\(\int \frac {x^2}{(1+2 x+\sqrt {-3-2 x+4 x^2})^2} \, dx\) [7]

Optimal result

Integrand size = 25, antiderivative size = 183 \[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {1}{64} \left (2 x+\sqrt {-3-2 x+4 x^2}\right )+\frac {4}{81 \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )}-\frac {2197}{3456 \left (1-2 \left (2 x+\sqrt {-3-2 x+4 x^2}\right )\right )^3}-\frac {1183}{3456 \left (1-2 \left (2 x+\sqrt {-3-2 x+4 x^2}\right )\right )^2}-\frac {377}{864 \left (1-2 \left (2 x+\sqrt {-3-2 x+4 x^2}\right )\right )}-\frac {56}{243} \log \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )+\frac {3341 \log \left (1-2 \left (2 x+\sqrt {-3-2 x+4 x^2}\right )\right )}{15552} \] Output:

1/32*x+1/64*(4*x^2-2*x-3)^(1/2)+4/(81+162*x+81*(4*x^2-2*x-3)^(1/2))-2197/3 
456/(1-4*x-2*(4*x^2-2*x-3)^(1/2))^3-1183/3456/(1-4*x-2*(4*x^2-2*x-3)^(1/2) 
)^2-377/(864-3456*x-1728*(4*x^2-2*x-3)^(1/2))-56/243*ln(1+2*x+(4*x^2-2*x-3 
)^(1/2))+3341/15552*ln(1-4*x-2*(4*x^2-2*x-3)^(1/2))
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.66 \[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {-\frac {6 \sqrt {-3-2 x+4 x^2} \left (346+159 x-204 x^2+288 x^3\right )}{2+3 x}+\frac {16 \left (-8+324 x+252 x^2-207 x^3+216 x^4\right )}{2+3 x}+243 \log \left (1-4 x+2 \sqrt {-3-2 x+4 x^2}\right )-3584 \log \left (-5-6 x+3 \sqrt {-3-2 x+4 x^2}\right )}{15552} \] Input:

Integrate[x^2/(1 + 2*x + Sqrt[-3 - 2*x + 4*x^2])^2,x]
 

Output:

((-6*Sqrt[-3 - 2*x + 4*x^2]*(346 + 159*x - 204*x^2 + 288*x^3))/(2 + 3*x) + 
 (16*(-8 + 324*x + 252*x^2 - 207*x^3 + 216*x^4))/(2 + 3*x) + 243*Log[1 - 4 
*x + 2*Sqrt[-3 - 2*x + 4*x^2]] - 3584*Log[-5 - 6*x + 3*Sqrt[-3 - 2*x + 4*x 
^2]])/15552
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.

\(\displaystyle \int \frac {x^2}{\left (\sqrt {4 x^2-2 x-3}+2 x+1\right )^2} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 x^2}{9}-\frac {1}{9} \sqrt {4 x^2-2 x-3} x-\frac {2 \sqrt {4 x^2-2 x-3}}{9 (3 x+2)}+\frac {2 \sqrt {4 x^2-2 x-3}}{27 (3 x+2)^2}+\frac {5}{54} \sqrt {4 x^2-2 x-3}-\frac {13 x}{54}-\frac {28}{81 (3 x+2)}+\frac {2}{81 (3 x+2)^2}+\frac {1}{6}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1549 \text {arctanh}\left (\frac {1-4 x}{2 \sqrt {4 x^2-2 x-3}}\right )}{15552}-\frac {28}{243} \text {arctanh}\left (\frac {11 x+7}{\sqrt {4 x^2-2 x-3}}\right )+\frac {2 x^3}{27}-\frac {13 x^2}{108}-\frac {1}{108} \left (4 x^2-2 x-3\right )^{3/2}-\frac {7}{864} (1-4 x) \sqrt {4 x^2-2 x-3}-\frac {2 \sqrt {4 x^2-2 x-3}}{81 (3 x+2)}-\frac {2}{27} \sqrt {4 x^2-2 x-3}+\frac {x}{6}-\frac {2}{243 (3 x+2)}-\frac {28}{243} \log (3 x+2)\)

Input:

Int[x^2/(1 + 2*x + Sqrt[-3 - 2*x + 4*x^2])^2,x]
 

Output:

x/6 - (13*x^2)/108 + (2*x^3)/27 - 2/(243*(2 + 3*x)) - (2*Sqrt[-3 - 2*x + 4 
*x^2])/27 - (7*(1 - 4*x)*Sqrt[-3 - 2*x + 4*x^2])/864 - (2*Sqrt[-3 - 2*x + 
4*x^2])/(81*(2 + 3*x)) - (-3 - 2*x + 4*x^2)^(3/2)/108 - (1549*ArcTanh[(1 - 
 4*x)/(2*Sqrt[-3 - 2*x + 4*x^2])])/15552 - (28*ArcTanh[(7 + 11*x)/Sqrt[-3 
- 2*x + 4*x^2]])/243 - (28*Log[2 + 3*x])/243
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13

\[\frac {x}{6}-\frac {2}{243 \left (2+3 x \right )}-\frac {28 \ln \left (2+3 x \right )}{243}-\frac {13 x^{2}}{108}+\frac {2 x^{3}}{27}+\frac {7 \left (8 x -2\right ) \sqrt {4 x^{2}-2 x -3}}{1728}-\frac {91 \ln \left (\frac {\left (4 x -1\right ) \sqrt {4}}{4}+\sqrt {4 x^{2}-2 x -3}\right ) \sqrt {4}}{3456}-\frac {2 \left (4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}\right )^{\frac {3}{2}}}{27 \left (x +\frac {2}{3}\right )}-\frac {28 \sqrt {36 \left (x +\frac {2}{3}\right )^{2}-66 x -43}}{243}+\frac {37 \ln \left (\frac {\left (4 x -1\right ) \sqrt {4}}{4}+\sqrt {4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}}\right ) \sqrt {4}}{486}+\frac {28 \,\operatorname {arctanh}\left (\frac {-21-33 x}{\sqrt {36 \left (x +\frac {2}{3}\right )^{2}-66 x -43}}\right )}{243}+\frac {\left (8 x -2\right ) \sqrt {4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}}}{27}-\frac {\left (4 x^{2}-2 x -3\right )^{\frac {3}{2}}}{108}\]

Input:

int(x^2/(1+2*x+(4*x^2-2*x-3)^(1/2))^2,x)
 

Output:

1/6*x-2/243/(2+3*x)-28/243*ln(2+3*x)-13/108*x^2+2/27*x^3+7/1728*(8*x-2)*(4 
*x^2-2*x-3)^(1/2)-91/3456*ln(1/4*(4*x-1)*4^(1/2)+(4*x^2-2*x-3)^(1/2))*4^(1 
/2)-2/27/(x+2/3)*(4*(x+2/3)^2-22/3*x-43/9)^(3/2)-28/243*(36*(x+2/3)^2-66*x 
-43)^(1/2)+37/486*ln(1/4*(4*x-1)*4^(1/2)+(4*(x+2/3)^2-22/3*x-43/9)^(1/2))* 
4^(1/2)+28/243*arctanh(9/2*(-14/3-22/3*x)/(36*(x+2/3)^2-66*x-43)^(1/2))+1/ 
27*(8*x-2)*(4*(x+2/3)^2-22/3*x-43/9)^(1/2)-1/108*(4*x^2-2*x-3)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {13824 \, x^{4} - 13248 \, x^{3} + 16128 \, x^{2} - 7168 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 7168 \, {\left (3 \, x + 2\right )} \log \left (-2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} - 1\right ) - 6196 \, {\left (3 \, x + 2\right )} \log \left (-4 \, x + 2 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 1\right ) - 7168 \, {\left (3 \, x + 2\right )} \log \left (-6 \, x + 3 \, \sqrt {4 \, x^{2} - 2 \, x - 3} - 5\right ) - 24 \, {\left (288 \, x^{3} - 204 \, x^{2} + 159 \, x + 346\right )} \sqrt {4 \, x^{2} - 2 \, x - 3} + 20283 \, x - 814}{62208 \, {\left (3 \, x + 2\right )}} \] Input:

integrate(x^2/(1+2*x+(4*x^2-2*x-3)^(1/2))^2,x, algorithm="fricas")
 

Output:

1/62208*(13824*x^4 - 13248*x^3 + 16128*x^2 - 7168*(3*x + 2)*log(3*x + 2) + 
 7168*(3*x + 2)*log(-2*x + sqrt(4*x^2 - 2*x - 3) - 1) - 6196*(3*x + 2)*log 
(-4*x + 2*sqrt(4*x^2 - 2*x - 3) + 1) - 7168*(3*x + 2)*log(-6*x + 3*sqrt(4* 
x^2 - 2*x - 3) - 5) - 24*(288*x^3 - 204*x^2 + 159*x + 346)*sqrt(4*x^2 - 2* 
x - 3) + 20283*x - 814)/(3*x + 2)
 

Sympy [F]

\[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\int \frac {x^{2}}{\left (2 x + \sqrt {4 x^{2} - 2 x - 3} + 1\right )^{2}}\, dx \] Input:

integrate(x**2/(1+2*x+(4*x**2-2*x-3)**(1/2))**2,x)
 

Output:

Integral(x**2/(2*x + sqrt(4*x**2 - 2*x - 3) + 1)**2, x)
 

Maxima [F]

\[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\int { \frac {x^{2}}{{\left (2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} + 1\right )}^{2}} \,d x } \] Input:

integrate(x^2/(1+2*x+(4*x^2-2*x-3)^(1/2))^2,x, algorithm="maxima")
 

Output:

integrate(x^2/(2*x + sqrt(4*x^2 - 2*x - 3) + 1)^2, x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {2}{27} \, x^{3} - \frac {13}{108} \, x^{2} - \frac {1}{864} \, {\left (4 \, {\left (8 \, x - 11\right )} x + 47\right )} \sqrt {4 \, x^{2} - 2 \, x - 3} + \frac {1}{6} \, x - \frac {4 \, {\left (22 \, x - 11 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 14\right )}}{243 \, {\left (3 \, {\left (2 \, x - \sqrt {4 \, x^{2} - 2 \, x - 3}\right )}^{2} + 16 \, x - 8 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 5\right )}} - \frac {2}{243 \, {\left (3 \, x + 2\right )}} - \frac {28}{243} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {28}{243} \, \log \left ({\left | -2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} - 1 \right |}\right ) - \frac {1549}{15552} \, \log \left ({\left | -4 \, x + 2 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 1 \right |}\right ) - \frac {28}{243} \, \log \left ({\left | -6 \, x + 3 \, \sqrt {4 \, x^{2} - 2 \, x - 3} - 5 \right |}\right ) \] Input:

integrate(x^2/(1+2*x+(4*x^2-2*x-3)^(1/2))^2,x, algorithm="giac")
 

Output:

2/27*x^3 - 13/108*x^2 - 1/864*(4*(8*x - 11)*x + 47)*sqrt(4*x^2 - 2*x - 3) 
+ 1/6*x - 4/243*(22*x - 11*sqrt(4*x^2 - 2*x - 3) + 14)/(3*(2*x - sqrt(4*x^ 
2 - 2*x - 3))^2 + 16*x - 8*sqrt(4*x^2 - 2*x - 3) + 5) - 2/243/(3*x + 2) - 
28/243*log(abs(3*x + 2)) + 28/243*log(abs(-2*x + sqrt(4*x^2 - 2*x - 3) - 1 
)) - 1549/15552*log(abs(-4*x + 2*sqrt(4*x^2 - 2*x - 3) + 1)) - 28/243*log( 
abs(-6*x + 3*sqrt(4*x^2 - 2*x - 3) - 5))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {x}{6}-\frac {28\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {2}{729\,\left (x+\frac {2}{3}\right )}-\frac {13\,x^2}{108}+\frac {2\,x^3}{27}+\int \frac {x^2\,\left (-8\,x^3+8\,x+3\right )}{2\,{\left (3\,x+2\right )}^2\,\sqrt {4\,x^2-2\,x-3}} \,d x \] Input:

int(x^2/(2*x + (4*x^2 - 2*x - 3)^(1/2) + 1)^2,x)
 

Output:

x/6 - (28*log(x + 2/3))/243 - 2/(729*(x + 2/3)) - (13*x^2)/108 + (2*x^3)/2 
7 + int((x^2*(8*x - 8*x^3 + 3))/(2*(3*x + 2)^2*(4*x^2 - 2*x - 3)^(1/2)), x 
)
 

Reduce [F]

\[ \int \frac {x^2}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\int \frac {x^{2}}{\left (1+2 x +\sqrt {4 x^{2}-2 x -3}\right )^{2}}d x \] Input:

int(x^2/(1+2*x+(4*x^2-2*x-3)^(1/2))^2,x)
 

Output:

int(x^2/(1+2*x+(4*x^2-2*x-3)^(1/2))^2,x)