Integrand size = 27, antiderivative size = 85 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\frac {2 (37+47 x)}{5 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {12 (701+836 x)}{25 \sqrt {2+5 x+3 x^2}}+\frac {104 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{25 \sqrt {5}} \]
-2/5*(37+47*x)/(3*x^2+5*x+2)^(3/2)+104/125*arctanh(1/10*(7+8*x)*5^(1/2)/(3 *x^2+5*x+2)^(1/2))*5^(1/2)+12/25*(701+836*x)/(3*x^2+5*x+2)^(1/2)
Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {2}{125} \left (\frac {5 \sqrt {2+5 x+3 x^2} \left (8227+30827 x+37698 x^2+15048 x^3\right )}{(1+x)^2 (2+3 x)^2}+104 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )\right ) \]
(2*((5*Sqrt[2 + 5*x + 3*x^2]*(8227 + 30827*x + 37698*x^2 + 15048*x^3))/((1 + x)^2*(2 + 3*x)^2) + 104*Sqrt[5]*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)]))/125
Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1235, 27, 1235, 27, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3) \left (3 x^2+5 x+2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1235 |
\(\displaystyle -\frac {2}{15} \int \frac {3 (188 x+269)}{(2 x+3) \left (3 x^2+5 x+2\right )^{3/2}}dx-\frac {2 (47 x+37)}{5 \left (3 x^2+5 x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{5} \int \frac {188 x+269}{(2 x+3) \left (3 x^2+5 x+2\right )^{3/2}}dx-\frac {2 (47 x+37)}{5 \left (3 x^2+5 x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 1235 |
\(\displaystyle -\frac {2}{5} \left (-\frac {2}{5} \int \frac {26}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\frac {6 (836 x+701)}{5 \sqrt {3 x^2+5 x+2}}\right )-\frac {2 (47 x+37)}{5 \left (3 x^2+5 x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{5} \left (-\frac {52}{5} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\frac {6 (836 x+701)}{5 \sqrt {3 x^2+5 x+2}}\right )-\frac {2 (47 x+37)}{5 \left (3 x^2+5 x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle -\frac {2}{5} \left (\frac {104}{5} \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )-\frac {6 (836 x+701)}{5 \sqrt {3 x^2+5 x+2}}\right )-\frac {2 (47 x+37)}{5 \left (3 x^2+5 x+2\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2}{5} \left (-\frac {52 \text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{5 \sqrt {5}}-\frac {6 (836 x+701)}{5 \sqrt {3 x^2+5 x+2}}\right )-\frac {2 (47 x+37)}{5 \left (3 x^2+5 x+2\right )^{3/2}}\) |
(-2*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)^(3/2)) - (2*((-6*(701 + 836*x))/(5*S qrt[2 + 5*x + 3*x^2]) - (52*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3* x^2])])/(5*Sqrt[5])))/5
3.26.21.3.1 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/(((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)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94
method | result | size |
trager | \(\frac {\frac {30096}{25} x^{3}+\frac {75396}{25} x^{2}+\frac {61654}{25} x +\frac {16454}{25}}{\left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}-\frac {104 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{125}\) | \(80\) |
default | \(\frac {5+6 x}{3 \left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}-\frac {8 \left (5+6 x \right )}{\sqrt {3 x^{2}+5 x +2}}+\frac {13}{15 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}-\frac {52 \left (5+6 x \right )}{15 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}+\frac {\frac {1872}{5}+\frac {11232 x}{25}}{\sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}+\frac {52}{25 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}-\frac {104 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{125}\) | \(144\) |
2/25*(15048*x^3+37698*x^2+30827*x+8227)/(3*x^2+5*x+2)^(3/2)-104/125*RootOf (_Z^2-5)*ln((-8*RootOf(_Z^2-5)*x+10*(3*x^2+5*x+2)^(1/2)-7*RootOf(_Z^2-5))/ (3+2*x))
Time = 0.37 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.47 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (26 \, \sqrt {5} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) + 5 \, {\left (15048 \, x^{3} + 37698 \, x^{2} + 30827 \, x + 8227\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}}{125 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]
2/125*(26*sqrt(5)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*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)) + 5*(15048*x^3 + 37698*x^2 + 30827*x + 8227)*sqrt(3*x^2 + 5*x + 2))/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)
\[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=- \int \frac {x}{18 x^{5} \sqrt {3 x^{2} + 5 x + 2} + 87 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 164 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 151 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 68 x \sqrt {3 x^{2} + 5 x + 2} + 12 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{18 x^{5} \sqrt {3 x^{2} + 5 x + 2} + 87 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 164 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 151 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 68 x \sqrt {3 x^{2} + 5 x + 2} + 12 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]
-Integral(x/(18*x**5*sqrt(3*x**2 + 5*x + 2) + 87*x**4*sqrt(3*x**2 + 5*x + 2) + 164*x**3*sqrt(3*x**2 + 5*x + 2) + 151*x**2*sqrt(3*x**2 + 5*x + 2) + 6 8*x*sqrt(3*x**2 + 5*x + 2) + 12*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-5/ (18*x**5*sqrt(3*x**2 + 5*x + 2) + 87*x**4*sqrt(3*x**2 + 5*x + 2) + 164*x** 3*sqrt(3*x**2 + 5*x + 2) + 151*x**2*sqrt(3*x**2 + 5*x + 2) + 68*x*sqrt(3*x **2 + 5*x + 2) + 12*sqrt(3*x**2 + 5*x + 2)), x)
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.19 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\frac {104}{125} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {10032 \, x}{25 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} + \frac {8412}{25 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {94 \, x}{5 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} - \frac {74}{5 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \]
-104/125*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs( 2*x + 3) - 2) + 10032/25*x/sqrt(3*x^2 + 5*x + 2) + 8412/25/sqrt(3*x^2 + 5* x + 2) - 94/5*x/(3*x^2 + 5*x + 2)^(3/2) - 74/5/(3*x^2 + 5*x + 2)^(3/2)
Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.20 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {104}{125} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {2 \, {\left ({\left (6 \, {\left (2508 \, x + 6283\right )} x + 30827\right )} x + 8227\right )}}{25 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \]
104/125*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))) + 2/25*((6*(2508*x + 6283)*x + 30827)*x + 8227)/(3*x^2 + 5*x + 2) ^(3/2)
Timed out. \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\int \frac {x-5}{\left (2\,x+3\right )\,{\left (3\,x^2+5\,x+2\right )}^{5/2}} \,d x \]