Integrand size = 27, antiderivative size = 100 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {\sqrt {3+2 x} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac {\sqrt {3+2 x} (2090+2529 x)}{6 \left (2+5 x+3 x^2\right )}+966 \text {arctanh}\left (\sqrt {3+2 x}\right )-1247 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
966*arctanh((3+2*x)^(1/2))-1247/5*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^( 1/2)-1/6*(121+139*x)*(3+2*x)^(1/2)/(3*x^2+5*x+2)^2+1/6*(2090+2529*x)*(3+2* x)^(1/2)/(3*x^2+5*x+2)
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {\sqrt {3+2 x} \left (1353+5123 x+6305 x^2+2529 x^3\right )}{2 \left (2+5 x+3 x^2\right )^2}+966 \text {arctanh}\left (\sqrt {3+2 x}\right )-1247 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
(Sqrt[3 + 2*x]*(1353 + 5123*x + 6305*x^2 + 2529*x^3))/(2*(2 + 5*x + 3*x^2) ^2) + 966*ArcTanh[Sqrt[3 + 2*x]] - 1247*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1233, 25, 1235, 27, 1197, 1480, 220}
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.
| \(\displaystyle \int \frac {(5-x) (2 x+3)^{3/2}}{\left (3 x^2+5 x+2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle \frac {1}{6} \int -\frac {703 x+1142}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )^2}dx-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{6} \int \frac {703 x+1142}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )^2}dx-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \int \frac {45 (281 x+603)}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {\sqrt {2 x+3} (2529 x+2090)}{3 x^2+5 x+2}\right )-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (9 \int \frac {281 x+603}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {\sqrt {2 x+3} (2529 x+2090)}{3 x^2+5 x+2}\right )-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {1}{6} \left (18 \int \frac {281 (2 x+3)+363}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}+\frac {\sqrt {2 x+3} (2529 x+2090)}{3 x^2+5 x+2}\right )-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {1}{6} \left (18 \left (1247 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-966 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )+\frac {\sqrt {2 x+3} (2529 x+2090)}{3 x^2+5 x+2}\right )-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{6} \left (18 \left (322 \text {arctanh}\left (\sqrt {2 x+3}\right )-\frac {1247 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}}\right )+\frac {\sqrt {2 x+3} (2529 x+2090)}{3 x^2+5 x+2}\right )-\frac {\sqrt {2 x+3} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
-1/6*(Sqrt[3 + 2*x]*(121 + 139*x))/(2 + 5*x + 3*x^2)^2 + ((Sqrt[3 + 2*x]*( 2090 + 2529*x))/(2 + 5*x + 3*x^2) + 18*(322*ArcTanh[Sqrt[3 + 2*x]] - (1247 *ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15]))/6
3.26.69.3.1 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[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 0])
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_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80
| method | result | size |
| risch | \(\frac {\left (2529 x^{3}+6305 x^{2}+5123 x +1353\right ) \sqrt {3+2 x}}{2 \left (3 x^{2}+5 x +2\right )^{2}}+483 \ln \left (\sqrt {3+2 x}+1\right )-\frac {1247 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{5}-483 \ln \left (\sqrt {3+2 x}-1\right )\) | \(80\) |
| trager | \(\frac {\left (2529 x^{3}+6305 x^{2}+5123 x +1353\right ) \sqrt {3+2 x}}{2 \left (3 x^{2}+5 x +2\right )^{2}}+\frac {1247 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{10}+483 \ln \left (\frac {\sqrt {3+2 x}+2+x}{1+x}\right )\) | \(101\) |
| derivativedivides | \(\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {68}{\sqrt {3+2 x}-1}-483 \ln \left (\sqrt {3+2 x}-1\right )-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {68}{\sqrt {3+2 x}+1}+483 \ln \left (\sqrt {3+2 x}+1\right )+\frac {1305 \left (3+2 x \right )^{\frac {3}{2}}-2345 \sqrt {3+2 x}}{\left (6 x +4\right )^{2}}-\frac {1247 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{5}\) | \(124\) |
| default | \(\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {68}{\sqrt {3+2 x}-1}-483 \ln \left (\sqrt {3+2 x}-1\right )-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {68}{\sqrt {3+2 x}+1}+483 \ln \left (\sqrt {3+2 x}+1\right )+\frac {1305 \left (3+2 x \right )^{\frac {3}{2}}-2345 \sqrt {3+2 x}}{\left (6 x +4\right )^{2}}-\frac {1247 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{5}\) | \(124\) |
| pseudoelliptic | \(\frac {-\frac {44892 \sqrt {15}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )}{5}-17388 \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \ln \left (\sqrt {3+2 x}-1\right )+17388 \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \ln \left (\sqrt {3+2 x}+1\right )+5058 \left (x^{3}+\frac {6305}{2529} x^{2}+\frac {5123}{2529} x +\frac {451}{843}\right ) \sqrt {3+2 x}}{\left (\sqrt {3+2 x}-1\right )^{2} \left (\sqrt {3+2 x}+1\right )^{2} \left (2+3 x \right )^{2}}\) | \(127\) |
1/2*(2529*x^3+6305*x^2+5123*x+1353)/(3*x^2+5*x+2)^2*(3+2*x)^(1/2)+483*ln(( 3+2*x)^(1/2)+1)-1247/5*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-483*ln ((3+2*x)^(1/2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (81) = 162\).
Time = 0.26 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.70 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {1247 \, \sqrt {5} \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 4830 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 4830 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + 5 \, {\left (2529 \, x^{3} + 6305 \, x^{2} + 5123 \, x + 1353\right )} \sqrt {2 \, x + 3}}{10 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]
1/10*(1247*sqrt(5)*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(-(sqrt (5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 4830*(9*x^4 + 30*x^3 + 3 7*x^2 + 20*x + 4)*log(sqrt(2*x + 3) + 1) - 4830*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) - 1) + 5*(2529*x^3 + 6305*x^2 + 5123*x + 1353 )*sqrt(2*x + 3))/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)
Time = 102.20 (sec) , antiderivative size = 405, normalized size of antiderivative = 4.05 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {551 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{5} - 3840 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right ) + 3400 \left (\begin {cases} \frac {\sqrt {15} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )^{2}}\right )}{375} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right ) - 483 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 483 \log {\left (\sqrt {2 x + 3} + 1 \right )} + \frac {68}{\sqrt {2 x + 3} + 1} - \frac {3}{\left (\sqrt {2 x + 3} + 1\right )^{2}} + \frac {68}{\sqrt {2 x + 3} - 1} + \frac {3}{\left (\sqrt {2 x + 3} - 1\right )^{2}} \]
551*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(1 5)/3))/5 - 3840*Piecewise((sqrt(15)*(-log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/4 + log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/4 - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 + 1 )) - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 - 1)))/75, (sqrt(2*x + 3) > -sqrt(15)/ 3) & (sqrt(2*x + 3) < sqrt(15)/3))) + 3400*Piecewise((sqrt(15)*(3*log(sqrt (15)*sqrt(2*x + 3)/5 - 1)/16 - 3*log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/16 + 3/ (16*(sqrt(15)*sqrt(2*x + 3)/5 + 1)) + 1/(16*(sqrt(15)*sqrt(2*x + 3)/5 + 1) **2) + 3/(16*(sqrt(15)*sqrt(2*x + 3)/5 - 1)) - 1/(16*(sqrt(15)*sqrt(2*x + 3)/5 - 1)**2))/375, (sqrt(2*x + 3) > -sqrt(15)/3) & (sqrt(2*x + 3) < sqrt( 15)/3))) - 483*log(sqrt(2*x + 3) - 1) + 483*log(sqrt(2*x + 3) + 1) + 68/(s qrt(2*x + 3) + 1) - 3/(sqrt(2*x + 3) + 1)**2 + 68/(sqrt(2*x + 3) - 1) + 3/ (sqrt(2*x + 3) - 1)**2
Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.33 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {1247}{10} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {2529 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 10151 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 13115 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 5445 \, \sqrt {2 \, x + 3}}{9 \, {\left (2 \, x + 3\right )}^{4} - 48 \, {\left (2 \, x + 3\right )}^{3} + 94 \, {\left (2 \, x + 3\right )}^{2} - 160 \, x - 215} + 483 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 483 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
1247/10*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + (2529*(2*x + 3)^(7/2) - 10151*(2*x + 3)^(5/2) + 13115*(2*x + 3)^( 3/2) - 5445*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 + 94*(2*x + 3)^ 2 - 160*x - 215) + 483*log(sqrt(2*x + 3) + 1) - 483*log(sqrt(2*x + 3) - 1)
Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.19 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {1247}{10} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {2529 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 10151 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 13115 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 5445 \, \sqrt {2 \, x + 3}}{{\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 483 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 483 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
1247/10*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3* sqrt(2*x + 3))) + (2529*(2*x + 3)^(7/2) - 10151*(2*x + 3)^(5/2) + 13115*(2 *x + 3)^(3/2) - 5445*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 + 483*lo g(sqrt(2*x + 3) + 1) - 483*log(abs(sqrt(2*x + 3) - 1))
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=966\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )+\frac {605\,\sqrt {2\,x+3}-\frac {13115\,{\left (2\,x+3\right )}^{3/2}}{9}+\frac {10151\,{\left (2\,x+3\right )}^{5/2}}{9}-281\,{\left (2\,x+3\right )}^{7/2}}{\frac {160\,x}{9}-\frac {94\,{\left (2\,x+3\right )}^2}{9}+\frac {16\,{\left (2\,x+3\right )}^3}{3}-{\left (2\,x+3\right )}^4+\frac {215}{9}}-\frac {1247\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{5} \]