Integrand size = 27, antiderivative size = 141 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {56399}{625 (3+2 x)^{5/2}}+\frac {102697}{1875 (3+2 x)^{3/2}}-\frac {24409}{3125 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2}+\frac {8852+9957 x}{50 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}+266 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {806841 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{3125} \]
56399/625/(3+2*x)^(5/2)+102697/1875/(3+2*x)^(3/2)-3/10*(37+47*x)/(3+2*x)^( 5/2)/(3*x^2+5*x+2)^2+1/50*(8852+9957*x)/(3+2*x)^(5/2)/(3*x^2+5*x+2)+266*ar ctanh((3+2*x)^(1/2))-806841/15625*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^( 1/2)-24409/3125/(3+2*x)^(1/2)
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.69 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {20250051+94082723 x+160041829 x^2+114099329 x^3+18312714 x^4-14906052 x^5-5272344 x^6}{18750 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2}+266 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {806841 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{3125} \]
(20250051 + 94082723*x + 160041829*x^2 + 114099329*x^3 + 18312714*x^4 - 14 906052*x^5 - 5272344*x^6)/(18750*(3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)^2) + 26 6*ArcTanh[Sqrt[3 + 2*x]] - (806841*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2* x]])/3125
Time = 0.37 (sec) , antiderivative size = 164, 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, 1235, 27, 1198, 1198, 1198, 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.
| \(\displaystyle \int \frac {5-x}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\frac {1}{10} \int \frac {1551 x+1772}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )^2}dx-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \int \frac {7 (9957 x+10907)}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )}dx+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \int \frac {9957 x+10907}{(2 x+3)^{7/2} \left (3 x^2+5 x+2\right )}dx+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \left (\frac {1}{5} \int \frac {24171 x+28921}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}dx+\frac {16114}{25 (2 x+3)^{5/2}}\right )+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \int \frac {44013 x+67763}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}dx+\frac {29342}{15 (2 x+3)^{3/2}}\right )+\frac {16114}{25 (2 x+3)^{5/2}}\right )+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {1}{5} \int \frac {108289-10461 x}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {6974}{5 \sqrt {2 x+3}}\right )+\frac {29342}{15 (2 x+3)^{3/2}}\right )+\frac {16114}{25 (2 x+3)^{5/2}}\right )+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {2}{5} \int \frac {247961-10461 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {6974}{5 \sqrt {2 x+3}}\right )+\frac {29342}{15 (2 x+3)^{3/2}}\right )+\frac {16114}{25 (2 x+3)^{5/2}}\right )+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {2}{5} \left (345789 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-356250 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-\frac {6974}{5 \sqrt {2 x+3}}\right )+\frac {29342}{15 (2 x+3)^{3/2}}\right )+\frac {16114}{25 (2 x+3)^{5/2}}\right )+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{5} \left (\frac {1}{5} \left (\frac {1}{5} \left (\frac {2}{5} \left (118750 \text {arctanh}\left (\sqrt {2 x+3}\right )-115263 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )-\frac {6974}{5 \sqrt {2 x+3}}\right )+\frac {29342}{15 (2 x+3)^{3/2}}\right )+\frac {16114}{25 (2 x+3)^{5/2}}\right )+\frac {9957 x+8852}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}\) |
(-3*(37 + 47*x))/(10*(3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)^2) + ((8852 + 9957* x)/(5*(3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)) + (7*(16114/(25*(3 + 2*x)^(5/2)) + (29342/(15*(3 + 2*x)^(3/2)) + (-6974/(5*Sqrt[3 + 2*x]) + (2*(118750*ArcT anh[Sqrt[3 + 2*x]] - 115263*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/5 )/5)/5))/5)/10
3.26.74.3.1 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]
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.67
| method | result | size |
| risch | \(-\frac {5272344 x^{6}+14906052 x^{5}-18312714 x^{4}-114099329 x^{3}-160041829 x^{2}-94082723 x -20250051}{18750 \left (3+2 x \right )^{\frac {5}{2}} \left (3 x^{2}+5 x +2\right )^{2}}+133 \ln \left (\sqrt {3+2 x}+1\right )-\frac {806841 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{15625}-133 \ln \left (\sqrt {3+2 x}-1\right )\) | \(95\) |
| trager | \(-\frac {5272344 x^{6}+14906052 x^{5}-18312714 x^{4}-114099329 x^{3}-160041829 x^{2}-94082723 x -20250051}{18750 \left (3+2 x \right )^{\frac {5}{2}} \left (3 x^{2}+5 x +2\right )^{2}}+133 \ln \left (\frac {\sqrt {3+2 x}+2+x}{1+x}\right )+\frac {567 \operatorname {RootOf}\left (\textit {\_Z}^{2}-30373935\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-30373935\right ) x +7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-30373935\right )-21345 \sqrt {3+2 x}}{2+3 x}\right )}{31250}\) | \(117\) |
| derivativedivides | \(\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {8}{\sqrt {3+2 x}-1}-133 \ln \left (\sqrt {3+2 x}-1\right )-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {8}{\sqrt {3+2 x}+1}+133 \ln \left (\sqrt {3+2 x}+1\right )-\frac {416}{625 \left (3+2 x \right )^{\frac {5}{2}}}-\frac {9824}{1875 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {137184}{3125 \sqrt {3+2 x}}+\frac {\frac {22599 \left (3+2 x \right )^{\frac {3}{2}}}{125}-\frac {196587 \sqrt {3+2 x}}{625}}{\left (6 x +4\right )^{2}}-\frac {806841 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{15625}\) | \(151\) |
| default | \(\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {8}{\sqrt {3+2 x}-1}-133 \ln \left (\sqrt {3+2 x}-1\right )-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {8}{\sqrt {3+2 x}+1}+133 \ln \left (\sqrt {3+2 x}+1\right )-\frac {416}{625 \left (3+2 x \right )^{\frac {5}{2}}}-\frac {9824}{1875 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {137184}{3125 \sqrt {3+2 x}}+\frac {\frac {22599 \left (3+2 x \right )^{\frac {3}{2}}}{125}-\frac {196587 \sqrt {3+2 x}}{625}}{\left (6 x +4\right )^{2}}-\frac {806841 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{15625}\) | \(151\) |
| pseudoelliptic | \(-\frac {12909456 \left (\sqrt {3+2 x}\, \sqrt {15}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \left (x +\frac {3}{2}\right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )+\frac {296875 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \left (x +\frac {3}{2}\right )^{2} \ln \left (\sqrt {3+2 x}-1\right )}{115263}-\frac {296875 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \left (x +\frac {3}{2}\right )^{2} \ln \left (\sqrt {3+2 x}+1\right )}{115263}-\frac {800209145 x^{2}}{174277656}-\frac {570496645 x^{3}}{174277656}-\frac {726695 x^{4}}{1383156}+\frac {98585 x^{5}}{230526}+\frac {17435 x^{6}}{115263}-\frac {67201945 x}{24896808}-\frac {33750085}{58092552}\right )}{15625 \left (3+2 x \right )^{\frac {5}{2}} \left (\sqrt {3+2 x}+1\right )^{2} \left (x +\frac {2}{3}\right )^{2} \left (\sqrt {3+2 x}-1\right )^{2}}\) | \(174\) |
-1/18750*(5272344*x^6+14906052*x^5-18312714*x^4-114099329*x^3-160041829*x^ 2-94082723*x-20250051)/(3+2*x)^(5/2)/(3*x^2+5*x+2)^2+133*ln((3+2*x)^(1/2)+ 1)-806841/15625*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-133*ln((3+2*x )^(1/2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (108) = 216\).
Time = 0.30 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.74 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {2420523 \, \sqrt {5} \sqrt {3} {\left (72 \, x^{7} + 564 \, x^{6} + 1862 \, x^{5} + 3355 \, x^{4} + 3560 \, x^{3} + 2223 \, x^{2} + 756 \, x + 108\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 12468750 \, {\left (72 \, x^{7} + 564 \, x^{6} + 1862 \, x^{5} + 3355 \, x^{4} + 3560 \, x^{3} + 2223 \, x^{2} + 756 \, x + 108\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 12468750 \, {\left (72 \, x^{7} + 564 \, x^{6} + 1862 \, x^{5} + 3355 \, x^{4} + 3560 \, x^{3} + 2223 \, x^{2} + 756 \, x + 108\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 5 \, {\left (5272344 \, x^{6} + 14906052 \, x^{5} - 18312714 \, x^{4} - 114099329 \, x^{3} - 160041829 \, x^{2} - 94082723 \, x - 20250051\right )} \sqrt {2 \, x + 3}}{93750 \, {\left (72 \, x^{7} + 564 \, x^{6} + 1862 \, x^{5} + 3355 \, x^{4} + 3560 \, x^{3} + 2223 \, x^{2} + 756 \, x + 108\right )}} \]
1/93750*(2420523*sqrt(5)*sqrt(3)*(72*x^7 + 564*x^6 + 1862*x^5 + 3355*x^4 + 3560*x^3 + 2223*x^2 + 756*x + 108)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 12468750*(72*x^7 + 564*x^6 + 1862*x^5 + 3355*x^4 + 3 560*x^3 + 2223*x^2 + 756*x + 108)*log(sqrt(2*x + 3) + 1) - 12468750*(72*x^ 7 + 564*x^6 + 1862*x^5 + 3355*x^4 + 3560*x^3 + 2223*x^2 + 756*x + 108)*log (sqrt(2*x + 3) - 1) - 5*(5272344*x^6 + 14906052*x^5 - 18312714*x^4 - 11409 9329*x^3 - 160041829*x^2 - 94082723*x - 20250051)*sqrt(2*x + 3))/(72*x^7 + 564*x^6 + 1862*x^5 + 3355*x^4 + 3560*x^3 + 2223*x^2 + 756*x + 108)
Time = 117.03 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.15 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {372033 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{15625} - \frac {351864 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{625} + \frac {33048 \left (\begin {cases} \frac {\sqrt {15} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )^{2}}\right )}{375} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{125} - 133 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 133 \log {\left (\sqrt {2 x + 3} + 1 \right )} + \frac {8}{\sqrt {2 x + 3} + 1} - \frac {3}{\left (\sqrt {2 x + 3} + 1\right )^{2}} + \frac {8}{\sqrt {2 x + 3} - 1} + \frac {3}{\left (\sqrt {2 x + 3} - 1\right )^{2}} - \frac {137184}{3125 \sqrt {2 x + 3}} - \frac {9824}{1875 \left (2 x + 3\right )^{\frac {3}{2}}} - \frac {416}{625 \left (2 x + 3\right )^{\frac {5}{2}}} \]
372033*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqr t(15)/3))/15625 - 351864*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)))/625 + 33048*Piecewise((sqrt(1 5)*(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(1 5)*sqrt(2*x + 3)/5 - 1)**2))/375, (sqrt(2*x + 3) > -sqrt(15)/3) & (sqrt(2* x + 3) < sqrt(15)/3)))/125 - 133*log(sqrt(2*x + 3) - 1) + 133*log(sqrt(2*x + 3) + 1) + 8/(sqrt(2*x + 3) + 1) - 3/(sqrt(2*x + 3) + 1)**2 + 8/(sqrt(2* x + 3) - 1) + 3/(sqrt(2*x + 3) - 1)**2 - 137184/(3125*sqrt(2*x + 3)) - 982 4/(1875*(2*x + 3)**(3/2)) - 416/(625*(2*x + 3)**(5/2))
Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.14 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {806841}{31250} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {659043 \, {\left (2 \, x + 3\right )}^{6} - 8136261 \, {\left (2 \, x + 3\right )}^{5} + 23916753 \, {\left (2 \, x + 3\right )}^{4} - 24720095 \, {\left (2 \, x + 3\right )}^{3} + 6945760 \, {\left (2 \, x + 3\right )}^{2} + 1457600 \, x + 2342400}{9375 \, {\left (9 \, {\left (2 \, x + 3\right )}^{\frac {13}{2}} - 48 \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} + 94 \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} - 80 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + 25 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}}\right )}} + 133 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 133 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
806841/31250*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt (2*x + 3))) - 1/9375*(659043*(2*x + 3)^6 - 8136261*(2*x + 3)^5 + 23916753* (2*x + 3)^4 - 24720095*(2*x + 3)^3 + 6945760*(2*x + 3)^2 + 1457600*x + 234 2400)/(9*(2*x + 3)^(13/2) - 48*(2*x + 3)^(11/2) + 94*(2*x + 3)^(9/2) - 80* (2*x + 3)^(7/2) + 25*(2*x + 3)^(5/2)) + 133*log(sqrt(2*x + 3) + 1) - 133*l og(sqrt(2*x + 3) - 1)
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {806841}{31250} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {202995 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 745077 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 831169 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 259087 \, \sqrt {2 \, x + 3}}{625 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} - \frac {32 \, {\left (12861 \, {\left (2 \, x + 3\right )}^{2} + 3070 \, x + 4800\right )}}{9375 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}}} + 133 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 133 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
806841/31250*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/625*(202995*(2*x + 3)^(7/2) - 745077*(2*x + 3)^(5 /2) + 831169*(2*x + 3)^(3/2) - 259087*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 - 32/9375*(12861*(2*x + 3)^2 + 3070*x + 4800)/(2*x + 3)^(5/2) + 1 33*log(sqrt(2*x + 3) + 1) - 133*log(abs(sqrt(2*x + 3) - 1))
Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.90 \[ \int \frac {5-x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )^3} \, dx=266\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {806841\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{15625}-\frac {\frac {58304\,x}{3375}+\frac {1389152\,{\left (2\,x+3\right )}^2}{16875}-\frac {4944019\,{\left (2\,x+3\right )}^3}{16875}+\frac {2657417\,{\left (2\,x+3\right )}^4}{9375}-\frac {301343\,{\left (2\,x+3\right )}^5}{3125}+\frac {24409\,{\left (2\,x+3\right )}^6}{3125}+\frac {31232}{1125}}{\frac {25\,{\left (2\,x+3\right )}^{5/2}}{9}-\frac {80\,{\left (2\,x+3\right )}^{7/2}}{9}+\frac {94\,{\left (2\,x+3\right )}^{9/2}}{9}-\frac {16\,{\left (2\,x+3\right )}^{11/2}}{3}+{\left (2\,x+3\right )}^{13/2}} \]
266*atanh((2*x + 3)^(1/2)) - (806841*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1 /2))/5))/15625 - ((58304*x)/3375 + (1389152*(2*x + 3)^2)/16875 - (4944019* (2*x + 3)^3)/16875 + (2657417*(2*x + 3)^4)/9375 - (301343*(2*x + 3)^5)/312 5 + (24409*(2*x + 3)^6)/3125 + 31232/1125)/((25*(2*x + 3)^(5/2))/9 - (80*( 2*x + 3)^(7/2))/9 + (94*(2*x + 3)^(9/2))/9 - (16*(2*x + 3)^(11/2))/3 + (2* x + 3)^(13/2))