Integrand size = 35, antiderivative size = 53 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=-\frac {1}{\left (1+x^2\right )^4}+\frac {2 (3+2 x)}{3 \left (1+x^2\right )^3}-\frac {3+8 x}{6 \left (1+x^2\right )^2}-\frac {1}{2 \left (1+x^2\right )} \] Output:
-1/(x^2+1)^4+2/3*(3+2*x)/(x^2+1)^3-1/6*(3+8*x)/(x^2+1)^2-1/(2*x^2+2)
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.64 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=-\frac {x^2 \left (3+8 x+12 x^2+8 x^3+3 x^4\right )}{6 \left (1+x^2\right )^4} \] Input:
Integrate[(-x - 4*x^2 - 5*x^3 + 5*x^5 + 4*x^6 + x^7)/(1 + x^2)^5,x]
Output:
-1/6*(x^2*(3 + 8*x + 12*x^2 + 8*x^3 + 3*x^4))/(1 + x^2)^4
Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {2342, 2335, 27, 2342, 2335, 27, 2345, 27, 241}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^7+4 x^6+5 x^5-5 x^3-4 x^2-x}{\left (x^2+1\right )^5} \, dx\) |
\(\Big \downarrow \) 2342 |
\(\displaystyle \int \frac {x \left (x^6+4 x^5+5 x^4-5 x^2-4 x-1\right )}{\left (x^2+1\right )^5}dx\) |
\(\Big \downarrow \) 2335 |
\(\displaystyle \frac {x^2}{\left (x^2+1\right )^4}-\frac {1}{8} \int \frac {8 \left (-x^5-4 x^4-4 x^3+4 x^2+3 x\right )}{\left (x^2+1\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2}{\left (x^2+1\right )^4}-\int \frac {-x^5-4 x^4-4 x^3+4 x^2+3 x}{\left (x^2+1\right )^4}dx\) |
\(\Big \downarrow \) 2342 |
\(\displaystyle \frac {x^2}{\left (x^2+1\right )^4}-\int \frac {x \left (-x^4-4 x^3-4 x^2+4 x+3\right )}{\left (x^2+1\right )^4}dx\) |
\(\Big \downarrow \) 2335 |
\(\displaystyle \frac {1}{6} \int -\frac {2 \left (-3 x^3-12 x^2+3 x+4\right )}{\left (x^2+1\right )^3}dx+\frac {x^2}{\left (x^2+1\right )^4}+\frac {(4-3 x) x}{3 \left (x^2+1\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} \int \frac {-3 x^3-12 x^2+3 x+4}{\left (x^2+1\right )^3}dx+\frac {x^2}{\left (x^2+1\right )^4}+\frac {(4-3 x) x}{3 \left (x^2+1\right )^3}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} \int \frac {12 x}{\left (x^2+1\right )^2}dx+\frac {3-8 x}{2 \left (x^2+1\right )^2}\right )+\frac {x^2}{\left (x^2+1\right )^4}+\frac {(4-3 x) x}{3 \left (x^2+1\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (3 \int \frac {x}{\left (x^2+1\right )^2}dx+\frac {3-8 x}{2 \left (x^2+1\right )^2}\right )+\frac {x^2}{\left (x^2+1\right )^4}+\frac {(4-3 x) x}{3 \left (x^2+1\right )^3}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {x^2}{\left (x^2+1\right )^4}+\frac {(4-3 x) x}{3 \left (x^2+1\right )^3}+\frac {1}{3} \left (\frac {3-8 x}{2 \left (x^2+1\right )^2}-\frac {3}{2 \left (x^2+1\right )}\right )\) |
Input:
Int[(-x - 4*x^2 - 5*x^3 + 5*x^5 + 4*x^6 + x^7)/(1 + x^2)^5,x]
Output:
x^2/(1 + x^2)^4 + ((4 - 3*x)*x)/(3*(1 + x^2)^3) + ((3 - 8*x)/(2*(1 + x^2)^ 2) - 3/(2*(1 + x^2)))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient [Pq, x, x]*(a + b*x^2)^p, x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[ Coeff[Pq, x, 0], 0] && !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(-\frac {x^{2} \left (3 x^{4}+8 x^{3}+12 x^{2}+8 x +3\right )}{6 \left (x^{2}+1\right )^{4}}\) | \(33\) |
default | \(\frac {-\frac {1}{2} x^{6}-\frac {4}{3} x^{5}-2 x^{4}-\frac {4}{3} x^{3}-\frac {1}{2} x^{2}}{\left (x^{2}+1\right )^{4}}\) | \(35\) |
norman | \(\frac {-\frac {1}{2} x^{6}-\frac {4}{3} x^{5}-2 x^{4}-\frac {4}{3} x^{3}-\frac {1}{2} x^{2}}{\left (x^{2}+1\right )^{4}}\) | \(35\) |
risch | \(\frac {-\frac {1}{2} x^{6}-\frac {4}{3} x^{5}-2 x^{4}-\frac {4}{3} x^{3}-\frac {1}{2} x^{2}}{\left (x^{2}+1\right )^{4}}\) | \(35\) |
parallelrisch | \(\frac {-3 x^{6}-8 x^{5}-12 x^{4}-8 x^{3}-3 x^{2}}{6 \left (x^{2}+1\right )^{4}}\) | \(36\) |
orering | \(-\frac {x \left (3 x^{4}+8 x^{3}+12 x^{2}+8 x +3\right ) \left (x^{7}+4 x^{6}+5 x^{5}-5 x^{3}-4 x^{2}-x \right )}{6 \left (x^{2}+1\right )^{4} \left (-1+x \right ) \left (1+x \right )^{5}}\) | \(68\) |
meijerg | \(-\frac {x^{2} \left (x^{6}+4 x^{4}+6 x^{2}+4\right )}{8 \left (x^{2}+1\right )^{4}}+\frac {x \left (-15 x^{6}-55 x^{4}-73 x^{2}+15\right )}{96 \left (x^{2}+1\right )^{4}}-\frac {5 x^{4} \left (x^{4}+4 x^{2}+6\right )}{24 \left (x^{2}+1\right )^{4}}+\frac {5 x^{6} \left (x^{2}+4\right )}{24 \left (x^{2}+1\right )^{4}}-\frac {x \left (-105 x^{6}+511 x^{4}+385 x^{2}+105\right )}{672 \left (x^{2}+1\right )^{4}}+\frac {x^{8}}{8 \left (x^{2}+1\right )^{4}}\) | \(134\) |
Input:
int((x^7+4*x^6+5*x^5-5*x^3-4*x^2-x)/(x^2+1)^5,x,method=_RETURNVERBOSE)
Output:
-1/6*x^2*(3*x^4+8*x^3+12*x^2+8*x+3)/(x^2+1)^4
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=-\frac {3 \, x^{6} + 8 \, x^{5} + 12 \, x^{4} + 8 \, x^{3} + 3 \, x^{2}}{6 \, {\left (x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1\right )}} \] Input:
integrate((x^7+4*x^6+5*x^5-5*x^3-4*x^2-x)/(x^2+1)^5,x, algorithm="fricas")
Output:
-1/6*(3*x^6 + 8*x^5 + 12*x^4 + 8*x^3 + 3*x^2)/(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1)
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=\frac {- 3 x^{6} - 8 x^{5} - 12 x^{4} - 8 x^{3} - 3 x^{2}}{6 x^{8} + 24 x^{6} + 36 x^{4} + 24 x^{2} + 6} \] Input:
integrate((x**7+4*x**6+5*x**5-5*x**3-4*x**2-x)/(x**2+1)**5,x)
Output:
(-3*x**6 - 8*x**5 - 12*x**4 - 8*x**3 - 3*x**2)/(6*x**8 + 24*x**6 + 36*x**4 + 24*x**2 + 6)
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=-\frac {3 \, x^{6} + 8 \, x^{5} + 12 \, x^{4} + 8 \, x^{3} + 3 \, x^{2}}{6 \, {\left (x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1\right )}} \] Input:
integrate((x^7+4*x^6+5*x^5-5*x^3-4*x^2-x)/(x^2+1)^5,x, algorithm="maxima")
Output:
-1/6*(3*x^6 + 8*x^5 + 12*x^4 + 8*x^3 + 3*x^2)/(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1)
Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.53 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=-\frac {3 \, {\left (x + \frac {1}{x}\right )}^{2} + 8 \, x + \frac {8}{x} + 6}{6 \, {\left (x + \frac {1}{x}\right )}^{4}} \] Input:
integrate((x^7+4*x^6+5*x^5-5*x^3-4*x^2-x)/(x^2+1)^5,x, algorithm="giac")
Output:
-1/6*(3*(x + 1/x)^2 + 8*x + 8/x + 6)/(x + 1/x)^4
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=\frac {\frac {4\,x}{3}+2}{{\left (x^2+1\right )}^3}-\frac {\frac {4\,x}{3}+\frac {1}{2}}{{\left (x^2+1\right )}^2}-\frac {1}{2\,\left (x^2+1\right )}-\frac {1}{{\left (x^2+1\right )}^4} \] Input:
int(-(x + 4*x^2 + 5*x^3 - 5*x^5 - 4*x^6 - x^7)/(x^2 + 1)^5,x)
Output:
((4*x)/3 + 2)/(x^2 + 1)^3 - ((4*x)/3 + 1/2)/(x^2 + 1)^2 - 1/(2*(x^2 + 1)) - 1/(x^2 + 1)^4
Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \frac {-x-4 x^2-5 x^3+5 x^5+4 x^6+x^7}{\left (1+x^2\right )^5} \, dx=\frac {3 x^{8}-32 x^{5}-30 x^{4}-32 x^{3}+3}{24 x^{8}+96 x^{6}+144 x^{4}+96 x^{2}+24} \] Input:
int((x^7+4*x^6+5*x^5-5*x^3-4*x^2-x)/(x^2+1)^5,x)
Output:
(3*x**8 - 32*x**5 - 30*x**4 - 32*x**3 + 3)/(24*(x**8 + 4*x**6 + 6*x**4 + 4 *x**2 + 1))