3.11.54 \(\int \frac {-16+32 x-8 x^2+16 x^3-x^4+2 x^5+e (48 x-12 x^3)}{16 x+8 x^3+x^5} \, dx\)

Optimal. Leaf size=23 \[ 2 x-3 e \left (1-\frac {4 x}{4+x^2}\right )-\log (x) \]

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Rubi [A]  time = 0.09, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {1594, 28, 1805, 1586, 43} \begin {gather*} \frac {12 e x}{x^2+4}+2 x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 32*x - 8*x^2 + 16*x^3 - x^4 + 2*x^5 + E*(48*x - 12*x^3))/(16*x + 8*x^3 + x^5),x]

[Out]

2*x + (12*E*x)/(4 + x^2) - Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1805

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16+32 x-8 x^2+16 x^3-x^4+2 x^5+e \left (48 x-12 x^3\right )}{x \left (16+8 x^2+x^4\right )} \, dx\\ &=\int \frac {-16+32 x-8 x^2+16 x^3-x^4+2 x^5+e \left (48 x-12 x^3\right )}{x \left (4+x^2\right )^2} \, dx\\ &=\frac {12 e x}{4+x^2}-\frac {1}{8} \int \frac {32-64 x+8 x^2-16 x^3}{x \left (4+x^2\right )} \, dx\\ &=\frac {12 e x}{4+x^2}-\frac {1}{8} \int \frac {8-16 x}{x} \, dx\\ &=\frac {12 e x}{4+x^2}-\frac {1}{8} \int \left (-16+\frac {8}{x}\right ) \, dx\\ &=2 x+\frac {12 e x}{4+x^2}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.83 \begin {gather*} 2 x+\frac {12 e x}{4+x^2}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 32*x - 8*x^2 + 16*x^3 - x^4 + 2*x^5 + E*(48*x - 12*x^3))/(16*x + 8*x^3 + x^5),x]

[Out]

2*x + (12*E*x)/(4 + x^2) - Log[x]

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fricas [A]  time = 0.71, size = 31, normalized size = 1.35 \begin {gather*} \frac {2 \, x^{3} + 12 \, x e - {\left (x^{2} + 4\right )} \log \relax (x) + 8 \, x}{x^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+48*x)*exp(1)+2*x^5-x^4+16*x^3-8*x^2+32*x-16)/(x^5+8*x^3+16*x),x, algorithm="fricas")

[Out]

(2*x^3 + 12*x*e - (x^2 + 4)*log(x) + 8*x)/(x^2 + 4)

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giac [A]  time = 0.29, size = 21, normalized size = 0.91 \begin {gather*} 2 \, x + \frac {12 \, x e}{x^{2} + 4} - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+48*x)*exp(1)+2*x^5-x^4+16*x^3-8*x^2+32*x-16)/(x^5+8*x^3+16*x),x, algorithm="giac")

[Out]

2*x + 12*x*e/(x^2 + 4) - log(abs(x))

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maple [A]  time = 0.05, size = 21, normalized size = 0.91

method	result	size
default	\(2 x -\ln \relax (x )+\frac {12 \,{\mathrm e} x}{x^{2}+4}\)	\(21\)
risch	\(2 x -\ln \relax (x )+\frac {12 \,{\mathrm e} x}{x^{2}+4}\)	\(21\)
norman	\(\frac {\left (8+12 \,{\mathrm e}\right ) x +2 x^{3}}{x^{2}+4}-\ln \relax (x )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*x^3+48*x)*exp(1)+2*x^5-x^4+16*x^3-8*x^2+32*x-16)/(x^5+8*x^3+16*x),x,method=_RETURNVERBOSE)

[Out]

2*x-ln(x)+12*exp(1)*x/(x^2+4)

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maxima [A]  time = 0.43, size = 20, normalized size = 0.87 \begin {gather*} 2 \, x + \frac {12 \, x e}{x^{2} + 4} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+48*x)*exp(1)+2*x^5-x^4+16*x^3-8*x^2+32*x-16)/(x^5+8*x^3+16*x),x, algorithm="maxima")

[Out]

2*x + 12*x*e/(x^2 + 4) - log(x)

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mupad [B]  time = 0.08, size = 20, normalized size = 0.87 \begin {gather*} 2\,x-\ln \relax (x)+\frac {12\,x\,\mathrm {e}}{x^2+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + exp(1)*(48*x - 12*x^3) - 8*x^2 + 16*x^3 - x^4 + 2*x^5 - 16)/(16*x + 8*x^3 + x^5),x)

[Out]

2*x - log(x) + (12*x*exp(1))/(x^2 + 4)

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sympy [A]  time = 0.22, size = 17, normalized size = 0.74 \begin {gather*} 2 x + \frac {12 e x}{x^{2} + 4} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x**3+48*x)*exp(1)+2*x**5-x**4+16*x**3-8*x**2+32*x-16)/(x**5+8*x**3+16*x),x)

[Out]

2*x + 12*E*x/(x**2 + 4) - log(x)

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