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

Optimal. Leaf size=87 \[ \frac{1-\frac{\sqrt{-x-1}}{\sqrt{x+3}}}{-\frac{3 (x+1)}{x+3}-\frac{2 \sqrt{-x-1}}{\sqrt{x+3}}+1}+\frac{\tan ^{-1}\left (\frac{1-\frac{3 \sqrt{-x-1}}{\sqrt{x+3}}}{\sqrt{2}}\right )}{\sqrt{2}} \]

[Out]

(1 - Sqrt[-1 - x]/Sqrt[3 + x])/(1 - (3*(1 + x))/(3 + x) - (2*Sqrt[-1 - x])/Sqrt[3 + x]) + ArcTan[(1 - (3*Sqrt[
-1 - x])/Sqrt[3 + x])/Sqrt[2]]/Sqrt[2]

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Rubi [A]  time = 0.0651296, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {12, 638, 618, 204} \[ \frac{1-\frac{\sqrt{-x-1}}{\sqrt{x+3}}}{-\frac{3 (x+1)}{x+3}-\frac{2 \sqrt{-x-1}}{\sqrt{x+3}}+1}+\frac{\tan ^{-1}\left (\frac{1-\frac{3 \sqrt{-x-1}}{\sqrt{x+3}}}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - Sqrt[-1 - x]/Sqrt[3 + x])/(1 - (3*(1 + x))/(3 + x) - (2*Sqrt[-1 - x])/Sqrt[3 + x]) + ArcTan[(1 - (3*Sqrt[
-1 - x])/Sqrt[3 + x])/Sqrt[2]]/Sqrt[2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (x+\sqrt{-3-4 x-x^2}\right )^2} \, dx &=2 \operatorname{Subst}\left (\int -\frac{2 x}{\left (1-2 x+3 x^2\right )^2} \, dx,x,\frac{\sqrt{-1-x}}{\sqrt{3+x}}\right )\\ &=-\left (4 \operatorname{Subst}\left (\int \frac{x}{\left (1-2 x+3 x^2\right )^2} \, dx,x,\frac{\sqrt{-1-x}}{\sqrt{3+x}}\right )\right )\\ &=\frac{1-\frac{\sqrt{-1-x}}{\sqrt{3+x}}}{1-\frac{3 (1+x)}{3+x}-\frac{2 \sqrt{-1-x}}{\sqrt{3+x}}}-\operatorname{Subst}\left (\int \frac{1}{1-2 x+3 x^2} \, dx,x,\frac{\sqrt{-1-x}}{\sqrt{3+x}}\right )\\ &=\frac{1-\frac{\sqrt{-1-x}}{\sqrt{3+x}}}{1-\frac{3 (1+x)}{3+x}-\frac{2 \sqrt{-1-x}}{\sqrt{3+x}}}+2 \operatorname{Subst}\left (\int \frac{1}{-8-x^2} \, dx,x,-2+\frac{6 \sqrt{-1-x}}{\sqrt{3+x}}\right )\\ &=\frac{1-\frac{\sqrt{-1-x}}{\sqrt{3+x}}}{1-\frac{3 (1+x)}{3+x}-\frac{2 \sqrt{-1-x}}{\sqrt{3+x}}}+\frac{\tan ^{-1}\left (\frac{1-\frac{3 \sqrt{-1-x}}{\sqrt{3+x}}}{\sqrt{2}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 1.62129, size = 881, normalized size = 10.13 \[ \frac{1}{16} \left (\frac{8 (x+3)}{2 x^2+4 x+3}+4 \sqrt{2} \tan ^{-1}\left (\sqrt{2} (x+1)\right )-\frac{2 i \left (-2 i+\sqrt{2}\right ) \tan ^{-1}\left (\frac{(x+2) \left (2 \left (9+2 i \sqrt{2}\right ) x^2+16 \left (2+i \sqrt{2}\right ) x+3 \left (5+4 i \sqrt{2}\right )\right )}{\left (8 i+6 \sqrt{2}\right ) x^3+\left (-6 \sqrt{1+2 i \sqrt{2}} \sqrt{-x^2-4 x-3}+8 \sqrt{2}+36 i\right ) x^2+\left (-12 \sqrt{1+2 i \sqrt{2}} \sqrt{-x^2-4 x-3}-5 \sqrt{2}+40 i\right ) x-9 \sqrt{1+2 i \sqrt{2}} \sqrt{-x^2-4 x-3}-6 \sqrt{2}+12 i}\right )}{\sqrt{1+2 i \sqrt{2}}}+\frac{2 \left (2 i+\sqrt{2}\right ) \tanh ^{-1}\left (\frac{(x+2) \left (2 \left (9 i+2 \sqrt{2}\right ) x^2+16 \left (2 i+\sqrt{2}\right ) x+3 \left (5 i+4 \sqrt{2}\right )\right )}{\left (-8 i+6 \sqrt{2}\right ) x^3+\left (-6 \sqrt{1-2 i \sqrt{2}} \sqrt{-x^2-4 x-3}+8 \sqrt{2}-36 i\right ) x^2-12 \sqrt{1-2 i \sqrt{2}} \sqrt{-x^2-4 x-3} x-5 \left (8 i+\sqrt{2}\right ) x-3 \left (3 \sqrt{1-2 i \sqrt{2}} \sqrt{-x^2-4 x-3}+2 \sqrt{2}+4 i\right )}\right )}{\sqrt{1-2 i \sqrt{2}}}-\frac{\left (2 i+\sqrt{2}\right ) \log \left (4 \left (2 x^2+4 x+3\right )^2\right )}{\sqrt{1-2 i \sqrt{2}}}-\frac{\left (-2 i+\sqrt{2}\right ) \log \left (4 \left (2 x^2+4 x+3\right )^2\right )}{\sqrt{1+2 i \sqrt{2}}}+\frac{\left (2 i+\sqrt{2}\right ) \log \left (\left (2 x^2+4 x+3\right ) \left (\left (2+2 i \sqrt{2}\right ) x^2+\left (-2 \sqrt{2-4 i \sqrt{2}} \sqrt{-x^2-4 x-3}+8 i \sqrt{2}+4\right ) x-2 \sqrt{2-4 i \sqrt{2}} \sqrt{-x^2-4 x-3}+6 i \sqrt{2}+3\right )\right )}{\sqrt{1-2 i \sqrt{2}}}+\frac{\left (-2 i+\sqrt{2}\right ) \log \left (\left (2 x^2+4 x+3\right ) \left (\left (2-2 i \sqrt{2}\right ) x^2-2 \left (\sqrt{2+4 i \sqrt{2}} \sqrt{-x^2-4 x-3}+4 i \sqrt{2}-2\right ) x-2 \sqrt{2+4 i \sqrt{2}} \sqrt{-x^2-4 x-3}-6 i \sqrt{2}+3\right )\right )}{\sqrt{1+2 i \sqrt{2}}}+\frac{8 (2 x+3) \sqrt{-x^2-4 x-3}}{2 x^2+4 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(3 + x))/(3 + 4*x + 2*x^2) + (8*(3 + 2*x)*Sqrt[-3 - 4*x - x^2])/(3 + 4*x + 2*x^2) + 4*Sqrt[2]*ArcTan[Sqrt[
2]*(1 + x)] - ((2*I)*(-2*I + Sqrt[2])*ArcTan[((2 + x)*(3*(5 + (4*I)*Sqrt[2]) + 16*(2 + I*Sqrt[2])*x + 2*(9 + (
2*I)*Sqrt[2])*x^2))/(12*I - 6*Sqrt[2] + (8*I + 6*Sqrt[2])*x^3 - 9*Sqrt[1 + (2*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2]
+ x*(40*I - 5*Sqrt[2] - 12*Sqrt[1 + (2*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2]) + x^2*(36*I + 8*Sqrt[2] - 6*Sqrt[1 +
(2*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2]))])/Sqrt[1 + (2*I)*Sqrt[2]] + (2*(2*I + Sqrt[2])*ArcTanh[((2 + x)*(3*(5*I
+ 4*Sqrt[2]) + 16*(2*I + Sqrt[2])*x + 2*(9*I + 2*Sqrt[2])*x^2))/(-5*(8*I + Sqrt[2])*x + (-8*I + 6*Sqrt[2])*x^
3 - 12*Sqrt[1 - (2*I)*Sqrt[2]]*x*Sqrt[-3 - 4*x - x^2] + x^2*(-36*I + 8*Sqrt[2] - 6*Sqrt[1 - (2*I)*Sqrt[2]]*Sqr
t[-3 - 4*x - x^2]) - 3*(4*I + 2*Sqrt[2] + 3*Sqrt[1 - (2*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2]))])/Sqrt[1 - (2*I)*Sq
rt[2]] - ((-2*I + Sqrt[2])*Log[4*(3 + 4*x + 2*x^2)^2])/Sqrt[1 + (2*I)*Sqrt[2]] - ((2*I + Sqrt[2])*Log[4*(3 + 4
*x + 2*x^2)^2])/Sqrt[1 - (2*I)*Sqrt[2]] + ((2*I + Sqrt[2])*Log[(3 + 4*x + 2*x^2)*(3 + (6*I)*Sqrt[2] + (2 + (2*
I)*Sqrt[2])*x^2 - 2*Sqrt[2 - (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2] + x*(4 + (8*I)*Sqrt[2] - 2*Sqrt[2 - (4*I)*Sqr
t[2]]*Sqrt[-3 - 4*x - x^2]))])/Sqrt[1 - (2*I)*Sqrt[2]] + ((-2*I + Sqrt[2])*Log[(3 + 4*x + 2*x^2)*(3 - (6*I)*Sq
rt[2] + (2 - (2*I)*Sqrt[2])*x^2 - 2*Sqrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2] - 2*x*(-2 + (4*I)*Sqrt[2] + S
qrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2]))])/Sqrt[1 + (2*I)*Sqrt[2]])/16

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Maple [B]  time = 0.079, size = 2407, normalized size = 27.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*(4*x+4)/(2*x^2+4*x+3)+1/4*2^(1/2)*arctan(1/4*(4*x+4)*2^(1/2))-1/2*(-4*x-6)/(2*x^2+4*x+3)+1/36*3^(1/2)*4^(
1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*(7*2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))+4*arctanh(3*x/(-3/
2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2)))/((x^2/(-3/2-x)^2-4)/(x/(-3/2-x)+1)^2)^(1/2)/(x/(-3/2-x)+1)+1/72*3^(1/2)*4^(
1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*(arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))*2^(1/2)*x^2/(-3/2-x)^2-8*arc
tanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))*x^2/(-3/2-x)^2+2*2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)
*2^(1/2))-16*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))-6*(3*x^2/(-3/2-x)^2-12)^(1/2))/((x^2/(-3/2-x)^2
-4)/(x/(-3/2-x)+1)^2)^(1/2)/(x/(-3/2-x)+1)/(x^2/(-3/2-x)^2+2)-2/9*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*
(2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))+arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2)))/((
x^2/(-3/2-x)^2-4)/(x/(-3/2-x)+1)^2)^(1/2)/(x/(-3/2-x)+1)-2/9*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*(3*ar
ctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))*2^(1/2)*x^6/(-3/2-x)^6+4*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-
12)^(1/2))*x^6/(-3/2-x)^6-2*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)/(x^2/(-3/2-x)^2-4))*x
^6/(-3/2-x)^6+2*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(-3/2-x)^2+4)/(x^2/(-3/2-x)^2-4))*x^6/(-3/2-x)^
6+(3*x^2/(-3/2-x)^2-12)^(1/2)*x^5/(-3/2-x)^5-(3*x^2/(-3/2-x)^2-12)^(3/2)*x^2/(-3/2-x)^2+(3*x^2/(-3/2-x)^2-12)^
(1/2)*x^4/(-3/2-x)^4-36*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))*2^(1/2)*x^2/(-3/2-x)^2-2*(3*x^2/(-3/2-
x)^2-12)^(1/2)*x^3/(-3/2-x)^3-48*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))*x^2/(-3/2-x)^2-8*(3*x^2/(-3
/2-x)^2-12)^(1/2)*x^2/(-3/2-x)^2+24*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)/(x^2/(-3/2-x)
^2-4))*x^2/(-3/2-x)^2-24*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(-3/2-x)^2+4)/(x^2/(-3/2-x)^2-4))*x^2/
(-3/2-x)^2-48*2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))-8*(3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)
-64*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))+16*(3*x^2/(-3/2-x)^2-12)^(1/2)+32*ln(((3*x^2/(-3/2-x)^2-
12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)/(x^2/(-3/2-x)^2-4))-32*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(
-3/2-x)^2+4)/(x^2/(-3/2-x)^2-4)))/((x^2/(-3/2-x)^2-4)/(x/(-3/2-x)+1)^2)^(1/2)/(x/(-3/2-x)+1)/(x^2/(-3/2-x)^2+2
)/((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)/((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(-3/2-
x)^2+4)+1/18*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*(11*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))*2
^(1/2)*x^6/(-3/2-x)^6+24*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))*x^6/(-3/2-x)^6-8*ln(((3*x^2/(-3/2-x
)^2-12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)/(x^2/(-3/2-x)^2-4))*x^6/(-3/2-x)^6+8*ln(((3*x^2/(-3/2-x)^2-12)^(1/2
)*x/(-3/2-x)-x^2/(-3/2-x)^2+4)/(x^2/(-3/2-x)^2-4))*x^6/(-3/2-x)^6+4*(3*x^2/(-3/2-x)^2-12)^(1/2)*x^5/(-3/2-x)^5
-(3*x^2/(-3/2-x)^2-12)^(3/2)*x^2/(-3/2-x)^2+(3*x^2/(-3/2-x)^2-12)^(1/2)*x^4/(-3/2-x)^4-132*arctan(1/6*(3*x^2/(
-3/2-x)^2-12)^(1/2)*2^(1/2))*2^(1/2)*x^2/(-3/2-x)^2-8*(3*x^2/(-3/2-x)^2-12)^(1/2)*x^3/(-3/2-x)^3-288*arctanh(3
*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))*x^2/(-3/2-x)^2-8*(3*x^2/(-3/2-x)^2-12)^(1/2)*x^2/(-3/2-x)^2+96*ln(((3
*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)/(x^2/(-3/2-x)^2-4))*x^2/(-3/2-x)^2-96*ln(((3*x^2/(-3/2-
x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(-3/2-x)^2+4)/(x^2/(-3/2-x)^2-4))*x^2/(-3/2-x)^2-176*2^(1/2)*arctan(1/6*(3*x^2/(
-3/2-x)^2-12)^(1/2)*2^(1/2))-32*(3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-384*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x
)^2-12)^(1/2))+16*(3*x^2/(-3/2-x)^2-12)^(1/2)+128*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)+x^2/(-3/2-x)^2-4)
/(x^2/(-3/2-x)^2-4))-128*ln(((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(-3/2-x)^2+4)/(x^2/(-3/2-x)^2-4)))/((x
^2/(-3/2-x)^2-4)/(x/(-3/2-x)+1)^2)^(1/2)/(x/(-3/2-x)+1)/(x^2/(-3/2-x)^2+2)/((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/
2-x)+x^2/(-3/2-x)^2-4)/((3*x^2/(-3/2-x)^2-12)^(1/2)*x/(-3/2-x)-x^2/(-3/2-x)^2+4)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x + \sqrt{-x^{2} - 4 \, x - 3}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.71634, size = 324, normalized size = 3.72 \begin{align*} \frac{2 \, \sqrt{2}{\left (2 \, x^{2} + 4 \, x + 3\right )} \arctan \left (\sqrt{2}{\left (x + 1\right )}\right ) - \sqrt{2}{\left (2 \, x^{2} + 4 \, x + 3\right )} \arctan \left (\frac{\sqrt{2}{\left (6 \, x^{2} + 20 \, x + 15\right )} \sqrt{-x^{2} - 4 \, x - 3}}{4 \,{\left (2 \, x^{3} + 11 \, x^{2} + 18 \, x + 9\right )}}\right ) + 4 \, \sqrt{-x^{2} - 4 \, x - 3}{\left (2 \, x + 3\right )} + 4 \, x + 12}{8 \,{\left (2 \, x^{2} + 4 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*(2*x^2 + 4*x + 3)*arctan(sqrt(2)*(x + 1)) - sqrt(2)*(2*x^2 + 4*x + 3)*arctan(1/4*sqrt(2)*(6*x^2
+ 20*x + 15)*sqrt(-x^2 - 4*x - 3)/(2*x^3 + 11*x^2 + 18*x + 9)) + 4*sqrt(-x^2 - 4*x - 3)*(2*x + 3) + 4*x + 12)
/(2*x^2 + 4*x + 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (x + \sqrt{- x^{2} - 4 x - 3}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13715, size = 355, normalized size = 4.08 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\sqrt{2}{\left (x + 1\right )}\right ) - \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\frac{3 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) - \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\frac{\sqrt{-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) + \frac{x + 3}{2 \,{\left (2 \, x^{2} + 4 \, x + 3\right )}} - \frac{\frac{10 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac{7 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} - \frac{2 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{3}}{{\left (x + 2\right )}^{3}} + 3}{3 \,{\left (\frac{8 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac{14 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + \frac{8 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{3}}{{\left (x + 2\right )}^{3}} + \frac{3 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{4}}{{\left (x + 2\right )}^{4}} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*arctan(sqrt(2)*(x + 1)) - 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1
)) - 1/4*sqrt(2)*arctan(1/2*sqrt(2)*((sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 1/2*(x + 3)/(2*x^2 + 4*x + 3)
- 1/3*(10*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 7*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 - 2*(sqrt(-x^2 - 4*x -
3) - 1)^3/(x + 2)^3 + 3)/(8*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 14*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 +
8*(sqrt(-x^2 - 4*x - 3) - 1)^3/(x + 2)^3 + 3*(sqrt(-x^2 - 4*x - 3) - 1)^4/(x + 2)^4 + 3)