Integrand size = 29, antiderivative size = 197 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {6 (37+47 x)}{5 (3+2 x)^{3/2} \sqrt {2+5 x+3 x^2}}-\frac {2516 \sqrt {2+5 x+3 x^2}}{75 (3+2 x)^{3/2}}-\frac {14876 \sqrt {2+5 x+3 x^2}}{375 \sqrt {3+2 x}}+\frac {7438 \sqrt {-2-5 x-3 x^2} E\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{125 \sqrt {3} \sqrt {2+5 x+3 x^2}}-\frac {1258 \sqrt {-2-5 x-3 x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right ),-\frac {2}{3}\right )}{25 \sqrt {3} \sqrt {2+5 x+3 x^2}} \]
-6/5*(37+47*x)/(3+2*x)^(3/2)/(3*x^2+5*x+2)^(1/2)+7438/375*EllipticE(3^(1/2 )*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1 /2)-1258/75*EllipticF(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1 /2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)-2516/75*(3*x^2+5*x+2)^(1/2)/(3+2*x)^(3/2)- 14876/375*(3*x^2+5*x+2)^(1/2)/(3+2*x)^(1/2)
Time = 31.31 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.90 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {2 \left (-20905-42025 x-18870 x^2+3719 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{5/2} \sqrt {\frac {2+3 x}{3+2 x}} E\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right )|\frac {3}{5}\right )-1832 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{5/2} \sqrt {\frac {2+3 x}{3+2 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right ),\frac {3}{5}\right )\right )}{375 (3+2 x)^{3/2} \sqrt {2+5 x+3 x^2}} \]
(2*(-20905 - 42025*x - 18870*x^2 + 3719*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(5/2)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] - 1832*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(5/2)*Sqrt[ (2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5]))/(37 5*(3 + 2*x)^(3/2)*Sqrt[2 + 5*x + 3*x^2])
Time = 0.37 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {1235, 1237, 27, 1237, 27, 1269, 1172, 27, 321, 327}
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.
| \(\displaystyle \int \frac {5-x}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\frac {2}{5} \int \frac {423 x+320}{(2 x+3)^{5/2} \sqrt {3 x^2+5 x+2}}dx-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}-\frac {2}{15} \int -\frac {1887 x+971}{2 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}dx\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \int \frac {1887 x+971}{(2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}dx+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {2}{5} \int \frac {3 (3719 x+4006)}{2 \sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \int \frac {3719 x+4006}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {3719}{2} \int \frac {\sqrt {2 x+3}}{\sqrt {3 x^2+5 x+2}}dx-\frac {3145}{2} \int \frac {1}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1172 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {3719 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {3} \sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {3145 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {3}}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\right )\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {3719 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {3145 \sqrt {-3 x^2-5 x-2} \int \frac {1}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 321 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {3719 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {3145 \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\right )\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 327 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{15} \left (\frac {7438 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {3719 \sqrt {-3 x^2-5 x-2} E\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {3145 \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\right )\right )+\frac {1258 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^{3/2}}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}\) |
(-6*(37 + 47*x))/(5*(3 + 2*x)^(3/2)*Sqrt[2 + 5*x + 3*x^2]) - (2*((1258*Sqr t[2 + 5*x + 3*x^2])/(15*(3 + 2*x)^(3/2)) + ((7438*Sqrt[2 + 5*x + 3*x^2])/( 5*Sqrt[3 + 2*x]) - (3*((3719*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[ 3]*Sqrt[1 + x]], -2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (3145*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])))/5)/15))/5
3.27.19.3.1 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]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c /(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0] && !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sy mbol] :> Simp[2*Rt[b^2 - 4*a*c, 2]*(d + e*x)^m*(Sqrt[(-c)*((a + b*x + c*x^2 )/(b^2 - 4*a*c))]/(c*Sqrt[a + b*x + c*x^2]*(2*c*((d + e*x)/(2*c*d - b*e - e *Rt[b^2 - 4*a*c, 2])))^m)) Subst[Int[(1 + 2*e*Rt[b^2 - 4*a*c, 2]*(x^2/(2* c*d - b*e - e*Rt[b^2 - 4*a*c, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^ 2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b, c, d, e }, x] && EqQ[m^2, 1/4]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[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[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[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]
Time = 0.36 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03
| method | result | size |
| default | \(-\frac {2 \left (1722 F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right ) \sqrt {15}\, x \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {3+2 x}+7438 \sqrt {15}\, E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right ) x \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {3+2 x}+2583 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {15}\, \sqrt {3+2 x}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )+11157 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {15}\, \sqrt {3+2 x}\, E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )+669420 x^{3}+2402880 x^{2}+2750205 x +982995\right )}{5625 \left (3+2 x \right )^{\frac {3}{2}} \sqrt {3 x^{2}+5 x +2}}\) | \(203\) |
| elliptic | \(\frac {\sqrt {\left (3+2 x \right ) \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {26 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}{75 \left (x +\frac {3}{2}\right )^{2}}-\frac {404 \left (6 x^{2}+10 x +4\right )}{75 \sqrt {\left (x +\frac {3}{2}\right ) \left (6 x^{2}+10 x +4\right )}}-\frac {2 \left (9+6 x \right ) \left (\frac {653}{125}+\frac {903 x}{125}\right )}{\sqrt {\left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right ) \left (9+6 x \right )}}-\frac {8012 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {45+30 x}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )}{1875 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}-\frac {7438 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {45+30 x}\, \left (\frac {E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )}{3}-F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )\right )}{1875 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}\right )}{\sqrt {3+2 x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(250\) |
-2/5625*(1722*EllipticF(1/5*(-20-30*x)^(1/2),1/2*10^(1/2))*15^(1/2)*x*(-20 -30*x)^(1/2)*(3+3*x)^(1/2)*(3+2*x)^(1/2)+7438*15^(1/2)*EllipticE(1/5*(-20- 30*x)^(1/2),1/2*10^(1/2))*x*(-20-30*x)^(1/2)*(3+3*x)^(1/2)*(3+2*x)^(1/2)+2 583*(-20-30*x)^(1/2)*(3+3*x)^(1/2)*15^(1/2)*(3+2*x)^(1/2)*EllipticF(1/5*(- 20-30*x)^(1/2),1/2*10^(1/2))+11157*(-20-30*x)^(1/2)*(3+3*x)^(1/2)*15^(1/2) *(3+2*x)^(1/2)*EllipticE(1/5*(-20-30*x)^(1/2),1/2*10^(1/2))+669420*x^3+240 2880*x^2+2750205*x+982995)/(3+2*x)^(3/2)/(3*x^2+5*x+2)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.64 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {1447 \, \sqrt {6} {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right ) - 66942 \, \sqrt {6} {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} {\rm weierstrassZeta}\left (\frac {19}{27}, -\frac {28}{729}, {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right )\right ) - 18 \, {\left (44628 \, x^{3} + 160192 \, x^{2} + 183347 \, x + 65533\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {2 \, x + 3}}{3375 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} \]
1/3375*(1447*sqrt(6)*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*weierstrassPIn verse(19/27, -28/729, x + 19/18) - 66942*sqrt(6)*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*weierstrassZeta(19/27, -28/729, weierstrassPInverse(19/27, - 28/729, x + 19/18)) - 18*(44628*x^3 + 160192*x^2 + 183347*x + 65533)*sqrt( 3*x^2 + 5*x + 2)*sqrt(2*x + 3))/(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)
\[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \frac {x}{12 x^{4} \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 56 x^{3} \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 95 x^{2} \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 69 x \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 18 \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{12 x^{4} \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 56 x^{3} \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 95 x^{2} \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 69 x \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2} + 18 \sqrt {2 x + 3} \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]
-Integral(x/(12*x**4*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 56*x**3*sqrt(2 *x + 3)*sqrt(3*x**2 + 5*x + 2) + 95*x**2*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 69*x*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 18*sqrt(2*x + 3)*sqrt(3* x**2 + 5*x + 2)), x) - Integral(-5/(12*x**4*sqrt(2*x + 3)*sqrt(3*x**2 + 5* x + 2) + 56*x**3*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 95*x**2*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 69*x*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 1 8*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2)), x)
\[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {x - 5}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (2 \, x + 3\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {x - 5}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (2 \, x + 3\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {x-5}{{\left (2\,x+3\right )}^{5/2}\,{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]