Integrand size = 16, antiderivative size = 313 \[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}+\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} c^2 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:
3/5*x*(-d^2*x^2+c^2)^(1/3)+2/5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*c^2*(c^(2 /3)-(-d^2*x^2+c^2)^(1/3))*((c^(4/3)+c^(2/3)*(-d^2*x^2+c^2)^(1/3)+(-d^2*x^2 +c^2)^(2/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))^2)^(1/2)*EllipticF (((1+3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2 +c^2)^(1/3)),2*I-I*3^(1/2))/d^2/x/(-c^(2/3)*(c^(2/3)-(-d^2*x^2+c^2)^(1/3)) /((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.18 \[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {x \sqrt [3]{c^2-d^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\sqrt [3]{1-\frac {d^2 x^2}{c^2}}} \] Input:
Integrate[(c^2 - d^2*x^2)^(1/3),x]
Output:
(x*(c^2 - d^2*x^2)^(1/3)*Hypergeometric2F1[-1/3, 1/2, 3/2, (d^2*x^2)/c^2]) /(1 - (d^2*x^2)/c^2)^(1/3)
Time = 0.42 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {211, 234, 760}
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.
\(\displaystyle \int \sqrt [3]{c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {2}{5} c^2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}-\frac {3 c^2 \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{5 d^2 x}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} c^2 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+\left (c^2-d^2 x^2\right )^{2/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\) |
Input:
Int[(c^2 - d^2*x^2)^(1/3),x]
Output:
(3*x*(c^2 - d^2*x^2)^(1/3))/5 + (2*3^(3/4)*Sqrt[2 - Sqrt[3]]*c^2*(c^(2/3) - (c^2 - d^2*x^2)^(1/3))*Sqrt[(c^(4/3) + c^(2/3)*(c^2 - d^2*x^2)^(1/3) + ( c^2 - d^2*x^2)^(2/3))/((1 - Sqrt[3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))^2]*E llipticF[ArcSin[((1 + Sqrt[3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))/((1 - Sqrt [3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(5*d^2*x*Sqrt[-(( c^(2/3)*(c^(2/3) - (c^2 - d^2*x^2)^(1/3)))/((1 - Sqrt[3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))^2)])
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
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 ] && NegQ[a]
\[\int \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}d x\]
Input:
int((-d^2*x^2+c^2)^(1/3),x)
Output:
int((-d^2*x^2+c^2)^(1/3),x)
\[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
Output:
integral((-d^2*x^2 + c^2)^(1/3), x)
Time = 0.46 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.10 \[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=c^{\frac {2}{3}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} \] Input:
integrate((-d**2*x**2+c**2)**(1/3),x)
Output:
c**(2/3)*x*hyper((-1/3, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2)
\[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(1/3), x)
\[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(1/3), x)
Time = 6.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.15 \[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {x\,{\left (c^2-d^2\,x^2\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {1}{2};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{1/3}} \] Input:
int((c^2 - d^2*x^2)^(1/3),x)
Output:
(x*(c^2 - d^2*x^2)^(1/3)*hypergeom([-1/3, 1/2], 3/2, (d^2*x^2)/c^2))/(1 - (d^2*x^2)/c^2)^(1/3)
\[ \int \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} x}{5}+\frac {2 \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}d x \right ) c^{2}}{5} \] Input:
int((-d^2*x^2+c^2)^(1/3),x)
Output:
(3*(c**2 - d**2*x**2)**(1/3)*x + 2*int((c**2 - d**2*x**2)**(1/3)/(c**2 - d **2*x**2),x)*c**2)/5