\(\int \frac {1}{(c+d x)^3 \sqrt [3]{c^2-d^2 x^2}} \, dx\) [262]

Optimal result

Integrand size = 24, antiderivative size = 708 \[ \int \frac {1}{(c+d x)^3 \sqrt [3]{c^2-d^2 x^2}} \, dx=\frac {15 x}{112 c^3 \sqrt [3]{c^2-d^2 x^2}}-\frac {3}{7 d (c+d x)^2 \sqrt [3]{c^2-d^2 x^2}}-\frac {3}{56 c d (c+d x) \sqrt [3]{c^2-d^2 x^2}}+\frac {15 x}{112 c^3 \left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}+\frac {15 \sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} E\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right )|-7+4 \sqrt {3}\right )}{224 c^{7/3} d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}-\frac {5\ 3^{3/4} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{56 \sqrt {2} c^{7/3} d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:

15/112*x/c^3/(-d^2*x^2+c^2)^(1/3)-3/7/d/(d*x+c)^2/(-d^2*x^2+c^2)^(1/3)-3/5 
6/c/d/(d*x+c)/(-d^2*x^2+c^2)^(1/3)+15/112*x/c^3/((1-3^(1/2))*c^(2/3)-(-d^2 
*x^2+c^2)^(1/3))+15/224*3^(1/4)*(1/2*6^(1/2)+1/2*2^(1/2))*(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)*EllipticE(((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))/c^(7/3)/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)-5/112*3^(3/4)*(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))*2^(1/2)/c^(7/3)/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 = 9.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.10 \[ \int \frac {1}{(c+d x)^3 \sqrt [3]{c^2-d^2 x^2}} \, dx=-\frac {3 (c-d x) \sqrt [3]{1+\frac {d x}{c}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {10}{3},\frac {5}{3},\frac {c-d x}{2 c}\right )}{16 \sqrt [3]{2} c^3 d \sqrt [3]{c^2-d^2 x^2}} \] Input:

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

Output:

(-3*(c - d*x)*(1 + (d*x)/c)^(1/3)*Hypergeometric2F1[2/3, 10/3, 5/3, (c - d 
*x)/(2*c)])/(16*2^(1/3)*c^3*d*(c^2 - d^2*x^2)^(1/3))
 

Rubi [C] (verified)

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

Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.09, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {506, 74}

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.

\(\displaystyle \int \frac {1}{(c+d x)^3 \sqrt [3]{c^2-d^2 x^2}} \, dx\)

\(\Big \downarrow \) 506

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

\(\Big \downarrow \) 74

\(\displaystyle -\frac {3 \sqrt [3]{1-\frac {2 c}{c+d x}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {8}{3},\frac {11}{3},\frac {2 c}{c+d x}\right )}{8 d (c+d x)^2 \sqrt [3]{c^2-d^2 x^2}}\)

Input:

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

Output:

(-3*(1 - (2*c)/(c + d*x))^(1/3)*Hypergeometric2F1[1/3, 8/3, 11/3, (2*c)/(c 
 + d*x)])/(8*d*(c + d*x)^2*(c^2 - d^2*x^2)^(1/3))
 

Defintions of rubi rules used

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 506

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(-(a + b*x^2)^p)*((1/(c + d*x))^(2*p)/(d*(1 - (c - 
d*q)/(c + d*x))^p*(1 - (c + d*q)/(c + d*x))^p))   Subst[Int[(1 - (c - d*q)* 
x)^p*((1 - (c + d*q)*x)^p/x^(n + 2*p + 2)), x], x, 1/(c + d*x)], x]] /; Fre 
eQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && NegQ[a/b]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(-d^2*x^2 + c^2)^(2/3)/(d^5*x^5 + 3*c*d^4*x^4 + 2*c^2*d^3*x^3 - 
2*c^3*d^2*x^2 - 3*c^4*d*x - c^5), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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