Integrand size = 27, antiderivative size = 102 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {(3+2 x)^{3/2} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac {25 \sqrt {3+2 x} (112+131 x)}{6 \left (2+5 x+3 x^2\right )}+1250 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {2905}{3} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
-1/6*(3+2*x)^(3/2)*(121+139*x)/(3*x^2+5*x+2)^2+1250*arctanh((3+2*x)^(1/2)) -2905/9*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+25/6*(112+131*x)*(3+2 *x)^(1/2)/(3*x^2+5*x+2)
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {\sqrt {3+2 x} \left (5237+19891 x+24497 x^2+9825 x^3\right )}{6 \left (2+5 x+3 x^2\right )^2}+1250 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {2905}{3} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
(Sqrt[3 + 2*x]*(5237 + 19891*x + 24497*x^2 + 9825*x^3))/(6*(2 + 5*x + 3*x^ 2)^2) + 1250*ArcTanh[Sqrt[3 + 2*x]] - (2905*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sq rt[3 + 2*x]])/3
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1233, 27, 1234, 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.
| \(\displaystyle \int \frac {(5-x) (2 x+3)^{5/2}}{\left (3 x^2+5 x+2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle \frac {1}{6} \int -\frac {25 \sqrt {2 x+3} (17 x+36)}{\left (3 x^2+5 x+2\right )^2}dx-\frac {(2 x+3)^{3/2} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {25}{6} \int \frac {\sqrt {2 x+3} (17 x+36)}{\left (3 x^2+5 x+2\right )^2}dx-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1234 |
| \(\displaystyle -\frac {25}{6} \left (-\int \frac {131 x+281}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle -\frac {25}{6} \left (-2 \int \frac {131 (2 x+3)+169}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle -\frac {25}{6} \left (-2 \left (581 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-450 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {25}{6} \left (-2 \left (150 \text {arctanh}\left (\sqrt {2 x+3}\right )-\frac {581 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}}\right )-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\) |
-1/6*((3 + 2*x)^(3/2)*(121 + 139*x))/(2 + 5*x + 3*x^2)^2 - (25*(-((Sqrt[3 + 2*x]*(112 + 131*x))/(2 + 5*x + 3*x^2)) - 2*(150*ArcTanh[Sqrt[3 + 2*x]] - (581*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15])))/6
3.26.68.3.1 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*(a + b*x + c*x^2)^(p + 1)*( (f*b - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g *(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)* (m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1 ] && GtQ[m, 0] && (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 = 0.47 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78
| method | result | size |
| risch | \(\frac {\left (9825 x^{3}+24497 x^{2}+19891 x +5237\right ) \sqrt {3+2 x}}{6 \left (3 x^{2}+5 x +2\right )^{2}}+625 \ln \left (\sqrt {3+2 x}+1\right )-\frac {2905 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{9}-625 \ln \left (\sqrt {3+2 x}-1\right )\) | \(80\) |
| trager | \(\frac {\left (9825 x^{3}+24497 x^{2}+19891 x +5237\right ) \sqrt {3+2 x}}{6 \left (3 x^{2}+5 x +2\right )^{2}}-625 \ln \left (\frac {-2-x +\sqrt {3+2 x}}{1+x}\right )+\frac {2905 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{18}\) | \(103\) |
| derivativedivides | \(\frac {1835 \left (3+2 x \right )^{\frac {3}{2}}-\frac {10025 \sqrt {3+2 x}}{3}}{\left (6 x +4\right )^{2}}-\frac {2905 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{9}-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {80}{\sqrt {3+2 x}+1}+625 \ln \left (\sqrt {3+2 x}+1\right )+\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {80}{\sqrt {3+2 x}-1}-625 \ln \left (\sqrt {3+2 x}-1\right )\) | \(124\) |
| default | \(\frac {1835 \left (3+2 x \right )^{\frac {3}{2}}-\frac {10025 \sqrt {3+2 x}}{3}}{\left (6 x +4\right )^{2}}-\frac {2905 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{9}-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {80}{\sqrt {3+2 x}+1}+625 \ln \left (\sqrt {3+2 x}+1\right )+\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {80}{\sqrt {3+2 x}-1}-625 \ln \left (\sqrt {3+2 x}-1\right )\) | \(124\) |
| pseudoelliptic | \(\frac {-11620 \sqrt {15}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )-22500 \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \ln \left (\sqrt {3+2 x}-1\right )+22500 \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \ln \left (\sqrt {3+2 x}+1\right )+6550 \left (x^{3}+\frac {187}{75} x^{2}+\frac {19891}{9825} x +\frac {5237}{9825}\right ) \sqrt {3+2 x}}{\left (2+3 x \right )^{2} \left (\sqrt {3+2 x}+1\right )^{2} \left (\sqrt {3+2 x}-1\right )^{2}}\) | \(127\) |
1/6*(9825*x^3+24497*x^2+19891*x+5237)/(3*x^2+5*x+2)^2*(3+2*x)^(1/2)+625*ln ((3+2*x)^(1/2)+1)-2905/9*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-625* 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 = 0.25 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.67 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {2905 \, \sqrt {5} \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 11250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 11250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + 3 \, {\left (9825 \, x^{3} + 24497 \, x^{2} + 19891 \, x + 5237\right )} \sqrt {2 \, x + 3}}{18 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]
1/18*(2905*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)) + 11250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) + 1) - 11250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) - 1) + 3*(9825*x^3 + 24497*x^2 + 19891*x + 5237)*sqrt(2*x + 3))/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)
Time = 102.47 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.98 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=141 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right ) - \frac {15800 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{3} + \frac {17000 \left (\begin {cases} \frac {\sqrt {15} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )^{2}}\right )}{375} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{3} - 625 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 625 \log {\left (\sqrt {2 x + 3} + 1 \right )} + \frac {80}{\sqrt {2 x + 3} + 1} - \frac {3}{\left (\sqrt {2 x + 3} + 1\right )^{2}} + \frac {80}{\sqrt {2 x + 3} - 1} + \frac {3}{\left (\sqrt {2 x + 3} - 1\right )^{2}} \]
141*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(1 5)/3)) - 15800*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)))/3 + 17000*Piecewise((sqrt(15)*(3*log(sq rt(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) < sqr t(15)/3)))/3 - 625*log(sqrt(2*x + 3) - 1) + 625*log(sqrt(2*x + 3) + 1) + 8 0/(sqrt(2*x + 3) + 1) - 3/(sqrt(2*x + 3) + 1)**2 + 80/(sqrt(2*x + 3) - 1) + 3/(sqrt(2*x + 3) - 1)**2
Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.31 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {2905}{18} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {9825 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 39431 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 50875 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 21125 \, \sqrt {2 \, x + 3}}{3 \, {\left (9 \, {\left (2 \, x + 3\right )}^{4} - 48 \, {\left (2 \, x + 3\right )}^{3} + 94 \, {\left (2 \, x + 3\right )}^{2} - 160 \, x - 215\right )}} + 625 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 625 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
2905/18*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/3*(9825*(2*x + 3)^(7/2) - 39431*(2*x + 3)^(5/2) + 50875*(2*x + 3)^(3/2) - 21125*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 + 94*(2*x + 3)^2 - 160*x - 215) + 625*log(sqrt(2*x + 3) + 1) - 625*log(sqrt(2*x + 3) - 1)
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.18 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {2905}{18} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {9825 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 39431 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 50875 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 21125 \, \sqrt {2 \, x + 3}}{3 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 625 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 625 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
2905/18*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3* sqrt(2*x + 3))) + 1/3*(9825*(2*x + 3)^(7/2) - 39431*(2*x + 3)^(5/2) + 5087 5*(2*x + 3)^(3/2) - 21125*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 + 6 25*log(sqrt(2*x + 3) + 1) - 625*log(abs(sqrt(2*x + 3) - 1))
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=1250\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )+\frac {\frac {21125\,\sqrt {2\,x+3}}{27}-\frac {50875\,{\left (2\,x+3\right )}^{3/2}}{27}+\frac {39431\,{\left (2\,x+3\right )}^{5/2}}{27}-\frac {3275\,{\left (2\,x+3\right )}^{7/2}}{9}}{\frac {160\,x}{9}-\frac {94\,{\left (2\,x+3\right )}^2}{9}+\frac {16\,{\left (2\,x+3\right )}^3}{3}-{\left (2\,x+3\right )}^4+\frac {215}{9}}-\frac {2905\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{9} \]