\(\int \frac {1}{x (1+2 x+\sqrt {-3-2 x+4 x^2})^3} \, dx\) [15]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {1}{x \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^3} \, dx=-\frac {1}{12 \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2}-\frac {25}{36 \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )}+\frac {59}{108} \log \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )-\frac {1}{8} \log \left (x-4 x^2-2 x \sqrt {-3-2 x+4 x^2}\right )-\frac {8}{27} \log \left (1-2 \left (2 x+\sqrt {-3-2 x+4 x^2}\right )\right ) \] Output:

-1/12/(1+2*x+(4*x^2-2*x-3)^(1/2))^2-25/(36+72*x+36*(4*x^2-2*x-3)^(1/2))+59 
/108*ln(1+2*x+(4*x^2-2*x-3)^(1/2))-1/8*ln(x-4*x^2-2*x*(4*x^2-2*x-3)^(1/2)) 
-8/27*ln(1-4*x-2*(4*x^2-2*x-3)^(1/2))
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^3} \, dx=\frac {1}{216} \left (\frac {3 (25+38 x)}{(2+3 x)^2}+\frac {3 (53+81 x) \sqrt {-3-2 x+4 x^2}}{(2+3 x)^2}-27 \log \left (x \left (-1+4 x-2 \sqrt {-3-2 x+4 x^2}\right )\right )+118 \log \left (-5-6 x+3 \sqrt {-3-2 x+4 x^2}\right )\right ) \] Input:

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

Output:

((3*(25 + 38*x))/(2 + 3*x)^2 + (3*(53 + 81*x)*Sqrt[-3 - 2*x + 4*x^2])/(2 + 
 3*x)^2 - 27*Log[x*(-1 + 4*x - 2*Sqrt[-3 - 2*x + 4*x^2])] + 118*Log[-5 - 6 
*x + 3*Sqrt[-3 - 2*x + 4*x^2]])/216
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.12, 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.

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {2 \sqrt {4 x^2-2 x-3}}{3 (3 x+2)^2}+\frac {\sqrt {4 x^2-2 x-3}}{12 (3 x+2)^3}-\frac {1}{8 x}+\frac {59}{72 (3 x+2)}-\frac {19}{36 (3 x+2)^2}+\frac {1}{36 (3 x+2)^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4}{27} \text {arctanh}\left (\frac {1-4 x}{2 \sqrt {4 x^2-2 x-3}}\right )+\frac {59}{216} \text {arctanh}\left (\frac {11 x+7}{\sqrt {4 x^2-2 x-3}}\right )+\frac {\sqrt {4 x^2-2 x-3} (11 x+7)}{24 (3 x+2)^2}+\frac {2 \sqrt {4 x^2-2 x-3}}{9 (3 x+2)}+\frac {19}{108 (3 x+2)}-\frac {1}{216 (3 x+2)^2}-\frac {\log (x)}{8}+\frac {59}{216} \log (3 x+2)\)

Input:

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

Output:

-1/216*1/(2 + 3*x)^2 + 19/(108*(2 + 3*x)) + (2*Sqrt[-3 - 2*x + 4*x^2])/(9* 
(2 + 3*x)) + ((7 + 11*x)*Sqrt[-3 - 2*x + 4*x^2])/(24*(2 + 3*x)^2) + (4*Arc 
Tanh[(1 - 4*x)/(2*Sqrt[-3 - 2*x + 4*x^2])])/27 + (59*ArcTanh[(7 + 11*x)/Sq 
rt[-3 - 2*x + 4*x^2]])/216 - Log[x]/8 + (59*Log[2 + 3*x])/216
 

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 = 66.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.27

method result size
default \(-\frac {1}{216 \left (2+3 x \right )^{2}}-\frac {\ln \left (x \right )}{8}+\frac {19}{108 \left (2+3 x \right )}+\frac {59 \ln \left (2+3 x \right )}{216}-\frac {\left (4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}\right )^{\frac {3}{2}}}{72 \left (x +\frac {2}{3}\right )^{2}}+\frac {5 \left (4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}\right )^{\frac {3}{2}}}{24 \left (x +\frac {2}{3}\right )}+\frac {59 \sqrt {36 \left (x +\frac {2}{3}\right )^{2}-66 x -43}}{216}-\frac {59 \,\operatorname {arctanh}\left (\frac {-21-33 x}{\sqrt {36 \left (x +\frac {2}{3}\right )^{2}-66 x -43}}\right )}{216}-\frac {5 \left (8 x -2\right ) \sqrt {4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}}}{48}-\frac {2 \ln \left (\frac {\left (4 x -1\right ) \sqrt {4}}{4}+\sqrt {4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}}\right ) \sqrt {4}}{27}\) \(166\)
trager \(\text {Expression too large to display}\) \(2056\)

Input:

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

Output:

-1/216/(2+3*x)^2-1/8*ln(x)+19/108/(2+3*x)+59/216*ln(2+3*x)-1/72/(x+2/3)^2* 
(4*(x+2/3)^2-22/3*x-43/9)^(3/2)+5/24/(x+2/3)*(4*(x+2/3)^2-22/3*x-43/9)^(3/ 
2)+59/216*(36*(x+2/3)^2-66*x-43)^(1/2)-59/216*arctanh(9/2*(-14/3-22/3*x)/( 
36*(x+2/3)^2-66*x-43)^(1/2))-5/48*(8*x-2)*(4*(x+2/3)^2-22/3*x-43/9)^(1/2)- 
2/27*ln(1/4*(4*x-1)*4^(1/2)+(4*(x+2/3)^2-22/3*x-43/9)^(1/2))*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^3} \, dx=\frac {486 \, x^{2} + 59 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 27 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (x\right ) - 59 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (-2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} - 1\right ) + 32 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (-4 \, x + 2 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 1\right ) + 59 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (-6 \, x + 3 \, \sqrt {4 \, x^{2} - 2 \, x - 3} - 5\right ) + 3 \, \sqrt {4 \, x^{2} - 2 \, x - 3} {\left (81 \, x + 53\right )} + 762 \, x + 291}{216 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

1/216*(486*x^2 + 59*(9*x^2 + 12*x + 4)*log(3*x + 2) - 27*(9*x^2 + 12*x + 4 
)*log(x) - 59*(9*x^2 + 12*x + 4)*log(-2*x + sqrt(4*x^2 - 2*x - 3) - 1) + 3 
2*(9*x^2 + 12*x + 4)*log(-4*x + 2*sqrt(4*x^2 - 2*x - 3) + 1) + 59*(9*x^2 + 
 12*x + 4)*log(-6*x + 3*sqrt(4*x^2 - 2*x - 3) - 5) + 3*sqrt(4*x^2 - 2*x - 
3)*(81*x + 53) + 762*x + 291)/(9*x^2 + 12*x + 4)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.56 \[ \int \frac {1}{x \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^3} \, dx=\frac {301 \, {\left (2 \, x - \sqrt {4 \, x^{2} - 2 \, x - 3}\right )}^{3} + 1164 \, {\left (2 \, x - \sqrt {4 \, x^{2} - 2 \, x - 3}\right )}^{2} + 2986 \, x - 1493 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 636}{36 \, {\left (3 \, {\left (2 \, x - \sqrt {4 \, x^{2} - 2 \, x - 3}\right )}^{2} + 16 \, x - 8 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 5\right )}^{2}} + \frac {38 \, x + 25}{72 \, {\left (3 \, x + 2\right )}^{2}} + \frac {59}{216} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x \right |}\right ) - \frac {59}{216} \, \log \left ({\left | -2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} - 1 \right |}\right ) + \frac {4}{27} \, \log \left ({\left | -4 \, x + 2 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 1 \right |}\right ) + \frac {59}{216} \, \log \left ({\left | -6 \, x + 3 \, \sqrt {4 \, x^{2} - 2 \, x - 3} - 5 \right |}\right ) \] Input:

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

Output:

1/36*(301*(2*x - sqrt(4*x^2 - 2*x - 3))^3 + 1164*(2*x - sqrt(4*x^2 - 2*x - 
 3))^2 + 2986*x - 1493*sqrt(4*x^2 - 2*x - 3) + 636)/(3*(2*x - sqrt(4*x^2 - 
 2*x - 3))^2 + 16*x - 8*sqrt(4*x^2 - 2*x - 3) + 5)^2 + 1/72*(38*x + 25)/(3 
*x + 2)^2 + 59/216*log(abs(3*x + 2)) - 1/8*log(abs(x)) - 59/216*log(abs(-2 
*x + sqrt(4*x^2 - 2*x - 3) - 1)) + 4/27*log(abs(-4*x + 2*sqrt(4*x^2 - 2*x 
- 3) + 1)) + 59/216*log(abs(-6*x + 3*sqrt(4*x^2 - 2*x - 3) - 5))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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