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

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

[Out]

-2*arctanh((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 = 0.11, 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, 206} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {2 \sqrt {3}-3} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {2 \sqrt {3}-3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 \tanh ^{-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 = 0.45, size = 267, normalized size = 6.36 \[ -\frac {2 \sqrt {6} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \left (4 i \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^2-x+1} \Pi \left (\frac {2 i \sqrt {3}}{-3+(2+i) \sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )+\sqrt {2 i x+\sqrt {3}-i} \left (\left (\sqrt {3}+(-2-i)\right ) x-i \sqrt {3}+(1+2 i)\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\left (-3+(2+i) \sqrt {3}\right ) \sqrt {-2 i x+\sqrt {3}+i} \sqrt {x^3+1}} \]

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

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fricas [B]  time = 0.51, size = 205, normalized size = 4.88 \[ \frac {1}{6} \, \sqrt {3} \sqrt {2 \, \sqrt {3} + 3} \log \left (\frac {x^{8} - 16 \, x^{7} + 112 \, x^{6} - 16 \, x^{5} + 112 \, x^{4} + 224 \, x^{3} + 64 \, x^{2} - 4 \, {\left (2 \, x^{6} - 18 \, x^{5} + 42 \, x^{4} - 8 \, x^{3} - \sqrt {3} {\left (x^{6} - 12 \, x^{5} + 18 \, x^{4} - 16 \, x^{3} - 12 \, x^{2} - 8\right )} + 24 \, x + 8\right )} \sqrt {x^{3} + 1} \sqrt {2 \, \sqrt {3} + 3} + 16 \, \sqrt {3} {\left (x^{7} - 2 \, x^{6} + 6 \, x^{5} + 5 \, x^{4} + 2 \, x^{3} + 6 \, x^{2} + 4 \, x + 4\right )} + 128 \, x + 112}{x^{8} + 8 \, x^{7} + 16 \, x^{6} - 16 \, x^{5} - 56 \, x^{4} + 32 \, x^{3} + 64 \, x^{2} - 64 \, x + 16}\right ) \]

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]

1/6*sqrt(3)*sqrt(2*sqrt(3) + 3)*log((x^8 - 16*x^7 + 112*x^6 - 16*x^5 + 112*x^4 + 224*x^3 + 64*x^2 - 4*(2*x^6 -
18*x^5 + 42*x^4 - 8*x^3 - sqrt(3)*(x^6 - 12*x^5 + 18*x^4 - 16*x^3 - 12*x^2 - 8) + 24*x + 8)*sqrt(x^3 + 1)*sqr
t(2*sqrt(3) + 3) + 16*sqrt(3)*(x^7 - 2*x^6 + 6*x^5 + 5*x^4 + 2*x^3 + 6*x^2 + 4*x + 4) + 128*x + 112)/(x^8 + 8*
x^7 + 16*x^6 - 16*x^5 - 56*x^4 + 32*x^3 + 64*x^2 - 64*x + 16))

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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 = 0.08, 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 \[ \int \frac {x + 1 + \sqrt {3}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x - \sqrt {3} + 1\right )}\, dx \]

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

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