Integrand size = 27, antiderivative size = 77 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{2 \sqrt {3}}+\frac {13 \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{2 \sqrt {5}} \]
-1/6*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+13/10*arctan h(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)^(1/2))*5^(1/2)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=\frac {13 \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )}{\sqrt {5}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{\sqrt {3}} \]
(13*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)])/Sqrt[5] - ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)]/Sqrt[3]
Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1269, 1092, 219, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3) \sqrt {3 x^2+5 x+2}} \, dx\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {13}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\frac {1}{2} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {13}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {13}{2} \int \frac {1}{(2 x+3) \sqrt {3 x^2+5 x+2}}dx-\frac {\text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle -13 \int \frac {1}{20-\frac {(8 x+7)^2}{3 x^2+5 x+2}}d\left (-\frac {8 x+7}{\sqrt {3 x^2+5 x+2}}\right )-\frac {\text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {13 \text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {5}}-\frac {\text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{2 \sqrt {3}}\) |
-1/2*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])]/Sqrt[3] + (13*Ar cTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(2*Sqrt[5])
3.25.100.3.1 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/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
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[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79
method | result | size |
default | \(-\frac {\ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{6}-\frac {13 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{10}\) | \(61\) |
trager | \(\frac {13 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{10}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{6}\) | \(92\) |
-1/6*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)-13/10*5^(1/2)*a rctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=\frac {1}{12} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + \frac {13}{20} \, \sqrt {5} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) \]
1/12*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120 *x + 49) + 13/20*sqrt(5)*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))
\[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=- \int \frac {x}{2 x \sqrt {3 x^{2} + 5 x + 2} + 3 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{2 x \sqrt {3 x^{2} + 5 x + 2} + 3 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]
-Integral(x/(2*x*sqrt(3*x**2 + 5*x + 2) + 3*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-5/(2*x*sqrt(3*x**2 + 5*x + 2) + 3*sqrt(3*x**2 + 5*x + 2)), x)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=-\frac {1}{6} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {13}{10} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) \]
-1/6*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 13/10*sqrt(5 )*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2)
Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.39 \[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=\frac {13}{10} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {1}{6} \, \sqrt {3} \log \left ({\left | -6 \, \sqrt {3} x - 5 \, \sqrt {3} + 6 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}\right ) \]
13/10*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))) + 1/6*sqrt(3)*log(abs(-6*sqrt(3)*x - 5*sqrt(3) + 6*sqrt(3*x^2 + 5*x + 2)))
Timed out. \[ \int \frac {5-x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx=-\int \frac {x-5}{\left (2\,x+3\right )\,\sqrt {3\,x^2+5\,x+2}} \,d x \]