3.1.93 \(\int \frac {1}{(2+5 x+3 x^2)^2} \, dx\) [93]

Optimal. Leaf size=34 \[ -\frac {5+6 x}{2+5 x+3 x^2}+6 \log (1+x)-6 \log (2+3 x) \]

[Out]

(-5-6*x)/(3*x^2+5*x+2)+6*ln(1+x)-6*ln(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {628, 630, 31} \begin {gather*} -\frac {6 x+5}{3 x^2+5 x+2}+6 \log (x+1)-6 \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 5*x + 3*x^2)^(-2),x]

[Out]

-((5 + 6*x)/(2 + 5*x + 3*x^2)) + 6*Log[1 + x] - 6*Log[2 + 3*x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 628

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 630

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

Rubi steps

\begin {align*} \int \frac {1}{\left (2+5 x+3 x^2\right )^2} \, dx &=-\frac {5+6 x}{2+5 x+3 x^2}-6 \int \frac {1}{2+5 x+3 x^2} \, dx\\ &=-\frac {5+6 x}{2+5 x+3 x^2}-18 \int \frac {1}{2+3 x} \, dx+18 \int \frac {1}{3+3 x} \, dx\\ &=-\frac {5+6 x}{2+5 x+3 x^2}+6 \log (1+x)-6 \log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 0.97 \begin {gather*} \frac {-5-6 x}{2+5 x+3 x^2}+6 \log (1+x)-6 \log (2+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 5*x + 3*x^2)^(-2),x]

[Out]

(-5 - 6*x)/(2 + 5*x + 3*x^2) + 6*Log[1 + x] - 6*Log[2 + 3*x]

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Maple [A]
time = 0.61, size = 32, normalized size = 0.94

method	result	size
default	\(-\frac {1}{x +1}+6 \ln \left (x +1\right )-\frac {3}{2+3 x}-6 \ln \left (2+3 x \right )\)	\(32\)
risch	\(\frac {-2 x -\frac {5}{3}}{x^{2}+\frac {5}{3} x +\frac {2}{3}}+6 \ln \left (x +1\right )-6 \ln \left (2+3 x \right )\)	\(32\)
norman	\(\frac {\frac {15}{2} x^{2}+\frac {13}{2} x}{3 x^{2}+5 x +2}+6 \ln \left (x +1\right )-6 \ln \left (2+3 x \right )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(x+1)+6*ln(x+1)-3/(2+3*x)-6*ln(2+3*x)

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Maxima [A]
time = 0.27, size = 34, normalized size = 1.00 \begin {gather*} -\frac {6 \, x + 5}{3 \, x^{2} + 5 \, x + 2} - 6 \, \log \left (3 \, x + 2\right ) + 6 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*x + 5)/(3*x^2 + 5*x + 2) - 6*log(3*x + 2) + 6*log(x + 1)

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Fricas [A]
time = 1.35, size = 53, normalized size = 1.56 \begin {gather*} -\frac {6 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 6 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (x + 1\right ) + 6 \, x + 5}{3 \, x^{2} + 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*(3*x^2 + 5*x + 2)*log(3*x + 2) - 6*(3*x^2 + 5*x + 2)*log(x + 1) + 6*x + 5)/(3*x^2 + 5*x + 2)

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Sympy [A]
time = 0.05, size = 31, normalized size = 0.91 \begin {gather*} \frac {- 6 x - 5}{3 x^{2} + 5 x + 2} - 6 \log {\left (x + \frac {2}{3} \right )} + 6 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x**2+5*x+2)**2,x)

[Out]

(-6*x - 5)/(3*x**2 + 5*x + 2) - 6*log(x + 2/3) + 6*log(x + 1)

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Giac [A]
time = 0.70, size = 36, normalized size = 1.06 \begin {gather*} -\frac {6 \, x + 5}{3 \, x^{2} + 5 \, x + 2} - 6 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + 6 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*x + 5)/(3*x^2 + 5*x + 2) - 6*log(abs(3*x + 2)) + 6*log(abs(x + 1))

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Mupad [B]
time = 0.20, size = 34, normalized size = 1.00 \begin {gather*} -6\,\ln \left (\frac {3\,x+2}{x+1}\right )-\frac {2\,\left (3\,x+\frac {5}{2}\right )}{3\,x^2+5\,x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x + 3*x^2 + 2)^2,x)

[Out]

- 6*log((3*x + 2)/(x + 1)) - (2*(3*x + 5/2))/(5*x + 3*x^2 + 2)

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