Integrand size = 27, antiderivative size = 119 \[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {6 (37+47 x)}{5 (3+2 x)^2 \sqrt {2+5 x+3 x^2}}-\frac {166 \sqrt {2+5 x+3 x^2}}{5 (3+2 x)^2}-\frac {864 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {483 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{25 \sqrt {5}} \]
483/125*arctanh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)^(1/2))*5^(1/2)-6/5*(37+ 47*x)/(3+2*x)^2/(3*x^2+5*x+2)^(1/2)-166/5*(3*x^2+5*x+2)^(1/2)/(3+2*x)^2-86 4/25*(3*x^2+5*x+2)^(1/2)/(3+2*x)
Time = 0.39 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71 \[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {2}{125} \left (-\frac {5 \sqrt {2+5 x+3 x^2} \left (3977+10988 x+9453 x^2+2592 x^3\right )}{(1+x) (3+2 x)^2 (2+3 x)}+483 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )\right ) \]
(2*((-5*Sqrt[2 + 5*x + 3*x^2]*(3977 + 10988*x + 9453*x^2 + 2592*x^3))/((1 + x)*(3 + 2*x)^2*(2 + 3*x)) + 483*Sqrt[5]*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5 ]/(1 + x)]))/125
Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1235, 1237, 27, 1228, 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.
| \(\displaystyle \int \frac {5-x}{(2 x+3)^3 \left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\frac {2}{5} \int \frac {564 x+431}{(2 x+3)^3 \sqrt {3 x^2+5 x+2}}dx-\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle -\frac {2}{5} \left (\frac {83 \sqrt {3 x^2+5 x+2}}{(2 x+3)^2}-\frac {1}{10} \int -\frac {15 (166 x+105)}{(2 x+3)^2 \sqrt {3 x^2+5 x+2}}dx\right )-\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2}{5} \left (\frac {3}{2} \int \frac {166 x+105}{(2 x+3)^2 \sqrt {3 x^2+5 x+2}}dx+\frac {83 \sqrt {3 x^2+5 x+2}}{(2 x+3)^2}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle -\frac {2}{5} \left (\frac {3}{2} \left (\frac {288 \sqrt {3 x^2+5 x+2}}{5 (2 x+3)}-\frac {161}{5} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )+\frac {83 \sqrt {3 x^2+5 x+2}}{(2 x+3)^2}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle -\frac {2}{5} \left (\frac {3}{2} \left (\frac {322}{5} \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )+\frac {288 \sqrt {3 x^2+5 x+2}}{5 (2 x+3)}\right )+\frac {83 \sqrt {3 x^2+5 x+2}}{(2 x+3)^2}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2}{5} \left (\frac {3}{2} \left (\frac {288 \sqrt {3 x^2+5 x+2}}{5 (2 x+3)}-\frac {161 \text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{5 \sqrt {5}}\right )+\frac {83 \sqrt {3 x^2+5 x+2}}{(2 x+3)^2}\right )-\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}\) |
(-6*(37 + 47*x))/(5*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2]) - (2*((83*Sqrt[2 + 5*x + 3*x^2])/(3 + 2*x)^2 + (3*((288*Sqrt[2 + 5*x + 3*x^2])/(5*(3 + 2*x)) - (161*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(5*Sqrt[5]))) /2))/5
3.26.13.3.1 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[(-(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]
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] )
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)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.35 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.57
| method | result | size |
| risch | \(-\frac {2 \left (2592 x^{3}+9453 x^{2}+10988 x +3977\right )}{25 \left (3+2 x \right )^{2} \sqrt {3 x^{2}+5 x +2}}-\frac {483 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{125}\) | \(68\) |
| trager | \(-\frac {2 \left (2592 x^{3}+9453 x^{2}+10988 x +3977\right )}{25 \left (3+2 x \right )^{2} \sqrt {3 x^{2}+5 x +2}}+\frac {483 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{125}\) | \(87\) |
| default | \(-\frac {5}{2 \left (x +\frac {3}{2}\right ) \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}+\frac {483}{50 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}-\frac {216 \left (5+6 x \right )}{25 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}-\frac {483 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{125}-\frac {13}{40 \left (x +\frac {3}{2}\right )^{2} \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}\) | \(111\) |
-2/25*(2592*x^3+9453*x^2+10988*x+3977)/(3+2*x)^2/(3*x^2+5*x+2)^(1/2)-483/1 25*5^(1/2)*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))
Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.05 \[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {483 \, \sqrt {5} {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) - 20 \, {\left (2592 \, x^{3} + 9453 \, x^{2} + 10988 \, x + 3977\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{250 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} \]
1/250*(483*sqrt(5)*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log((4*sqrt(5)*s qrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2 + 12*x + 9)) - 20*(2592*x^3 + 9453*x^2 + 10988*x + 3977)*sqrt(3*x^2 + 5*x + 2))/(12*x^ 4 + 56*x^3 + 95*x^2 + 69*x + 18)
\[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \frac {x}{24 x^{5} \sqrt {3 x^{2} + 5 x + 2} + 148 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 358 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 423 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 243 x \sqrt {3 x^{2} + 5 x + 2} + 54 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{24 x^{5} \sqrt {3 x^{2} + 5 x + 2} + 148 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 358 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 423 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 243 x \sqrt {3 x^{2} + 5 x + 2} + 54 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]
-Integral(x/(24*x**5*sqrt(3*x**2 + 5*x + 2) + 148*x**4*sqrt(3*x**2 + 5*x + 2) + 358*x**3*sqrt(3*x**2 + 5*x + 2) + 423*x**2*sqrt(3*x**2 + 5*x + 2) + 243*x*sqrt(3*x**2 + 5*x + 2) + 54*sqrt(3*x**2 + 5*x + 2)), x) - Integral(- 5/(24*x**5*sqrt(3*x**2 + 5*x + 2) + 148*x**4*sqrt(3*x**2 + 5*x + 2) + 358* x**3*sqrt(3*x**2 + 5*x + 2) + 423*x**2*sqrt(3*x**2 + 5*x + 2) + 243*x*sqrt (3*x**2 + 5*x + 2) + 54*sqrt(3*x**2 + 5*x + 2)), x)
Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.32 \[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {483}{125} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) - \frac {1296 \, x}{25 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {1677}{50 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {13}{10 \, {\left (4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{2} + 12 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + 9 \, \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}} - \frac {5}{2 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + 3 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} \]
-483/125*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs( 2*x + 3) - 2) - 1296/25*x/sqrt(3*x^2 + 5*x + 2) - 1677/50/sqrt(3*x^2 + 5*x + 2) - 13/10/(4*sqrt(3*x^2 + 5*x + 2)*x^2 + 12*sqrt(3*x^2 + 5*x + 2)*x + 9*sqrt(3*x^2 + 5*x + 2)) - 5/(2*sqrt(3*x^2 + 5*x + 2)*x + 3*sqrt(3*x^2 + 5 *x + 2))
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (97) = 194\).
Time = 0.31 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.89 \[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {483}{125} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) - \frac {6 \, {\left (903 \, x + 653\right )}}{125 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {2 \, {\left (2442 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{3} + 9999 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 35473 \, \sqrt {3} x + 12979 \, \sqrt {3} - 35473 \, \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}}{125 \, {\left (2 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 6 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} + 11\right )}^{2}} \]
483/125*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))) - 6/125*(903*x + 653)/sqrt(3*x^2 + 5*x + 2) - 2/125*(2442*(sqrt(3 )*x - sqrt(3*x^2 + 5*x + 2))^3 + 9999*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5* x + 2))^2 + 35473*sqrt(3)*x + 12979*sqrt(3) - 35473*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)^2
Timed out. \[ \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {x-5}{{\left (2\,x+3\right )}^3\,{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]