x-4 x^3+\frac {x \left (1+(1+e)^2 x\right )}{1+x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.12 \begin {gather*} -x-4 x^3+\frac {-1-2 e-e^2+x}{1+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + 4*E*x + 2*E^2*x - 15*x^2 - 25*x^4 - 12*x^6)/(1 + 2*x^2 + x^4),x]

[Out]

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

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fricas [A]  time = 0.58, size = 27, normalized size = 1.08 \begin {gather*} -\frac {4 \, x^{5} + 5 \, x^{3} + e^{2} + 2 \, e + 1}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)^2+4*x*exp(1)-12*x^6-25*x^4-15*x^2+2*x)/(x^4+2*x^2+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 28, normalized size = 1.12 \begin {gather*} -4 \, x^{3} - x + \frac {x - e^{2} - 2 \, e - 1}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)^2+4*x*exp(1)-12*x^6-25*x^4-15*x^2+2*x)/(x^4+2*x^2+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 29, normalized size = 1.16

method	result	size
default	\(-4 x^{3}-x +\frac {x -{\mathrm e}^{2}-2 \,{\mathrm e}-1}{x^{2}+1}\)	\(29\)
risch	\(-4 x^{3}-x +\frac {x -{\mathrm e}^{2}-2 \,{\mathrm e}-1}{x^{2}+1}\)	\(29\)
gosper	\(-\frac {4 x^{5}+5 x^{3}+{\mathrm e}^{2}+2 \,{\mathrm e}+1}{x^{2}+1}\)	\(30\)
norman	\(\frac {-4 x^{5}-1-5 x^{3}-{\mathrm e}^{2}-2 \,{\mathrm e}}{x^{2}+1}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(1)^2+4*x*exp(1)-12*x^6-25*x^4-15*x^2+2*x)/(x^4+2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 28, normalized size = 1.12 \begin {gather*} -4 \, x^{3} - x + \frac {x - e^{2} - 2 \, e - 1}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)^2+4*x*exp(1)-12*x^6-25*x^4-15*x^2+2*x)/(x^4+2*x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 27, normalized size = 1.08 \begin {gather*} -\frac {4\,x^5+5\,x^3+2\,\mathrm {e}+{\mathrm {e}}^2+1}{x^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 4*x*exp(1) + 2*x*exp(2) - 15*x^2 - 25*x^4 - 12*x^6)/(2*x^2 + x^4 + 1),x)

[Out]

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

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sympy [A]  time = 0.34, size = 24, normalized size = 0.96 \begin {gather*} - 4 x^{3} - x - \frac {- x + 1 + 2 e + e^{2}}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)**2+4*x*exp(1)-12*x**6-25*x**4-15*x**2+2*x)/(x**4+2*x**2+1),x)

[Out]

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

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