3.22.49 \(\int \frac {(5-x) (3+2 x)}{(2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=36 \[ -\frac {139 x+121}{3 \left (3 x^2+5 x+2\right )}+47 \log (x+1)-47 \log (3 x+2) \]

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {777, 616, 31} \begin {gather*} -\frac {139 x+121}{3 \left (3 x^2+5 x+2\right )}+47 \log (x+1)-47 \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((5 - x)*(3 + 2*x))/(2 + 5*x + 3*x^2)^2,x]

[Out]

-(121 + 139*x)/(3*(2 + 5*x + 3*x^2)) + 47*Log[1 + x] - 47*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 616

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]

Rule 777

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

Rubi steps

\begin {align*} \int \frac {(5-x) (3+2 x)}{\left (2+5 x+3 x^2\right )^2} \, dx &=-\frac {121+139 x}{3 \left (2+5 x+3 x^2\right )}-47 \int \frac {1}{2+5 x+3 x^2} \, dx\\ &=-\frac {121+139 x}{3 \left (2+5 x+3 x^2\right )}-141 \int \frac {1}{2+3 x} \, dx+141 \int \frac {1}{3+3 x} \, dx\\ &=-\frac {121+139 x}{3 \left (2+5 x+3 x^2\right )}+47 \log (1+x)-47 \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.94 \begin {gather*} -\frac {139 x+121}{9 x^2+15 x+6}+47 \log (x+1)-47 \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((5 - x)*(3 + 2*x))/(2 + 5*x + 3*x^2)^2,x]

[Out]

-((121 + 139*x)/(6 + 15*x + 9*x^2)) + 47*Log[1 + x] - 47*Log[2 + 3*x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(5-x) (3+2 x)}{\left (2+5 x+3 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((5 - x)*(3 + 2*x))/(2 + 5*x + 3*x^2)^2,x]

[Out]

IntegrateAlgebraic[((5 - x)*(3 + 2*x))/(2 + 5*x + 3*x^2)^2, x]

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fricas [A]  time = 0.38, size = 53, normalized size = 1.47 \begin {gather*} -\frac {141 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 141 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (x + 1\right ) + 139 \, x + 121}{3 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(141*(3*x^2 + 5*x + 2)*log(3*x + 2) - 141*(3*x^2 + 5*x + 2)*log(x + 1) + 139*x + 121)/(3*x^2 + 5*x + 2)

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giac [A]  time = 0.16, size = 36, normalized size = 1.00 \begin {gather*} -\frac {139 \, x + 121}{3 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} - 47 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + 47 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(139*x + 121)/(3*x^2 + 5*x + 2) - 47*log(abs(3*x + 2)) + 47*log(abs(x + 1))

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maple [A]  time = 0.05, size = 32, normalized size = 0.89 \begin {gather*} -47 \ln \left (3 x +2\right )+47 \ln \left (x +1\right )-\frac {85}{3 \left (3 x +2\right )}-\frac {6}{x +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)*(2*x+3)/(3*x^2+5*x+2)^2,x)

[Out]

-85/3/(3*x+2)-47*ln(3*x+2)-6/(x+1)+47*ln(x+1)

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maxima [A]  time = 0.61, size = 34, normalized size = 0.94 \begin {gather*} -\frac {139 \, x + 121}{3 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} - 47 \, \log \left (3 \, x + 2\right ) + 47 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(139*x + 121)/(3*x^2 + 5*x + 2) - 47*log(3*x + 2) + 47*log(x + 1)

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mupad [B]  time = 2.27, size = 26, normalized size = 0.72 \begin {gather*} 94\,\mathrm {atanh}\left (6\,x+5\right )-\frac {\frac {139\,x}{9}+\frac {121}{9}}{x^2+\frac {5\,x}{3}+\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x + 3)*(x - 5))/(5*x + 3*x^2 + 2)^2,x)

[Out]

94*atanh(6*x + 5) - ((139*x)/9 + 121/9)/((5*x)/3 + x^2 + 2/3)

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sympy [A]  time = 0.13, size = 29, normalized size = 0.81 \begin {gather*} - \frac {139 x + 121}{9 x^{2} + 15 x + 6} - 47 \log {\left (x + \frac {2}{3} \right )} + 47 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(139*x + 121)/(9*x**2 + 15*x + 6) - 47*log(x + 2/3) + 47*log(x + 1)

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