Integrand size = 27, antiderivative size = 105 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=-\frac {(8+x) \sqrt {2+5 x+3 x^2}}{2 (3+2 x)}+\frac {43 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{8 \sqrt {3}}-\frac {57 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{8 \sqrt {5}} \]
43/24*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)-57/40*arcta nh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)^(1/2))*5^(1/2)-1/2*(8+x)*(3*x^2+5*x+ 2)^(1/2)/(3+2*x)
Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=-\frac {(8+x) \sqrt {2+5 x+3 x^2}}{6+4 x}-\frac {57 \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )}{4 \sqrt {5}}+\frac {43 \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{4 \sqrt {3}} \]
-(((8 + x)*Sqrt[2 + 5*x + 3*x^2])/(6 + 4*x)) - (57*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)])/(4*Sqrt[5]) + (43*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/( 1 + x)])/(4*Sqrt[3])
Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1230, 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.
| \(\displaystyle \int \frac {(5-x) \sqrt {3 x^2+5 x+2}}{(2 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 1230 |
| \(\displaystyle -\frac {1}{8} \int -\frac {2 (43 x+36)}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\frac {\sqrt {3 x^2+5 x+2} (x+8)}{2 (2 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {43 x+36}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\frac {(x+8) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {1}{4} \left (\frac {43}{2} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx-\frac {57}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )-\frac {(x+8) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {1}{4} \left (43 \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}-\frac {57}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )-\frac {(x+8) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{4} \left (\frac {43 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}}-\frac {57}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )-\frac {(x+8) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {1}{4} \left (57 \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )+\frac {43 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}}\right )-\frac {(x+8) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{4} \left (\frac {43 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}}-\frac {57 \text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {5}}\right )-\frac {(x+8) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)}\) |
-1/2*((8 + x)*Sqrt[2 + 5*x + 3*x^2])/(3 + 2*x) + ((43*ArcTanh[(5 + 6*x)/(2 *Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(2*Sqrt[3]) - (57*ArcTanh[(7 + 8*x)/(2*S qrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(2*Sqrt[5]))/4
3.25.12.3.1 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)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 1] || EqQ[p, 1] || (IntegerQ[p] && !RationalQ[m])) && NeQ[m, -1] && !ILtQ [m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
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 = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(-\frac {3 x^{3}+29 x^{2}+42 x +16}{2 \left (3+2 x \right ) \sqrt {3 x^{2}+5 x +2}}+\frac {43 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{24}+\frac {57 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{40}\) | \(97\) |
| trager | \(-\frac {\left (8+x \right ) \sqrt {3 x^{2}+5 x +2}}{2 \left (3+2 x \right )}-\frac {43 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}+5 x +2}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{24}-\frac {57 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{40}\) | \(116\) |
| default | \(-\frac {57 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}{40}+\frac {43 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}\right ) \sqrt {3}}{24}+\frac {57 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{40}-\frac {13 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{10 \left (x +\frac {3}{2}\right )}+\frac {13 \left (5+6 x \right ) \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}{20}\) | \(121\) |
-1/2*(3*x^3+29*x^2+42*x+16)/(3+2*x)/(3*x^2+5*x+2)^(1/2)+43/24*ln(1/3*(5/2+ 3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)+57/40*5^(1/2)*arctanh(2/5*(-7/2- 4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))
Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.21 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=\frac {215 \, \sqrt {3} {\left (2 \, x + 3\right )} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + 171 \, \sqrt {5} {\left (2 \, x + 3\right )} \log \left (-\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} - 124 \, x^{2} - 212 \, x - 89}{4 \, x^{2} + 12 \, x + 9}\right ) - 120 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (x + 8\right )}}{240 \, {\left (2 \, x + 3\right )}} \]
1/240*(215*sqrt(3)*(2*x + 3)*log(4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49) + 171*sqrt(5)*(2*x + 3)*log(-(4*sqrt(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) - 124*x^2 - 212*x - 89)/(4*x^2 + 12*x + 9)) - 120*sq rt(3*x^2 + 5*x + 2)*(x + 8))/(2*x + 3)
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=- \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{4 x^{2} + 12 x + 9}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{4 x^{2} + 12 x + 9}\, dx \]
-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(4*x**2 + 12*x + 9), x) - Integral(x*s qrt(3*x**2 + 5*x + 2)/(4*x**2 + 12*x + 9), x)
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=\frac {43}{24} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) + \frac {57}{40} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) - \frac {1}{4} \, \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {13 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{4 \, {\left (2 \, x + 3\right )}} \]
43/24*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) + 57/40*sqrt( 5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) - 1/4*sqrt(3*x^2 + 5*x + 2) - 13/4*sqrt(3*x^2 + 5*x + 2)/(2*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (81) = 162\).
Time = 0.47 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.77 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=-\frac {43}{24} \, \sqrt {3} \log \left (\frac {{\left | -2 \, \sqrt {3} + 2 \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {2 \, \sqrt {5}}{2 \, x + 3} \right |}}{{\left | 2 \, \sqrt {3} + 2 \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {2 \, \sqrt {5}}{2 \, x + 3} \right |}}\right ) \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) + \frac {57}{40} \, \sqrt {5} \log \left ({\left | \sqrt {5} {\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )} - 4 \right |}\right ) \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) - \frac {13}{8} \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) + \frac {4 \, {\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) - 3 \, \sqrt {5} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}{4 \, {\left ({\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )}^{2} - 3\right )}} \]
-43/24*sqrt(3)*log(abs(-2*sqrt(3) + 2*sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + 2*sqrt(5)/(2*x + 3))/abs(2*sqrt(3) + 2*sqrt(-8/(2*x + 3) + 5/(2*x + 3 )^2 + 3) + 2*sqrt(5)/(2*x + 3)))*sgn(1/(2*x + 3)) + 57/40*sqrt(5)*log(abs( sqrt(5)*(sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x + 3)) - 4)) *sgn(1/(2*x + 3)) - 13/8*sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3)*sgn(1/(2*x + 3)) + 1/4*(4*(sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x + 3 ))*sgn(1/(2*x + 3)) - 3*sqrt(5)*sgn(1/(2*x + 3)))/((sqrt(-8/(2*x + 3) + 5/ (2*x + 3)^2 + 3) + sqrt(5)/(2*x + 3))^2 - 3)
Timed out. \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx=-\int \frac {\left (x-5\right )\,\sqrt {3\,x^2+5\,x+2}}{{\left (2\,x+3\right )}^2} \,d x \]