3.175 \(\int \frac {1}{\sqrt {1+t^3}} \, dt\)

Optimal. Leaf size=103 \[ \frac {2 \sqrt {2+\sqrt {3}} (t+1) \sqrt {\frac {t^2-t+1}{\left (t+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {t-\sqrt {3}+1}{t+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {t+1}{\left (t+\sqrt {3}+1\right )^2}} \sqrt {t^3+1}} \]

[Out]

2/3*(1+t)*EllipticF((1+t-3^(1/2))/(1+t+3^(1/2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((t^2-t+1)/(1+t+3^(1/
2))^2)^(1/2)*3^(3/4)/(t^3+1)^(1/2)/((1+t)/(1+t+3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {218} \[ \frac {2 \sqrt {2+\sqrt {3}} (t+1) \sqrt {\frac {t^2-t+1}{\left (t+\sqrt {3}+1\right )^2}} \text {EllipticF}\left (\sin ^{-1}\left (\frac {t-\sqrt {3}+1}{t+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {t+1}{\left (t+\sqrt {3}+1\right )^2}} \sqrt {t^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + t^3],t]

[Out]

(2*Sqrt[2 + Sqrt[3]]*(1 + t)*Sqrt[(1 - t + t^2)/(1 + Sqrt[3] + t)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + t)/(1 + S
qrt[3] + t)], -7 - 4*Sqrt[3]])/(3^(1/4)*Sqrt[(1 + t)/(1 + Sqrt[3] + t)^2]*Sqrt[1 + t^3])

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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1+t^3}} \, dt &=\frac {2 \sqrt {2+\sqrt {3}} (1+t) \sqrt {\frac {1-t+t^2}{\left (1+\sqrt {3}+t\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+t}{1+\sqrt {3}+t}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1+t}{\left (1+\sqrt {3}+t\right )^2}} \sqrt {1+t^3}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 17, normalized size = 0.17 \[ t \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-t^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + t^3],t]

[Out]

t*Hypergeometric2F1[1/3, 1/2, 4/3, -t^3]

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {t^{3} + 1}}, t\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(t^3 + 1), t)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: integrate(1/sqrt(t^3+1),t)

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maple [A]  time = 0.13, size = 116, normalized size = 1.13 \[ \frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {t +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {t -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {t -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {t +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {t^{3}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3/2-1/2*I*3^(1/2))*((t+1)/(3/2-1/2*I*3^(1/2)))^(1/2)*((t-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((t
-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(t^3+1)^(1/2)*EllipticF(((t+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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {t^{3} + 1}}\,{d t} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(t^3 + 1), t)

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mupad [B]  time = 0.29, size = 155, normalized size = 1.50 \[ \frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {t-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {t+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-t+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {t+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{\sqrt {t^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,t-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(t^3 + 1)^(1/2),t)

[Out]

((3^(1/2)*1i + 3)*((t + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((t + 1)/((3^(1/2)*1i)/2 + 3/2))^(
1/2)*(((3^(1/2)*1i)/2 - t + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticF(asin(((t + 1)/((3^(1/2)*1i)/2 + 3/2))
^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(t^3 - t*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1
/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)

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sympy [A]  time = 0.63, size = 27, normalized size = 0.26 \[ \frac {t \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {t^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

t*gamma(1/3)*hyper((1/3, 1/2), (4/3,), t**3*exp_polar(I*pi))/(3*gamma(4/3))

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