Integrand size = 24, antiderivative size = 378 \[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {12}{11} c^3 x \sqrt [3]{c^2-d^2 x^2}-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}-\frac {15 c (11 c+4 d x) \left (c^2-d^2 x^2\right )^{4/3}}{154 d}+\frac {8\ 3^{3/4} \sqrt {2-\sqrt {3}} c^5 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{11 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:
12/11*c^3*x*(-d^2*x^2+c^2)^(1/3)-3/14*(d*x+c)^2*(-d^2*x^2+c^2)^(4/3)/d-15/ 154*c*(4*d*x+11*c)*(-d^2*x^2+c^2)^(4/3)/d+8/11*3^(3/4)*(1/2*6^(1/2)-1/2*2^ (1/2))*c^5*(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 = 5.71 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.28 \[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {1}{154} \sqrt [3]{c^2-d^2 x^2} \left (3 \left (-\frac {66 c^4}{d}-42 c^3 x+55 c^2 d x^2+42 c d^2 x^3+11 d^3 x^4\right )+\frac {280 c^3 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\sqrt [3]{1-\frac {d^2 x^2}{c^2}}}\right ) \] Input:
Integrate[(c + d*x)^3*(c^2 - d^2*x^2)^(1/3),x]
Output:
((c^2 - d^2*x^2)^(1/3)*(3*((-66*c^4)/d - 42*c^3*x + 55*c^2*d*x^2 + 42*c*d^ 2*x^3 + 11*d^3*x^4) + (280*c^3*x*Hypergeometric2F1[-1/3, 1/2, 3/2, (d^2*x^ 2)/c^2])/(1 - (d^2*x^2)/c^2)^(1/3)))/154
Time = 0.62 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {497, 27, 497, 27, 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.
| \(\displaystyle \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle -\frac {3 \int -\frac {20}{3} c d^2 (c+d x)^2 \sqrt [3]{c^2-d^2 x^2}dx}{14 d^2}-\frac {3 \left (c^2-d^2 x^2\right )^{4/3} (c+d x)^2}{14 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {10}{7} c \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle \frac {10}{7} c \left (-\frac {3 \int -\frac {14}{3} c d^2 (c+d x) \sqrt [3]{c^2-d^2 x^2}dx}{11 d^2}-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\right )-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {10}{7} c \left (\frac {14}{11} c \int (c+d x) \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\right )-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {10}{7} c \left (\frac {14}{11} c \left (c \int \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\right )-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {10}{7} c \left (\frac {14}{11} c \left (c \left (\frac {2}{5} c^2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\right )-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
| \(\Big \downarrow \) 234 |
| \(\displaystyle \frac {10}{7} c \left (\frac {14}{11} c \left (c \left (\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}-\frac {3 c^2 \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{5 d^2 x}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\right )-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
| \(\Big \downarrow \) 760 |
| \(\displaystyle \frac {10}{7} c \left (\frac {14}{11} c \left (c \left (\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} c^2 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+\left (c^2-d^2 x^2\right )^{2/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\right )-\frac {3 (c+d x)^2 \left (c^2-d^2 x^2\right )^{4/3}}{14 d}\) |
Input:
Int[(c + d*x)^3*(c^2 - d^2*x^2)^(1/3),x]
Output:
(-3*(c + d*x)^2*(c^2 - d^2*x^2)^(4/3))/(14*d) + (10*c*((-3*(c + d*x)*(c^2 - d^2*x^2)^(4/3))/(11*d) + (14*c*((-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]*Ell ipticF[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)]))))/11))/7
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b *(n + 2*p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n , p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, 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]
\[\int \left (d x +c \right )^{3} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}d x\]
Input:
int((d*x+c)^3*(-d^2*x^2+c^2)^(1/3),x)
Output:
int((d*x+c)^3*(-d^2*x^2+c^2)^(1/3),x)
\[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{3} \,d x } \] Input:
integrate((d*x+c)^3*(-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
Output:
integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(-d^2*x^2 + c^2)^(1/3), x)
Time = 1.68 (sec) , antiderivative size = 654, normalized size of antiderivative = 1.73 \[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=c^{\frac {11}{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 )} + c^{\frac {5}{3}} d^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + 3 c^{2} 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 ) + d^{3} \left (\begin {cases} \frac {9 c^{\frac {26}{3}} \sqrt [3]{-1 + \frac {d^{2} x^{2}}{c^{2}}} e^{\frac {i \pi }{3}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} - \frac {9 c^{\frac {26}{3}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} - \frac {6 c^{\frac {20}{3}} d^{2} x^{2} \sqrt [3]{-1 + \frac {d^{2} x^{2}}{c^{2}}} e^{\frac {i \pi }{3}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} + \frac {9 c^{\frac {20}{3}} d^{2} x^{2}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} - \frac {15 c^{\frac {14}{3}} d^{4} x^{4} \sqrt [3]{-1 + \frac {d^{2} x^{2}}{c^{2}}} e^{\frac {i \pi }{3}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} + \frac {12 c^{\frac {8}{3}} d^{6} x^{6} \sqrt [3]{-1 + \frac {d^{2} x^{2}}{c^{2}}} e^{\frac {i \pi }{3}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} & \text {for}\: \left |{\frac {d^{2} x^{2}}{c^{2}}}\right | > 1 \\\frac {9 c^{\frac {26}{3}} \sqrt [3]{1 - \frac {d^{2} x^{2}}{c^{2}}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} - \frac {9 c^{\frac {26}{3}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} - \frac {6 c^{\frac {20}{3}} d^{2} x^{2} \sqrt [3]{1 - \frac {d^{2} x^{2}}{c^{2}}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} + \frac {9 c^{\frac {20}{3}} d^{2} x^{2}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} - \frac {15 c^{\frac {14}{3}} d^{4} x^{4} \sqrt [3]{1 - \frac {d^{2} x^{2}}{c^{2}}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} + \frac {12 c^{\frac {8}{3}} d^{6} x^{6} \sqrt [3]{1 - \frac {d^{2} x^{2}}{c^{2}}}}{- 56 c^{4} d^{4} + 56 c^{2} d^{6} x^{2}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((d*x+c)**3*(-d**2*x**2+c**2)**(1/3),x)
Output:
c**(11/3)*x*hyper((-1/3, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + c**(5/3)*d**2*x**3*hyper((-1/3, 3/2), (5/2,), d**2*x**2*exp_polar(2*I*pi) /c**2) + 3*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)) + d**3*Piecewise((9*c**(26/3)*(-1 + d**2*x**2/c**2)**(1/3)*exp(I*pi/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) - 9 *c**(26/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) - 6*c**(20/3)*d**2*x**2*(-1 + d**2*x**2/c**2)**(1/3)*exp(I*pi/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) + 9*c**(20/3)*d**2*x**2/(-56*c**4*d**4 + 56*c**2*d**6*x**2) - 15*c**(14/3) *d**4*x**4*(-1 + d**2*x**2/c**2)**(1/3)*exp(I*pi/3)/(-56*c**4*d**4 + 56*c* *2*d**6*x**2) + 12*c**(8/3)*d**6*x**6*(-1 + d**2*x**2/c**2)**(1/3)*exp(I*p i/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2), Abs(d**2*x**2/c**2) > 1), (9*c** (26/3)*(1 - d**2*x**2/c**2)**(1/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) - 9 *c**(26/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) - 6*c**(20/3)*d**2*x**2*(1 - d**2*x**2/c**2)**(1/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) + 9*c**(20/3) *d**2*x**2/(-56*c**4*d**4 + 56*c**2*d**6*x**2) - 15*c**(14/3)*d**4*x**4*(1 - d**2*x**2/c**2)**(1/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2) + 12*c**(8/3 )*d**6*x**6*(1 - d**2*x**2/c**2)**(1/3)/(-56*c**4*d**4 + 56*c**2*d**6*x**2 ), True))
\[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{3} \,d x } \] Input:
integrate((d*x+c)^3*(-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c)^3, x)
\[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{3} \,d x } \] Input:
integrate((d*x+c)^3*(-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c)^3, x)
Timed out. \[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\int {\left (c^2-d^2\,x^2\right )}^{1/3}\,{\left (c+d\,x\right )}^3 \,d x \] Input:
int((c^2 - d^2*x^2)^(1/3)*(c + d*x)^3,x)
Output:
int((c^2 - d^2*x^2)^(1/3)*(c + d*x)^3, x)
\[ \int (c+d x)^3 \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {-198 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c^{4}+42 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c^{3} d x +165 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c^{2} d^{2} x^{2}+126 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c \,d^{3} x^{3}+33 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} d^{4} x^{4}+112 \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}d x \right ) c^{5} d}{154 d} \] Input:
int((d*x+c)^3*(-d^2*x^2+c^2)^(1/3),x)
Output:
( - 198*(c**2 - d**2*x**2)**(1/3)*c**4 + 42*(c**2 - d**2*x**2)**(1/3)*c**3 *d*x + 165*(c**2 - d**2*x**2)**(1/3)*c**2*d**2*x**2 + 126*(c**2 - d**2*x** 2)**(1/3)*c*d**3*x**3 + 33*(c**2 - d**2*x**2)**(1/3)*d**4*x**4 + 112*int(( c**2 - d**2*x**2)**(1/3)/(c**2 - d**2*x**2),x)*c**5*d)/(154*d)