Integrand size = 38, antiderivative size = 120 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\frac {761}{256} (5+2 x)^2 \sqrt {3-x+2 x^2}-\frac {105}{128} (5+2 x)^3 \sqrt {3-x+2 x^2}+\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}-\frac {(19227+4676 x) \sqrt {3-x+2 x^2}}{2048}-\frac {85429 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}} \]
-85429/8192*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+761/256*(5+2*x)^2*(2*x^ 2-x+3)^(1/2)-105/128*(5+2*x)^3*(2*x^2-x+3)^(1/2)+1/16*(5+2*x)^4*(2*x^2-x+3 )^(1/2)-1/2048*(19227+4676*x)*(2*x^2-x+3)^(1/2)
Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.58 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\frac {4 \sqrt {3-x+2 x^2} \left (2973-6916 x+352 x^2+7040 x^3+2048 x^4\right )-85429 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{8192} \]
(4*Sqrt[3 - x + 2*x^2]*(2973 - 6916*x + 352*x^2 + 7040*x^3 + 2048*x^4) - 8 5429*Sqrt[2]*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/8192
Time = 0.42 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {2184, 27, 2184, 27, 2184, 27, 1225, 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.
\(\displaystyle \int \frac {(2 x+5) \left (5 x^4-x^3+3 x^2+x+2\right )}{\sqrt {2 x^2-x+3}} \, dx\) |
\(\Big \downarrow \) 2184 |
\(\displaystyle \frac {1}{160} \int -\frac {5 (2 x+5) \left (840 x^3+1116 x^2+878 x+1011\right )}{\sqrt {2 x^2-x+3}}dx+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} (2 x+5)^4 \sqrt {2 x^2-x+3}-\frac {1}{32} \int \frac {(2 x+5) \left (840 x^3+1116 x^2+878 x+1011\right )}{\sqrt {2 x^2-x+3}}dx\) |
\(\Big \downarrow \) 2184 |
\(\displaystyle \frac {1}{32} \left (-\frac {1}{64} \int -\frac {8 (2 x+5) \left (9132 x^2+2636 x+8187\right )}{\sqrt {2 x^2-x+3}}dx-\frac {105}{4} \sqrt {2 x^2-x+3} (2 x+5)^3\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{32} \left (\frac {1}{8} \int \frac {(2 x+5) \left (9132 x^2+2636 x+8187\right )}{\sqrt {2 x^2-x+3}}dx-\frac {105}{4} (2 x+5)^3 \sqrt {2 x^2-x+3}\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 2184 |
\(\displaystyle \frac {1}{32} \left (\frac {1}{8} \left (\frac {1}{24} \int \frac {12 (1915-2338 x) (2 x+5)}{\sqrt {2 x^2-x+3}}dx+761 \sqrt {2 x^2-x+3} (2 x+5)^2\right )-\frac {105}{4} (2 x+5)^3 \sqrt {2 x^2-x+3}\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{32} \left (\frac {1}{8} \left (\frac {1}{2} \int \frac {(1915-2338 x) (2 x+5)}{\sqrt {2 x^2-x+3}}dx+761 \sqrt {2 x^2-x+3} (2 x+5)^2\right )-\frac {105}{4} (2 x+5)^3 \sqrt {2 x^2-x+3}\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{32} \left (\frac {1}{8} \left (\frac {1}{2} \left (\frac {85429}{8} \int \frac {1}{\sqrt {2 x^2-x+3}}dx-\frac {1}{4} (4676 x+19227) \sqrt {2 x^2-x+3}\right )+761 \sqrt {2 x^2-x+3} (2 x+5)^2\right )-\frac {105}{4} (2 x+5)^3 \sqrt {2 x^2-x+3}\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{32} \left (\frac {1}{8} \left (\frac {1}{2} \left (\frac {85429 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{8 \sqrt {46}}-\frac {1}{4} (4676 x+19227) \sqrt {2 x^2-x+3}\right )+761 \sqrt {2 x^2-x+3} (2 x+5)^2\right )-\frac {105}{4} (2 x+5)^3 \sqrt {2 x^2-x+3}\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{32} \left (\frac {1}{8} \left (\frac {1}{2} \left (\frac {85429 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{8 \sqrt {2}}-\frac {1}{4} (4676 x+19227) \sqrt {2 x^2-x+3}\right )+761 \sqrt {2 x^2-x+3} (2 x+5)^2\right )-\frac {105}{4} (2 x+5)^3 \sqrt {2 x^2-x+3}\right )+\frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4\) |
((5 + 2*x)^4*Sqrt[3 - x + 2*x^2])/16 + ((-105*(5 + 2*x)^3*Sqrt[3 - x + 2*x ^2])/4 + (761*(5 + 2*x)^2*Sqrt[3 - x + 2*x^2] + (-1/4*((19227 + 4676*x)*Sq rt[3 - x + 2*x^2]) + (85429*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(8*Sqrt[2]))/2)/ 8)/32
3.4.44.3.1 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_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c *d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !(IGt Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 1.50 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42
method | result | size |
risch | \(\frac {\left (2048 x^{4}+7040 x^{3}+352 x^{2}-6916 x +2973\right ) \sqrt {2 x^{2}-x +3}}{2048}+\frac {85429 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{8192}\) | \(50\) |
trager | \(\left (x^{4}+\frac {55}{16} x^{3}+\frac {11}{64} x^{2}-\frac {1729}{512} x +\frac {2973}{2048}\right ) \sqrt {2 x^{2}-x +3}-\frac {85429 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )+4 \sqrt {2 x^{2}-x +3}\right )}{8192}\) | \(72\) |
default | \(\frac {85429 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{8192}+\frac {11 x^{2} \sqrt {2 x^{2}-x +3}}{64}-\frac {1729 x \sqrt {2 x^{2}-x +3}}{512}+\frac {2973 \sqrt {2 x^{2}-x +3}}{2048}+x^{4} \sqrt {2 x^{2}-x +3}+\frac {55 x^{3} \sqrt {2 x^{2}-x +3}}{16}\) | \(95\) |
1/2048*(2048*x^4+7040*x^3+352*x^2-6916*x+2973)*(2*x^2-x+3)^(1/2)+85429/819 2*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\frac {1}{2048} \, {\left (2048 \, x^{4} + 7040 \, x^{3} + 352 \, x^{2} - 6916 \, x + 2973\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {85429}{16384} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \]
1/2048*(2048*x^4 + 7040*x^3 + 352*x^2 - 6916*x + 2973)*sqrt(2*x^2 - x + 3) + 85429/16384*sqrt(2)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x ^2 + 16*x - 25)
Time = 0.64 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.50 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\sqrt {2 x^{2} - x + 3} \left (x^{4} + \frac {55 x^{3}}{16} + \frac {11 x^{2}}{64} - \frac {1729 x}{512} + \frac {2973}{2048}\right ) + \frac {85429 \sqrt {2} \operatorname {asinh}{\left (\frac {4 \sqrt {23} \left (x - \frac {1}{4}\right )}{23} \right )}}{8192} \]
sqrt(2*x**2 - x + 3)*(x**4 + 55*x**3/16 + 11*x**2/64 - 1729*x/512 + 2973/2 048) + 85429*sqrt(2)*asinh(4*sqrt(23)*(x - 1/4)/23)/8192
Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.80 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\sqrt {2 \, x^{2} - x + 3} x^{4} + \frac {55}{16} \, \sqrt {2 \, x^{2} - x + 3} x^{3} + \frac {11}{64} \, \sqrt {2 \, x^{2} - x + 3} x^{2} - \frac {1729}{512} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {85429}{8192} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {2973}{2048} \, \sqrt {2 \, x^{2} - x + 3} \]
sqrt(2*x^2 - x + 3)*x^4 + 55/16*sqrt(2*x^2 - x + 3)*x^3 + 11/64*sqrt(2*x^2 - x + 3)*x^2 - 1729/512*sqrt(2*x^2 - x + 3)*x + 85429/8192*sqrt(2)*arcsin h(1/23*sqrt(23)*(4*x - 1)) + 2973/2048*sqrt(2*x^2 - x + 3)
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.57 \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\frac {1}{2048} \, {\left (4 \, {\left (8 \, {\left (4 \, {\left (16 \, x + 55\right )} x + 11\right )} x - 1729\right )} x + 2973\right )} \sqrt {2 \, x^{2} - x + 3} - \frac {85429}{8192} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \]
1/2048*(4*(8*(4*(16*x + 55)*x + 11)*x - 1729)*x + 2973)*sqrt(2*x^2 - x + 3 ) - 85429/8192*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)
Timed out. \[ \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx=\int \frac {\left (2\,x+5\right )\,\left (5\,x^4-x^3+3\,x^2+x+2\right )}{\sqrt {2\,x^2-x+3}} \,d x \]