Integrand size = 31, antiderivative size = 158 \[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=-2 \sqrt {1+x+x^2}+\frac {1}{4} (1+2 x) \sqrt {1+x+x^2}-2 \sqrt {4+2 x+x^2}+\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {11}{2} \text {arcsinh}\left (\frac {1+x}{\sqrt {3}}\right )+\frac {43}{8} \text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right )-2 \sqrt {7} \text {arctanh}\left (\frac {1+5 x}{2 \sqrt {7} \sqrt {1+x+x^2}}\right )+2 \sqrt {7} \text {arctanh}\left (\frac {1-2 x}{\sqrt {7} \sqrt {4+2 x+x^2}}\right ) \]
11/2*arcsinh(1/3*(1+x)*3^(1/2))+43/8*arcsinh(1/3*(1+2*x)*3^(1/2))-2*arctan h(1/14*(1+5*x)*7^(1/2)/(x^2+x+1)^(1/2))*7^(1/2)+2*arctanh(1/7*(1-2*x)*7^(1 /2)/(x^2+2*x+4)^(1/2))*7^(1/2)-2*(x^2+x+1)^(1/2)+1/4*(1+2*x)*(x^2+x+1)^(1/ 2)-2*(x^2+2*x+4)^(1/2)+1/2*(1+x)*(x^2+2*x+4)^(1/2)
Time = 5.51 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.01 \[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\frac {1}{8} \left (-14 \sqrt {1+x+x^2}+4 x \sqrt {1+x+x^2}-12 \sqrt {4+2 x+x^2}+4 x \sqrt {4+2 x+x^2}-32 \sqrt {7} \text {arctanh}\left (\frac {3+x-\sqrt {1+x+x^2}}{\sqrt {7}}\right )-32 \sqrt {7} \text {arctanh}\left (\frac {3+x-\sqrt {4+2 x+x^2}}{\sqrt {7}}\right )-43 \log \left (-1-2 x+2 \sqrt {1+x+x^2}\right )-44 \log \left (-1-x+\sqrt {4+2 x+x^2}\right )\right ) \]
(-14*Sqrt[1 + x + x^2] + 4*x*Sqrt[1 + x + x^2] - 12*Sqrt[4 + 2*x + x^2] + 4*x*Sqrt[4 + 2*x + x^2] - 32*Sqrt[7]*ArcTanh[(3 + x - Sqrt[1 + x + x^2])/S qrt[7]] - 32*Sqrt[7]*ArcTanh[(3 + x - Sqrt[4 + 2*x + x^2])/Sqrt[7]] - 43*L og[-1 - 2*x + 2*Sqrt[1 + x + x^2]] - 44*Log[-1 - x + Sqrt[4 + 2*x + x^2]]) /8
Time = 0.62 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {x+1}{\sqrt {x^2+2 x+4}-\sqrt {x^2+x+1}} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {x}{\sqrt {x^2+x+1}-\sqrt {x^2+2 x+4}}-\frac {1}{\sqrt {x^2+x+1}-\sqrt {x^2+2 x+4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {11}{2} \text {arcsinh}\left (\frac {x+1}{\sqrt {3}}\right )+\frac {43}{8} \text {arcsinh}\left (\frac {2 x+1}{\sqrt {3}}\right )-2 \sqrt {7} \text {arctanh}\left (\frac {5 x+1}{2 \sqrt {7} \sqrt {x^2+x+1}}\right )+2 \sqrt {7} \text {arctanh}\left (\frac {1-2 x}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )+\frac {1}{2} \sqrt {x^2+2 x+4} (x+1)+\frac {1}{4} (2 x+1) \sqrt {x^2+x+1}-2 \sqrt {x^2+x+1}-2 \sqrt {x^2+2 x+4}\) |
-2*Sqrt[1 + x + x^2] + ((1 + 2*x)*Sqrt[1 + x + x^2])/4 - 2*Sqrt[4 + 2*x + x^2] + ((1 + x)*Sqrt[4 + 2*x + x^2])/2 + (11*ArcSinh[(1 + x)/Sqrt[3]])/2 + (43*ArcSinh[(1 + 2*x)/Sqrt[3]])/8 - 2*Sqrt[7]*ArcTanh[(1 + 5*x)/(2*Sqrt[7 ]*Sqrt[1 + x + x^2])] + 2*Sqrt[7]*ArcTanh[(1 - 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])]
3.3.91.3.1 Defintions of rubi rules used
Time = 0.02 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.89
method | result | size |
default | \(-2 \sqrt {\left (3+x \right )^{2}-5 x -8}+\frac {43 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{8}+2 \sqrt {7}\, \operatorname {arctanh}\left (\frac {\left (-1-5 x \right ) \sqrt {7}}{14 \sqrt {\left (3+x \right )^{2}-5 x -8}}\right )-2 \sqrt {\left (3+x \right )^{2}-4 x -5}+\frac {11 \,\operatorname {arcsinh}\left (\frac {\left (1+x \right ) \sqrt {3}}{3}\right )}{2}+2 \sqrt {7}\, \operatorname {arctanh}\left (\frac {\left (2-4 x \right ) \sqrt {7}}{14 \sqrt {\left (3+x \right )^{2}-4 x -5}}\right )+\frac {\left (1+2 x \right ) \sqrt {x^{2}+x +1}}{4}+\frac {\left (2 x +2\right ) \sqrt {x^{2}+2 x +4}}{4}\) | \(140\) |
-2*((3+x)^2-5*x-8)^(1/2)+43/8*arcsinh(2/3*3^(1/2)*(x+1/2))+2*7^(1/2)*arcta nh(1/14*(-1-5*x)*7^(1/2)/((3+x)^2-5*x-8)^(1/2))-2*((3+x)^2-4*x-5)^(1/2)+11 /2*arcsinh(1/3*(1+x)*3^(1/2))+2*7^(1/2)*arctanh(1/14*(2-4*x)*7^(1/2)/((3+x )^2-4*x-5)^(1/2))+1/4*(1+2*x)*(x^2+x+1)^(1/2)+1/4*(2*x+2)*(x^2+2*x+4)^(1/2 )
Time = 0.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.98 \[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\frac {1}{4} \, \sqrt {x^{2} + x + 1} {\left (2 \, x - 7\right )} + \frac {1}{2} \, \sqrt {x^{2} + 2 \, x + 4} {\left (x - 3\right )} + 2 \, \sqrt {7} \log \left (\frac {2 \, \sqrt {7} {\left (5 \, x + 1\right )} + 2 \, \sqrt {x^{2} + x + 1} {\left (5 \, \sqrt {7} - 14\right )} - 25 \, x - 5}{x + 3}\right ) + 2 \, \sqrt {7} \log \left (\frac {\sqrt {7} {\left (2 \, x - 1\right )} + \sqrt {x^{2} + 2 \, x + 4} {\left (2 \, \sqrt {7} - 7\right )} - 4 \, x + 2}{x + 3}\right ) - \frac {11}{2} \, \log \left (-x + \sqrt {x^{2} + 2 \, x + 4} - 1\right ) - \frac {43}{8} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]
1/4*sqrt(x^2 + x + 1)*(2*x - 7) + 1/2*sqrt(x^2 + 2*x + 4)*(x - 3) + 2*sqrt (7)*log((2*sqrt(7)*(5*x + 1) + 2*sqrt(x^2 + x + 1)*(5*sqrt(7) - 14) - 25*x - 5)/(x + 3)) + 2*sqrt(7)*log((sqrt(7)*(2*x - 1) + sqrt(x^2 + 2*x + 4)*(2 *sqrt(7) - 7) - 4*x + 2)/(x + 3)) - 11/2*log(-x + sqrt(x^2 + 2*x + 4) - 1) - 43/8*log(-2*x + 2*sqrt(x^2 + x + 1) - 1)
\[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\int \frac {x + 1}{- \sqrt {x^{2} + x + 1} + \sqrt {x^{2} + 2 x + 4}}\, dx \]
\[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\int { \frac {x + 1}{\sqrt {x^{2} + 2 \, x + 4} - \sqrt {x^{2} + x + 1}} \,d x } \]
\[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\int { \frac {x + 1}{\sqrt {x^{2} + 2 \, x + 4} - \sqrt {x^{2} + x + 1}} \,d x } \]
Timed out. \[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\int -\frac {x+1}{\sqrt {x^2+x+1}-\sqrt {x^2+2\,x+4}} \,d x \]
\[ \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx=\int \frac {x +1}{\sqrt {x^{2}+2 x +4}-\sqrt {x^{2}+x +1}}d x \]