\(\int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx\) [243]

Optimal result

Integrand size = 24, antiderivative size = 347 \[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {42}{55} c^2 x \sqrt [3]{c^2-d^2 x^2}-\frac {3 (11 c+4 d x) \left (c^2-d^2 x^2\right )^{4/3}}{44 d}+\frac {28\ 3^{3/4} \sqrt {2-\sqrt {3}} c^4 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{55 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:

42/55*c^2*x*(-d^2*x^2+c^2)^(1/3)-3/44*(4*d*x+11*c)*(-d^2*x^2+c^2)^(4/3)/d+ 
28/55*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*c^4*(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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.61 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.27 \[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {1}{44} \sqrt [3]{c^2-d^2 x^2} \left (-\frac {33 c^3}{d}+33 c d x^2+12 x \left (-c^2+d^2 x^2\right )+\frac {56 c^2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\sqrt [3]{1-\frac {d^2 x^2}{c^2}}}\right ) \] Input:

Integrate[(c + d*x)^2*(c^2 - d^2*x^2)^(1/3),x]
 

Output:

((c^2 - d^2*x^2)^(1/3)*((-33*c^3)/d + 33*c*d*x^2 + 12*x*(-c^2 + d^2*x^2) + 
 (56*c^2*x*Hypergeometric2F1[-1/3, 1/2, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2 
)/c^2)^(1/3)))/44
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.

\(\displaystyle \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx\)

\(\Big \downarrow \) 497

\(\displaystyle -\frac {3 \int -\frac {14}{3} c d^2 (c+d x) \sqrt [3]{c^2-d^2 x^2}dx}{11 d^2}-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {14}{11} c \int (c+d x) \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {14}{11} c \left (c \int \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {14}{11} c \left (c \left (\frac {2}{5} c^2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {14}{11} c \left (c \left (\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}-\frac {3 c^2 \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{5 d^2 x}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {14}{11} c \left (c \left (\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} c^2 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+\left (c^2-d^2 x^2\right )^{2/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}+\frac {3}{5} x \sqrt [3]{c^2-d^2 x^2}\right )-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{8 d}\right )-\frac {3 (c+d x) \left (c^2-d^2 x^2\right )^{4/3}}{11 d}\)

Input:

Int[(c + d*x)^2*(c^2 - d^2*x^2)^(1/3),x]
 

Output:

(-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]*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)]))))/11
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 211

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])
 

rule 234

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]
 

rule 455

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]
 

rule 497

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]
 

rule 760

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]
 

Maple [F]

\[\int \left (d x +c \right )^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}d x\]

Input:

int((d*x+c)^2*(-d^2*x^2+c^2)^(1/3),x)
 

Output:

int((d*x+c)^2*(-d^2*x^2+c^2)^(1/3),x)
 

Fricas [F]

\[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{2} \,d x } \] Input:

integrate((d*x+c)^2*(-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
 

Output:

integral((d^2*x^2 + 2*c*d*x + c^2)*(-d^2*x^2 + c^2)^(1/3), x)
 

Sympy [A] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.32 \[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=c^{\frac {8}{3}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + \frac {c^{\frac {2}{3}} d^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{3} + 2 c d \left (\begin {cases} \frac {x^{2} \sqrt [3]{c^{2}}}{2} & \text {for}\: d^{2} = 0 \\- \frac {3 \left (c^{2} - d^{2} x^{2}\right )^{\frac {4}{3}}}{8 d^{2}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((d*x+c)**2*(-d**2*x**2+c**2)**(1/3),x)
 

Output:

c**(8/3)*x*hyper((-1/3, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + 
c**(2/3)*d**2*x**3*hyper((-1/3, 3/2), (5/2,), d**2*x**2*exp_polar(2*I*pi)/ 
c**2)/3 + 2*c*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))
 

Maxima [F]

\[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{2} \,d x } \] Input:

integrate((d*x+c)^2*(-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c)^2, x)
 

Giac [F]

\[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{2} \,d x } \] Input:

integrate((d*x+c)^2*(-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
 

Output:

integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\int {\left (c^2-d^2\,x^2\right )}^{1/3}\,{\left (c+d\,x\right )}^2 \,d x \] Input:

int((c^2 - d^2*x^2)^(1/3)*(c + d*x)^2,x)
 

Output:

int((c^2 - d^2*x^2)^(1/3)*(c + d*x)^2, x)
 

Reduce [F]

\[ \int (c+d x)^2 \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {-165 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c^{3}+108 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c^{2} d x +165 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} c \,d^{2} x^{2}+60 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}} d^{3} x^{3}+112 \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}d x \right ) c^{4} d}{220 d} \] Input:

int((d*x+c)^2*(-d^2*x^2+c^2)^(1/3),x)
                                                                                    
                                                                                    
 

Output:

( - 165*(c**2 - d**2*x**2)**(1/3)*c**3 + 108*(c**2 - d**2*x**2)**(1/3)*c** 
2*d*x + 165*(c**2 - d**2*x**2)**(1/3)*c*d**2*x**2 + 60*(c**2 - d**2*x**2)* 
*(1/3)*d**3*x**3 + 112*int((c**2 - d**2*x**2)**(1/3)/(c**2 - d**2*x**2),x) 
*c**4*d)/(220*d)