Integrand size = 27, antiderivative size = 119 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=\frac {153 (7+8 x) \sqrt {2+5 x+3 x^2}}{800 (3+2 x)^2}-\frac {13 \left (2+5 x+3 x^2\right )^{3/2}}{20 (3+2 x)^4}-\frac {4 \left (2+5 x+3 x^2\right )^{3/2}}{5 (3+2 x)^3}-\frac {153 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{1600 \sqrt {5}} \]
-13/20*(3*x^2+5*x+2)^(3/2)/(3+2*x)^4-4/5*(3*x^2+5*x+2)^(3/2)/(3+2*x)^3-153 /8000*arctanh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)^(1/2))*5^(1/2)+153/800*(7 +8*x)*(3*x^2+5*x+2)^(1/2)/(3+2*x)^2
Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=\frac {\frac {5 \sqrt {2+5 x+3 x^2} \left (4759+9108 x+5252 x^2+1056 x^3\right )}{(3+2 x)^4}-153 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )}{4000} \]
((5*Sqrt[2 + 5*x + 3*x^2]*(4759 + 9108*x + 5252*x^2 + 1056*x^3))/(3 + 2*x) ^4 - 153*Sqrt[5]*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)])/4000
Time = 0.28 (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 = {1237, 27, 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.
| \(\displaystyle \int \frac {(5-x) \sqrt {3 x^2+5 x+2}}{(2 x+3)^5} \, dx\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle -\frac {1}{20} \int -\frac {3 (41-26 x) \sqrt {3 x^2+5 x+2}}{2 (2 x+3)^4}dx-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{20 (2 x+3)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{40} \int \frac {(41-26 x) \sqrt {3 x^2+5 x+2}}{(2 x+3)^4}dx-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{20 (2 x+3)^4}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {3}{40} \left (51 \int \frac {\sqrt {3 x^2+5 x+2}}{(2 x+3)^3}dx-\frac {32 \left (3 x^2+5 x+2\right )^{3/2}}{3 (2 x+3)^3}\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{20 (2 x+3)^4}\) |
| \(\Big \downarrow \) 1152 |
| \(\displaystyle \frac {3}{40} \left (51 \left (\frac {(8 x+7) \sqrt {3 x^2+5 x+2}}{20 (2 x+3)^2}-\frac {1}{40} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx\right )-\frac {32 \left (3 x^2+5 x+2\right )^{3/2}}{3 (2 x+3)^3}\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{20 (2 x+3)^4}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {3}{40} \left (51 \left (\frac {1}{20} \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )+\frac {\sqrt {3 x^2+5 x+2} (8 x+7)}{20 (2 x+3)^2}\right )-\frac {32 \left (3 x^2+5 x+2\right )^{3/2}}{3 (2 x+3)^3}\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{20 (2 x+3)^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{40} \left (51 \left (\frac {(8 x+7) \sqrt {3 x^2+5 x+2}}{20 (2 x+3)^2}-\frac {\text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{40 \sqrt {5}}\right )-\frac {32 \left (3 x^2+5 x+2\right )^{3/2}}{3 (2 x+3)^3}\right )-\frac {13 \left (3 x^2+5 x+2\right )^{3/2}}{20 (2 x+3)^4}\) |
(-13*(2 + 5*x + 3*x^2)^(3/2))/(20*(3 + 2*x)^4) + (3*((-32*(2 + 5*x + 3*x^2 )^(3/2))/(3*(3 + 2*x)^3) + 51*(((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] ))))/40
3.25.15.3.1 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[((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]
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 = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66
| method | result | size |
| risch | \(\frac {3168 x^{5}+21036 x^{4}+55696 x^{3}+70321 x^{2}+42011 x +9518}{800 \left (3+2 x \right )^{4} \sqrt {3 x^{2}+5 x +2}}+\frac {153 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{8000}\) | \(78\) |
| trager | \(\frac {\left (1056 x^{3}+5252 x^{2}+9108 x +4759\right ) \sqrt {3 x^{2}+5 x +2}}{800 \left (3+2 x \right )^{4}}-\frac {153 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{8000}\) | \(87\) |
| default | \(-\frac {\left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{10 \left (x +\frac {3}{2}\right )^{3}}-\frac {153 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{800 \left (x +\frac {3}{2}\right )^{2}}-\frac {153 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{500 \left (x +\frac {3}{2}\right )}-\frac {153 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}{8000}+\frac {153 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{8000}+\frac {153 \left (5+6 x \right ) \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}}{1000}-\frac {13 \left (3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}\right )^{\frac {3}{2}}}{320 \left (x +\frac {3}{2}\right )^{4}}\) | \(153\) |
1/800*(3168*x^5+21036*x^4+55696*x^3+70321*x^2+42011*x+9518)/(3+2*x)^4/(3*x ^2+5*x+2)^(1/2)+153/8000*5^(1/2)*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x+3/2 )^2-16*x-19)^(1/2))
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.06 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=\frac {153 \, \sqrt {5} {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )} \log \left (-\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} - 124 \, x^{2} - 212 \, x - 89}{4 \, x^{2} + 12 \, x + 9}\right ) + 20 \, {\left (1056 \, x^{3} + 5252 \, x^{2} + 9108 \, x + 4759\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{16000 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \]
1/16000*(153*sqrt(5)*(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)*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*(1056*x^3 + 5252*x^2 + 9108*x + 4759)*sqrt(3*x^2 + 5*x + 2))/(1 6*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=- \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243}\, dx \]
-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(32*x**5 + 240*x**4 + 720*x**3 + 1080* x**2 + 810*x + 243), x) - Integral(x*sqrt(3*x**2 + 5*x + 2)/(32*x**5 + 240 *x**4 + 720*x**3 + 1080*x**2 + 810*x + 243), x)
Time = 0.30 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.44 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=\frac {153}{8000} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {459}{800} \, \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {13 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{20 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} - \frac {4 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{5 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} - \frac {153 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{200 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {153 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{200 \, {\left (2 \, x + 3\right )}} \]
153/8000*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs( 2*x + 3) - 2) + 459/800*sqrt(3*x^2 + 5*x + 2) - 13/20*(3*x^2 + 5*x + 2)^(3 /2)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81) - 4/5*(3*x^2 + 5*x + 2)^(3/2) /(8*x^3 + 36*x^2 + 54*x + 27) - 153/200*(3*x^2 + 5*x + 2)^(3/2)/(4*x^2 + 1 2*x + 9) - 153/200*sqrt(3*x^2 + 5*x + 2)/(2*x + 3)
Time = 0.30 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.54 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=-\frac {3}{8000} \, \sqrt {5} {\left (44 \, \sqrt {5} \sqrt {3} + 51 \, \log \left (-\sqrt {5} \sqrt {3} + 4\right )\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) + \frac {153}{8000} \, \sqrt {5} \log \left ({\left | \sqrt {5} {\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )} - 4 \right |}\right ) \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) - \frac {1}{1600} \, {\left (\frac {5 \, {\left (\frac {2 \, {\left (\frac {65 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}{2 \, x + 3} - 24 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )\right )}}{2 \, x + 3} - 25 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )\right )}}{2 \, x + 3} - 132 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )\right )} \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} \]
-3/8000*sqrt(5)*(44*sqrt(5)*sqrt(3) + 51*log(-sqrt(5)*sqrt(3) + 4))*sgn(1/ (2*x + 3)) + 153/8000*sqrt(5)*log(abs(sqrt(5)*(sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x + 3)) - 4))*sgn(1/(2*x + 3)) - 1/1600*(5*(2*(65 *sgn(1/(2*x + 3))/(2*x + 3) - 24*sgn(1/(2*x + 3)))/(2*x + 3) - 25*sgn(1/(2 *x + 3)))/(2*x + 3) - 132*sgn(1/(2*x + 3)))*sqrt(-8/(2*x + 3) + 5/(2*x + 3 )^2 + 3)
Timed out. \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^5} \, dx=-\int \frac {\left (x-5\right )\,\sqrt {3\,x^2+5\,x+2}}{{\left (2\,x+3\right )}^5} \,d x \]