Integrand size = 27, antiderivative size = 117 \[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {2 (3+2 x)^3 (121+139 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {1664}{27} (3+2 x)^2 \sqrt {2+5 x+3 x^2}+\frac {10}{81} (3369+1438 x) \sqrt {2+5 x+3 x^2}+\frac {6265 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{81 \sqrt {3}} \]
6265/243*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)-2/3*(3+2 *x)^3*(121+139*x)/(3*x^2+5*x+2)^(1/2)+1664/27*(3+2*x)^2*(3*x^2+5*x+2)^(1/2 )+10/81*(3369+1438*x)*(3*x^2+5*x+2)^(1/2)
Time = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.71 \[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {2}{243} \left (\frac {3 \sqrt {2+5 x+3 x^2} \left (9591+6920 x-3331 x^2-102 x^3+72 x^4\right )}{(1+x) (2+3 x)}-6265 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )\right ) \]
(-2*((3*Sqrt[2 + 5*x + 3*x^2]*(9591 + 6920*x - 3331*x^2 - 102*x^3 + 72*x^4 ))/((1 + x)*(2 + 3*x)) - 6265*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)]))/243
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1233, 1236, 27, 1225, 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.
| \(\displaystyle \int \frac {(5-x) (2 x+3)^4}{\left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle \frac {2}{3} \int \frac {(2 x+3)^2 (832 x+723)}{\sqrt {3 x^2+5 x+2}}dx-\frac {2 (2 x+3)^3 (139 x+121)}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {2}{3} \left (\frac {1}{9} \int \frac {5 (2 x+3) (1438 x+1325)}{\sqrt {3 x^2+5 x+2}}dx+\frac {832}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {2 (2 x+3)^3 (139 x+121)}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} \left (\frac {5}{9} \int \frac {(2 x+3) (1438 x+1325)}{\sqrt {3 x^2+5 x+2}}dx+\frac {832}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {2 (2 x+3)^3 (139 x+121)}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {2}{3} \left (\frac {5}{9} \left (\frac {1253}{6} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx+\frac {1}{3} \sqrt {3 x^2+5 x+2} (1438 x+3369)\right )+\frac {832}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {2 (2 x+3)^3 (139 x+121)}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {2}{3} \left (\frac {5}{9} \left (\frac {1253}{3} \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}+\frac {1}{3} \sqrt {3 x^2+5 x+2} (1438 x+3369)\right )+\frac {832}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {2 (2 x+3)^3 (139 x+121)}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2}{3} \left (\frac {5}{9} \left (\frac {1253 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{6 \sqrt {3}}+\frac {1}{3} \sqrt {3 x^2+5 x+2} (1438 x+3369)\right )+\frac {832}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {2 (2 x+3)^3 (139 x+121)}{3 \sqrt {3 x^2+5 x+2}}\) |
(-2*(3 + 2*x)^3*(121 + 139*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (2*((832*(3 + 2 *x)^2*Sqrt[2 + 5*x + 3*x^2])/9 + (5*(((3369 + 1438*x)*Sqrt[2 + 5*x + 3*x^2 ])/3 + (1253*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(6*Sqrt [3])))/9))/3
3.26.6.3.1 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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
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 + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !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*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(-\frac {2 \left (72 x^{4}-102 x^{3}-3331 x^{2}+6920 x +9591\right )}{81 \sqrt {3 x^{2}+5 x +2}}+\frac {6265 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{243}\) | \(65\) |
| trager | \(-\frac {2 \left (72 x^{4}-102 x^{3}-3331 x^{2}+6920 x +9591\right )}{81 \sqrt {3 x^{2}+5 x +2}}+\frac {6265 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{243}\) | \(77\) |
| default | \(-\frac {2525 \left (5+6 x \right )}{162 \sqrt {3 x^{2}+5 x +2}}-\frac {25739}{162 \sqrt {3 x^{2}+5 x +2}}-\frac {16 x^{4}}{9 \sqrt {3 x^{2}+5 x +2}}+\frac {68 x^{3}}{27 \sqrt {3 x^{2}+5 x +2}}+\frac {6662 x^{2}}{81 \sqrt {3 x^{2}+5 x +2}}-\frac {6265 x}{81 \sqrt {3 x^{2}+5 x +2}}+\frac {6265 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{243}\) | \(130\) |
-2/81*(72*x^4-102*x^3-3331*x^2+6920*x+9591)/(3*x^2+5*x+2)^(1/2)+6265/243*l n(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)
Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.83 \[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {6265 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) - 12 \, {\left (72 \, x^{4} - 102 \, x^{3} - 3331 \, x^{2} + 6920 \, x + 9591\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{486 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]
1/486*(6265*sqrt(3)*(3*x^2 + 5*x + 2)*log(4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)* (6*x + 5) + 72*x^2 + 120*x + 49) - 12*(72*x^4 - 102*x^3 - 3331*x^2 + 6920* x + 9591)*sqrt(3*x^2 + 5*x + 2))/(3*x^2 + 5*x + 2)
\[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \left (- \frac {999 x}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \left (- \frac {864 x^{2}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \left (- \frac {264 x^{3}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {16 x^{4}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \frac {16 x^{5}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {405}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]
-Integral(-999*x/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-864*x**2/(3*x**2*sqrt(3*x** 2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-264*x**3/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5* x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x) - Integral(16*x**4/(3*x**2*sqrt(3*x **2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x ) - Integral(16*x**5/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5* x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-405/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x)
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {16 \, x^{4}}{9 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} + \frac {68 \, x^{3}}{27 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} + \frac {6662 \, x^{2}}{81 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} + \frac {6265}{243} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) - \frac {13840 \, x}{81 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {6394}{27 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} \]
-16/9*x^4/sqrt(3*x^2 + 5*x + 2) + 68/27*x^3/sqrt(3*x^2 + 5*x + 2) + 6662/8 1*x^2/sqrt(3*x^2 + 5*x + 2) + 6265/243*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) - 13840/81*x/sqrt(3*x^2 + 5*x + 2) - 6394/27/sqrt(3*x^ 2 + 5*x + 2)
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.57 \[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {6265}{243} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) - \frac {2 \, {\left ({\left ({\left (6 \, {\left (12 \, x - 17\right )} x - 3331\right )} x + 6920\right )} x + 9591\right )}}{81 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} \]
-6265/243*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5)) - 2/81*(((6*(12*x - 17)*x - 3331)*x + 6920)*x + 9591)/sqrt(3*x^2 + 5* x + 2)
Timed out. \[ \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {{\left (2\,x+3\right )}^4\,\left (x-5\right )}{{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]