Integrand size = 27, antiderivative size = 128 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {7451}{75 (3+2 x)^{3/2}}+\frac {6853}{125 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{10 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}+\frac {9146+10551 x}{50 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+310 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {45603}{125} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
7451/75/(3+2*x)^(3/2)-3/10*(37+47*x)/(3+2*x)^(3/2)/(3*x^2+5*x+2)^2+1/50*(9 146+10551*x)/(3+2*x)^(3/2)/(3*x^2+5*x+2)+310*arctanh((3+2*x)^(1/2))-45603/ 625*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+6853/125/(3+2*x)^(1/2)
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.71 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=310 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {\frac {5 \left (1057511+5278129 x+10168583 x^2+9453447 x^3+4247856 x^4+740124 x^5\right )}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2}-273618 \sqrt {15} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{3750} \]
310*ArcTanh[Sqrt[3 + 2*x]] + ((5*(1057511 + 5278129*x + 10168583*x^2 + 945 3447*x^3 + 4247856*x^4 + 740124*x^5))/((3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)^2 ) - 273618*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/3750
Time = 0.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {1235, 1235, 27, 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.
| \(\displaystyle \int \frac {5-x}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\frac {1}{10} \int \frac {1269 x+1550}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2}dx-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \int \frac {5 (10551 x+12101)}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}dx+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{10} \left (\int \frac {10551 x+12101}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}dx+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \int \frac {22353 x+30103}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}dx+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {14902}{15 (2 x+3)^{3/2}}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (\frac {1}{5} \int \frac {20559 x+59309}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {13706}{5 \sqrt {2 x+3}}\right )+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {14902}{15 (2 x+3)^{3/2}}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (\frac {2}{5} \int \frac {20559 (2 x+3)+56941}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}+\frac {13706}{5 \sqrt {2 x+3}}\right )+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {14902}{15 (2 x+3)^{3/2}}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (\frac {2}{5} \left (136809 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-116250 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )+\frac {13706}{5 \sqrt {2 x+3}}\right )+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {14902}{15 (2 x+3)^{3/2}}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (\frac {2}{5} \left (38750 \text {arctanh}\left (\sqrt {2 x+3}\right )-45603 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )+\frac {13706}{5 \sqrt {2 x+3}}\right )+\frac {10551 x+9146}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}+\frac {14902}{15 (2 x+3)^{3/2}}\right )-\frac {3 (47 x+37)}{10 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )^2}\) |
(-3*(37 + 47*x))/(10*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)^2) + (14902/(15*(3 + 2*x)^(3/2)) + (9146 + 10551*x)/(5*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) + ( 13706/(5*Sqrt[3 + 2*x]) + (2*(38750*ArcTanh[Sqrt[3 + 2*x]] - 45603*Sqrt[3/ 5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/5)/5)/10
3.26.73.3.1 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 = 0.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.70
| method | result | size |
| risch | \(\frac {740124 x^{5}+4247856 x^{4}+9453447 x^{3}+10168583 x^{2}+5278129 x +1057511}{750 \left (3 x^{2}+5 x +2\right )^{2} \left (3+2 x \right )^{\frac {3}{2}}}+155 \ln \left (\sqrt {3+2 x}+1\right )-\frac {45603 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{625}-155 \ln \left (\sqrt {3+2 x}-1\right )\) | \(90\) |
| trager | \(\frac {\left (740124 x^{5}+4247856 x^{4}+9453447 x^{3}+10168583 x^{2}+5278129 x +1057511\right ) \sqrt {3+2 x}}{750 \left (6 x^{3}+19 x^{2}+19 x +6\right )^{2}}-155 \ln \left (\frac {-2-x +\sqrt {3+2 x}}{1+x}\right )+\frac {81 \operatorname {RootOf}\left (\textit {\_Z}^{2}-4754535\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-4754535\right ) x +7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-4754535\right )-8445 \sqrt {3+2 x}}{2+3 x}\right )}{1250}\) | \(119\) |
| derivativedivides | \(\frac {\frac {171801 \left (3+2 x \right )^{\frac {3}{2}}}{625}-\frac {60021 \sqrt {3+2 x}}{125}}{\left (6 x +4\right )^{2}}-\frac {45603 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{625}-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {20}{\sqrt {3+2 x}+1}+155 \ln \left (\sqrt {3+2 x}+1\right )+\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {20}{\sqrt {3+2 x}-1}-155 \ln \left (\sqrt {3+2 x}-1\right )-\frac {416}{375 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {9824}{625 \sqrt {3+2 x}}\) | \(142\) |
| default | \(\frac {\frac {171801 \left (3+2 x \right )^{\frac {3}{2}}}{625}-\frac {60021 \sqrt {3+2 x}}{125}}{\left (6 x +4\right )^{2}}-\frac {45603 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{625}-\frac {3}{\left (\sqrt {3+2 x}+1\right )^{2}}+\frac {20}{\sqrt {3+2 x}+1}+155 \ln \left (\sqrt {3+2 x}+1\right )+\frac {3}{\left (\sqrt {3+2 x}-1\right )^{2}}+\frac {20}{\sqrt {3+2 x}-1}-155 \ln \left (\sqrt {3+2 x}-1\right )-\frac {416}{375 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {9824}{625 \sqrt {3+2 x}}\) | \(142\) |
| pseudoelliptic | \(-\frac {3283416 \left (\sqrt {3+2 x}\, \sqrt {15}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \left (x +\frac {3}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )+\frac {96875 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \left (x +\frac {3}{2}\right ) \ln \left (\sqrt {3+2 x}-1\right )}{45603}-\frac {96875 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right )^{2} \left (1+x \right )^{2} \left (x +\frac {3}{2}\right ) \ln \left (\sqrt {3+2 x}+1\right )}{45603}-\frac {34265 x^{5}}{45603}-\frac {196660 x^{4}}{45603}-\frac {5251915 x^{3}}{547236}-\frac {50842915 x^{2}}{4925124}-\frac {26390645 x}{4925124}-\frac {5287555}{4925124}\right )}{625 \left (3+2 x \right )^{\frac {3}{2}} \left (\sqrt {3+2 x}-1\right )^{2} \left (2+3 x \right )^{2} \left (\sqrt {3+2 x}+1\right )^{2}}\) | \(165\) |
1/750*(740124*x^5+4247856*x^4+9453447*x^3+10168583*x^2+5278129*x+1057511)/ (3*x^2+5*x+2)^2/(3+2*x)^(3/2)+155*ln((3+2*x)^(1/2)+1)-45603/625*arctanh(1/ 5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-155*ln((3+2*x)^(1/2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (99) = 198\).
Time = 0.30 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.72 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {136809 \, \sqrt {5} \sqrt {3} {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 581250 \, {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 581250 \, {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + 5 \, {\left (740124 \, x^{5} + 4247856 \, x^{4} + 9453447 \, x^{3} + 10168583 \, x^{2} + 5278129 \, x + 1057511\right )} \sqrt {2 \, x + 3}}{3750 \, {\left (36 \, x^{6} + 228 \, x^{5} + 589 \, x^{4} + 794 \, x^{3} + 589 \, x^{2} + 228 \, x + 36\right )}} \]
1/3750*(136809*sqrt(5)*sqrt(3)*(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589 *x^2 + 228*x + 36)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2 )) + 581250*(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589*x^2 + 228*x + 36)* log(sqrt(2*x + 3) + 1) - 581250*(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 58 9*x^2 + 228*x + 36)*log(sqrt(2*x + 3) - 1) + 5*(740124*x^5 + 4247856*x^4 + 9453447*x^3 + 10168583*x^2 + 5278129*x + 1057511)*sqrt(2*x + 3))/(36*x^6 + 228*x^5 + 589*x^4 + 794*x^3 + 589*x^2 + 228*x + 36)
Time = 115.56 (sec) , antiderivative size = 432, normalized size of antiderivative = 3.38 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {104463 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{3125} - \frac {106272 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{125} + \frac {11016 \left (\begin {cases} \frac {\sqrt {15} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )^{2}}\right )}{375} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{25} - 155 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 155 \log {\left (\sqrt {2 x + 3} + 1 \right )} + \frac {20}{\sqrt {2 x + 3} + 1} - \frac {3}{\left (\sqrt {2 x + 3} + 1\right )^{2}} + \frac {20}{\sqrt {2 x + 3} - 1} + \frac {3}{\left (\sqrt {2 x + 3} - 1\right )^{2}} - \frac {9824}{625 \sqrt {2 x + 3}} - \frac {416}{375 \left (2 x + 3\right )^{\frac {3}{2}}} \]
104463*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqr t(15)/3))/3125 - 106272*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) > -s qrt(15)/3) & (sqrt(2*x + 3) < sqrt(15)/3)))/125 + 11016*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)))/25 - 155*log(sqrt(2*x + 3) - 1) + 155*log(sqrt(2*x + 3) + 1) + 20/(sqrt(2*x + 3) + 1) - 3/(sqrt(2*x + 3) + 1)**2 + 20/(sqrt(2* x + 3) - 1) + 3/(sqrt(2*x + 3) - 1)**2 - 9824/(625*sqrt(2*x + 3)) - 416/(3 75*(2*x + 3)**(3/2))
Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.19 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {45603}{1250} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {185031 \, {\left (2 \, x + 3\right )}^{5} - 651537 \, {\left (2 \, x + 3\right )}^{4} + 619101 \, {\left (2 \, x + 3\right )}^{3} - 10115 \, {\left (2 \, x + 3\right )}^{2} - 228160 \, x - 352640}{375 \, {\left (9 \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} - 48 \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} + 94 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 80 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 25 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}\right )}} + 155 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 155 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
45603/1250*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2 *x + 3))) + 1/375*(185031*(2*x + 3)^5 - 651537*(2*x + 3)^4 + 619101*(2*x + 3)^3 - 10115*(2*x + 3)^2 - 228160*x - 352640)/(9*(2*x + 3)^(11/2) - 48*(2 *x + 3)^(9/2) + 94*(2*x + 3)^(7/2) - 80*(2*x + 3)^(5/2) + 25*(2*x + 3)^(3/ 2)) + 155*log(sqrt(2*x + 3) + 1) - 155*log(sqrt(2*x + 3) - 1)
Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.05 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=\frac {45603}{1250} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {64 \, {\left (921 \, x + 1414\right )}}{1875 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}} + \frac {396801 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 1551207 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 1922011 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 737605 \, \sqrt {2 \, x + 3}}{625 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 155 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 155 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
45603/1250*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 64/1875*(921*x + 1414)/(2*x + 3)^(3/2) + 1/625*(39680 1*(2*x + 3)^(7/2) - 1551207*(2*x + 3)^(5/2) + 1922011*(2*x + 3)^(3/2) - 73 7605*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 + 155*log(sqrt(2*x + 3) + 1) - 155*log(abs(sqrt(2*x + 3) - 1))
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.92 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^3} \, dx=310\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {45603\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{625}-\frac {\frac {45632\,x}{675}+\frac {2023\,{\left (2\,x+3\right )}^2}{675}-\frac {68789\,{\left (2\,x+3\right )}^3}{375}+\frac {24131\,{\left (2\,x+3\right )}^4}{125}-\frac {6853\,{\left (2\,x+3\right )}^5}{125}+\frac {70528}{675}}{\frac {25\,{\left (2\,x+3\right )}^{3/2}}{9}-\frac {80\,{\left (2\,x+3\right )}^{5/2}}{9}+\frac {94\,{\left (2\,x+3\right )}^{7/2}}{9}-\frac {16\,{\left (2\,x+3\right )}^{9/2}}{3}+{\left (2\,x+3\right )}^{11/2}} \]
310*atanh((2*x + 3)^(1/2)) - (45603*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/ 2))/5))/625 - ((45632*x)/675 + (2023*(2*x + 3)^2)/675 - (68789*(2*x + 3)^3 )/375 + (24131*(2*x + 3)^4)/125 - (6853*(2*x + 3)^5)/125 + 70528/675)/((25 *(2*x + 3)^(3/2))/9 - (80*(2*x + 3)^(5/2))/9 + (94*(2*x + 3)^(7/2))/9 - (1 6*(2*x + 3)^(9/2))/3 + (2*x + 3)^(11/2))