Integrand size = 27, antiderivative size = 94 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=\frac {47 (7+8 x) \sqrt {2+5 x+3 x^2}}{200 (3+2 x)^2}-\frac {13 \left (2+5 x+3 x^2\right )^{3/2}}{15 (3+2 x)^3}-\frac {47 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{400 \sqrt {5}} \]
-13/15*(3*x^2+5*x+2)^(3/2)/(3+2*x)^3-47/2000*arctanh(1/10*(7+8*x)*5^(1/2)/ (3*x^2+5*x+2)^(1/2))*5^(1/2)+47/200*(7+8*x)*(3*x^2+5*x+2)^(1/2)/(3+2*x)^2
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=\frac {\frac {5 \sqrt {2+5 x+3 x^2} \left (1921+2758 x+696 x^2\right )}{(3+2 x)^3}-141 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )}{3000} \]
((5*Sqrt[2 + 5*x + 3*x^2]*(1921 + 2758*x + 696*x^2))/(3 + 2*x)^3 - 141*Sqr t[5]*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)])/3000
Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1228, 1152, 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.
\(\displaystyle \int \frac {(5-x) \sqrt {3 x^2+5 x+2}}{(2 x+3)^4} \, dx\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {47}{10} \int \frac {\sqrt {3 x^2+5 x+2}}{(2 x+3)^3}dx-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{15 (2 x+3)^3}\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle \frac {47}{10} \left (\frac {(8 x+7) \sqrt {3 x^2+5 x+2}}{20 (2 x+3)^2}-\frac {1}{40} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{15 (2 x+3)^3}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {47}{10} \left (\frac {1}{20} \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )+\frac {\sqrt {3 x^2+5 x+2} (8 x+7)}{20 (2 x+3)^2}\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{15 (2 x+3)^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {47}{10} \left (\frac {(8 x+7) \sqrt {3 x^2+5 x+2}}{20 (2 x+3)^2}-\frac {\text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{40 \sqrt {5}}\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{15 (2 x+3)^3}\) |
(-13*(2 + 5*x + 3*x^2)^(3/2))/(15*(3 + 2*x)^3) + (47*(((7 + 8*x)*Sqrt[2 + 5*x + 3*x^2])/(20*(3 + 2*x)^2) - ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])]/(40*Sqrt[5])))/10
3.25.14.3.1 Defintions of rubi rules used
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[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b *x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a *c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
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[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {2088 x^{4}+11754 x^{3}+20945 x^{2}+15121 x +3842}{600 \left (3+2 x \right )^{3} \sqrt {3 x^{2}+5 x +2}}+\frac {47 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{2000}\) | \(73\) |
trager | \(\frac {\left (696 x^{2}+2758 x +1921\right ) \sqrt {3 x^{2}+5 x +2}}{600 \left (3+2 x \right )^{3}}+\frac {47 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{2000}\) | \(82\) |
default | \(-\frac {47 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{200 \left (x +\frac {3}{2}\right )^{2}}-\frac {47 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{125 \left (x +\frac {3}{2}\right )}-\frac {47 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}{2000}+\frac {47 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{2000}+\frac {47 \left (5+6 x \right ) \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}{250}-\frac {13 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{120 \left (x +\frac {3}{2}\right )^{3}}\) | \(132\) |
1/600*(2088*x^4+11754*x^3+20945*x^2+15121*x+3842)/(3+2*x)^3/(3*x^2+5*x+2)^ (1/2)+47/2000*5^(1/2)*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19 )^(1/2))
Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=\frac {141 \, \sqrt {5} {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )} \log \left (-\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} - 124 \, x^{2} - 212 \, x - 89}{4 \, x^{2} + 12 \, x + 9}\right ) + 20 \, {\left (696 \, x^{2} + 2758 \, x + 1921\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{12000 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} \]
1/12000*(141*sqrt(5)*(8*x^3 + 36*x^2 + 54*x + 27)*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)) + 20*( 696*x^2 + 2758*x + 1921)*sqrt(3*x^2 + 5*x + 2))/(8*x^3 + 36*x^2 + 54*x + 2 7)
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=- \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81}\, dx \]
-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(16*x**4 + 96*x**3 + 216*x**2 + 216*x + 81), x) - Integral(x*sqrt(3*x**2 + 5*x + 2)/(16*x**4 + 96*x**3 + 216*x** 2 + 216*x + 81), x)
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.44 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=\frac {47}{2000} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {141}{200} \, \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {13 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{15 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} - \frac {47 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{50 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {47 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{50 \, {\left (2 \, x + 3\right )}} \]
47/2000*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2 *x + 3) - 2) + 141/200*sqrt(3*x^2 + 5*x + 2) - 13/15*(3*x^2 + 5*x + 2)^(3/ 2)/(8*x^3 + 36*x^2 + 54*x + 27) - 47/50*(3*x^2 + 5*x + 2)^(3/2)/(4*x^2 + 1 2*x + 9) - 47/50*sqrt(3*x^2 + 5*x + 2)/(2*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (76) = 152\).
Time = 0.32 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.73 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=-\frac {47}{2000} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) - \frac {1236 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{5} - 4830 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{4} - 90290 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{3} - 144945 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} - 287985 \, \sqrt {3} x - 69339 \, \sqrt {3} + 287985 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{600 \, {\left (2 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 6 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} + 11\right )}^{3}} \]
-47/2000*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))) - 1/600*(1236*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^5 - 4830*sqrt( 3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^4 - 90290*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^3 - 144945*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 - 2879 85*sqrt(3)*x - 69339*sqrt(3) + 287985*sqrt(3*x^2 + 5*x + 2))/(2*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 6*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2) ) + 11)^3
Timed out. \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^4} \, dx=-\int \frac {\left (x-5\right )\,\sqrt {3\,x^2+5\,x+2}}{{\left (2\,x+3\right )}^4} \,d x \]