Integrand size = 30, antiderivative size = 47 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=-\frac {2 (73+367 x)}{621 \left (2-x+3 x^2\right )^{3/2}}-\frac {4 (3889-4290 x)}{14283 \sqrt {2-x+3 x^2}} \] Output:
1/621*(-146-734*x)/(3*x^2-x+2)^(3/2)-4/14283*(3889-4290*x)/(3*x^2-x+2)^(1/ 2)
Time = 0.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.70 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=\frac {2 \left (-1915+1833 x-3546 x^2+2860 x^3\right )}{1587 \left (2-x+3 x^2\right )^{3/2}} \] Input:
Integrate[((1 + 2*x)*(1 + 3*x + 4*x^2))/(2 - x + 3*x^2)^(5/2),x]
Output:
(2*(-1915 + 1833*x - 3546*x^2 + 2860*x^3))/(1587*(2 - x + 3*x^2)^(3/2))
Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2191, 27, 1158}
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.
\(\displaystyle \int \frac {(2 x+1) \left (4 x^2+3 x+1\right )}{\left (3 x^2-x+2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2}{69} \int \frac {828 x+577}{9 \left (3 x^2-x+2\right )^{3/2}}dx-\frac {2 (367 x+73)}{621 \left (3 x^2-x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{621} \int \frac {828 x+577}{\left (3 x^2-x+2\right )^{3/2}}dx-\frac {2 (367 x+73)}{621 \left (3 x^2-x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 1158 |
\(\displaystyle -\frac {4 (3889-4290 x)}{14283 \sqrt {3 x^2-x+2}}-\frac {2 (367 x+73)}{621 \left (3 x^2-x+2\right )^{3/2}}\) |
Input:
Int[((1 + 2*x)*(1 + 3*x + 4*x^2))/(2 - x + 3*x^2)^(5/2),x]
Output:
(-2*(73 + 367*x))/(621*(2 - x + 3*x^2)^(3/2)) - (4*(3889 - 4290*x))/(14283 *Sqrt[2 - x + 3*x^2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbo l] :> Simp[-2*((b*d - 2*a*e + (2*c*d - b*e)*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c, d, e}, x]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ (p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int [(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* (2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 2 - 4*a*c, 0] && LtQ[p, -1]
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(\frac {\frac {5720}{1587} x^{3}-\frac {2364}{529} x^{2}+\frac {1222}{529} x -\frac {3830}{1587}}{\left (3 x^{2}-x +2\right )^{\frac {3}{2}}}\) | \(30\) |
trager | \(\frac {\frac {5720}{1587} x^{3}-\frac {2364}{529} x^{2}+\frac {1222}{529} x -\frac {3830}{1587}}{\left (3 x^{2}-x +2\right )^{\frac {3}{2}}}\) | \(30\) |
risch | \(\frac {\frac {5720}{1587} x^{3}-\frac {2364}{529} x^{2}+\frac {1222}{529} x -\frac {3830}{1587}}{\left (3 x^{2}-x +2\right )^{\frac {3}{2}}}\) | \(30\) |
orering | \(\frac {\frac {5720}{1587} x^{3}-\frac {2364}{529} x^{2}+\frac {1222}{529} x -\frac {3830}{1587}}{\left (3 x^{2}-x +2\right )^{\frac {3}{2}}}\) | \(30\) |
default | \(\frac {\frac {715 x}{621}-\frac {715}{3726}}{\left (3 x^{2}-x +2\right )^{\frac {3}{2}}}+\frac {\frac {5720 x}{4761}-\frac {2860}{14283}}{\sqrt {3 x^{2}-x +2}}-\frac {295}{162 \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}-\frac {13 x}{9 \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}-\frac {8 x^{2}}{3 \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}\) | \(86\) |
Input:
int((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/1587/(3*x^2-x+2)^(3/2)*(2860*x^3-3546*x^2+1833*x-1915)
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (2860 \, x^{3} - 3546 \, x^{2} + 1833 \, x - 1915\right )} \sqrt {3 \, x^{2} - x + 2}}{1587 \, {\left (9 \, x^{4} - 6 \, x^{3} + 13 \, x^{2} - 4 \, x + 4\right )}} \] Input:
integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x, algorithm="fricas")
Output:
2/1587*(2860*x^3 - 3546*x^2 + 1833*x - 1915)*sqrt(3*x^2 - x + 2)/(9*x^4 - 6*x^3 + 13*x^2 - 4*x + 4)
\[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=\int \frac {\left (2 x + 1\right ) \left (4 x^{2} + 3 x + 1\right )}{\left (3 x^{2} - x + 2\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((1+2*x)*(4*x**2+3*x+1)/(3*x**2-x+2)**(5/2),x)
Output:
Integral((2*x + 1)*(4*x**2 + 3*x + 1)/(3*x**2 - x + 2)**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.62 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=\frac {5720 \, x}{4761 \, \sqrt {3 \, x^{2} - x + 2}} - \frac {8 \, x^{2}}{3 \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}}} - \frac {2860}{14283 \, \sqrt {3 \, x^{2} - x + 2}} - \frac {182 \, x}{621 \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}}} - \frac {1250}{621 \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}}} \] Input:
integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x, algorithm="maxima")
Output:
5720/4761*x/sqrt(3*x^2 - x + 2) - 8/3*x^2/(3*x^2 - x + 2)^(3/2) - 2860/142 83/sqrt(3*x^2 - x + 2) - 182/621*x/(3*x^2 - x + 2)^(3/2) - 1250/621/(3*x^2 - x + 2)^(3/2)
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.60 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left (2 \, {\left (1430 \, x - 1773\right )} x + 1833\right )} x - 1915\right )}}{1587 \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}}} \] Input:
integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x, algorithm="giac")
Output:
2/1587*((2*(1430*x - 1773)*x + 1833)*x - 1915)/(3*x^2 - x + 2)^(3/2)
Time = 18.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=-\frac {442\,x-5720\,x\,\left (3\,x^2-x+2\right )+15556\,x^2+11490}{\sqrt {3\,x^2-x+2}\,\left (14283\,x^2-4761\,x+9522\right )} \] Input:
int(((2*x + 1)*(3*x + 4*x^2 + 1))/(3*x^2 - x + 2)^(5/2),x)
Output:
-(442*x - 5720*x*(3*x^2 - x + 2) + 15556*x^2 + 11490)/((3*x^2 - x + 2)^(1/ 2)*(14283*x^2 - 4761*x + 9522))
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.43 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx=\frac {51480 \sqrt {3 x^{2}-x +2}\, x^{3}-63828 \sqrt {3 x^{2}-x +2}\, x^{2}+32994 \sqrt {3 x^{2}-x +2}\, x -34470 \sqrt {3 x^{2}-x +2}+18072 \sqrt {3}\, x^{4}-12048 \sqrt {3}\, x^{3}+26104 \sqrt {3}\, x^{2}-8032 \sqrt {3}\, x +8032 \sqrt {3}}{128547 x^{4}-85698 x^{3}+185679 x^{2}-57132 x +57132} \] Input:
int((1+2*x)*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x)
Output:
(2*(25740*sqrt(3*x**2 - x + 2)*x**3 - 31914*sqrt(3*x**2 - x + 2)*x**2 + 16 497*sqrt(3*x**2 - x + 2)*x - 17235*sqrt(3*x**2 - x + 2) + 9036*sqrt(3)*x** 4 - 6024*sqrt(3)*x**3 + 13052*sqrt(3)*x**2 - 4016*sqrt(3)*x + 4016*sqrt(3) ))/(14283*(9*x**4 - 6*x**3 + 13*x**2 - 4*x + 4))