Integrand size = 27, antiderivative size = 107 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=\frac {(121+124 x) \sqrt {2+5 x+3 x^2}}{40 (3+2 x)^2}-\frac {1}{8} \sqrt {3} \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )+\frac {27 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{80 \sqrt {5}} \]
-1/8*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+27/400*arcta nh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)^(1/2))*5^(1/2)+1/40*(121+124*x)*(3*x ^2+5*x+2)^(1/2)/(3+2*x)^2
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.87 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=\frac {1}{200} \left (\frac {5 (121+124 x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2}+27 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )-50 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )\right ) \]
((5*(121 + 124*x)*Sqrt[2 + 5*x + 3*x^2])/(3 + 2*x)^2 + 27*Sqrt[5]*ArcTanh[ Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)] - 50*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/200
Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1229, 27, 1269, 1092, 219, 1154, 219}
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.
| \(\displaystyle \int \frac {(5-x) \sqrt {3 x^2+5 x+2}}{(2 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 1229 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {1}{80} \int \frac {3 (20 x+21)}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {3}{80} \int \frac {20 x+21}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {3}{80} \left (10 \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx-9 \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {3}{80} \left (20 \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}-9 \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {3}{80} \left (\frac {10 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{\sqrt {3}}-9 \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {3}{80} \left (18 \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )+\frac {10 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{\sqrt {3}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(124 x+121) \sqrt {3 x^2+5 x+2}}{40 (2 x+3)^2}-\frac {3}{80} \left (\frac {10 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{\sqrt {3}}-\frac {9 \text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{\sqrt {5}}\right )\) |
((121 + 124*x)*Sqrt[2 + 5*x + 3*x^2])/(40*(3 + 2*x)^2) - (3*((10*ArcTanh[( 5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/Sqrt[3] - (9*ArcTanh[(7 + 8*x )/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/Sqrt[5]))/80
3.25.13.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, 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/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)))*((d*g - e*f*(m + 2))*(c* d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x), x] - Simp[p/(e^2*(m + 1 )*(m + 2)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2 )^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1) - c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c *(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1) - b*(d*g*( m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g }, x] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] && !ILtQ[m + 2*p + 3, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.91
| method | result | size |
| risch | \(\frac {372 x^{3}+983 x^{2}+853 x +242}{40 \left (3+2 x \right )^{2} \sqrt {3 x^{2}+5 x +2}}-\frac {\ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{8}-\frac {27 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{400}\) | \(97\) |
| trager | \(\frac {\left (121+124 x \right ) \sqrt {3 x^{2}+5 x +2}}{40 \left (3+2 x \right )^{2}}+\frac {27 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{400}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{8}\) | \(118\) |
| default | \(-\frac {21 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{50 \left (x +\frac {3}{2}\right )}+\frac {27 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}{400}-\frac {\ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}\right ) \sqrt {3}}{8}-\frac {27 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{400}+\frac {21 \left (5+6 x \right ) \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}{100}-\frac {13 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{40 \left (x +\frac {3}{2}\right )^{2}}\) | \(142\) |
1/40*(372*x^3+983*x^2+853*x+242)/(3+2*x)^2/(3*x^2+5*x+2)^(1/2)-1/8*ln(1/3* (5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)-27/400*5^(1/2)*arctanh(2/5* (-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.34 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=\frac {50 \, \sqrt {3} {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + 27 \, \sqrt {5} {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) + 20 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (124 \, x + 121\right )}}{800 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} \]
1/800*(50*sqrt(3)*(4*x^2 + 12*x + 9)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)* (6*x + 5) + 72*x^2 + 120*x + 49) + 27*sqrt(5)*(4*x^2 + 12*x + 9)*log((4*sq rt(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2 + 12* x + 9)) + 20*sqrt(3*x^2 + 5*x + 2)*(124*x + 121))/(4*x^2 + 12*x + 9)
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=- \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} + 36 x^{2} + 54 x + 27}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} + 36 x^{2} + 54 x + 27}\, dx \]
-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(8*x**3 + 36*x**2 + 54*x + 27), x) - I ntegral(x*sqrt(3*x**2 + 5*x + 2)/(8*x**3 + 36*x**2 + 54*x + 27), x)
Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=-\frac {1}{8} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {27}{400} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {39}{40} \, \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {13 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{10 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {21 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{20 \, {\left (2 \, x + 3\right )}} \]
-1/8*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 27/400*sqrt( 5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) + 39/40*sqrt(3*x^2 + 5*x + 2) - 13/10*(3*x^2 + 5*x + 2)^(3/2)/(4*x^2 + 12* x + 9) - 21/20*sqrt(3*x^2 + 5*x + 2)/(2*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (83) = 166\).
Time = 0.32 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.24 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=\frac {27}{400} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {1}{8} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) + \frac {886 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{3} + 2897 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 9039 \, \sqrt {3} x + 3037 \, \sqrt {3} - 9039 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{40 \, {\left (2 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 6 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} + 11\right )}^{2}} \]
27/400*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) + 1/8*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2 )) - 5)) + 1/40*(886*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^3 + 2897*sqrt(3)* (sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 9039*sqrt(3)*x + 3037*sqrt(3) - 90 39*sqrt(3*x^2 + 5*x + 2))/(2*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 6*sqr t(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) + 11)^2
Timed out. \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx=-\int \frac {\left (x-5\right )\,\sqrt {3\,x^2+5\,x+2}}{{\left (2\,x+3\right )}^3} \,d x \]