Integrand size = 27, antiderivative size = 111 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {2114}{125 (3+2 x)^{5/2}}-\frac {7042}{375 (3+2 x)^{3/2}}-\frac {24626}{625 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}+14 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {15876}{625} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
-2114/125/(3+2*x)^(5/2)-7042/375/(3+2*x)^(3/2)-3/5*(37+47*x)/(3+2*x)^(5/2) /(3*x^2+5*x+2)+14*arctanh((3+2*x)^(1/2))+15876/3125*arctanh(1/5*15^(1/2)*( 3+2*x)^(1/2))*15^(1/2)-24626/625/(3+2*x)^(1/2)
Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=14 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {-\frac {5 \left (1646109+5977997 x+7782530 x^2+4348428 x^3+886536 x^4\right )}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}+47628 \sqrt {15} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{9375} \]
14*ArcTanh[Sqrt[3 + 2*x]] + ((-5*(1646109 + 5977997*x + 7782530*x^2 + 4348 428*x^3 + 886536*x^4))/((3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)) + 47628*Sqrt[15 ]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/9375
Time = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1235, 27, 1198, 1198, 1198, 1197, 25, 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.
| \(\displaystyle \int \frac {5-x}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\frac {1}{5} \int \frac {7 (141 x+136)}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )}dx-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {7}{5} \int \frac {141 x+136}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )}dx-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \int \frac {453 x+428}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}dx+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \int \frac {1509 x+1384}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}dx+\frac {1006}{15 (2 x+3)^{3/2}}\right )+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {1}{5} \int \frac {5277 x+4652}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {3518}{5 \sqrt {2 x+3}}\right )+\frac {1006}{15 (2 x+3)^{3/2}}\right )+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {2}{5} \int -\frac {6527-5277 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}+\frac {3518}{5 \sqrt {2 x+3}}\right )+\frac {1006}{15 (2 x+3)^{3/2}}\right )+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {3518}{5 \sqrt {2 x+3}}-\frac {2}{5} \int \frac {6527-5277 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}\right )+\frac {1006}{15 (2 x+3)^{3/2}}\right )+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {2}{5} \left (3402 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}+1875 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )+\frac {3518}{5 \sqrt {2 x+3}}\right )+\frac {1006}{15 (2 x+3)^{3/2}}\right )+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {2}{5} \left (-625 \text {arctanh}\left (\sqrt {2 x+3}\right )-1134 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )+\frac {3518}{5 \sqrt {2 x+3}}\right )+\frac {1006}{15 (2 x+3)^{3/2}}\right )+\frac {302}{25 (2 x+3)^{5/2}}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\) |
(-3*(37 + 47*x))/(5*(3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)) - (7*(302/(25*(3 + 2*x)^(5/2)) + (1006/(15*(3 + 2*x)^(3/2)) + (3518/(5*Sqrt[3 + 2*x]) + (2*(- 625*ArcTanh[Sqrt[3 + 2*x]] - 1134*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x ]]))/5)/5)/5))/5
3.26.65.3.1 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), x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c *d^2 - b*d*e + a*e^2))), x] + Simp[1/(c*d^2 - b*d*e + a*e^2) Int[(d + e*x )^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] && LtQ[m, -1 ]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.41 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95
| method | result | size |
| trager | \(-\frac {886536 x^{4}+4348428 x^{3}+7782530 x^{2}+5977997 x +1646109}{1875 \left (3+2 x \right )^{\frac {5}{2}} \left (3 x^{2}+5 x +2\right )}+7 \ln \left (\frac {\sqrt {3+2 x}+2+x}{1+x}\right )-\frac {7938 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{3125}\) | \(106\) |
| derivativedivides | \(-\frac {918 \sqrt {3+2 x}}{625 \left (\frac {4}{3}+2 x \right )}+\frac {15876 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{3125}-\frac {6}{\sqrt {3+2 x}-1}-7 \ln \left (\sqrt {3+2 x}-1\right )-\frac {6}{\sqrt {3+2 x}+1}+7 \ln \left (\sqrt {3+2 x}+1\right )-\frac {104}{125 \left (3+2 x \right )^{\frac {5}{2}}}-\frac {1624}{375 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {16208}{625 \sqrt {3+2 x}}\) | \(113\) |
| default | \(-\frac {918 \sqrt {3+2 x}}{625 \left (\frac {4}{3}+2 x \right )}+\frac {15876 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{3125}-\frac {6}{\sqrt {3+2 x}-1}-7 \ln \left (\sqrt {3+2 x}-1\right )-\frac {6}{\sqrt {3+2 x}+1}+7 \ln \left (\sqrt {3+2 x}+1\right )-\frac {104}{125 \left (3+2 x \right )^{\frac {5}{2}}}-\frac {1624}{375 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {16208}{625 \sqrt {3+2 x}}\) | \(113\) |
| pseudoelliptic | \(\frac {571536 \sqrt {3+2 x}\, \sqrt {15}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \left (x +\frac {3}{2}\right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )-787500 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \left (x +\frac {3}{2}\right )^{2} \ln \left (\sqrt {3+2 x}-1\right )+787500 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \left (x +\frac {3}{2}\right )^{2} \ln \left (\sqrt {3+2 x}+1\right )-4432680 x^{4}-21742140 x^{3}-38912650 x^{2}-29889985 x -8230545}{\left (3+2 x \right )^{\frac {5}{2}} \left (28125 x^{2}+46875 x +18750\right )}\) | \(137\) |
-1/1875*(886536*x^4+4348428*x^3+7782530*x^2+5977997*x+1646109)/(3+2*x)^(5/ 2)/(3*x^2+5*x+2)+7*ln(((3+2*x)^(1/2)+2+x)/(1+x))-7938/3125*RootOf(_Z^2-15) *ln((-3*RootOf(_Z^2-15)*x+15*(3+2*x)^(1/2)-7*RootOf(_Z^2-15))/(2+3*x))
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.75 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {23814 \, \sqrt {5} \sqrt {3} {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) + 65625 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 65625 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 5 \, {\left (886536 \, x^{4} + 4348428 \, x^{3} + 7782530 \, x^{2} + 5977997 \, x + 1646109\right )} \sqrt {2 \, x + 3}}{9375 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )}} \]
1/9375*(23814*sqrt(5)*sqrt(3)*(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243* x + 54)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x + 7)/(3*x + 2)) + 65625*( 24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(sqrt(2*x + 3) + 1) - 65625*(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(sqrt(2*x + 3) - 1) - 5*(886536*x^4 + 4348428*x^3 + 7782530*x^2 + 5977997*x + 1646109 )*sqrt(2*x + 3))/(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)
Time = 47.34 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.23 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=- \frac {7479 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{3125} + \frac {5508 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{125} - 7 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 7 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} - \frac {16208}{625 \sqrt {2 x + 3}} - \frac {1624}{375 \left (2 x + 3\right )^{\frac {3}{2}}} - \frac {104}{125 \left (2 x + 3\right )^{\frac {5}{2}}} \]
-7479*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt (15)/3))/3125 + 5508*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)))/125 - 7*log(sqrt(2*x + 3) - 1) + 7*log(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) - 1) - 16208/(625*sqrt(2*x + 3)) - 1624/(375*(2*x + 3)**(3/2)) - 104/(125*(2*x + 3)**(5/2))
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {7938}{3125} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {2 \, {\left (110817 \, {\left (2 \, x + 3\right )}^{4} - 242697 \, {\left (2 \, x + 3\right )}^{3} + 91420 \, {\left (2 \, x + 3\right )}^{2} + 28120 \, x + 46080\right )}}{1875 \, {\left (3 \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} - 8 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + 5 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}}\right )}} + 7 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 7 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
-7938/3125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2 *x + 3))) - 2/1875*(110817*(2*x + 3)^4 - 242697*(2*x + 3)^3 + 91420*(2*x + 3)^2 + 28120*x + 46080)/(3*(2*x + 3)^(9/2) - 8*(2*x + 3)^(7/2) + 5*(2*x + 3)^(5/2)) + 7*log(sqrt(2*x + 3) + 1) - 7*log(sqrt(2*x + 3) - 1)
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {7938}{3125} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {6 \, {\left (4209 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 6709 \, \sqrt {2 \, x + 3}\right )}}{625 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - \frac {16 \, {\left (3039 \, {\left (2 \, x + 3\right )}^{2} + 1015 \, x + 1620\right )}}{1875 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}}} + 7 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 7 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
-7938/3125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 6/625*(4209*(2*x + 3)^(3/2) - 6709*sqrt(2*x + 3))/(3* (2*x + 3)^2 - 16*x - 19) - 16/1875*(3039*(2*x + 3)^2 + 1015*x + 1620)/(2*x + 3)^(5/2) + 7*log(sqrt(2*x + 3) + 1) - 7*log(abs(sqrt(2*x + 3) - 1))
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^2} \, dx=14\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )+\frac {15876\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{3125}-\frac {\frac {11248\,x}{1125}+\frac {36568\,{\left (2\,x+3\right )}^2}{1125}-\frac {161798\,{\left (2\,x+3\right )}^3}{1875}+\frac {24626\,{\left (2\,x+3\right )}^4}{625}+\frac {2048}{125}}{\frac {5\,{\left (2\,x+3\right )}^{5/2}}{3}-\frac {8\,{\left (2\,x+3\right )}^{7/2}}{3}+{\left (2\,x+3\right )}^{9/2}} \]