Integrand size = 25, antiderivative size = 82 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=-\frac {81}{512} (1-4 x) \sqrt {3-x+2 x^2}+\frac {73}{96} \left (3-x+2 x^2\right )^{3/2}+\frac {5}{8} x \left (3-x+2 x^2\right )^{3/2}-\frac {1863 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{1024 \sqrt {2}} \] Output:
-81/512*(1-4*x)*(2*x^2-x+3)^(1/2)+73/96*(2*x^2-x+3)^(3/2)+5/8*x*(2*x^2-x+3 )^(3/2)-1863/2048*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)
Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\frac {4 \sqrt {3-x+2 x^2} \left (3261+2684 x+1376 x^2+1920 x^3\right )-5589 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{6144} \] Input:
Integrate[Sqrt[3 - x + 2*x^2]*(2 + 3*x + 5*x^2),x]
Output:
(4*Sqrt[3 - x + 2*x^2]*(3261 + 2684*x + 1376*x^2 + 1920*x^3) - 5589*Sqrt[2 ]*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/6144
Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2192, 27, 1160, 1087, 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.
\(\displaystyle \int \sqrt {2 x^2-x+3} \left (5 x^2+3 x+2\right ) \, dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{8} \int \frac {1}{2} (73 x+2) \sqrt {2 x^2-x+3}dx+\frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} \int (73 x+2) \sqrt {2 x^2-x+3}dx+\frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{16} \left (\frac {81}{4} \int \sqrt {2 x^2-x+3}dx+\frac {73}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{16} \left (\frac {81}{4} \left (\frac {23}{16} \int \frac {1}{\sqrt {2 x^2-x+3}}dx-\frac {1}{8} (1-4 x) \sqrt {2 x^2-x+3}\right )+\frac {73}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{16} \left (\frac {81}{4} \left (\frac {1}{16} \sqrt {\frac {23}{2}} \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)-\frac {1}{8} (1-4 x) \sqrt {2 x^2-x+3}\right )+\frac {73}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{16} \left (\frac {81}{4} \left (\frac {23 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{16 \sqrt {2}}-\frac {1}{8} (1-4 x) \sqrt {2 x^2-x+3}\right )+\frac {73}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}\) |
Input:
Int[Sqrt[3 - x + 2*x^2]*(2 + 3*x + 5*x^2),x]
Output:
(5*x*(3 - x + 2*x^2)^(3/2))/8 + ((73*(3 - x + 2*x^2)^(3/2))/6 + (81*(-1/8* ((1 - 4*x)*Sqrt[3 - x + 2*x^2]) + (23*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(16*Sq rt[2])))/4)/16
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[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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 = 1.96 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.55
method | result | size |
risch | \(\frac {\left (1920 x^{3}+1376 x^{2}+2684 x +3261\right ) \sqrt {2 x^{2}-x +3}}{1536}+\frac {1863 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{2048}\) | \(45\) |
default | \(\frac {81 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{512}+\frac {1863 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{2048}+\frac {73 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{96}+\frac {5 x \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{8}\) | \(64\) |
trager | \(\left (\frac {5}{4} x^{3}+\frac {43}{48} x^{2}+\frac {671}{384} x +\frac {1087}{512}\right ) \sqrt {2 x^{2}-x +3}+\frac {1863 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {2 x^{2}-x +3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )}{2048}\) | \(71\) |
Input:
int((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2),x,method=_RETURNVERBOSE)
Output:
1/1536*(1920*x^3+1376*x^2+2684*x+3261)*(2*x^2-x+3)^(1/2)+1863/2048*2^(1/2) *arcsinh(4/23*23^(1/2)*(x-1/4))
Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\frac {1}{1536} \, {\left (1920 \, x^{3} + 1376 \, x^{2} + 2684 \, x + 3261\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1863}{4096} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \] Input:
integrate((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2),x, algorithm="fricas")
Output:
1/1536*(1920*x^3 + 1376*x^2 + 2684*x + 3261)*sqrt(2*x^2 - x + 3) + 1863/40 96*sqrt(2)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25)
Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\sqrt {2 x^{2} - x + 3} \cdot \left (\frac {5 x^{3}}{4} + \frac {43 x^{2}}{48} + \frac {671 x}{384} + \frac {1087}{512}\right ) + \frac {1863 \sqrt {2} \operatorname {asinh}{\left (\frac {4 \sqrt {23} \left (x - \frac {1}{4}\right )}{23} \right )}}{2048} \] Input:
integrate((2*x**2-x+3)**(1/2)*(5*x**2+3*x+2),x)
Output:
sqrt(2*x**2 - x + 3)*(5*x**3/4 + 43*x**2/48 + 671*x/384 + 1087/512) + 1863 *sqrt(2)*asinh(4*sqrt(23)*(x - 1/4)/23)/2048
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\frac {5}{8} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {73}{96} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {81}{128} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {1863}{2048} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {81}{512} \, \sqrt {2 \, x^{2} - x + 3} \] Input:
integrate((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2),x, algorithm="maxima")
Output:
5/8*(2*x^2 - x + 3)^(3/2)*x + 73/96*(2*x^2 - x + 3)^(3/2) + 81/128*sqrt(2* x^2 - x + 3)*x + 1863/2048*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 81/5 12*sqrt(2*x^2 - x + 3)
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\frac {1}{1536} \, {\left (4 \, {\left (8 \, {\left (60 \, x + 43\right )} x + 671\right )} x + 3261\right )} \sqrt {2 \, x^{2} - x + 3} - \frac {1863}{2048} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \] Input:
integrate((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2),x, algorithm="giac")
Output:
1/1536*(4*(8*(60*x + 43)*x + 671)*x + 3261)*sqrt(2*x^2 - x + 3) - 1863/204 8*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)
Time = 16.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.45 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\frac {23\,\sqrt {2}\,\ln \left (\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (2\,x-\frac {1}{2}\right )}{2}\right )}{256}+\frac {\left (\frac {x}{2}-\frac {1}{8}\right )\,\sqrt {2\,x^2-x+3}}{8}+\frac {73\,\sqrt {2\,x^2-x+3}\,\left (32\,x^2-4\,x+45\right )}{1536}+\frac {5\,x\,{\left (2\,x^2-x+3\right )}^{3/2}}{8}+\frac {1679\,\sqrt {2}\,\ln \left (2\,\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (4\,x-1\right )}{2}\right )}{2048} \] Input:
int((2*x^2 - x + 3)^(1/2)*(3*x + 5*x^2 + 2),x)
Output:
(23*2^(1/2)*log((2*x^2 - x + 3)^(1/2) + (2^(1/2)*(2*x - 1/2))/2))/256 + (( x/2 - 1/8)*(2*x^2 - x + 3)^(1/2))/8 + (73*(2*x^2 - x + 3)^(1/2)*(32*x^2 - 4*x + 45))/1536 + (5*x*(2*x^2 - x + 3)^(3/2))/8 + (1679*2^(1/2)*log(2*(2*x ^2 - x + 3)^(1/2) + (2^(1/2)*(4*x - 1))/2))/2048
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx=\frac {5 \sqrt {2 x^{2}-x +3}\, x^{3}}{4}+\frac {43 \sqrt {2 x^{2}-x +3}\, x^{2}}{48}+\frac {671 \sqrt {2 x^{2}-x +3}\, x}{384}+\frac {1087 \sqrt {2 x^{2}-x +3}}{512}+\frac {1863 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right )}{2048} \] Input:
int((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2),x)
Output:
(7680*sqrt(2*x**2 - x + 3)*x**3 + 5504*sqrt(2*x**2 - x + 3)*x**2 + 10736*s qrt(2*x**2 - x + 3)*x + 13044*sqrt(2*x**2 - x + 3) + 5589*sqrt(2)*log((2*s qrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23)))/6144