3.101.49 \(\int \frac {4+32 x+15 x^2-150 x^3-125 x^4}{1+6 x+27 x^2+68 x^3+135 x^4+150 x^5+125 x^6} \, dx\)

Optimal. Leaf size=18 \[ \frac {x}{\left (x+\frac {1}{x+\log \left (e^{2+4 x}\right )}\right )^2} \]

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Rubi [B]  time = 0.13, antiderivative size = 62, normalized size of antiderivative = 3.44, number of steps used = 11, number of rules used = 5, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2074, 638, 614, 618, 204} \begin {gather*} -\frac {15 (5 x+1)}{8 \left (5 x^2+2 x+1\right )}+\frac {115 x+31}{8 \left (5 x^2+2 x+1\right )}-\frac {5 x+2}{\left (5 x^2+2 x+1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 32*x + 15*x^2 - 150*x^3 - 125*x^4)/(1 + 6*x + 27*x^2 + 68*x^3 + 135*x^4 + 150*x^5 + 125*x^6),x]

[Out]

-((2 + 5*x)/(1 + 2*x + 5*x^2)^2) - (15*(1 + 5*x))/(8*(1 + 2*x + 5*x^2)) + (31 + 115*x)/(8*(1 + 2*x + 5*x^2))

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 614

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

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 (\frac {4 (-3+5 x)}{\left (1+2 x+5 x^2\right )^3}+\frac {21-10 x}{\left (1+2 x+5 x^2\right )^2}-\frac {5}{1+2 x+5 x^2}\right ) \, dx\\ &=4 \int \frac {-3+5 x}{\left (1+2 x+5 x^2\right )^3} \, dx-5 \int \frac {1}{1+2 x+5 x^2} \, dx+\int \frac {21-10 x}{\left (1+2 x+5 x^2\right )^2} \, dx\\ &=-\frac {2+5 x}{\left (1+2 x+5 x^2\right )^2}+\frac {31+115 x}{8 \left (1+2 x+5 x^2\right )}+10 \operatorname {Subst}\left (\int \frac {1}{-16-x^2} \, dx,x,2+10 x\right )+\frac {115}{8} \int \frac {1}{1+2 x+5 x^2} \, dx-15 \int \frac {1}{\left (1+2 x+5 x^2\right )^2} \, dx\\ &=-\frac {2+5 x}{\left (1+2 x+5 x^2\right )^2}-\frac {15 (1+5 x)}{8 \left (1+2 x+5 x^2\right )}+\frac {31+115 x}{8 \left (1+2 x+5 x^2\right )}-\frac {5}{2} \tan ^{-1}\left (\frac {1}{2} (1+5 x)\right )-\frac {75}{8} \int \frac {1}{1+2 x+5 x^2} \, dx-\frac {115}{4} \operatorname {Subst}\left (\int \frac {1}{-16-x^2} \, dx,x,2+10 x\right )\\ &=-\frac {2+5 x}{\left (1+2 x+5 x^2\right )^2}-\frac {15 (1+5 x)}{8 \left (1+2 x+5 x^2\right )}+\frac {31+115 x}{8 \left (1+2 x+5 x^2\right )}+\frac {75}{16} \tan ^{-1}\left (\frac {1}{2} (1+5 x)\right )+\frac {75}{4} \operatorname {Subst}\left (\int \frac {1}{-16-x^2} \, dx,x,2+10 x\right )\\ &=-\frac {2+5 x}{\left (1+2 x+5 x^2\right )^2}-\frac {15 (1+5 x)}{8 \left (1+2 x+5 x^2\right )}+\frac {31+115 x}{8 \left (1+2 x+5 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.17 \begin {gather*} \frac {x (2+5 x)^2}{\left (1+2 x+5 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 32*x + 15*x^2 - 150*x^3 - 125*x^4)/(1 + 6*x + 27*x^2 + 68*x^3 + 135*x^4 + 150*x^5 + 125*x^6),x]

[Out]

(x*(2 + 5*x)^2)/(1 + 2*x + 5*x^2)^2

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fricas [B]  time = 0.78, size = 37, normalized size = 2.06 \begin {gather*} \frac {25 \, x^{3} + 20 \, x^{2} + 4 \, x}{25 \, x^{4} + 20 \, x^{3} + 14 \, x^{2} + 4 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-125*x^4-150*x^3+15*x^2+32*x+4)/(125*x^6+150*x^5+135*x^4+68*x^3+27*x^2+6*x+1),x, algorithm="fricas"
)

[Out]

(25*x^3 + 20*x^2 + 4*x)/(25*x^4 + 20*x^3 + 14*x^2 + 4*x + 1)

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giac [A]  time = 0.17, size = 27, normalized size = 1.50 \begin {gather*} \frac {25 \, x^{3} + 20 \, x^{2} + 4 \, x}{{\left (5 \, x^{2} + 2 \, x + 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-125*x^4-150*x^3+15*x^2+32*x+4)/(125*x^6+150*x^5+135*x^4+68*x^3+27*x^2+6*x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 28, normalized size = 1.56

method	result	size
norman	\(\frac {25 x^{3}+20 x^{2}+4 x}{\left (5 x^{2}+2 x +1\right )^{2}}\)	\(28\)
default	\(-\frac {25 \left (-x^{3}-\frac {4}{5} x^{2}-\frac {4}{25} x \right )}{\left (5 x^{2}+2 x +1\right )^{2}}\)	\(29\)
gosper	\(\frac {x \left (5 x +2\right )^{2}}{25 x^{4}+20 x^{3}+14 x^{2}+4 x +1}\)	\(32\)
risch	\(\frac {x^{3}+\frac {4}{5} x^{2}+\frac {4}{25} x}{x^{4}+\frac {4}{5} x^{3}+\frac {14}{25} x^{2}+\frac {4}{25} x +\frac {1}{25}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-125*x^4-150*x^3+15*x^2+32*x+4)/(125*x^6+150*x^5+135*x^4+68*x^3+27*x^2+6*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.36, size = 37, normalized size = 2.06 \begin {gather*} \frac {25 \, x^{3} + 20 \, x^{2} + 4 \, x}{25 \, x^{4} + 20 \, x^{3} + 14 \, x^{2} + 4 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-125*x^4-150*x^3+15*x^2+32*x+4)/(125*x^6+150*x^5+135*x^4+68*x^3+27*x^2+6*x+1),x, algorithm="maxima"
)

[Out]

(25*x^3 + 20*x^2 + 4*x)/(25*x^4 + 20*x^3 + 14*x^2 + 4*x + 1)

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mupad [B]  time = 7.71, size = 21, normalized size = 1.17 \begin {gather*} \frac {x\,{\left (5\,x+2\right )}^2}{{\left (5\,x^2+2\,x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + 15*x^2 - 150*x^3 - 125*x^4 + 4)/(6*x + 27*x^2 + 68*x^3 + 135*x^4 + 150*x^5 + 125*x^6 + 1),x)

[Out]

(x*(5*x + 2)^2)/(2*x + 5*x^2 + 1)^2

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sympy [B]  time = 0.12, size = 36, normalized size = 2.00 \begin {gather*} - \frac {- 25 x^{3} - 20 x^{2} - 4 x}{25 x^{4} + 20 x^{3} + 14 x^{2} + 4 x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-125*x**4-150*x**3+15*x**2+32*x+4)/(125*x**6+150*x**5+135*x**4+68*x**3+27*x**2+6*x+1),x)

[Out]

-(-25*x**3 - 20*x**2 - 4*x)/(25*x**4 + 20*x**3 + 14*x**2 + 4*x + 1)

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