Integrand size = 25, antiderivative size = 59 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\frac {39}{16} \sqrt {3-x+2 x^2}+\frac {5}{4} x \sqrt {3-x+2 x^2}+\frac {17 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \] Output:
39/16*(2*x^2-x+3)^(1/2)+5/4*x*(2*x^2-x+3)^(1/2)+17/64*arcsinh(1/23*(1-4*x) *23^(1/2))*2^(1/2)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\frac {1}{64} \left (4 (39+20 x) \sqrt {3-x+2 x^2}+17 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )\right ) \] Input:
Integrate[(2 + 3*x + 5*x^2)/Sqrt[3 - x + 2*x^2],x]
Output:
(4*(39 + 20*x)*Sqrt[3 - x + 2*x^2] + 17*Sqrt[2]*Log[1 - 4*x + 2*Sqrt[6 - 2 *x + 4*x^2]])/64
Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2192, 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.
\(\displaystyle \int \frac {5 x^2+3 x+2}{\sqrt {2 x^2-x+3}} \, dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{4} \int -\frac {14-39 x}{2 \sqrt {2 x^2-x+3}}dx+\frac {5}{4} \sqrt {2 x^2-x+3} x\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{4} x \sqrt {2 x^2-x+3}-\frac {1}{8} \int \frac {14-39 x}{\sqrt {2 x^2-x+3}}dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{8} \left (\frac {39}{2} \sqrt {2 x^2-x+3}-\frac {17}{4} \int \frac {1}{\sqrt {2 x^2-x+3}}dx\right )+\frac {5}{4} \sqrt {2 x^2-x+3} x\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{8} \left (\frac {39}{2} \sqrt {2 x^2-x+3}-\frac {17 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{4 \sqrt {46}}\right )+\frac {5}{4} \sqrt {2 x^2-x+3} x\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{8} \left (\frac {39}{2} \sqrt {2 x^2-x+3}-\frac {17 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{4 \sqrt {2}}\right )+\frac {5}{4} \sqrt {2 x^2-x+3} x\) |
Input:
Int[(2 + 3*x + 5*x^2)/Sqrt[3 - x + 2*x^2],x]
Output:
(5*x*Sqrt[3 - x + 2*x^2])/4 + ((39*Sqrt[3 - x + 2*x^2])/2 - (17*ArcSinh[(- 1 + 4*x)/Sqrt[23]])/(4*Sqrt[2]))/8
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 = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 1.89 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.59
method | result | size |
risch | \(\frac {\left (20 x +39\right ) \sqrt {2 x^{2}-x +3}}{16}-\frac {17 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}\) | \(35\) |
default | \(-\frac {17 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}+\frac {39 \sqrt {2 x^{2}-x +3}}{16}+\frac {5 x \sqrt {2 x^{2}-x +3}}{4}\) | \(45\) |
trager | \(\left (\frac {5 x}{4}+\frac {39}{16}\right ) \sqrt {2 x^{2}-x +3}-\frac {17 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {2 x^{2}-x +3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )}{64}\) | \(61\) |
Input:
int((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/16*(20*x+39)*(2*x^2-x+3)^(1/2)-17/64*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/ 4))
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\frac {1}{16} \, \sqrt {2 \, x^{2} - x + 3} {\left (20 \, x + 39\right )} + \frac {17}{128} \, \sqrt {2} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \] Input:
integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x, algorithm="fricas")
Output:
1/16*sqrt(2*x^2 - x + 3)*(20*x + 39) + 17/128*sqrt(2)*log(4*sqrt(2)*sqrt(2 *x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25)
Time = 0.38 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\left (\frac {5 x}{4} + \frac {39}{16}\right ) \sqrt {2 x^{2} - x + 3} - \frac {17 \sqrt {2} \operatorname {asinh}{\left (\frac {4 \sqrt {23} \left (x - \frac {1}{4}\right )}{23} \right )}}{64} \] Input:
integrate((5*x**2+3*x+2)/(2*x**2-x+3)**(1/2),x)
Output:
(5*x/4 + 39/16)*sqrt(2*x**2 - x + 3) - 17*sqrt(2)*asinh(4*sqrt(23)*(x - 1/ 4)/23)/64
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\frac {5}{4} \, \sqrt {2 \, x^{2} - x + 3} x - \frac {17}{64} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {39}{16} \, \sqrt {2 \, x^{2} - x + 3} \] Input:
integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x, algorithm="maxima")
Output:
5/4*sqrt(2*x^2 - x + 3)*x - 17/64*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) + 39/16*sqrt(2*x^2 - x + 3)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\frac {1}{16} \, \sqrt {2 \, x^{2} - x + 3} {\left (20 \, x + 39\right )} + \frac {17}{64} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \] Input:
integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x, algorithm="giac")
Output:
1/16*sqrt(2*x^2 - x + 3)*(20*x + 39) + 17/64*sqrt(2)*log(-2*sqrt(2)*(sqrt( 2)*x - sqrt(2*x^2 - x + 3)) + 1)
Timed out. \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\int \frac {5\,x^2+3\,x+2}{\sqrt {2\,x^2-x+3}} \,d x \] Input:
int((3*x + 5*x^2 + 2)/(2*x^2 - x + 3)^(1/2),x)
Output:
int((3*x + 5*x^2 + 2)/(2*x^2 - x + 3)^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int \frac {2+3 x+5 x^2}{\sqrt {3-x+2 x^2}} \, dx=\frac {5 \sqrt {2 x^{2}-x +3}\, x}{4}+\frac {39 \sqrt {2 x^{2}-x +3}}{16}-\frac {17 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right )}{64} \] Input:
int((5*x^2+3*x+2)/(2*x^2-x+3)^(1/2),x)
Output:
(80*sqrt(2*x**2 - x + 3)*x + 156*sqrt(2*x**2 - x + 3) - 17*sqrt(2)*log((2* sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23)))/64