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

Optimal. Leaf size=42 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {3+2 \sqrt {3}}} \]

[Out]

-2*arctan((1+x)*(3+2*3^(1/2))^(1/2)/(x^3+1)^(1/2))/(3+2*3^(1/2))^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2140, 203} \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {3+2 \sqrt {3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 2140

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> With[{k = Simplify[(d*e
+ 2*c*f)/(c*f)]}, Dist[((1 + k)*e)/d, Subst[Int[1/(1 + (3 + 2*k)*a*x^2), x], x, (1 + ((1 + k)*d*x)/c)/Sqrt[a +
b*x^3]], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && EqQ[b^2*c^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6
, 0] && EqQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^3), 0]

Rubi steps

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

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Mathematica [C]  time = 0.44, size = 269, normalized size = 6.40 \[ \frac {2 \sqrt {6} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \left (4 \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^2-x+1} \Pi \left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )+\sqrt {2 i x+\sqrt {3}-i} \left (\left ((1+2 i)+i \sqrt {3}\right ) x-\sqrt {3}-(2+i)\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^3+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[6]*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*(Sqrt[-I + Sqrt[3] + (2*I)*x]*((-2 - I) - Sqrt[3] + ((1 + 2*I) +
I*Sqrt[3])*x)*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])] +
4*Sqrt[I + Sqrt[3] - (2*I)*x]*Sqrt[1 - x + x^2]*EllipticPi[(2*Sqrt[3])/(3*I + (1 + 2*I)*Sqrt[3]), ArcSin[Sqrt[
I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])]))/((3*I + (1 + 2*I)*Sqrt[3])*Sqrt[I +
Sqrt[3] - (2*I)*x]*Sqrt[1 + x^3])

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fricas [A]  time = 0.47, size = 50, normalized size = 1.19 \[ \frac {1}{3} \, \sqrt {3} \sqrt {2 \, \sqrt {3} - 3} \arctan \left (\frac {{\left (\sqrt {3} {\left (x^{2} - 4 \, x - 2\right )} - 6 \, x - 6\right )} \sqrt {2 \, \sqrt {3} - 3}}{6 \, \sqrt {x^{3} + 1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*sqrt(2*sqrt(3) - 3)*arctan(1/6*(sqrt(3)*(x^2 - 4*x - 2) - 6*x - 6)*sqrt(2*sqrt(3) - 3)/sqrt(x^3 +
1))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%{[1,-1]:[1,0,-3]%%},[2]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%
} Error: Bad Argument Value

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maple [C]  time = 0.03, size = 245, normalized size = 5.83 \[ \frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3/2-1/2*I*3^(1/2))*((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x
-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)*EllipticF(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3
/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))-4*(3/2-1/2*I*3^(1/2))*((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2
-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)*E
llipticPi(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),1/3*(-3/2+1/2*I*3^(1/2))*3^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I
*3^(1/2)))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x - \sqrt {3} + 1}{\sqrt {x^{3} + 1} {\left (x + \sqrt {3} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.02 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x - \sqrt {3} + 1}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 1 + \sqrt {3}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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