\(\int \frac {1}{(1+2 x+\sqrt {-3-2 x+4 x^2})^2} \, dx\) [9]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {1}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {1}{9 \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )}-\frac {13}{18 \left (3-2 \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )\right )}-\frac {13}{27} \log \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )+\frac {13}{27} \log \left (3-2 \left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )\right ) \] Output:

1/(9+18*x+9*(4*x^2-2*x-3)^(1/2))-13/(18-72*x-36*(4*x^2-2*x-3)^(1/2))-13/27 
*ln(1+2*x+(4*x^2-2*x-3)^(1/2))+13/27*ln(1-4*x-2*(4*x^2-2*x-3)^(1/2))
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {-1+24 x+36 x^2-3 (5+6 x) \sqrt {-3-2 x+4 x^2}-26 (2+3 x) \log \left (-5-6 x+3 \sqrt {-3-2 x+4 x^2}\right )}{54 (2+3 x)} \] Input:

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

Output:

(-1 + 24*x + 36*x^2 - 3*(5 + 6*x)*Sqrt[-3 - 2*x + 4*x^2] - 26*(2 + 3*x)*Lo 
g[-5 - 6*x + 3*Sqrt[-3 - 2*x + 4*x^2]])/(54*(2 + 3*x))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2541, 27, 1195, 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.

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

\(\Big \downarrow \) 2541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1195

\(\displaystyle -\int \left (-\frac {26}{27 \left (2 \left (2 x+\sqrt {4 x^2-2 x-3}+1\right )-3\right )}+\frac {13}{9 \left (2 \left (2 x+\sqrt {4 x^2-2 x-3}+1\right )-3\right )^2}+\frac {13}{27 \left (2 x+\sqrt {4 x^2-2 x-3}+1\right )}+\frac {1}{9 \left (2 x+\sqrt {4 x^2-2 x-3}+1\right )^2}\right )d\left (2 x+\sqrt {4 x^2-2 x-3}+1\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{9 \left (\sqrt {4 x^2-2 x-3}+2 x+1\right )}-\frac {13}{18 \left (3-2 \left (\sqrt {4 x^2-2 x-3}+2 x+1\right )\right )}-\frac {13}{27} \log \left (\sqrt {4 x^2-2 x-3}+2 x+1\right )+\frac {13}{27} \log \left (3-2 \left (\sqrt {4 x^2-2 x-3}+2 x+1\right )\right )\)

Input:

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

Output:

1/(9*(1 + 2*x + Sqrt[-3 - 2*x + 4*x^2])) - 13/(18*(3 - 2*(1 + 2*x + Sqrt[- 
3 - 2*x + 4*x^2]))) - (13*Log[1 + 2*x + Sqrt[-3 - 2*x + 4*x^2]])/27 + (13* 
Log[3 - 2*(1 + 2*x + Sqrt[-3 - 2*x + 4*x^2])])/27
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 2009

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

rule 2541

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2   Subst[Int[(g + h*x^n)^p*((d 
^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x 
)^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e 
, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.69

method	result	size
trager	\(\frac {\left (17+24 x \right ) x}{72+108 x}-\frac {\left (5+6 x \right ) \sqrt {4 x^{2}-2 x -3}}{18 \left (2+3 x \right )}+\frac {13 \ln \left (-\frac {5+6 x +3 \sqrt {4 x^{2}-2 x -3}}{2+3 x}\right )}{27}\)	\(74\)
default	\(-\frac {1}{54 \left (2+3 x \right )}-\frac {13 \ln \left (2+3 x \right )}{54}+\frac {2 x}{9}-\frac {\left (4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}\right )^{\frac {3}{2}}}{6 \left (x +\frac {2}{3}\right )}-\frac {13 \sqrt {36 \left (x +\frac {2}{3}\right )^{2}-66 x -43}}{54}+\frac {13 \ln \left (\frac {\left (4 x -1\right ) \sqrt {4}}{4}+\sqrt {4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}}\right ) \sqrt {4}}{108}+\frac {13 \,\operatorname {arctanh}\left (\frac {-21-33 x}{\sqrt {36 \left (x +\frac {2}{3}\right )^{2}-66 x -43}}\right )}{54}+\frac {\left (8 x -2\right ) \sqrt {4 \left (x +\frac {2}{3}\right )^{2}-\frac {22 x}{3}-\frac {43}{9}}}{12}\)	\(135\)

Input:

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

Output:

1/36*(17+24*x)*x/(2+3*x)-1/18*(5+6*x)/(2+3*x)*(4*x^2-2*x-3)^(1/2)+13/27*ln 
(-(5+6*x+3*(4*x^2-2*x-3)^(1/2))/(2+3*x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {72 \, x^{2} - 26 \, {\left (3 \, x + 2\right )} \log \left (48 \, x^{2} - \sqrt {4 \, x^{2} - 2 \, x - 3} {\left (24 \, x + 7\right )} + 2 \, x - 23\right ) - 26 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 26 \, {\left (3 \, x + 2\right )} \log \left (-2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} - 1\right ) - 6 \, \sqrt {4 \, x^{2} - 2 \, x - 3} {\left (6 \, x + 5\right )} + 45 \, x - 4}{108 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

1/108*(72*x^2 - 26*(3*x + 2)*log(48*x^2 - sqrt(4*x^2 - 2*x - 3)*(24*x + 7) 
 + 2*x - 23) - 26*(3*x + 2)*log(3*x + 2) + 26*(3*x + 2)*log(-2*x + sqrt(4* 
x^2 - 2*x - 3) - 1) - 6*sqrt(4*x^2 - 2*x - 3)*(6*x + 5) + 45*x - 4)/(3*x + 
 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.56 \[ \int \frac {1}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {2}{9} \, x - \frac {1}{9} \, \sqrt {4 \, x^{2} - 2 \, x - 3} - \frac {22 \, x - 11 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 14}{27 \, {\left (3 \, {\left (2 \, x - \sqrt {4 \, x^{2} - 2 \, x - 3}\right )}^{2} + 16 \, x - 8 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 5\right )}} - \frac {1}{54 \, {\left (3 \, x + 2\right )}} - \frac {13}{54} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {13}{54} \, \log \left ({\left | -2 \, x + \sqrt {4 \, x^{2} - 2 \, x - 3} - 1 \right |}\right ) - \frac {13}{54} \, \log \left ({\left | -4 \, x + 2 \, \sqrt {4 \, x^{2} - 2 \, x - 3} + 1 \right |}\right ) - \frac {13}{54} \, \log \left ({\left | -6 \, x + 3 \, \sqrt {4 \, x^{2} - 2 \, x - 3} - 5 \right |}\right ) \] Input:

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

Output:

2/9*x - 1/9*sqrt(4*x^2 - 2*x - 3) - 1/27*(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) - 1/54/(3*x + 2) - 13/54*log(abs(3*x + 2)) + 13/54*log(abs(-2*x + sqr 
t(4*x^2 - 2*x - 3) - 1)) - 13/54*log(abs(-4*x + 2*sqrt(4*x^2 - 2*x - 3) + 
1)) - 13/54*log(abs(-6*x + 3*sqrt(4*x^2 - 2*x - 3) - 5))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {2\,x}{9}-\frac {13\,\ln \left (x+\frac {2}{3}\right )}{54}+\int \frac {-8\,x^3+8\,x+3}{2\,{\left (3\,x+2\right )}^2\,\sqrt {4\,x^2-2\,x-3}} \,d x-\frac {1}{162\,\left (x+\frac {2}{3}\right )} \] Input:

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

Output:

(2*x)/9 - (13*log(x + 2/3))/54 + int((8*x - 8*x^3 + 3)/(2*(3*x + 2)^2*(4*x 
^2 - 2*x - 3)^(1/2)), x) - 1/(162*(x + 2/3))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (1+2 x+\sqrt {-3-2 x+4 x^2}\right )^2} \, dx=\frac {-36 \sqrt {4 x^{2}-2 x -3}\, x -30 \sqrt {4 x^{2}-2 x -3}+156 \,\mathrm {log}\left (3 \sqrt {4 x^{2}-2 x -3}+6 x +5\right ) x +104 \,\mathrm {log}\left (3 \sqrt {4 x^{2}-2 x -3}+6 x +5\right )-156 \,\mathrm {log}\left (3 x +2\right ) x -104 \,\mathrm {log}\left (3 x +2\right )+72 x^{2}+51 x}{324 x +216} \] Input:

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

Output:

( - 36*sqrt(4*x**2 - 2*x - 3)*x - 30*sqrt(4*x**2 - 2*x - 3) + 156*log(3*sq 
rt(4*x**2 - 2*x - 3) + 6*x + 5)*x + 104*log(3*sqrt(4*x**2 - 2*x - 3) + 6*x 
 + 5) - 156*log(3*x + 2)*x - 104*log(3*x + 2) + 72*x**2 + 51*x)/(108*(3*x 
+ 2))