Integrand size = 50, antiderivative size = 60 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=\frac {2}{\left (3+x+x^4\right )^3}-\frac {3 x}{\left (3+x+x^4\right )^3}+\frac {5 x^2}{\left (3+x+x^4\right )^3}+\frac {x^4}{\left (3+x+x^4\right )^3}-\frac {5 x^6}{\left (3+x+x^4\right )^3} \]
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.45 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=\frac {2-3 x+5 x^2+x^4-5 x^6}{\left (3+x+x^4\right )^3} \]
Integrate[-((15 - 36*x + 5*x^2 + 12*x^3 - 34*x^4 + 140*x^5 + 15*x^6 + 8*x^ 7 - 30*x^9)/(3 + x + x^4)^4),x]
Time = 0.43 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {25, 2527, 27, 2527, 27, 2527, 27, 2527, 27, 2021}
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.
\(\displaystyle \int -\frac {-30 x^9+8 x^7+15 x^6+140 x^5-34 x^4+12 x^3+5 x^2-36 x+15}{\left (x^4+x+3\right )^4} \, dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {-30 x^9+8 x^7+15 x^6+140 x^5-34 x^4+12 x^3+5 x^2-36 x+15}{\left (x^4+x+3\right )^4}dx\) |
\(\Big \downarrow \) 2527 |
\(\displaystyle \frac {1}{6} \int -\frac {6 \left (8 x^7+50 x^5-34 x^4+12 x^3+5 x^2-36 x+15\right )}{\left (x^4+x+3\right )^4}dx-\frac {5 x^6}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {8 x^7+50 x^5-34 x^4+12 x^3+5 x^2-36 x+15}{\left (x^4+x+3\right )^4}dx-\frac {5 x^6}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 2527 |
\(\displaystyle \frac {1}{8} \int -\frac {8 \left (50 x^5-33 x^4+24 x^3+5 x^2-36 x+15\right )}{\left (x^4+x+3\right )^4}dx+\frac {x^4}{\left (x^4+x+3\right )^3}-\frac {5 x^6}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {50 x^5-33 x^4+24 x^3+5 x^2-36 x+15}{\left (x^4+x+3\right )^4}dx+\frac {x^4}{\left (x^4+x+3\right )^3}-\frac {5 x^6}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 2527 |
\(\displaystyle \frac {1}{10} \int -\frac {30 \left (-11 x^4+8 x^3-2 x+5\right )}{\left (x^4+x+3\right )^4}dx+\frac {x^4}{\left (x^4+x+3\right )^3}-\frac {5 x^6}{\left (x^4+x+3\right )^3}+\frac {5 x^2}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -3 \int \frac {-11 x^4+8 x^3-2 x+5}{\left (x^4+x+3\right )^4}dx+\frac {x^4}{\left (x^4+x+3\right )^3}-\frac {5 x^6}{\left (x^4+x+3\right )^3}+\frac {5 x^2}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 2527 |
\(\displaystyle -3 \left (\frac {x}{\left (x^4+x+3\right )^3}-\frac {1}{11} \int -\frac {22 \left (4 x^3+1\right )}{\left (x^4+x+3\right )^4}dx\right )+\frac {x^4}{\left (x^4+x+3\right )^3}-\frac {5 x^6}{\left (x^4+x+3\right )^3}+\frac {5 x^2}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -3 \left (2 \int \frac {4 x^3+1}{\left (x^4+x+3\right )^4}dx+\frac {x}{\left (x^4+x+3\right )^3}\right )+\frac {x^4}{\left (x^4+x+3\right )^3}-\frac {5 x^6}{\left (x^4+x+3\right )^3}+\frac {5 x^2}{\left (x^4+x+3\right )^3}\) |
\(\Big \downarrow \) 2021 |
\(\displaystyle \frac {x^4}{\left (x^4+x+3\right )^3}-3 \left (\frac {x}{\left (x^4+x+3\right )^3}-\frac {2}{3 \left (x^4+x+3\right )^3}\right )-\frac {5 x^6}{\left (x^4+x+3\right )^3}+\frac {5 x^2}{\left (x^4+x+3\right )^3}\) |
Int[-((15 - 36*x + 5*x^2 + 12*x^3 - 34*x^4 + 140*x^5 + 15*x^6 + 8*x^7 - 30 *x^9)/(3 + x + x^4)^4),x]
(5*x^2)/(3 + x + x^4)^3 + x^4/(3 + x + x^4)^3 - (5*x^6)/(3 + x + x^4)^3 - 3*(-2/(3*(3 + x + x^4)^3) + x/(3 + x + x^4)^3)
3.5.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x ]}, Simp[Coeff[Pp, x, p]*x^(p - q + 1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x, q]*Pp , Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; Free Q[m, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && NeQ[m, -1]
Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x ]}, Simp[Coeff[Pm, x, m]*x^(m - n + 1)*(Qn^(p + 1)/((m + n*p + 1)*Coeff[Qn, x, n])), x] + Simp[1/((m + n*p + 1)*Coeff[Qn, x, n]) Int[ExpandToSum[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p, x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[p, -1]
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {-5 x^{6}+x^{4}+5 x^{2}-3 x +2}{\left (x^{4}+x +3\right )^{3}}\) | \(28\) |
norman | \(\frac {-5 x^{6}+x^{4}+5 x^{2}-3 x +2}{\left (x^{4}+x +3\right )^{3}}\) | \(28\) |
risch | \(\frac {-5 x^{6}+x^{4}+5 x^{2}-3 x +2}{\left (x^{4}+x +3\right )^{3}}\) | \(28\) |
parallelrisch | \(\frac {-5 x^{6}+x^{4}+5 x^{2}-3 x +2}{\left (x^{4}+x +3\right )^{3}}\) | \(28\) |
gosper | \(-\frac {5 x^{6}-x^{4}-5 x^{2}+3 x -2}{\left (x^{4}+x +3\right )^{3}}\) | \(31\) |
int((30*x^9-8*x^7-15*x^6-140*x^5+34*x^4-12*x^3-5*x^2+36*x-15)/(x^4+x+3)^4, x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=-\frac {5 \, x^{6} - x^{4} - 5 \, x^{2} + 3 \, x - 2}{x^{12} + 3 \, x^{9} + 9 \, x^{8} + 3 \, x^{6} + 18 \, x^{5} + 27 \, x^{4} + x^{3} + 9 \, x^{2} + 27 \, x + 27} \]
integrate((30*x^9-8*x^7-15*x^6-140*x^5+34*x^4-12*x^3-5*x^2+36*x-15)/(x^4+x +3)^4,x, algorithm="fricas")
-(5*x^6 - x^4 - 5*x^2 + 3*x - 2)/(x^12 + 3*x^9 + 9*x^8 + 3*x^6 + 18*x^5 + 27*x^4 + x^3 + 9*x^2 + 27*x + 27)
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=\frac {- 5 x^{6} + x^{4} + 5 x^{2} - 3 x + 2}{x^{12} + 3 x^{9} + 9 x^{8} + 3 x^{6} + 18 x^{5} + 27 x^{4} + x^{3} + 9 x^{2} + 27 x + 27} \]
(-5*x**6 + x**4 + 5*x**2 - 3*x + 2)/(x**12 + 3*x**9 + 9*x**8 + 3*x**6 + 18 *x**5 + 27*x**4 + x**3 + 9*x**2 + 27*x + 27)
Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=-\frac {5 \, x^{6} - x^{4} - 5 \, x^{2} + 3 \, x - 2}{x^{12} + 3 \, x^{9} + 9 \, x^{8} + 3 \, x^{6} + 18 \, x^{5} + 27 \, x^{4} + x^{3} + 9 \, x^{2} + 27 \, x + 27} \]
integrate((30*x^9-8*x^7-15*x^6-140*x^5+34*x^4-12*x^3-5*x^2+36*x-15)/(x^4+x +3)^4,x, algorithm="maxima")
-(5*x^6 - x^4 - 5*x^2 + 3*x - 2)/(x^12 + 3*x^9 + 9*x^8 + 3*x^6 + 18*x^5 + 27*x^4 + x^3 + 9*x^2 + 27*x + 27)
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.50 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=-\frac {5 \, x^{6} - x^{4} - 5 \, x^{2} + 3 \, x - 2}{{\left (x^{4} + x + 3\right )}^{3}} \]
integrate((30*x^9-8*x^7-15*x^6-140*x^5+34*x^4-12*x^3-5*x^2+36*x-15)/(x^4+x +3)^4,x, algorithm="giac")
Time = 9.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.45 \[ \int -\frac {15-36 x+5 x^2+12 x^3-34 x^4+140 x^5+15 x^6+8 x^7-30 x^9}{\left (3+x+x^4\right )^4} \, dx=\frac {-5\,x^6+x^4+5\,x^2-3\,x+2}{{\left (x^4+x+3\right )}^3} \]