Integrand size = 24, antiderivative size = 77 \[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=-\frac {13 \sqrt {2+3 x^2}}{70 (3+2 x)^2}-\frac {281 \sqrt {2+3 x^2}}{2450 (3+2 x)}-\frac {291 \text {arctanh}\left (\frac {4-9 x}{\sqrt {35} \sqrt {2+3 x^2}}\right )}{1225 \sqrt {35}} \]
-291/42875*arctanh(1/35*(4-9*x)*35^(1/2)/(3*x^2+2)^(1/2))*35^(1/2)-13/70*( 3*x^2+2)^(1/2)/(3+2*x)^2-281/2450*(3*x^2+2)^(1/2)/(3+2*x)
Time = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {-\frac {35 (649+281 x) \sqrt {2+3 x^2}}{(3+2 x)^2}+582 \sqrt {35} \text {arctanh}\left (\frac {3 \sqrt {3}+2 \sqrt {3} x-2 \sqrt {2+3 x^2}}{\sqrt {35}}\right )}{42875} \]
((-35*(649 + 281*x)*Sqrt[2 + 3*x^2])/(3 + 2*x)^2 + 582*Sqrt[35]*ArcTanh[(3 *Sqrt[3] + 2*Sqrt[3]*x - 2*Sqrt[2 + 3*x^2])/Sqrt[35]])/42875
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {688, 25, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3)^3 \sqrt {3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 688 |
\(\displaystyle -\frac {1}{70} \int -\frac {82-39 x}{(2 x+3)^2 \sqrt {3 x^2+2}}dx-\frac {13 \sqrt {3 x^2+2}}{70 (2 x+3)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{70} \int \frac {82-39 x}{(2 x+3)^2 \sqrt {3 x^2+2}}dx-\frac {13 \sqrt {3 x^2+2}}{70 (2 x+3)^2}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle \frac {1}{70} \left (\frac {582}{35} \int \frac {1}{(2 x+3) \sqrt {3 x^2+2}}dx-\frac {281 \sqrt {3 x^2+2}}{35 (2 x+3)}\right )-\frac {13 \sqrt {3 x^2+2}}{70 (2 x+3)^2}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {1}{70} \left (-\frac {582}{35} \int \frac {1}{35-\frac {(4-9 x)^2}{3 x^2+2}}d\frac {4-9 x}{\sqrt {3 x^2+2}}-\frac {281 \sqrt {3 x^2+2}}{35 (2 x+3)}\right )-\frac {13 \sqrt {3 x^2+2}}{70 (2 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{70} \left (-\frac {582 \text {arctanh}\left (\frac {4-9 x}{\sqrt {35} \sqrt {3 x^2+2}}\right )}{35 \sqrt {35}}-\frac {281 \sqrt {3 x^2+2}}{35 (2 x+3)}\right )-\frac {13 \sqrt {3 x^2+2}}{70 (2 x+3)^2}\) |
(-13*Sqrt[2 + 3*x^2])/(70*(3 + 2*x)^2) + ((-281*Sqrt[2 + 3*x^2])/(35*(3 + 2*x)) - (582*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(35*Sqrt[35])) /70
3.15.4.3.1 Defintions of rubi rules used
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 = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-\frac {843 x^{3}+1947 x^{2}+562 x +1298}{1225 \left (3+2 x \right )^{2} \sqrt {3 x^{2}+2}}-\frac {291 \sqrt {35}\, \operatorname {arctanh}\left (\frac {2 \left (4-9 x \right ) \sqrt {35}}{35 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-36 x -19}}\right )}{42875}\) | \(65\) |
trager | \(-\frac {\left (281 x +649\right ) \sqrt {3 x^{2}+2}}{1225 \left (3+2 x \right )^{2}}+\frac {291 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) \ln \left (\frac {9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) x +35 \sqrt {3 x^{2}+2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right )}{3+2 x}\right )}{42875}\) | \(71\) |
default | \(-\frac {281 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-9 x -\frac {19}{4}}}{4900 \left (x +\frac {3}{2}\right )}-\frac {291 \sqrt {35}\, \operatorname {arctanh}\left (\frac {2 \left (4-9 x \right ) \sqrt {35}}{35 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-36 x -19}}\right )}{42875}-\frac {13 \sqrt {3 \left (x +\frac {3}{2}\right )^{2}-9 x -\frac {19}{4}}}{280 \left (x +\frac {3}{2}\right )^{2}}\) | \(74\) |
-1/1225*(843*x^3+1947*x^2+562*x+1298)/(3+2*x)^2/(3*x^2+2)^(1/2)-291/42875* 35^(1/2)*arctanh(2/35*(4-9*x)*35^(1/2)/(12*(x+3/2)^2-36*x-19)^(1/2))
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16 \[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {291 \, \sqrt {35} {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (-\frac {\sqrt {35} \sqrt {3 \, x^{2} + 2} {\left (9 \, x - 4\right )} + 93 \, x^{2} - 36 \, x + 43}{4 \, x^{2} + 12 \, x + 9}\right ) - 70 \, \sqrt {3 \, x^{2} + 2} {\left (281 \, x + 649\right )}}{85750 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} \]
1/85750*(291*sqrt(35)*(4*x^2 + 12*x + 9)*log(-(sqrt(35)*sqrt(3*x^2 + 2)*(9 *x - 4) + 93*x^2 - 36*x + 43)/(4*x^2 + 12*x + 9)) - 70*sqrt(3*x^2 + 2)*(28 1*x + 649))/(4*x^2 + 12*x + 9)
\[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=- \int \frac {x}{8 x^{3} \sqrt {3 x^{2} + 2} + 36 x^{2} \sqrt {3 x^{2} + 2} + 54 x \sqrt {3 x^{2} + 2} + 27 \sqrt {3 x^{2} + 2}}\, dx - \int \left (- \frac {5}{8 x^{3} \sqrt {3 x^{2} + 2} + 36 x^{2} \sqrt {3 x^{2} + 2} + 54 x \sqrt {3 x^{2} + 2} + 27 \sqrt {3 x^{2} + 2}}\right )\, dx \]
-Integral(x/(8*x**3*sqrt(3*x**2 + 2) + 36*x**2*sqrt(3*x**2 + 2) + 54*x*sqr t(3*x**2 + 2) + 27*sqrt(3*x**2 + 2)), x) - Integral(-5/(8*x**3*sqrt(3*x**2 + 2) + 36*x**2*sqrt(3*x**2 + 2) + 54*x*sqrt(3*x**2 + 2) + 27*sqrt(3*x**2 + 2)), x)
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {291}{42875} \, \sqrt {35} \operatorname {arsinh}\left (\frac {3 \, \sqrt {6} x}{2 \, {\left | 2 \, x + 3 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 3 \right |}}\right ) - \frac {13 \, \sqrt {3 \, x^{2} + 2}}{70 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {281 \, \sqrt {3 \, x^{2} + 2}}{2450 \, {\left (2 \, x + 3\right )}} \]
291/42875*sqrt(35)*arcsinh(3/2*sqrt(6)*x/abs(2*x + 3) - 2/3*sqrt(6)/abs(2* x + 3)) - 13/70*sqrt(3*x^2 + 2)/(4*x^2 + 12*x + 9) - 281/2450*sqrt(3*x^2 + 2)/(2*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (62) = 124\).
Time = 0.31 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.38 \[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {291}{42875} \, \sqrt {35} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {35} - 3 \, \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {35} + 3 \, \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) - \frac {1164 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{3} + 6463 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} - 17904 \, \sqrt {3} x + 2248 \, \sqrt {3} + 17904 \, \sqrt {3 \, x^{2} + 2}}{4900 \, {\left ({\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} + 3 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )} - 2\right )}^{2}} \]
291/42875*sqrt(35)*log(-abs(-2*sqrt(3)*x - sqrt(35) - 3*sqrt(3) + 2*sqrt(3 *x^2 + 2))/(2*sqrt(3)*x - sqrt(35) + 3*sqrt(3) - 2*sqrt(3*x^2 + 2))) - 1/4 900*(1164*(sqrt(3)*x - sqrt(3*x^2 + 2))^3 + 6463*sqrt(3)*(sqrt(3)*x - sqrt (3*x^2 + 2))^2 - 17904*sqrt(3)*x + 2248*sqrt(3) + 17904*sqrt(3*x^2 + 2))/( (sqrt(3)*x - sqrt(3*x^2 + 2))^2 + 3*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 2)) - 2)^2
Time = 10.83 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {5-x}{(3+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {291\,\sqrt {35}\,\ln \left (x+\frac {3}{2}\right )}{42875}-\frac {291\,\sqrt {35}\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {35}\,\sqrt {x^2+\frac {2}{3}}}{9}-\frac {4}{9}\right )}{42875}-\frac {281\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{4900\,\left (x+\frac {3}{2}\right )}-\frac {13\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{280\,\left (x^2+3\,x+\frac {9}{4}\right )} \]