Integrand size = 20, antiderivative size = 103 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=-\frac {35 (5+6 x) \sqrt {2+5 x+3 x^2}}{1152}+\frac {35}{144} (5+6 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {1}{15} \left (2+5 x+3 x^2\right )^{5/2}+\frac {35 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{2304 \sqrt {3}} \]
35/144*(5+6*x)*(3*x^2+5*x+2)^(3/2)-1/15*(3*x^2+5*x+2)^(5/2)+35/6912*arctan h(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)-35/1152*(5+6*x)*(3*x^2+ 5*x+2)^(1/2)
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.69 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=\frac {-3 \sqrt {2+5 x+3 x^2} \left (-11589-43070 x-48792 x^2-13680 x^3+3456 x^4\right )+175 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{17280} \]
(-3*Sqrt[2 + 5*x + 3*x^2]*(-11589 - 43070*x - 48792*x^2 - 13680*x^3 + 3456 *x^4) + 175*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/17280
Time = 0.22 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1160, 1087, 1087, 1092, 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.
\(\displaystyle \int (5-x) \left (3 x^2+5 x+2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {35}{6} \int \left (3 x^2+5 x+2\right )^{3/2}dx-\frac {1}{15} \left (3 x^2+5 x+2\right )^{5/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {35}{6} \left (\frac {1}{24} (6 x+5) \left (3 x^2+5 x+2\right )^{3/2}-\frac {1}{16} \int \sqrt {3 x^2+5 x+2}dx\right )-\frac {1}{15} \left (3 x^2+5 x+2\right )^{5/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {35}{6} \left (\frac {1}{16} \left (\frac {1}{24} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx-\frac {1}{12} (6 x+5) \sqrt {3 x^2+5 x+2}\right )+\frac {1}{24} (6 x+5) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} \left (3 x^2+5 x+2\right )^{5/2}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {35}{6} \left (\frac {1}{16} \left (\frac {1}{12} \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}-\frac {1}{12} (6 x+5) \sqrt {3 x^2+5 x+2}\right )+\frac {1}{24} (6 x+5) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} \left (3 x^2+5 x+2\right )^{5/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {35}{6} \left (\frac {1}{16} \left (\frac {\text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{24 \sqrt {3}}-\frac {1}{12} (6 x+5) \sqrt {3 x^2+5 x+2}\right )+\frac {1}{24} (6 x+5) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} \left (3 x^2+5 x+2\right )^{5/2}\) |
-1/15*(2 + 5*x + 3*x^2)^(5/2) + (35*(((5 + 6*x)*(2 + 5*x + 3*x^2)^(3/2))/2 4 + (-1/12*((5 + 6*x)*Sqrt[2 + 5*x + 3*x^2]) + ArcTanh[(5 + 6*x)/(2*Sqrt[3 ]*Sqrt[2 + 5*x + 3*x^2])]/(24*Sqrt[3]))/16))/6
3.25.22.3.1 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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {\left (3456 x^{4}-13680 x^{3}-48792 x^{2}-43070 x -11589\right ) \sqrt {3 x^{2}+5 x +2}}{5760}+\frac {35 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{6912}\) | \(65\) |
trager | \(\left (-\frac {3}{5} x^{4}+\frac {19}{8} x^{3}+\frac {2033}{240} x^{2}+\frac {4307}{576} x +\frac {3863}{1920}\right ) \sqrt {3 x^{2}+5 x +2}-\frac {35 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}+5 x +2}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{6912}\) | \(76\) |
default | \(\frac {35 \left (5+6 x \right ) \left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}{144}-\frac {35 \left (5+6 x \right ) \sqrt {3 x^{2}+5 x +2}}{1152}+\frac {35 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{6912}-\frac {\left (3 x^{2}+5 x +2\right )^{\frac {5}{2}}}{15}\) | \(83\) |
-1/5760*(3456*x^4-13680*x^3-48792*x^2-43070*x-11589)*(3*x^2+5*x+2)^(1/2)+3 5/6912*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.71 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=-\frac {1}{5760} \, {\left (3456 \, x^{4} - 13680 \, x^{3} - 48792 \, x^{2} - 43070 \, x - 11589\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {35}{13824} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) \]
-1/5760*(3456*x^4 - 13680*x^3 - 48792*x^2 - 43070*x - 11589)*sqrt(3*x^2 + 5*x + 2) + 35/13824*sqrt(3)*log(4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49)
Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=\sqrt {3 x^{2} + 5 x + 2} \left (- \frac {3 x^{4}}{5} + \frac {19 x^{3}}{8} + \frac {2033 x^{2}}{240} + \frac {4307 x}{576} + \frac {3863}{1920}\right ) + \frac {35 \sqrt {3} \log {\left (6 x + 2 \sqrt {3} \sqrt {3 x^{2} + 5 x + 2} + 5 \right )}}{6912} \]
sqrt(3*x**2 + 5*x + 2)*(-3*x**4/5 + 19*x**3/8 + 2033*x**2/240 + 4307*x/576 + 3863/1920) + 35*sqrt(3)*log(6*x + 2*sqrt(3)*sqrt(3*x**2 + 5*x + 2) + 5) /6912
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=-\frac {1}{15} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}} + \frac {35}{24} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x + \frac {175}{144} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} - \frac {35}{192} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + \frac {35}{6912} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) - \frac {175}{1152} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]
-1/15*(3*x^2 + 5*x + 2)^(5/2) + 35/24*(3*x^2 + 5*x + 2)^(3/2)*x + 175/144* (3*x^2 + 5*x + 2)^(3/2) - 35/192*sqrt(3*x^2 + 5*x + 2)*x + 35/6912*sqrt(3) *log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) - 175/1152*sqrt(3*x^2 + 5* x + 2)
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.67 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=-\frac {1}{5760} \, {\left (2 \, {\left (12 \, {\left (6 \, {\left (24 \, x - 95\right )} x - 2033\right )} x - 21535\right )} x - 11589\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {35}{6912} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) \]
-1/5760*(2*(12*(6*(24*x - 95)*x - 2033)*x - 21535)*x - 11589)*sqrt(3*x^2 + 5*x + 2) - 35/6912*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5 *x + 2)) - 5))
Time = 11.59 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.26 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{3/2} \, dx=\frac {35\,\sqrt {3}\,\ln \left (\sqrt {3\,x^2+5\,x+2}+\frac {\sqrt {3}\,\left (3\,x+\frac {5}{2}\right )}{3}\right )}{6912}-\frac {5\,\left (6\,x+5\right )\,\sqrt {3\,x^2+5\,x+2}}{1152}+\frac {5\,\left (3\,x+\frac {5}{2}\right )\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{12}-\frac {5\,\left (\frac {x}{2}+\frac {5}{12}\right )\,\sqrt {3\,x^2+5\,x+2}}{16}+\frac {5\,x\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{24}+\frac {25\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{144}-\frac {{\left (3\,x^2+5\,x+2\right )}^{5/2}}{15} \]
(35*3^(1/2)*log((5*x + 3*x^2 + 2)^(1/2) + (3^(1/2)*(3*x + 5/2))/3))/6912 - (5*(6*x + 5)*(5*x + 3*x^2 + 2)^(1/2))/1152 + (5*(3*x + 5/2)*(5*x + 3*x^2 + 2)^(3/2))/12 - (5*(x/2 + 5/12)*(5*x + 3*x^2 + 2)^(1/2))/16 + (5*x*(5*x + 3*x^2 + 2)^(3/2))/24 + (25*(5*x + 3*x^2 + 2)^(3/2))/144 - (5*x + 3*x^2 + 2)^(5/2)/15