Integrand size = 27, antiderivative size = 51 \[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=\frac {1}{10} \arctan \left (\frac {5 (2+x)}{2 \sqrt {-7+2 x+5 x^2}}\right )+\frac {1}{5} \text {arctanh}\left (\frac {5 (1+x)}{\sqrt {-7+2 x+5 x^2}}\right ) \]
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=\frac {1}{10} \arctan \left (\frac {5+\frac {5 x}{2}}{\sqrt {-7+2 x+5 x^2}}\right )+\frac {1}{5} \text {arctanh}\left (\frac {5+5 x}{\sqrt {-7+2 x+5 x^2}}\right ) \]
ArcTan[(5 + (5*x)/2)/Sqrt[-7 + 2*x + 5*x^2]]/10 + ArcTanh[(5 + 5*x)/Sqrt[- 7 + 2*x + 5*x^2]]/5
Time = 0.22 (sec) , antiderivative size = 51, 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 = {1317, 27, 1362, 216, 220}
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.
| \(\displaystyle \int \frac {1}{\sqrt {5 x^2+2 x-7} \left (5 x^2+12 x+8\right )} \, dx\) |
| \(\Big \downarrow \) 1317 |
| \(\displaystyle \frac {1}{50} \int -\frac {50 (x+1)}{\sqrt {5 x^2+2 x-7} \left (5 x^2+12 x+8\right )}dx-\frac {1}{50} \int -\frac {50 (x+2)}{\sqrt {5 x^2+2 x-7} \left (5 x^2+12 x+8\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x+2}{\sqrt {5 x^2+2 x-7} \left (5 x^2+12 x+8\right )}dx-\int \frac {x+1}{\sqrt {5 x^2+2 x-7} \left (5 x^2+12 x+8\right )}dx\) |
| \(\Big \downarrow \) 1362 |
| \(\displaystyle -32 \int \frac {1}{\frac {6400 (x+1)^2}{5 x^2+2 x-7}-256}d\frac {8 (x+1)}{\sqrt {5 x^2+2 x-7}}-8 \int \frac {1}{\frac {400 (x+2)^2}{5 x^2+2 x-7}+64}d\left (-\frac {2 (x+2)}{\sqrt {5 x^2+2 x-7}}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{10} \arctan \left (\frac {5 (x+2)}{2 \sqrt {5 x^2+2 x-7}}\right )-32 \int \frac {1}{\frac {6400 (x+1)^2}{5 x^2+2 x-7}-256}d\frac {8 (x+1)}{\sqrt {5 x^2+2 x-7}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{10} \arctan \left (\frac {5 (x+2)}{2 \sqrt {5 x^2+2 x-7}}\right )+\frac {1}{5} \text {arctanh}\left (\frac {5 (x+1)}{\sqrt {5 x^2+2 x-7}}\right )\) |
ArcTan[(5*(2 + x))/(2*Sqrt[-7 + 2*x + 5*x^2])]/10 + ArcTanh[(5*(1 + x))/Sq rt[-7 + 2*x + 5*x^2]]/5
3.2.21.3.1 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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[1/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)* (x_)^2]), x_Symbol] :> With[{q = Rt[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f) , 2]}, Simp[1/(2*q) Int[(c*d - a*f + q + (c*e - b*f)*x)/((a + b*x + c*x^2 )*Sqrt[d + e*x + f*x^2]), x], x] - Simp[1/(2*q) Int[(c*d - a*f - q + (c*e - b*f)*x)/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && NeQ[c*e - b*f, 0] && NegQ[b^2 - 4*a*c]
Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[-2*g*(g*b - 2*a*h) Subst[I nt[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - (b*d - a*e)*x^2, x], x], x, Simp[ g*b - 2*a*h - (b*h - 2*g*c)*x, x]/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b , c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && Ne Q[b*d - a*e, 0] && EqQ[h^2*(b*d - a*e) - 2*g*h*(c*d - a*f) + g^2*(c*e - b*f ), 0]
Leaf count of result is larger than twice the leaf count of optimal. \(143\) vs. \(2(41)=82\).
Time = 0.77 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.82
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {4 \left (2+x \right )^{2}}{\left (-1-x \right )^{2}}+9}\, \left (2 \,\operatorname {arctanh}\left (\frac {\sqrt {-\frac {4 \left (2+x \right )^{2}}{\left (-1-x \right )^{2}}+9}}{5}\right )+\arctan \left (\frac {5 \sqrt {-\frac {4 \left (2+x \right )^{2}}{\left (-1-x \right )^{2}}+9}\, \left (2+x \right )}{2 \left (\frac {4 \left (2+x \right )^{2}}{\left (-1-x \right )^{2}}-9\right ) \left (-1-x \right )}\right )\right )}{10 \sqrt {-\frac {\frac {4 \left (2+x \right )^{2}}{\left (-1-x \right )^{2}}-9}{\left (1+\frac {2+x}{-1-x}\right )^{2}}}\, \left (1+\frac {2+x}{-1-x}\right )}\) | \(144\) |
| trager | \(\operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \ln \left (-\frac {-129600 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x +8750 \sqrt {5 x^{2}+2 x -7}\, \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+18630 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -155 \sqrt {5 x^{2}+2 x -7}-30330 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+1105 x +5729}{20 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -5 x -4}\right )+\frac {\ln \left (\frac {129600 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x +8750 \sqrt {5 x^{2}+2 x -7}\, \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-33210 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -1595 \sqrt {5 x^{2}+2 x -7}-30330 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+353 x +337}{20 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x +x +4}\right )}{5}-\ln \left (\frac {129600 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x +8750 \sqrt {5 x^{2}+2 x -7}\, \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-33210 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -1595 \sqrt {5 x^{2}+2 x -7}-30330 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+353 x +337}{20 \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x +x +4}\right ) \operatorname {RootOf}\left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )\) | \(356\) |
-1/10*(-4*(2+x)^2/(-1-x)^2+9)^(1/2)*(2*arctanh(1/5*(-4*(2+x)^2/(-1-x)^2+9) ^(1/2))+arctan(5/2*(-4*(2+x)^2/(-1-x)^2+9)^(1/2)/(4*(2+x)^2/(-1-x)^2-9)*(2 +x)/(-1-x)))/(-(4*(2+x)^2/(-1-x)^2-9)/(1+(2+x)/(-1-x))^2)^(1/2)/(1+(2+x)/( -1-x))
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (41) = 82\).
Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 3.02 \[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=\frac {1}{20} \, \arctan \left (\frac {27 \, x^{2} + 20 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 2\right )} + 36 \, x}{31 \, x^{2} + 16 \, x - 56}\right ) + \frac {1}{20} \, \arctan \left (-\frac {27 \, x^{2} - 20 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 2\right )} + 36 \, x}{31 \, x^{2} + 16 \, x - 56}\right ) + \frac {1}{20} \, \log \left (\frac {15 \, x^{2} + 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 1\right )} + 26 \, x + 9}{x^{2}}\right ) - \frac {1}{20} \, \log \left (\frac {15 \, x^{2} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 1\right )} + 26 \, x + 9}{x^{2}}\right ) \]
1/20*arctan((27*x^2 + 20*sqrt(5*x^2 + 2*x - 7)*(x + 2) + 36*x)/(31*x^2 + 1 6*x - 56)) + 1/20*arctan(-(27*x^2 - 20*sqrt(5*x^2 + 2*x - 7)*(x + 2) + 36* x)/(31*x^2 + 16*x - 56)) + 1/20*log((15*x^2 + 5*sqrt(5*x^2 + 2*x - 7)*(x + 1) + 26*x + 9)/x^2) - 1/20*log((15*x^2 - 5*sqrt(5*x^2 + 2*x - 7)*(x + 1) + 26*x + 9)/x^2)
\[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=\int \frac {1}{\sqrt {\left (x - 1\right ) \left (5 x + 7\right )} \left (5 x^{2} + 12 x + 8\right )}\, dx \]
\[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=\int { \frac {1}{{\left (5 \, x^{2} + 12 \, x + 8\right )} \sqrt {5 \, x^{2} + 2 \, x - 7}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (41) = 82\).
Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 4.02 \[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=-\frac {1}{10} \, \arctan \left (-\frac {5 \, \sqrt {5} x + 6 \, \sqrt {5} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} + 5}{2 \, {\left (\sqrt {5} + 5\right )}}\right ) - \frac {1}{10} \, \arctan \left (\frac {5 \, \sqrt {5} x + 6 \, \sqrt {5} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} - 5}{2 \, {\left (\sqrt {5} - 5\right )}}\right ) + \frac {1}{10} \, \log \left (5 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )}^{2} + 2 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )} {\left (6 \, \sqrt {5} + 5\right )} + 20 \, \sqrt {5} + 65\right ) - \frac {1}{10} \, \log \left (5 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )}^{2} + 2 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )} {\left (6 \, \sqrt {5} - 5\right )} - 20 \, \sqrt {5} + 65\right ) \]
-1/10*arctan(-1/2*(5*sqrt(5)*x + 6*sqrt(5) - 5*sqrt(5*x^2 + 2*x - 7) + 5)/ (sqrt(5) + 5)) - 1/10*arctan(1/2*(5*sqrt(5)*x + 6*sqrt(5) - 5*sqrt(5*x^2 + 2*x - 7) - 5)/(sqrt(5) - 5)) + 1/10*log(5*(sqrt(5)*x - sqrt(5*x^2 + 2*x - 7))^2 + 2*(sqrt(5)*x - sqrt(5*x^2 + 2*x - 7))*(6*sqrt(5) + 5) + 20*sqrt(5 ) + 65) - 1/10*log(5*(sqrt(5)*x - sqrt(5*x^2 + 2*x - 7))^2 + 2*(sqrt(5)*x - sqrt(5*x^2 + 2*x - 7))*(6*sqrt(5) - 5) - 20*sqrt(5) + 65)
Timed out. \[ \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx=\int \frac {1}{\sqrt {5\,x^2+2\,x-7}\,\left (5\,x^2+12\,x+8\right )} \,d x \]