Integrand size = 13, antiderivative size = 172 \[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\frac {2}{5} t \sqrt {4+t^3}-\frac {8\ 2^{2/3} \sqrt {2+\sqrt {3}} \left (2^{2/3}+t\right ) \sqrt {\frac {2 \sqrt [3]{2}-2^{2/3} t+t^2}{\left (2^{2/3} \left (1+\sqrt {3}\right )+t\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{2/3} \left (1-\sqrt {3}\right )+t}{2^{2/3} \left (1+\sqrt {3}\right )+t}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {2^{2/3}+t}{\left (2^{2/3} \left (1+\sqrt {3}\right )+t\right )^2}} \sqrt {4+t^3}} \] Output:
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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.73 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.20 \[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\frac {2}{5} t \left (\sqrt {4+t^3}-2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {t^3}{4}\right )\right ) \] Input:
Integrate[t^3/Sqrt[4 + t^3],t]
Output:
(2*t*(Sqrt[4 + t^3] - 2*Hypergeometric2F1[1/3, 1/2, 4/3, -1/4*t^3]))/5
Time = 0.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {843, 759}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {t^3}{\sqrt {t^3+4}} \, dt\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {2}{5} t \sqrt {t^3+4}-\frac {8}{5} \int \frac {1}{\sqrt {t^3+4}}dt\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {2}{5} t \sqrt {t^3+4}-\frac {8\ 2^{2/3} \sqrt {2+\sqrt {3}} \left (t+2^{2/3}\right ) \sqrt {\frac {t^2-2^{2/3} t+2 \sqrt [3]{2}}{\left (t+2^{2/3} \left (1+\sqrt {3}\right )\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {t+2^{2/3} \left (1-\sqrt {3}\right )}{t+2^{2/3} \left (1+\sqrt {3}\right )}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {t+2^{2/3}}{\left (t+2^{2/3} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {t^3+4}}\) |
Input:
Int[t^3/Sqrt[4 + t^3],t]
Output:
(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[ArcSi n[(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])
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((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]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
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] - Simp[ 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]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.10
| method | result | size |
| meijerg | \(\frac {t^{4} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -\frac {t^{3}}{4}\right )}{8}\) | \(17\) |
| default | \(\frac {2 t \sqrt {t^{3}+4}}{5}+\frac {8 i \sqrt {3}\, 2^{\frac {2}{3}} \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \sqrt {\frac {2^{\frac {2}{3}}+t}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\, \sqrt {-i \left (t -\frac {2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {6}\, \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}}{6}, \sqrt {\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\right )}{15 \sqrt {t^{3}+4}}\) | \(168\) |
| risch | \(\frac {2 t \sqrt {t^{3}+4}}{5}+\frac {8 i \sqrt {3}\, 2^{\frac {2}{3}} \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \sqrt {\frac {2^{\frac {2}{3}}+t}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\, \sqrt {-i \left (t -\frac {2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {6}\, \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}}{6}, \sqrt {\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\right )}{15 \sqrt {t^{3}+4}}\) | \(168\) |
| elliptic | \(\frac {2 t \sqrt {t^{3}+4}}{5}+\frac {8 i \sqrt {3}\, 2^{\frac {2}{3}} \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \sqrt {\frac {2^{\frac {2}{3}}+t}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\, \sqrt {-i \left (t -\frac {2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {6}\, \sqrt {i \left (t -\frac {2^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \sqrt {3}\, 2^{\frac {1}{3}}}}{6}, \sqrt {\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{\frac {3 \,2^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{2}}}\right )}{15 \sqrt {t^{3}+4}}\) | \(168\) |
Input:
int(t^3/(t^3+4)^(1/2),t,method=_RETURNVERBOSE)
Output:
1/8*t^4*hypergeom([1/2,4/3],[7/3],-1/4*t^3)
Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.10 \[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\frac {2}{5} \, \sqrt {t^{3} + 4} t - \frac {16}{5} \, {\rm weierstrassPInverse}\left (0, -16, t\right ) \] Input:
integrate(t^3/(t^3+4)^(1/2),t, algorithm="fricas")
Output:
2/5*sqrt(t^3 + 4)*t - 16/5*weierstrassPInverse(0, -16, t)
Time = 0.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.18 \[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\frac {t^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {t^{3} e^{i \pi }}{4}} \right )}}{6 \Gamma \left (\frac {7}{3}\right )} \] Input:
integrate(t**3/(t**3+4)**(1/2),t)
Output:
t**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), t**3*exp_polar(I*pi)/4)/(6*gamma (7/3))
\[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\int { \frac {t^{3}}{\sqrt {t^{3} + 4}} \,d t } \] Input:
integrate(t^3/(t^3+4)^(1/2),t, algorithm="maxima")
Output:
integrate(t^3/sqrt(t^3 + 4), t)
\[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\int { \frac {t^{3}}{\sqrt {t^{3} + 4}} \,d t } \] Input:
integrate(t^3/(t^3+4)^(1/2),t, algorithm="giac")
Output:
integrate(t^3/sqrt(t^3 + 4), t)
Time = 0.04 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.75 \[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\frac {2\,t\,\sqrt {t^3+4}}{5}-\frac {16\,\sqrt {-\frac {t-2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\,\sqrt {-\frac {t+2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2^{2/3}-2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\,\sqrt {\frac {t+2^{2/3}}{2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\,\left (2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {t+2^{2/3}}{2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}}\right )\middle |\frac {2^{2/3}+2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2^{2/3}-2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )}{5\,\sqrt {t^3+\left (2^{2/3}+2^{2/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-2^{2/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,t^2+\left (2\,2^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-2\,2^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-2\,2^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,t-4\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int(t^3/(t^3 + 4)^(1/2),t)
Output:
(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))
\[ \int \frac {t^3}{\sqrt {4+t^3}} \, dt=\frac {2 \sqrt {t^{3}+4}\, t}{5}-\frac {8 \left (\int \frac {\sqrt {t^{3}+4}}{t^{3}+4}d t \right )}{5} \] Input:
int(t^3/(t^3+4)^(1/2),t)
Output:
(2*(sqrt(t**3 + 4)*t - 4*int(sqrt(t**3 + 4)/(t**3 + 4),t)))/5