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

Optimal. Leaf size=146 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}+\frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}} \]

[Out]

1/3*(1+x)*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((x^2-x+1)/(1+x+3^(1/
2))^2)^(1/2)*3^(1/4)/(x^3+1)^(1/2)/((1+x)/(1+x+3^(1/2))^2)^(1/2)+arctan((1+x)*(3+2*3^(1/2))^(1/2)/(x^3+1)^(1/2
))/(9+6*3^(1/2))^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2135, 218, 2140, 203} \[ \frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \text {EllipticF}\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (x+1)}{\sqrt {x^3+1}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[3 + 2*Sqrt[3]]*(1 + x))/Sqrt[1 + x^3]]/Sqrt[3*(3 + 2*Sqrt[3])] + (Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[
(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^
(3/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^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 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 2135

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(-6*a*d^3)/(c*(b*c^3 - 28*a*d^3)), In
t[1/Sqrt[a + b*x^3], x], x] + Dist[1/(c*(b*c^3 - 28*a*d^3)), Int[Simp[c*(b*c^3 - 22*a*d^3) + 6*a*d^4*x, x]/((c
+ d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2*c^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 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}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx &=-\frac {\int \frac {6 \left (1-\sqrt {3}\right )+6 x}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx}{12 \sqrt {3}}+\frac {\int \frac {1}{\sqrt {1+x^3}} \, dx}{2 \sqrt {3}}\\ &=\frac {\sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}+\frac {\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 )}{\sqrt {3}}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {3}} (1+x)}{\sqrt {1+x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}+\frac {\sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 136, normalized size = 0.93 \[ -\frac {4 \sqrt {2} \sqrt {\frac {i (x+1)}{\sqrt {3}+3 i}} \sqrt {x^2-x+1} \Pi \left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}};\sin ^{-1}\left (\frac {\sqrt {-2 i x+\sqrt {3}+i}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {x^3+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*Sqrt[2]*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*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)*Sqr
t[3])*Sqrt[1 + x^3])

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fricas [F]  time = 1.31, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{3} + 1} {\left (x - \sqrt {3} + 1\right )}}{x^{5} + 2 \, x^{4} - 2 \, x^{3} + x^{2} + 2 \, x - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^3 + 1)*(x - sqrt(3) + 1)/(x^5 + 2*x^4 - 2*x^3 + x^2 + 2*x - 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Unable to divide,
perhaps due to rounding error%%%{1,[2]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 132, normalized size = 0.90 \[ \frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \EllipticPi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(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)*3^(1/2)*EllipticPi(((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 \[ \int \frac {1}{\sqrt {x^{3} + 1} {\left (x + \sqrt {3} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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