3.202 \(\int \frac{1}{(2+3 x+x^2)^5} \, dx\)

Optimal. Leaf size=87 \[ \frac{35 (2 x+3)}{x^2+3 x+2}-\frac{35 (2 x+3)}{6 \left (x^2+3 x+2\right )^2}+\frac{7 (2 x+3)}{6 \left (x^2+3 x+2\right )^3}-\frac{2 x+3}{4 \left (x^2+3 x+2\right )^4}+70 \log (x+1)-70 \log (x+2) \]

[Out]

-(3 + 2*x)/(4*(2 + 3*x + x^2)^4) + (7*(3 + 2*x))/(6*(2 + 3*x + x^2)^3) - (35*(3 + 2*x))/(6*(2 + 3*x + x^2)^2)
+ (35*(3 + 2*x))/(2 + 3*x + x^2) + 70*Log[1 + x] - 70*Log[2 + x]

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Rubi [A]  time = 0.0232128, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {614, 616, 31} \[ \frac{35 (2 x+3)}{x^2+3 x+2}-\frac{35 (2 x+3)}{6 \left (x^2+3 x+2\right )^2}+\frac{7 (2 x+3)}{6 \left (x^2+3 x+2\right )^3}-\frac{2 x+3}{4 \left (x^2+3 x+2\right )^4}+70 \log (x+1)-70 \log (x+2) \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x + x^2)^(-5),x]

[Out]

-(3 + 2*x)/(4*(2 + 3*x + x^2)^4) + (7*(3 + 2*x))/(6*(2 + 3*x + x^2)^3) - (35*(3 + 2*x))/(6*(2 + 3*x + x^2)^2)
+ (35*(3 + 2*x))/(2 + 3*x + x^2) + 70*Log[1 + x] - 70*Log[2 + x]

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 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 31

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

Rubi steps

\begin{align*} \int \frac{1}{\left (2+3 x+x^2\right )^5} \, dx &=-\frac{3+2 x}{4 \left (2+3 x+x^2\right )^4}-\frac{7}{2} \int \frac{1}{\left (2+3 x+x^2\right )^4} \, dx\\ &=-\frac{3+2 x}{4 \left (2+3 x+x^2\right )^4}+\frac{7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}+\frac{35}{3} \int \frac{1}{\left (2+3 x+x^2\right )^3} \, dx\\ &=-\frac{3+2 x}{4 \left (2+3 x+x^2\right )^4}+\frac{7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}-\frac{35 (3+2 x)}{6 \left (2+3 x+x^2\right )^2}-35 \int \frac{1}{\left (2+3 x+x^2\right )^2} \, dx\\ &=-\frac{3+2 x}{4 \left (2+3 x+x^2\right )^4}+\frac{7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}-\frac{35 (3+2 x)}{6 \left (2+3 x+x^2\right )^2}+\frac{35 (3+2 x)}{2+3 x+x^2}+70 \int \frac{1}{2+3 x+x^2} \, dx\\ &=-\frac{3+2 x}{4 \left (2+3 x+x^2\right )^4}+\frac{7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}-\frac{35 (3+2 x)}{6 \left (2+3 x+x^2\right )^2}+\frac{35 (3+2 x)}{2+3 x+x^2}+70 \int \frac{1}{1+x} \, dx-70 \int \frac{1}{2+x} \, dx\\ &=-\frac{3+2 x}{4 \left (2+3 x+x^2\right )^4}+\frac{7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}-\frac{35 (3+2 x)}{6 \left (2+3 x+x^2\right )^2}+\frac{35 (3+2 x)}{2+3 x+x^2}+70 \log (1+x)-70 \log (2+x)\\ \end{align*}

Mathematica [A]  time = 0.0246254, size = 87, normalized size = 1. \[ \frac{-2 x-3}{4 \left (x^2+3 x+2\right )^4}+\frac{35 (2 x+3)}{x^2+3 x+2}-\frac{35 (2 x+3)}{6 \left (x^2+3 x+2\right )^2}+\frac{7 (2 x+3)}{6 \left (x^2+3 x+2\right )^3}+70 \log (x+1)-70 \log (x+2) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x + x^2)^(-5),x]

[Out]

(-3 - 2*x)/(4*(2 + 3*x + x^2)^4) + (7*(3 + 2*x))/(6*(2 + 3*x + x^2)^3) - (35*(3 + 2*x))/(6*(2 + 3*x + x^2)^2)
+ (35*(3 + 2*x))/(2 + 3*x + x^2) + 70*Log[1 + x] - 70*Log[2 + x]

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Maple [A]  time = 0.01, size = 70, normalized size = 0.8 \begin{align*}{\frac{1}{4\, \left ( 2+x \right ) ^{4}}}+{\frac{5}{3\, \left ( 2+x \right ) ^{3}}}+{\frac{15}{2\, \left ( 2+x \right ) ^{2}}}+35\, \left ( 2+x \right ) ^{-1}-70\,\ln \left ( 2+x \right ) -{\frac{1}{4\, \left ( 1+x \right ) ^{4}}}+{\frac{5}{3\, \left ( 1+x \right ) ^{3}}}-{\frac{15}{2\, \left ( 1+x \right ) ^{2}}}+35\, \left ( 1+x \right ) ^{-1}+70\,\ln \left ( 1+x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/(2+x)^4+5/3/(2+x)^3+15/2/(2+x)^2+35/(2+x)-70*ln(2+x)-1/4/(1+x)^4+5/3/(1+x)^3-15/2/(1+x)^2+35/(1+x)+70*ln(1
+x)

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Maxima [A]  time = 0.928019, size = 122, normalized size = 1.4 \begin{align*} \frac{840 \, x^{7} + 8820 \, x^{6} + 38920 \, x^{5} + 93450 \, x^{4} + 131768 \, x^{3} + 109116 \, x^{2} + 49176 \, x + 9315}{12 \,{\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )}} - 70 \, \log \left (x + 2\right ) + 70 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(840*x^7 + 8820*x^6 + 38920*x^5 + 93450*x^4 + 131768*x^3 + 109116*x^2 + 49176*x + 9315)/(x^8 + 12*x^7 + 6
2*x^6 + 180*x^5 + 321*x^4 + 360*x^3 + 248*x^2 + 96*x + 16) - 70*log(x + 2) + 70*log(x + 1)

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Fricas [B]  time = 1.82358, size = 481, normalized size = 5.53 \begin{align*} \frac{840 \, x^{7} + 8820 \, x^{6} + 38920 \, x^{5} + 93450 \, x^{4} + 131768 \, x^{3} + 109116 \, x^{2} - 840 \,{\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )} \log \left (x + 2\right ) + 840 \,{\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )} \log \left (x + 1\right ) + 49176 \, x + 9315}{12 \,{\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(840*x^7 + 8820*x^6 + 38920*x^5 + 93450*x^4 + 131768*x^3 + 109116*x^2 - 840*(x^8 + 12*x^7 + 62*x^6 + 180*
x^5 + 321*x^4 + 360*x^3 + 248*x^2 + 96*x + 16)*log(x + 2) + 840*(x^8 + 12*x^7 + 62*x^6 + 180*x^5 + 321*x^4 + 3
60*x^3 + 248*x^2 + 96*x + 16)*log(x + 1) + 49176*x + 9315)/(x^8 + 12*x^7 + 62*x^6 + 180*x^5 + 321*x^4 + 360*x^
3 + 248*x^2 + 96*x + 16)

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Sympy [A]  time = 0.193934, size = 88, normalized size = 1.01 \begin{align*} \frac{840 x^{7} + 8820 x^{6} + 38920 x^{5} + 93450 x^{4} + 131768 x^{3} + 109116 x^{2} + 49176 x + 9315}{12 x^{8} + 144 x^{7} + 744 x^{6} + 2160 x^{5} + 3852 x^{4} + 4320 x^{3} + 2976 x^{2} + 1152 x + 192} + 70 \log{\left (x + 1 \right )} - 70 \log{\left (x + 2 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(840*x**7 + 8820*x**6 + 38920*x**5 + 93450*x**4 + 131768*x**3 + 109116*x**2 + 49176*x + 9315)/(12*x**8 + 144*x
**7 + 744*x**6 + 2160*x**5 + 3852*x**4 + 4320*x**3 + 2976*x**2 + 1152*x + 192) + 70*log(x + 1) - 70*log(x + 2)

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Giac [A]  time = 1.05658, size = 84, normalized size = 0.97 \begin{align*} \frac{840 \, x^{7} + 8820 \, x^{6} + 38920 \, x^{5} + 93450 \, x^{4} + 131768 \, x^{3} + 109116 \, x^{2} + 49176 \, x + 9315}{12 \,{\left (x^{2} + 3 \, x + 2\right )}^{4}} - 70 \, \log \left ({\left | x + 2 \right |}\right ) + 70 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(840*x^7 + 8820*x^6 + 38920*x^5 + 93450*x^4 + 131768*x^3 + 109116*x^2 + 49176*x + 9315)/(x^2 + 3*x + 2)^4
- 70*log(abs(x + 2)) + 70*log(abs(x + 1))