Integrand size = 38, antiderivative size = 86 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {-53+373 x}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {6055-28981 x}{3174 \sqrt {3-x+2 x^2}}+\frac {5}{4} \sqrt {3-x+2 x^2}-\frac {71 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{8 \sqrt {2}} \]
1/69*(-53+373*x)/(2*x^2-x+3)^(3/2)-71/16*arcsinh(1/23*(1-4*x)*23^(1/2))*2^ (1/2)+1/3174*(6055-28981*x)/(2*x^2-x+3)^(1/2)+5/4*(2*x^2-x+3)^(1/2)
Time = 0.64 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {102869-199290 x+185337 x^2-147664 x^3+31740 x^4}{6348 \left (3-x+2 x^2\right )^{3/2}}-\frac {71 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{8 \sqrt {2}} \]
(102869 - 199290*x + 185337*x^2 - 147664*x^3 + 31740*x^4)/(6348*(3 - x + 2 *x^2)^(3/2)) - (71*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/(8*Sqrt[2])
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {2191, 27, 2191, 27, 1160, 1090, 222}
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.
\(\displaystyle \int \frac {(2 x+5) \left (5 x^4-x^3+3 x^2+x+2\right )}{\left (2 x^2-x+3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2}{69} \int -\frac {-690 x^3-1932 x^2+233}{4 \left (2 x^2-x+3\right )^{3/2}}dx-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{138} \int \frac {-690 x^3-1932 x^2+233}{\left (2 x^2-x+3\right )^{3/2}}dx-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {1}{138} \left (\frac {6055-28981 x}{23 \sqrt {2 x^2-x+3}}-\frac {2}{23} \int -\frac {1587 (10 x+33)}{4 \sqrt {2 x^2-x+3}}dx\right )-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{138} \left (\frac {69}{2} \int \frac {10 x+33}{\sqrt {2 x^2-x+3}}dx+\frac {6055-28981 x}{23 \sqrt {2 x^2-x+3}}\right )-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{138} \left (\frac {69}{2} \left (\frac {71}{2} \int \frac {1}{\sqrt {2 x^2-x+3}}dx+5 \sqrt {2 x^2-x+3}\right )+\frac {6055-28981 x}{23 \sqrt {2 x^2-x+3}}\right )-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{138} \left (\frac {69}{2} \left (\frac {71 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{2 \sqrt {46}}+5 \sqrt {2 x^2-x+3}\right )+\frac {6055-28981 x}{23 \sqrt {2 x^2-x+3}}\right )-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{138} \left (\frac {69}{2} \left (\frac {71 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{2 \sqrt {2}}+5 \sqrt {2 x^2-x+3}\right )+\frac {6055-28981 x}{23 \sqrt {2 x^2-x+3}}\right )-\frac {53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
-1/69*(53 - 373*x)/(3 - x + 2*x^2)^(3/2) + ((6055 - 28981*x)/(23*Sqrt[3 - x + 2*x^2]) + (69*(5*Sqrt[3 - x + 2*x^2] + (71*ArcSinh[(-1 + 4*x)/Sqrt[23] ])/(2*Sqrt[2])))/2)/138
3.4.59.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
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]
Timed out.
hanged
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.36 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {112677 \, \sqrt {2} {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (31740 \, x^{4} - 147664 \, x^{3} + 185337 \, x^{2} - 199290 \, x + 102869\right )} \sqrt {2 \, x^{2} - x + 3}}{50784 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \]
1/50784*(112677*sqrt(2)*(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)*log(-4*sqrt(2)* sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 8*(31740*x^4 - 14766 4*x^3 + 185337*x^2 - 199290*x + 102869)*sqrt(2*x^2 - x + 3))/(4*x^4 - 4*x^ 3 + 13*x^2 - 6*x + 9)
\[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\int \frac {\left (2 x + 5\right ) \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{\left (2 x^{2} - x + 3\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (69) = 138\).
Time = 0.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.35 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {5 \, x^{4}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {71}{12696} \, x {\left (\frac {284 \, x}{\sqrt {2 \, x^{2} - x + 3}} - \frac {3174 \, x^{2}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {71}{\sqrt {2 \, x^{2} - x + 3}} + \frac {805 \, x}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {3243}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}\right )} + \frac {71}{16} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {5041}{6348} \, \sqrt {2 \, x^{2} - x + 3} - \frac {10007 \, x}{3174 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {59 \, x^{2}}{2 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {2959}{2116 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {807 \, x}{92 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {7603}{276 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
5*x^4/(2*x^2 - x + 3)^(3/2) + 71/12696*x*(284*x/sqrt(2*x^2 - x + 3) - 3174 *x^2/(2*x^2 - x + 3)^(3/2) - 71/sqrt(2*x^2 - x + 3) + 805*x/(2*x^2 - x + 3 )^(3/2) - 3243/(2*x^2 - x + 3)^(3/2)) + 71/16*sqrt(2)*arcsinh(1/23*sqrt(23 )*(4*x - 1)) - 5041/6348*sqrt(2*x^2 - x + 3) - 10007/3174*x/sqrt(2*x^2 - x + 3) + 59/2*x^2/(2*x^2 - x + 3)^(3/2) - 2959/2116/sqrt(2*x^2 - x + 3) - 8 07/92*x/(2*x^2 - x + 3)^(3/2) + 7603/276/(2*x^2 - x + 3)^(3/2)
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.77 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=-\frac {71}{16} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left ({\left (4 \, {\left (7935 \, x - 36916\right )} x + 185337\right )} x - 199290\right )} x + 102869}{6348 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
-71/16*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/6 348*(((4*(7935*x - 36916)*x + 185337)*x - 199290)*x + 102869)/(2*x^2 - x + 3)^(3/2)
Timed out. \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\int \frac {\left (2\,x+5\right )\,\left (5\,x^4-x^3+3\,x^2+x+2\right )}{{\left (2\,x^2-x+3\right )}^{5/2}} \,d x \]