\(\int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{(1+x^2)^2 (2+x+x^2)^2} \, dx\) [277]

Optimal result

Integrand size = 44, antiderivative size = 33 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {3}{1+x^2}+\frac {1}{2+x+x^2}+\log \left (1+x^2\right )-\log \left (2+x+x^2\right ) \]

[Out]

-3/(x^2+1)+1/(x^2+x+2)+ln(x^2+1)-ln(x^2+x+2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6874, 267, 266, 643, 642} \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {3}{x^2+1}+\frac {1}{x^2+x+2}+\log \left (x^2+1\right )-\log \left (x^2+x+2\right ) \]

[In]

Int[(-3 + 25*x + 23*x^2 + 32*x^3 + 15*x^4 + 7*x^5 + x^6)/((1 + x^2)^2*(2 + x + x^2)^2),x]

[Out]

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

Rule 266

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {3}{1+x^2}+\frac {1}{2+x+x^2}+\log \left (1+x^2\right )-\log \left (2+x+x^2\right ) \]

[In]

Integrate[(-3 + 25*x + 23*x^2 + 32*x^3 + 15*x^4 + 7*x^5 + x^6)/((1 + x^2)^2*(2 + x + x^2)^2),x]

[Out]

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

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03

method	result	size
default	\(-\frac {3}{x^{2}+1}+\frac {1}{x^{2}+x +2}+\ln \left (x^{2}+1\right )-\ln \left (x^{2}+x +2\right )\)	\(34\)
norman	\(\frac {-2 x^{2}-3 x -5}{\left (x^{2}+x +2\right ) \left (x^{2}+1\right )}-\ln \left (x^{2}+x +2\right )+\ln \left (x^{2}+1\right )\)	\(43\)
risch	\(\frac {-2 x^{2}-3 x -5}{\left (x^{2}+x +2\right ) \left (x^{2}+1\right )}-\ln \left (x^{2}+x +2\right )+\ln \left (x^{2}+1\right )\)	\(43\)
parallelrisch	\(\frac {\ln \left (x^{2}+1\right ) x^{4}-\ln \left (x^{2}+x +2\right ) x^{4}-5+\ln \left (x^{2}+1\right ) x^{3}-\ln \left (x^{2}+x +2\right ) x^{3}+3 \ln \left (x^{2}+1\right ) x^{2}-3 \ln \left (x^{2}+x +2\right ) x^{2}+\ln \left (x^{2}+1\right ) x -\ln \left (x^{2}+x +2\right ) x -2 x^{2}+2 \ln \left (x^{2}+1\right )-2 \ln \left (x^{2}+x +2\right )-3 x}{\left (x^{2}+1\right ) \left (x^{2}+x +2\right )}\)	\(129\)

[In]

int((x^6+7*x^5+15*x^4+32*x^3+23*x^2+25*x-3)/(x^2+1)^2/(x^2+x+2)^2,x,method=_RETURNVERBOSE)

[Out]

-3/(x^2+1)+1/(x^2+x+2)+ln(x^2+1)-ln(x^2+x+2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (33) = 66\).

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.18 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {2 \, x^{2} + {\left (x^{4} + x^{3} + 3 \, x^{2} + x + 2\right )} \log \left (x^{2} + x + 2\right ) - {\left (x^{4} + x^{3} + 3 \, x^{2} + x + 2\right )} \log \left (x^{2} + 1\right ) + 3 \, x + 5}{x^{4} + x^{3} + 3 \, x^{2} + x + 2} \]

[In]

integrate((x^6+7*x^5+15*x^4+32*x^3+23*x^2+25*x-3)/(x^2+1)^2/(x^2+x+2)^2,x, algorithm="fricas")

[Out]

-(2*x^2 + (x^4 + x^3 + 3*x^2 + x + 2)*log(x^2 + x + 2) - (x^4 + x^3 + 3*x^2 + x + 2)*log(x^2 + 1) + 3*x + 5)/(
x^4 + x^3 + 3*x^2 + x + 2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=\frac {- 2 x^{2} - 3 x - 5}{x^{4} + x^{3} + 3 x^{2} + x + 2} + \log {\left (x^{2} + 1 \right )} - \log {\left (x^{2} + x + 2 \right )} \]

[In]

integrate((x**6+7*x**5+15*x**4+32*x**3+23*x**2+25*x-3)/(x**2+1)**2/(x**2+x+2)**2,x)

[Out]

(-2*x**2 - 3*x - 5)/(x**4 + x**3 + 3*x**2 + x + 2) + log(x**2 + 1) - log(x**2 + x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {2 \, x^{2} + 3 \, x + 5}{x^{4} + x^{3} + 3 \, x^{2} + x + 2} - \log \left (x^{2} + x + 2\right ) + \log \left (x^{2} + 1\right ) \]

[In]

integrate((x^6+7*x^5+15*x^4+32*x^3+23*x^2+25*x-3)/(x^2+1)^2/(x^2+x+2)^2,x, algorithm="maxima")

[Out]

-(2*x^2 + 3*x + 5)/(x^4 + x^3 + 3*x^2 + x + 2) - log(x^2 + x + 2) + log(x^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {2 \, x^{2} + 3 \, x + 5}{x^{4} + x^{3} + 3 \, x^{2} + x + 2} - \log \left (x^{2} + x + 2\right ) + \log \left (x^{2} + 1\right ) \]

[In]

integrate((x^6+7*x^5+15*x^4+32*x^3+23*x^2+25*x-3)/(x^2+1)^2/(x^2+x+2)^2,x, algorithm="giac")

[Out]

-(2*x^2 + 3*x + 5)/(x^4 + x^3 + 3*x^2 + x + 2) - log(x^2 + x + 2) + log(x^2 + 1)

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {-3+25 x+23 x^2+32 x^3+15 x^4+7 x^5+x^6}{\left (1+x^2\right )^2 \left (2+x+x^2\right )^2} \, dx=-\frac {2\,x^2+3\,x+5}{x^4+x^3+3\,x^2+x+2}+\mathrm {atan}\left (\frac {\frac {x\,224{}\mathrm {i}}{11}+\frac {224}{11}{}\mathrm {i}}{44\,x^2+16\,x+60}-\frac {3}{11}{}\mathrm {i}\right )\,2{}\mathrm {i} \]

[In]

int((25*x + 23*x^2 + 32*x^3 + 15*x^4 + 7*x^5 + x^6 - 3)/((x^2 + 1)^2*(x + x^2 + 2)^2),x)

[Out]

atan(((x*224i)/11 + 224i/11)/(16*x + 44*x^2 + 60) - 3i/11)*2i - (3*x + 2*x^2 + 5)/(x + 3*x^2 + x^3 + x^4 + 2)