\(\int \frac {1}{(2+3 x+x^2)^5} \, dx\) [202]

Optimal result

Integrand size = 10, antiderivative size = 87 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {-3-2 x}{4 \left (2+3 x+x^2\right )^4}+\frac {7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}-\frac {35 (3+2 x)}{6 \left (2+3 x+x^2\right )^2}+\frac {35 (3+2 x)}{2+3 x+x^2}+70 \log (1+x)-70 \log (2+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {628, 630, 31} \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {35 (2 x+3)}{x^2+3 x+2}-\frac {35 (2 x+3)}{6 \left (x^2+3 x+2\right )^2}+\frac {7 (2 x+3)}{6 \left (x^2+3 x+2\right )^3}-\frac {2 x+3}{4 \left (x^2+3 x+2\right )^4}+70 \log (x+1)-70 \log (x+2) \]

[In]

Int[(2 + 3*x + x^2)^(-5),x]

[Out]

-1/4*(3 + 2*x)/(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 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*} \text {integral}& = -\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] (verified)

Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {-3-2 x}{4 \left (2+3 x+x^2\right )^4}+\frac {7 (3+2 x)}{6 \left (2+3 x+x^2\right )^3}-\frac {35 (3+2 x)}{6 \left (2+3 x+x^2\right )^2}+\frac {35 (3+2 x)}{2+3 x+x^2}+70 \log (1+x)-70 \log (2+x) \]

[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]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69

method	result	size
norman	\(\frac {70 x^{7}+735 x^{6}+4098 x +9093 x^{2}+\frac {9730}{3} x^{5}+\frac {15575}{2} x^{4}+\frac {32942}{3} x^{3}+\frac {3105}{4}}{\left (x^{2}+3 x +2\right )^{4}}+70 \ln \left (1+x \right )-70 \ln \left (2+x \right )\)	\(60\)
risch	\(\frac {70 x^{7}+735 x^{6}+4098 x +9093 x^{2}+\frac {9730}{3} x^{5}+\frac {15575}{2} x^{4}+\frac {32942}{3} x^{3}+\frac {3105}{4}}{\left (x^{2}+3 x +2\right )^{4}}+70 \ln \left (1+x \right )-70 \ln \left (2+x \right )\)	\(60\)
default	\(\frac {1}{4 \left (2+x \right )^{4}}+\frac {5}{3 \left (2+x \right )^{3}}+\frac {15}{2 \left (2+x \right )^{2}}+\frac {35}{2+x}-70 \ln \left (2+x \right )-\frac {1}{4 \left (1+x \right )^{4}}+\frac {5}{3 \left (1+x \right )^{3}}-\frac {15}{2 \left (1+x \right )^{2}}+\frac {35}{1+x}+70 \ln \left (1+x \right )\)	\(70\)
parallelrisch	\(\frac {9315+8820 x^{6}+49176 x +38920 x^{5}+13440 \ln \left (1+x \right )-13440 \ln \left (2+x \right )+93450 x^{4}+131768 x^{3}+109116 x^{2}+840 x^{7}-80640 \ln \left (2+x \right ) x +52080 \ln \left (1+x \right ) x^{6}-52080 \ln \left (2+x \right ) x^{6}+151200 \ln \left (1+x \right ) x^{5}+840 \ln \left (1+x \right ) x^{8}-840 \ln \left (2+x \right ) x^{8}+10080 \ln \left (1+x \right ) x^{7}-10080 \ln \left (2+x \right ) x^{7}-151200 \ln \left (2+x \right ) x^{5}-269640 \ln \left (2+x \right ) x^{4}+269640 \ln \left (1+x \right ) x^{4}+302400 \ln \left (1+x \right ) x^{3}-302400 \ln \left (2+x \right ) x^{3}-208320 \ln \left (2+x \right ) x^{2}+208320 \ln \left (1+x \right ) x^{2}+80640 \ln \left (1+x \right ) x}{12 \left (x^{2}+3 x +2\right )^{4}}\)	\(200\)

[In]

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

[Out]

(70*x^7+735*x^6+4098*x+9093*x^2+9730/3*x^5+15575/2*x^4+32942/3*x^3+3105/4)/(x^2+3*x+2)^4+70*ln(1+x)-70*ln(2+x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (81) = 162\).

Time = 0.25 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {840 \, x^{7} + 8820 \, x^{6} + 38920 \, x^{5} + 93450 \, x^{4} + 131768 \, x^{3} + 109116 \, x^{2} - 840 \, {\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )} \log \left (x + 2\right ) + 840 \, {\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )} \log \left (x + 1\right ) + 49176 \, x + 9315}{12 \, {\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {840 x^{7} + 8820 x^{6} + 38920 x^{5} + 93450 x^{4} + 131768 x^{3} + 109116 x^{2} + 49176 x + 9315}{12 x^{8} + 144 x^{7} + 744 x^{6} + 2160 x^{5} + 3852 x^{4} + 4320 x^{3} + 2976 x^{2} + 1152 x + 192} + 70 \log {\left (x + 1 \right )} - 70 \log {\left (x + 2 \right )} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {840 \, x^{7} + 8820 \, x^{6} + 38920 \, x^{5} + 93450 \, x^{4} + 131768 \, x^{3} + 109116 \, x^{2} + 49176 \, x + 9315}{12 \, {\left (x^{8} + 12 \, x^{7} + 62 \, x^{6} + 180 \, x^{5} + 321 \, x^{4} + 360 \, x^{3} + 248 \, x^{2} + 96 \, x + 16\right )}} - 70 \, \log \left (x + 2\right ) + 70 \, \log \left (x + 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=\frac {840 \, x^{7} + 8820 \, x^{6} + 38920 \, x^{5} + 93450 \, x^{4} + 131768 \, x^{3} + 109116 \, x^{2} + 49176 \, x + 9315}{12 \, {\left (x^{2} + 3 \, x + 2\right )}^{4}} - 70 \, \log \left ({\left | x + 2 \right |}\right ) + 70 \, \log \left ({\left | x + 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (2+3 x+x^2\right )^5} \, dx=70\,\ln \left (\frac {x+1}{x+2}\right )+70\,\left (x+\frac {3}{2}\right )\,\left (\frac {1}{x^2+3\,x+2}-\frac {1}{6\,{\left (x^2+3\,x+2\right )}^2}+\frac {1}{30\,{\left (x^2+3\,x+2\right )}^3}-\frac {1}{140\,{\left (x^2+3\,x+2\right )}^4}\right ) \]

[In]

int(1/(3*x + x^2 + 2)^5,x)

[Out]

70*log((x + 1)/(x + 2)) + 70*(x + 3/2)*(1/(3*x + x^2 + 2) - 1/(6*(3*x + x^2 + 2)^2) + 1/(30*(3*x + x^2 + 2)^3)
- 1/(140*(3*x + x^2 + 2)^4))