Integrand size = 40, antiderivative size = 172 \[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\frac {(141051019-23482924 x) \sqrt {3-x+2 x^2}}{65536}+\frac {(500141-123060 x) \left (3-x+2 x^2\right )^{3/2}}{12288}+\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}+\frac {1622009981 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{131072 \sqrt {2}}-\frac {99009 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{8 \sqrt {2}} \]
1/12288*(500141-123060*x)*(2*x^2-x+3)^(3/2)+3505/896*(2*x^2-x+3)^(5/2)-311 /448*(5+2*x)*(2*x^2-x+3)^(5/2)+5/112*(5+2*x)^2*(2*x^2-x+3)^(5/2)+162200998 1/262144*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-99009/16*arctanh(1/24*(17- 22*x)*2^(1/2)/(2*x^2-x+3)^(1/2))*2^(1/2)+1/65536*(141051019-23482924*x)*(2 *x^2-x+3)^(1/2)
Time = 0.69 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\frac {4 \sqrt {3-x+2 x^2} \left (3149403255-609499532 x+159973408 x^2-46476672 x^3+14493696 x^4-3710976 x^5+983040 x^6\right )+68130865152 \sqrt {2} \text {arctanh}\left (\frac {1}{6} \left (5+2 x-\sqrt {6-2 x+4 x^2}\right )\right )+34062209601 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{5505024} \]
(4*Sqrt[3 - x + 2*x^2]*(3149403255 - 609499532*x + 159973408*x^2 - 4647667 2*x^3 + 14493696*x^4 - 3710976*x^5 + 983040*x^6) + 68130865152*Sqrt[2]*Arc Tanh[(5 + 2*x - Sqrt[6 - 2*x + 4*x^2])/6] + 34062209601*Sqrt[2]*Log[1 - 4* x + 2*Sqrt[6 - 2*x + 4*x^2]])/5505024
Time = 0.59 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.11, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2184, 2184, 27, 2184, 27, 1231, 27, 1231, 27, 1269, 1090, 222, 1154, 219}
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.
| \(\displaystyle \int \frac {\left (2 x^2-x+3\right )^{3/2} \left (5 x^4-x^3+3 x^2+x+2\right )}{2 x+5} \, dx\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {1}{224} \int \frac {\left (2 x^2-x+3\right )^{3/2} \left (-7464 x^3-14508 x^2-9926 x+573\right )}{2 x+5}dx+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{96} \int -\frac {24 \left (-70100 x^2-85940 x+17923\right ) \left (2 x^2-x+3\right )^{3/2}}{2 x+5}dx-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{224} \left (-\frac {1}{4} \int \frac {\left (-70100 x^2-85940 x+17923\right ) \left (2 x^2-x+3\right )^{3/2}}{2 x+5}dx-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (3505 \left (2 x^2-x+3\right )^{5/2}-\frac {1}{40} \int -\frac {140 (7397-20510 x) \left (2 x^2-x+3\right )^{3/2}}{2 x+5}dx\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \int \frac {(7397-20510 x) \left (2 x^2-x+3\right )^{3/2}}{2 x+5}dx+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}-\frac {1}{64} \int -\frac {2 (4441417-11741462 x) \sqrt {2 x^2-x+3}}{2 x+5}dx\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \int \frac {(4441417-11741462 x) \sqrt {2 x^2-x+3}}{2 x+5}dx+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{8} (141051019-23482924 x) \sqrt {2 x^2-x+3}-\frac {1}{32} \int -\frac {2 (1622930831-3244019962 x)}{(2 x+5) \sqrt {2 x^2-x+3}}dx\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{16} \int \frac {1622930831-3244019962 x}{(2 x+5) \sqrt {2 x^2-x+3}}dx+\frac {1}{8} \sqrt {2 x^2-x+3} (141051019-23482924 x)\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{16} \left (9732980736 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-1622009981 \int \frac {1}{\sqrt {2 x^2-x+3}}dx\right )+\frac {1}{8} \sqrt {2 x^2-x+3} (141051019-23482924 x)\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{16} \left (9732980736 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-\frac {1622009981 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{\sqrt {46}}\right )+\frac {1}{8} \sqrt {2 x^2-x+3} (141051019-23482924 x)\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{16} \left (9732980736 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-\frac {1622009981 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {2}}\right )+\frac {1}{8} \sqrt {2 x^2-x+3} (141051019-23482924 x)\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{16} \left (-19465961472 \int \frac {1}{288-\frac {(17-22 x)^2}{2 x^2-x+3}}d\frac {17-22 x}{\sqrt {2 x^2-x+3}}-\frac {1622009981 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {2}}\right )+\frac {1}{8} \sqrt {2 x^2-x+3} (141051019-23482924 x)\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{224} \left (\frac {1}{4} \left (\frac {7}{2} \left (\frac {1}{32} \left (\frac {1}{16} \left (-\frac {1622009981 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {2}}-811081728 \sqrt {2} \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )\right )+\frac {1}{8} \sqrt {2 x^2-x+3} (141051019-23482924 x)\right )+\frac {1}{48} (500141-123060 x) \left (2 x^2-x+3\right )^{3/2}\right )+3505 \left (2 x^2-x+3\right )^{5/2}\right )-\frac {311}{2} (2 x+5) \left (2 x^2-x+3\right )^{5/2}\right )+\frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}\) |
(5*(5 + 2*x)^2*(3 - x + 2*x^2)^(5/2))/112 + ((-311*(5 + 2*x)*(3 - x + 2*x^ 2)^(5/2))/2 + (3505*(3 - x + 2*x^2)^(5/2) + (7*(((500141 - 123060*x)*(3 - x + 2*x^2)^(3/2))/48 + (((141051019 - 23482924*x)*Sqrt[3 - x + 2*x^2])/8 + ((-1622009981*ArcSinh[(-1 + 4*x)/Sqrt[23]])/Sqrt[2] - 811081728*Sqrt[2]*A rcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/16)/32))/2)/4)/224
3.4.36.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ (c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c ^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x ] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || !R ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Integer Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
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]))
Timed out.
hanged
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.78 \[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\frac {1}{1376256} \, {\left (983040 \, x^{6} - 3710976 \, x^{5} + 14493696 \, x^{4} - 46476672 \, x^{3} + 159973408 \, x^{2} - 609499532 \, x + 3149403255\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1622009981}{524288} \, \sqrt {2} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + \frac {99009}{32} \, \sqrt {2} \log \left (-\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) \]
1/1376256*(983040*x^6 - 3710976*x^5 + 14493696*x^4 - 46476672*x^3 + 159973 408*x^2 - 609499532*x + 3149403255)*sqrt(2*x^2 - x + 3) + 1622009981/52428 8*sqrt(2)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25 ) + 99009/32*sqrt(2)*log(-(24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) + 10 60*x^2 - 1036*x + 1153)/(4*x^2 + 20*x + 25))
\[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\int \frac {\left (2 x^{2} - x + 3\right )^{\frac {3}{2}} \cdot \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{2 x + 5}\, dx \]
Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91 \[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\frac {5}{28} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} x^{2} - \frac {111}{224} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} x + \frac {1395}{896} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} - \frac {10255}{1024} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {500141}{12288} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {5870731}{16384} \, \sqrt {2 \, x^{2} - x + 3} x - \frac {1622009981}{262144} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) + \frac {99009}{16} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) + \frac {141051019}{65536} \, \sqrt {2 \, x^{2} - x + 3} \]
5/28*(2*x^2 - x + 3)^(5/2)*x^2 - 111/224*(2*x^2 - x + 3)^(5/2)*x + 1395/89 6*(2*x^2 - x + 3)^(5/2) - 10255/1024*(2*x^2 - x + 3)^(3/2)*x + 500141/1228 8*(2*x^2 - x + 3)^(3/2) - 5870731/16384*sqrt(2*x^2 - x + 3)*x - 1622009981 /262144*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) + 99009/16*sqrt(2 )*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) + 1 41051019/65536*sqrt(2*x^2 - x + 3)
Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.81 \[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\frac {1}{1376256} \, {\left (4 \, {\left (8 \, {\left (12 \, {\left (16 \, {\left (4 \, {\left (40 \, x - 151\right )} x + 2359\right )} x - 121033\right )} x + 4999169\right )} x - 152374883\right )} x + 3149403255\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1622009981}{262144} \, \sqrt {2} \log \left (-4 \, \sqrt {2} x + \sqrt {2} + 4 \, \sqrt {2 \, x^{2} - x + 3}\right ) - \frac {99009}{16} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {99009}{16} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) \]
1/1376256*(4*(8*(12*(16*(4*(40*x - 151)*x + 2359)*x - 121033)*x + 4999169) *x - 152374883)*x + 3149403255)*sqrt(2*x^2 - x + 3) + 1622009981/262144*sq rt(2)*log(-4*sqrt(2)*x + sqrt(2) + 4*sqrt(2*x^2 - x + 3)) - 99009/16*sqrt( 2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 99009/16*sqr t(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3)))
Timed out. \[ \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx=\int \frac {{\left (2\,x^2-x+3\right )}^{3/2}\,\left (5\,x^4-x^3+3\,x^2+x+2\right )}{2\,x+5} \,d x \]