Integrand size = 24, antiderivative size = 121 \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=-\frac {13 \sqrt {2+3 x^2}}{140 (3+2 x)^4}-\frac {97 \sqrt {2+3 x^2}}{2100 (3+2 x)^3}-\frac {87 \sqrt {2+3 x^2}}{4900 (3+2 x)^2}-\frac {991 \sqrt {2+3 x^2}}{171500 (3+2 x)}+\frac {27 \text {arctanh}\left (\frac {4-9 x}{\sqrt {35} \sqrt {2+3 x^2}}\right )}{42875 \sqrt {35}} \]
27/1500625*arctanh(1/35*(4-9*x)*35^(1/2)/(3*x^2+2)^(1/2))*35^(1/2)-13/140* (3*x^2+2)^(1/2)/(3+2*x)^4-97/2100*(3*x^2+2)^(1/2)/(3+2*x)^3-87/4900*(3*x^2 +2)^(1/2)/(3+2*x)^2-991/171500*(3*x^2+2)^(1/2)/(3+2*x)
Time = 0.66 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=\frac {-\frac {35 \sqrt {2+3 x^2} \left (70389+79423 x+35892 x^2+5946 x^3\right )}{(3+2 x)^4}-162 \sqrt {35} \text {arctanh}\left (\frac {3 \sqrt {3}+2 \sqrt {3} x-2 \sqrt {2+3 x^2}}{\sqrt {35}}\right )}{4501875} \]
((-35*Sqrt[2 + 3*x^2]*(70389 + 79423*x + 35892*x^2 + 5946*x^3))/(3 + 2*x)^ 4 - 162*Sqrt[35]*ArcTanh[(3*Sqrt[3] + 2*Sqrt[3]*x - 2*Sqrt[2 + 3*x^2])/Sqr t[35]])/4501875
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {688, 25, 688, 27, 688, 27, 679, 488, 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.
| \(\displaystyle \int \frac {5-x}{(2 x+3)^5 \sqrt {3 x^2+2}} \, dx\) |
| \(\Big \downarrow \) 688 |
| \(\displaystyle -\frac {1}{140} \int -\frac {164-117 x}{(2 x+3)^4 \sqrt {3 x^2+2}}dx-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{140} \int \frac {164-117 x}{(2 x+3)^4 \sqrt {3 x^2+2}}dx-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 688 |
| \(\displaystyle \frac {1}{140} \left (-\frac {1}{105} \int -\frac {42 (72-97 x)}{(2 x+3)^3 \sqrt {3 x^2+2}}dx-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{140} \left (\frac {2}{5} \int \frac {72-97 x}{(2 x+3)^3 \sqrt {3 x^2+2}}dx-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 688 |
| \(\displaystyle \frac {1}{140} \left (\frac {2}{5} \left (-\frac {1}{70} \int -\frac {5 (104-261 x)}{(2 x+3)^2 \sqrt {3 x^2+2}}dx-\frac {87 \sqrt {3 x^2+2}}{14 (2 x+3)^2}\right )-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{140} \left (\frac {2}{5} \left (\frac {1}{14} \int \frac {104-261 x}{(2 x+3)^2 \sqrt {3 x^2+2}}dx-\frac {87 \sqrt {3 x^2+2}}{14 (2 x+3)^2}\right )-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 679 |
| \(\displaystyle \frac {1}{140} \left (\frac {2}{5} \left (\frac {1}{14} \left (-\frac {108}{35} \int \frac {1}{(2 x+3) \sqrt {3 x^2+2}}dx-\frac {991 \sqrt {3 x^2+2}}{35 (2 x+3)}\right )-\frac {87 \sqrt {3 x^2+2}}{14 (2 x+3)^2}\right )-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 488 |
| \(\displaystyle \frac {1}{140} \left (\frac {2}{5} \left (\frac {1}{14} \left (\frac {108}{35} \int \frac {1}{35-\frac {(4-9 x)^2}{3 x^2+2}}d\frac {4-9 x}{\sqrt {3 x^2+2}}-\frac {991 \sqrt {3 x^2+2}}{35 (2 x+3)}\right )-\frac {87 \sqrt {3 x^2+2}}{14 (2 x+3)^2}\right )-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{140} \left (\frac {2}{5} \left (\frac {1}{14} \left (\frac {108 \text {arctanh}\left (\frac {4-9 x}{\sqrt {35} \sqrt {3 x^2+2}}\right )}{35 \sqrt {35}}-\frac {991 \sqrt {3 x^2+2}}{35 (2 x+3)}\right )-\frac {87 \sqrt {3 x^2+2}}{14 (2 x+3)^2}\right )-\frac {97 \sqrt {3 x^2+2}}{15 (2 x+3)^3}\right )-\frac {13 \sqrt {3 x^2+2}}{140 (2 x+3)^4}\) |
(-13*Sqrt[2 + 3*x^2])/(140*(3 + 2*x)^4) + ((-97*Sqrt[2 + 3*x^2])/(15*(3 + 2*x)^3) + (2*((-87*Sqrt[2 + 3*x^2])/(14*(3 + 2*x)^2) + ((-991*Sqrt[2 + 3*x ^2])/(35*(3 + 2*x)) + (108*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/ (35*Sqrt[35]))/14))/5)/140
3.15.6.3.1 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/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/( (m + 1)*(c*d^2 + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.77 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.62
| method | result | size |
| risch | \(-\frac {17838 x^{5}+107676 x^{4}+250161 x^{3}+282951 x^{2}+158846 x +140778}{128625 \left (3+2 x \right )^{4} \sqrt {3 x^{2}+2}}+\frac {27 \sqrt {35}\, \operatorname {arctanh}\left (\frac {2 \left (4-9 x \right ) \sqrt {35}}{35 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-36 x -19}}\right )}{1500625}\) | \(75\) |
| trager | \(-\frac {\left (5946 x^{3}+35892 x^{2}+79423 x +70389\right ) \sqrt {3 x^{2}+2}}{128625 \left (3+2 x \right )^{4}}+\frac {27 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) \ln \left (\frac {-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) x +35 \sqrt {3 x^{2}+2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right )}{3+2 x}\right )}{1500625}\) | \(81\) |
| default | \(-\frac {97 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-9 x -\frac {19}{4}}}{16800 \left (x +\frac {3}{2}\right )^{3}}-\frac {87 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-9 x -\frac {19}{4}}}{19600 \left (x +\frac {3}{2}\right )^{2}}-\frac {991 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-9 x -\frac {19}{4}}}{343000 \left (x +\frac {3}{2}\right )}+\frac {27 \sqrt {35}\, \operatorname {arctanh}\left (\frac {2 \left (4-9 x \right ) \sqrt {35}}{35 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-36 x -19}}\right )}{1500625}-\frac {13 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-9 x -\frac {19}{4}}}{2240 \left (x +\frac {3}{2}\right )^{4}}\) | \(116\) |
-1/128625*(17838*x^5+107676*x^4+250161*x^3+282951*x^2+158846*x+140778)/(3+ 2*x)^4/(3*x^2+2)^(1/2)+27/1500625*35^(1/2)*arctanh(2/35*(4-9*x)*35^(1/2)/( 12*(x+3/2)^2-36*x-19)^(1/2))
Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98 \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=\frac {81 \, \sqrt {35} {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )} \log \left (\frac {\sqrt {35} \sqrt {3 \, x^{2} + 2} {\left (9 \, x - 4\right )} - 93 \, x^{2} + 36 \, x - 43}{4 \, x^{2} + 12 \, x + 9}\right ) - 70 \, {\left (5946 \, x^{3} + 35892 \, x^{2} + 79423 \, x + 70389\right )} \sqrt {3 \, x^{2} + 2}}{9003750 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \]
1/9003750*(81*sqrt(35)*(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)*log((sqrt( 35)*sqrt(3*x^2 + 2)*(9*x - 4) - 93*x^2 + 36*x - 43)/(4*x^2 + 12*x + 9)) - 70*(5946*x^3 + 35892*x^2 + 79423*x + 70389)*sqrt(3*x^2 + 2))/(16*x^4 + 96* x^3 + 216*x^2 + 216*x + 81)
Timed out. \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.13 \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=-\frac {27}{1500625} \, \sqrt {35} \operatorname {arsinh}\left (\frac {3 \, \sqrt {6} x}{2 \, {\left | 2 \, x + 3 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 3 \right |}}\right ) - \frac {13 \, \sqrt {3 \, x^{2} + 2}}{140 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} - \frac {97 \, \sqrt {3 \, x^{2} + 2}}{2100 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} - \frac {87 \, \sqrt {3 \, x^{2} + 2}}{4900 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {991 \, \sqrt {3 \, x^{2} + 2}}{171500 \, {\left (2 \, x + 3\right )}} \]
-27/1500625*sqrt(35)*arcsinh(3/2*sqrt(6)*x/abs(2*x + 3) - 2/3*sqrt(6)/abs( 2*x + 3)) - 13/140*sqrt(3*x^2 + 2)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81 ) - 97/2100*sqrt(3*x^2 + 2)/(8*x^3 + 36*x^2 + 54*x + 27) - 87/4900*sqrt(3* x^2 + 2)/(4*x^2 + 12*x + 9) - 991/171500*sqrt(3*x^2 + 2)/(2*x + 3)
Time = 0.29 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.58 \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=\frac {1}{12005000} \, \sqrt {35} {\left (991 \, \sqrt {35} \sqrt {3} - 216 \, \log \left (\sqrt {35} \sqrt {3} - 9\right )\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) - \frac {1}{1029000} \, {\left (\frac {35 \, {\left (\frac {7 \, {\left (\frac {97}{\mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )} + \frac {195}{{\left (2 \, x + 3\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}\right )}}{2 \, x + 3} + \frac {261}{\mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}\right )}}{2 \, x + 3} + \frac {2973}{\mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}\right )} \sqrt {-\frac {18}{2 \, x + 3} + \frac {35}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {27 \, \sqrt {35} \log \left (\sqrt {35} {\left (\sqrt {-\frac {18}{2 \, x + 3} + \frac {35}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {35}}{2 \, x + 3}\right )} - 9\right )}{1500625 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )} \]
1/12005000*sqrt(35)*(991*sqrt(35)*sqrt(3) - 216*log(sqrt(35)*sqrt(3) - 9)) *sgn(1/(2*x + 3)) - 1/1029000*(35*(7*(97/sgn(1/(2*x + 3)) + 195/((2*x + 3) *sgn(1/(2*x + 3))))/(2*x + 3) + 261/sgn(1/(2*x + 3)))/(2*x + 3) + 2973/sgn (1/(2*x + 3)))*sqrt(-18/(2*x + 3) + 35/(2*x + 3)^2 + 3) + 27/1500625*sqrt( 35)*log(sqrt(35)*(sqrt(-18/(2*x + 3) + 35/(2*x + 3)^2 + 3) + sqrt(35)/(2*x + 3)) - 9)/sgn(1/(2*x + 3))
Time = 0.21 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.21 \[ \int \frac {5-x}{(3+2 x)^5 \sqrt {2+3 x^2}} \, dx=\frac {\sqrt {35}\,\left (\frac {2808\,\ln \left (x+\frac {3}{2}\right )}{42875}-\frac {2808\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {35}\,\sqrt {x^2+\frac {2}{3}}}{9}-\frac {4}{9}\right )}{42875}\right )}{560}-\frac {\sqrt {35}\,\left (\frac {324\,\ln \left (x+\frac {3}{2}\right )}{8575}-\frac {324\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {35}\,\sqrt {x^2+\frac {2}{3}}}{9}-\frac {4}{9}\right )}{8575}\right )}{280}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {18252}{42875\,\left (x+\frac {3}{2}\right )}+\frac {702}{1225\,{\left (x+\frac {3}{2}\right )}^2}+\frac {117}{175\,{\left (x+\frac {3}{2}\right )}^3}+\frac {39}{70\,{\left (x+\frac {3}{2}\right )}^4}\right )}{96}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {636}{8575\,\left (x+\frac {3}{2}\right )}+\frac {18}{245\,{\left (x+\frac {3}{2}\right )}^2}+\frac {2}{35\,{\left (x+\frac {3}{2}\right )}^3}\right )}{48} \]
(35^(1/2)*((2808*log(x + 3/2))/42875 - (2808*log(x - (3^(1/2)*35^(1/2)*(x^ 2 + 2/3)^(1/2))/9 - 4/9))/42875))/560 - (35^(1/2)*((324*log(x + 3/2))/8575 - (324*log(x - (3^(1/2)*35^(1/2)*(x^2 + 2/3)^(1/2))/9 - 4/9))/8575))/280 - (3^(1/2)*(x^2 + 2/3)^(1/2)*(18252/(42875*(x + 3/2)) + 702/(1225*(x + 3/2 )^2) + 117/(175*(x + 3/2)^3) + 39/(70*(x + 3/2)^4)))/96 + (3^(1/2)*(x^2 + 2/3)^(1/2)*(636/(8575*(x + 3/2)) + 18/(245*(x + 3/2)^2) + 2/(35*(x + 3/2)^ 3)))/48