3.41 \(\int \frac {t^3}{\sqrt {4+t^3}} \, dt\)

Optimal. Leaf size=172 \[ \frac {2}{5} t \sqrt {t^3+4}-\frac {8\ 2^{2/3} \sqrt {2+\sqrt {3}} \left (t+2^{2/3}\right ) \sqrt {\frac {t^2-2^{2/3} t+2 \sqrt [3]{2}}{\left (t+2^{2/3} \left (1+\sqrt {3}\right )\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {t+2^{2/3} \left (1-\sqrt {3}\right )}{t+2^{2/3} \left (1+\sqrt {3}\right )}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {t+2^{2/3}}{\left (t+2^{2/3} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {t^3+4}} \]

[Out]

2/5*t*(t^3+4)^(1/2)-8/15*2^(2/3)*(2^(2/3)+t)*EllipticF((t+2^(2/3)*(1-3^(1/2)))/(t+2^(2/3)*(1+3^(1/2))),I*3^(1/
2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((2*2^(1/3)-2^(2/3)*t+t^2)/(t+2^(2/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/(t^3+4)^
(1/2)/((2^(2/3)+t)/(t+2^(2/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {321, 218} \[ \frac {2}{5} t \sqrt {t^3+4}-\frac {8\ 2^{2/3} \sqrt {2+\sqrt {3}} \left (t+2^{2/3}\right ) \sqrt {\frac {t^2-2^{2/3} t+2 \sqrt [3]{2}}{\left (t+2^{2/3} \left (1+\sqrt {3}\right )\right )^2}} \text {EllipticF}\left (\sin ^{-1}\left (\frac {t+2^{2/3} \left (1-\sqrt {3}\right )}{t+2^{2/3} \left (1+\sqrt {3}\right )}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {t+2^{2/3}}{\left (t+2^{2/3} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {t^3+4}} \]

Antiderivative was successfully verified.

[In]

Int[t^3/Sqrt[4 + t^3],t]

[Out]

(2*t*Sqrt[4 + t^3])/5 - (8*2^(2/3)*Sqrt[2 + Sqrt[3]]*(2^(2/3) + t)*Sqrt[(2*2^(1/3) - 2^(2/3)*t + t^2)/(2^(2/3)
*(1 + Sqrt[3]) + t)^2]*EllipticF[ArcSin[(2^(2/3)*(1 - Sqrt[3]) + t)/(2^(2/3)*(1 + Sqrt[3]) + t)], -7 - 4*Sqrt[
3]])/(5*3^(1/4)*Sqrt[(2^(2/3) + t)/(2^(2/3)*(1 + Sqrt[3]) + t)^2]*Sqrt[4 + 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]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {t^3}{\sqrt {4+t^3}} \, dt &=\frac {2}{5} t \sqrt {4+t^3}-\frac {8}{5} \int \frac {1}{\sqrt {4+t^3}} \, dt\\ &=\frac {2}{5} t \sqrt {4+t^3}-\frac {8\ 2^{2/3} \sqrt {2+\sqrt {3}} \left (2^{2/3}+t\right ) \sqrt {\frac {2 \sqrt [3]{2}-2^{2/3} t+t^2}{\left (2^{2/3} \left (1+\sqrt {3}\right )+t\right )^2}} F\left (\sin ^{-1}\left (\frac {2^{2/3} \left (1-\sqrt {3}\right )+t}{2^{2/3} \left (1+\sqrt {3}\right )+t}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {2^{2/3}+t}{\left (2^{2/3} \left (1+\sqrt {3}\right )+t\right )^2}} \sqrt {4+t^3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.20 \[ \frac {2}{5} t \left (\sqrt {t^3+4}-2 \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {t^3}{4}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[t^3/Sqrt[4 + t^3],t]

[Out]

(2*t*(Sqrt[4 + t^3] - 2*Hypergeometric2F1[1/3, 1/2, 4/3, -1/4*t^3]))/5

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {t^{3}}{\sqrt {t^{3} + 4}}, t\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t^3/(t^3+4)^(1/2),t, algorithm="fricas")

[Out]

integral(t^3/sqrt(t^3 + 4), 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(t^3/(t^3+4)^(1/2),t, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2/5*t*sqrt(t^3+4)+integrate(-8/5/sqrt(t^
3+4),t)

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maple [A]  time = 0.18, size = 168, normalized size = 0.98 \[ \frac {2 \sqrt {t^{3}+4}\, t}{5}+\frac {8 i \sqrt {3}\, 2^{\frac {2}{3}} \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \sqrt {\frac {t +2^{\frac {2}{3}}}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\, \sqrt {-i \left (t -\frac {2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \EllipticF \left (\frac {\sqrt {6}\, \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}}{6}, \sqrt {\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\right )}{15 \sqrt {t^{3}+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(t^3/(t^3+4)^(1/2),t)

[Out]

2/5*t*(t^3+4)^(1/2)+8/15*I*3^(1/2)*2^(2/3)*(I*(t-1/2*2^(2/3)-1/2*I*3^(1/2)*2^(2/3))*3^(1/2)*2^(1/3))^(1/2)*((2
^(2/3)+t)/(3/2*2^(2/3)+1/2*I*3^(1/2)*2^(2/3)))^(1/2)*(-I*(t-1/2*2^(2/3)+1/2*I*3^(1/2)*2^(2/3))*3^(1/2)*2^(1/3)
)^(1/2)/(t^3+4)^(1/2)*EllipticF(1/6*6^(1/2)*(I*(t-1/2*2^(2/3)-1/2*I*3^(1/2)*2^(2/3))*3^(1/2)*2^(1/3))^(1/2),(I
*3^(1/2)*2^(2/3)/(3/2*2^(2/3)+1/2*I*3^(1/2)*2^(2/3)))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {t^{3}}{\sqrt {t^{3} + 4}}\,{d t} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t^3/(t^3+4)^(1/2),t, algorithm="maxima")

[Out]

integrate(t^3/sqrt(t^3 + 4), t)

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mupad [B]  time = 0.08, size = 301, normalized size = 1.75 \[ \frac {2\,t\,\sqrt {t^3+4}}{5}-\frac {16\,\sqrt {-\frac {t-2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\,\sqrt {-\frac {t+2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2^{2/3}-2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\,\sqrt {\frac {t+2^{2/3}}{2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\,\left (2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {t+2^{2/3}}{2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\right )\middle |\frac {2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2^{2/3}-2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )}{5\,\sqrt {t^3+\left (2^{2/3}+2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,t^2+\left (2\,2^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-2\,2^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-2\,2^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,t-4\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(t^3/(t^3 + 4)^(1/2),t)

[Out]

(2*t*(t^3 + 4)^(1/2))/5 - (16*(-(t - 2^(2/3)*((3^(1/2)*1i)/2 + 1/2))/(2^(2/3) + 2^(2/3)*((3^(1/2)*1i)/2 + 1/2)
))^(1/2)*(-(t + 2^(2/3)*((3^(1/2)*1i)/2 - 1/2))/(2^(2/3) - 2^(2/3)*((3^(1/2)*1i)/2 - 1/2)))^(1/2)*((t + 2^(2/3
))/(2^(2/3) + 2^(2/3)*((3^(1/2)*1i)/2 + 1/2)))^(1/2)*(2^(2/3) + 2^(2/3)*((3^(1/2)*1i)/2 + 1/2))*ellipticF(asin
(((t + 2^(2/3))/(2^(2/3) + 2^(2/3)*((3^(1/2)*1i)/2 + 1/2)))^(1/2)), (2^(2/3) + 2^(2/3)*((3^(1/2)*1i)/2 + 1/2))
/(2^(2/3) - 2^(2/3)*((3^(1/2)*1i)/2 - 1/2))))/(5*(t^2*(2^(2/3) + 2^(2/3)*((3^(1/2)*1i)/2 - 1/2) - 2^(2/3)*((3^
(1/2)*1i)/2 + 1/2)) - 4*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + t^3 - t*(2*2^(1/3)*((3^(1/2)*1i)/2 + 1
/2) - 2*2^(1/3)*((3^(1/2)*1i)/2 - 1/2) + 2*2^(1/3)*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2)))^(1/2))

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sympy [A]  time = 0.72, size = 31, normalized size = 0.18 \[ \frac {t^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {t^{3} e^{i \pi }}{4}} \right )}}{6 \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t**3/(t**3+4)**(1/2),t)

[Out]

t**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), t**3*exp_polar(I*pi)/4)/(6*gamma(7/3))

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