3.1.94 \(\int \frac {1}{(2+5 x-3 x^2)^2} \, dx\) [94]

Optimal. Leaf size=42 \[ -\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (1+3 x) \]

[Out]

1/49*(-5+6*x)/(-3*x^2+5*x+2)-6/343*ln(2-x)+6/343*ln(1+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 42, 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 {5-6 x}{49 \left (-3 x^2+5 x+2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (3 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 5*x - 3*x^2)^(-2),x]

[Out]

-1/49*(5 - 6*x)/(2 + 5*x - 3*x^2) - (6*Log[2 - x])/343 + (6*Log[1 + 3*x])/343

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}{49 \left (2+5 x-3 x^2\right )}+\frac {6}{49} \int \frac {1}{2+5 x-3 x^2} \, dx\\ &=-\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}-\frac {18}{343} \int \frac {1}{-1-3 x} \, dx+\frac {18}{343} \int \frac {1}{6-3 x} \, dx\\ &=-\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (1+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 42, normalized size = 1.00 \begin {gather*} \frac {5-6 x}{49 \left (-2-5 x+3 x^2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (1+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 5*x - 3*x^2)^(-2),x]

[Out]

(5 - 6*x)/(49*(-2 - 5*x + 3*x^2)) - (6*Log[2 - x])/343 + (6*Log[1 + 3*x])/343

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Maple [A]
time = 0.53, size = 32, normalized size = 0.76

method	result	size
default	\(-\frac {1}{49 \left (x -2\right )}-\frac {6 \ln \left (x -2\right )}{343}-\frac {3}{49 \left (3 x +1\right )}+\frac {6 \ln \left (3 x +1\right )}{343}\)	\(32\)
risch	\(\frac {-\frac {2 x}{49}+\frac {5}{147}}{x^{2}-\frac {5}{3} x -\frac {2}{3}}-\frac {6 \ln \left (x -2\right )}{343}+\frac {6 \ln \left (3 x +1\right )}{343}\)	\(32\)
norman	\(\frac {\frac {15}{98} x^{2}-\frac {37}{98} x}{3 x^{2}-5 x -2}-\frac {6 \ln \left (x -2\right )}{343}+\frac {6 \ln \left (3 x +1\right )}{343}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/49/(x-2)-6/343*ln(x-2)-3/49/(3*x+1)+6/343*ln(3*x+1)

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Maxima [A]
time = 0.28, size = 34, normalized size = 0.81 \begin {gather*} -\frac {6 \, x - 5}{49 \, {\left (3 \, x^{2} - 5 \, x - 2\right )}} + \frac {6}{343} \, \log \left (3 \, x + 1\right ) - \frac {6}{343} \, \log \left (x - 2\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]

-1/49*(6*x - 5)/(3*x^2 - 5*x - 2) + 6/343*log(3*x + 1) - 6/343*log(x - 2)

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Fricas [A]
time = 2.29, size = 53, normalized size = 1.26 \begin {gather*} \frac {6 \, {\left (3 \, x^{2} - 5 \, x - 2\right )} \log \left (3 \, x + 1\right ) - 6 \, {\left (3 \, x^{2} - 5 \, x - 2\right )} \log \left (x - 2\right ) - 42 \, x + 35}{343 \, {\left (3 \, x^{2} - 5 \, x - 2\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="fricas")

[Out]

1/343*(6*(3*x^2 - 5*x - 2)*log(3*x + 1) - 6*(3*x^2 - 5*x - 2)*log(x - 2) - 42*x + 35)/(3*x^2 - 5*x - 2)

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Sympy [A]
time = 0.06, size = 32, normalized size = 0.76 \begin {gather*} \frac {5 - 6 x}{147 x^{2} - 245 x - 98} - \frac {6 \log {\left (x - 2 \right )}}{343} + \frac {6 \log {\left (x + \frac {1}{3} \right )}}{343} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5 - 6*x)/(147*x**2 - 245*x - 98) - 6*log(x - 2)/343 + 6*log(x + 1/3)/343

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Giac [A]
time = 0.66, size = 36, normalized size = 0.86 \begin {gather*} -\frac {6 \, x - 5}{49 \, {\left (3 \, x^{2} - 5 \, x - 2\right )}} + \frac {6}{343} \, \log \left ({\left | 3 \, x + 1 \right |}\right ) - \frac {6}{343} \, \log \left ({\left | x - 2 \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]

-1/49*(6*x - 5)/(3*x^2 - 5*x - 2) + 6/343*log(abs(3*x + 1)) - 6/343*log(abs(x - 2))

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Mupad [B]
time = 0.08, size = 34, normalized size = 0.81 \begin {gather*} \frac {6\,\ln \left (\frac {3\,x+1}{x-2}\right )}{343}+\frac {2\,\left (3\,x-\frac {5}{2}\right )}{49\,\left (-3\,x^2+5\,x+2\right )} \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 + 1)/(x - 2)))/343 + (2*(3*x - 5/2))/(49*(5*x - 3*x^2 + 2))

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