Integrand size = 22, antiderivative size = 337 \[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3}{5} c x \sqrt [3]{c^2-d^2 x^2}-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}+\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} c^3 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:
3/5*c*x*(-d^2*x^2+c^2)^(1/3)-3/8*(-d^2*x^2+c^2)^(4/3)/d+2/5*3^(3/4)*(1/2*6 ^(1/2)-1/2*2^(1/2))*c^3*(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 = 5.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.24 \[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}+\frac {c x \sqrt [3]{c^2-d^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\sqrt [3]{1-\frac {d^2 x^2}{c^2}}} \] Input:
Integrate[(c + d*x)*(c^2 - d^2*x^2)^(1/3),x]
Output:
(-3*(c^2 - d^2*x^2)^(4/3))/(8*d) + (c*x*(c^2 - d^2*x^2)^(1/3)*Hypergeometr ic2F1[-1/3, 1/2, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^(1/3)
Time = 0.47 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {455, 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.
| \(\displaystyle \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle c \int \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle c \left (\frac {2}{5} c^2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\) |
| \(\Big \downarrow \) 234 |
| \(\displaystyle c \left (\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}-\frac {3 c^2 \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{5 d^2 x}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\) |
| \(\Big \downarrow \) 760 |
| \(\displaystyle c \left (\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} c^2 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+\left (c^2-d^2 x^2\right )^{2/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\) |
Input:
Int[(c + d*x)*(c^2 - d^2*x^2)^(1/3),x]
Output:
(-3*(c^2 - d^2*x^2)^(4/3))/(8*d) + c*((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]*EllipticF[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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
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]
\[\int \left (d x +c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}d x\]
Input:
int((d*x+c)*(-d^2*x^2+c^2)^(1/3),x)
Output:
int((d*x+c)*(-d^2*x^2+c^2)^(1/3),x)
\[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
Output:
integral((-d^2*x^2 + c^2)^(1/3)*(d*x + c), x)
Time = 1.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.20 \[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=c^{\frac {5}{3}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + d \left (\begin {cases} \frac {x^{2} \sqrt [3]{c^{2}}}{2} & \text {for}\: d^{2} = 0 \\- \frac {3 \left (c^{2} - d^{2} x^{2}\right )^{\frac {4}{3}}}{8 d^{2}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((d*x+c)*(-d**2*x**2+c**2)**(1/3),x)
Output:
c**(5/3)*x*hyper((-1/3, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + d*Piecewise((x**2*(c**2)**(1/3)/2, Eq(d**2, 0)), (-3*(c**2 - d**2*x**2)**( 4/3)/(8*d**2), True))
\[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c), x)
\[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c), x)
Time = 6.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.20 \[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {c\,x\,{\left (c^2-d^2\,x^2\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {1}{2};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{1/3}}-\frac {3\,{\left (c^2-d^2\,x^2\right )}^{4/3}}{8\,d} \] Input:
int((c^2 - d^2*x^2)^(1/3)*(c + d*x),x)
Output:
(c*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) - (3*(c^2 - d^2*x^2)^(4/3))/(8*d)
\[ \int (c+d x) \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {-15 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c^{2}+24 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c d x +15 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} d^{2} x^{2}+16 \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}d x \right ) c^{3} d}{40 d} \] Input:
int((d*x+c)*(-d^2*x^2+c^2)^(1/3),x)
Output:
( - 15*(c**2 - d**2*x**2)**(1/3)*c**2 + 24*(c**2 - d**2*x**2)**(1/3)*c*d*x + 15*(c**2 - d**2*x**2)**(1/3)*d**2*x**2 + 16*int((c**2 - d**2*x**2)**(1/ 3)/(c**2 - d**2*x**2),x)*c**3*d)/(40*d)