3.23.74 \(\int \frac {-512+1568 x+1630 x^2+622 x^3+110 x^4+8 x^5}{256 x+224 x^2+81 x^3+14 x^4+x^5} \, dx\)

Optimal. Leaf size=29 \[ 2 \left (4 x+\log \left (\frac {e^{\frac {x}{x-(4+x)^2}}}{4 x}\right )\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 21, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 4, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2074, 638, 618, 204} \begin {gather*} -\frac {2 x}{x^2+7 x+16}+8 x-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-512 + 1568*x + 1630*x^2 + 622*x^3 + 110*x^4 + 8*x^5)/(256*x + 224*x^2 + 81*x^3 + 14*x^4 + x^5),x]

[Out]

8*x - (2*x)/(16 + 7*x + x^2) - 2*Log[x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (8-\frac {2}{x}-\frac {2 (32+7 x)}{\left (16+7 x+x^2\right )^2}+\frac {2}{16+7 x+x^2}\right ) \, dx\\ &=8 x-2 \log (x)-2 \int \frac {32+7 x}{\left (16+7 x+x^2\right )^2} \, dx+2 \int \frac {1}{16+7 x+x^2} \, dx\\ &=8 x-\frac {2 x}{16+7 x+x^2}-2 \log (x)-2 \int \frac {1}{16+7 x+x^2} \, dx-4 \operatorname {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,7+2 x\right )\\ &=8 x-\frac {2 x}{16+7 x+x^2}+\frac {4 \tan ^{-1}\left (\frac {7+2 x}{\sqrt {15}}\right )}{\sqrt {15}}-2 \log (x)+4 \operatorname {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,7+2 x\right )\\ &=8 x-\frac {2 x}{16+7 x+x^2}-2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.79 \begin {gather*} 2 \left (4 x-\frac {x}{16+7 x+x^2}-\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-512 + 1568*x + 1630*x^2 + 622*x^3 + 110*x^4 + 8*x^5)/(256*x + 224*x^2 + 81*x^3 + 14*x^4 + x^5),x]

[Out]

2*(4*x - x/(16 + 7*x + x^2) - Log[x])

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fricas [A]  time = 1.05, size = 38, normalized size = 1.31 \begin {gather*} \frac {2 \, {\left (4 \, x^{3} + 28 \, x^{2} - {\left (x^{2} + 7 \, x + 16\right )} \log \relax (x) + 63 \, x\right )}}{x^{2} + 7 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^5+110*x^4+622*x^3+1630*x^2+1568*x-512)/(x^5+14*x^4+81*x^3+224*x^2+256*x),x, algorithm="fricas")

[Out]

2*(4*x^3 + 28*x^2 - (x^2 + 7*x + 16)*log(x) + 63*x)/(x^2 + 7*x + 16)

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giac [A]  time = 0.17, size = 22, normalized size = 0.76 \begin {gather*} 8 \, x - \frac {2 \, x}{x^{2} + 7 \, x + 16} - 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^5+110*x^4+622*x^3+1630*x^2+1568*x-512)/(x^5+14*x^4+81*x^3+224*x^2+256*x),x, algorithm="giac")

[Out]

8*x - 2*x/(x^2 + 7*x + 16) - 2*log(abs(x))

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maple [A]  time = 0.04, size = 22, normalized size = 0.76

method	result	size
default	\(8 x -\frac {2 x}{x^{2}+7 x +16}-2 \ln \relax (x )\)	\(22\)
risch	\(8 x -\frac {2 x}{x^{2}+7 x +16}-2 \ln \relax (x )\)	\(22\)
norman	\(\frac {8 x^{3}-266 x -896}{x^{2}+7 x +16}-2 \ln \relax (x )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^5+110*x^4+622*x^3+1630*x^2+1568*x-512)/(x^5+14*x^4+81*x^3+224*x^2+256*x),x,method=_RETURNVERBOSE)

[Out]

8*x-2*x/(x^2+7*x+16)-2*ln(x)

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maxima [A]  time = 0.34, size = 21, normalized size = 0.72 \begin {gather*} 8 \, x - \frac {2 \, x}{x^{2} + 7 \, x + 16} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^5+110*x^4+622*x^3+1630*x^2+1568*x-512)/(x^5+14*x^4+81*x^3+224*x^2+256*x),x, algorithm="maxima")

[Out]

8*x - 2*x/(x^2 + 7*x + 16) - 2*log(x)

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mupad [B]  time = 1.27, size = 21, normalized size = 0.72 \begin {gather*} 8\,x-2\,\ln \relax (x)-\frac {2\,x}{x^2+7\,x+16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1568*x + 1630*x^2 + 622*x^3 + 110*x^4 + 8*x^5 - 512)/(256*x + 224*x^2 + 81*x^3 + 14*x^4 + x^5),x)

[Out]

8*x - 2*log(x) - (2*x)/(7*x + x^2 + 16)

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sympy [A]  time = 0.11, size = 19, normalized size = 0.66 \begin {gather*} 8 x - \frac {2 x}{x^{2} + 7 x + 16} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**5+110*x**4+622*x**3+1630*x**2+1568*x-512)/(x**5+14*x**4+81*x**3+224*x**2+256*x),x)

[Out]

8*x - 2*x/(x**2 + 7*x + 16) - 2*log(x)

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