\(\int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx\) [278]

Optimal result

Integrand size = 26, antiderivative size = 294 \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=-\frac {\sqrt [3]{c^2-d^2 x^2}}{2 d (c+d x)^{7/3}}+\frac {\sqrt [3]{c^2-d^2 x^2}}{12 c d (c+d x)^{4/3}}+\frac {\sqrt [3]{c^2-d^2 x^2} \arctan \left (\frac {\sqrt [3]{c}+2^{2/3} \sqrt [3]{c-d x}}{\sqrt {3} \sqrt [3]{c}}\right )}{6\ 2^{2/3} \sqrt {3} c^{5/3} d \sqrt [3]{c-d x} \sqrt [3]{c+d x}}+\frac {\sqrt [3]{c^2-d^2 x^2} \log (c+d x)}{36\ 2^{2/3} c^{5/3} d \sqrt [3]{c-d x} \sqrt [3]{c+d x}}-\frac {\sqrt [3]{c^2-d^2 x^2} \log \left (\sqrt [3]{2} \sqrt [3]{c}-\sqrt [3]{c-d x}\right )}{12\ 2^{2/3} c^{5/3} d \sqrt [3]{c-d x} \sqrt [3]{c+d x}} \] Output:

-1/2*(-d^2*x^2+c^2)^(1/3)/d/(d*x+c)^(7/3)+1/12*(-d^2*x^2+c^2)^(1/3)/c/d/(d 
*x+c)^(4/3)+1/36*(-d^2*x^2+c^2)^(1/3)*arctan(1/3*(c^(1/3)+2^(2/3)*(-d*x+c) 
^(1/3))*3^(1/2)/c^(1/3))*2^(1/3)*3^(1/2)/c^(5/3)/d/(-d*x+c)^(1/3)/(d*x+c)^ 
(1/3)+1/72*(-d^2*x^2+c^2)^(1/3)*ln(d*x+c)*2^(1/3)/c^(5/3)/d/(-d*x+c)^(1/3) 
/(d*x+c)^(1/3)-1/24*(-d^2*x^2+c^2)^(1/3)*ln(2^(1/3)*c^(1/3)-(-d*x+c)^(1/3) 
)*2^(1/3)/c^(5/3)/d/(-d*x+c)^(1/3)/(d*x+c)^(1/3)
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\frac {\frac {6 c^{2/3} (-5 c+d x) \sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{7/3}}-2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{c+d x}}{\sqrt [3]{c} \sqrt [3]{c+d x}+2^{2/3} \sqrt [3]{c^2-d^2 x^2}}\right )-2 \sqrt [3]{2} \log \left (-2 \sqrt [3]{c} \sqrt [3]{c+d x}+2^{2/3} \sqrt [3]{c^2-d^2 x^2}\right )+\sqrt [3]{2} \log \left (2 c^{2/3} (c+d x)^{2/3}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c+d x} \sqrt [3]{c^2-d^2 x^2}+\sqrt [3]{2} \left (c^2-d^2 x^2\right )^{2/3}\right )}{72 c^{5/3} d} \] Input:

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

Output:

((6*c^(2/3)*(-5*c + d*x)*(c^2 - d^2*x^2)^(1/3))/(c + d*x)^(7/3) - 2*2^(1/3 
)*Sqrt[3]*ArcTan[(Sqrt[3]*c^(1/3)*(c + d*x)^(1/3))/(c^(1/3)*(c + d*x)^(1/3 
) + 2^(2/3)*(c^2 - d^2*x^2)^(1/3))] - 2*2^(1/3)*Log[-2*c^(1/3)*(c + d*x)^( 
1/3) + 2^(2/3)*(c^2 - d^2*x^2)^(1/3)] + 2^(1/3)*Log[2*c^(2/3)*(c + d*x)^(2 
/3) + 2^(2/3)*c^(1/3)*(c + d*x)^(1/3)*(c^2 - d^2*x^2)^(1/3) + 2^(1/3)*(c^2 
 - d^2*x^2)^(2/3)])/(72*c^(5/3)*d)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.82, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {474, 473, 27, 51, 52, 69, 16, 1082, 217}

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.

\(\displaystyle \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx\)

\(\Big \downarrow \) 474

\(\displaystyle \frac {\sqrt [3]{\frac {d x}{c}+1} \int \frac {\sqrt [3]{c^2-d^2 x^2}}{\left (\frac {d x}{c}+1\right )^{10/3}}dx}{c^3 \sqrt [3]{c+d x}}\)

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 51

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{6} c \int \frac {1}{(c+d x)^2 \left (c^2-c d x\right )^{2/3}}dx-\frac {\sqrt [3]{c^2-c d x}}{2 d (c+d x)^2}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{6} c \left (\frac {\int \frac {1}{(c+d x) \left (c^2-c d x\right )^{2/3}}dx}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )-\frac {\sqrt [3]{c^2-c d x}}{2 d (c+d x)^2}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 69

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{6} c \left (\frac {-\frac {3 \int \frac {1}{\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}}d\sqrt [3]{c^2-c d x}}{2\ 2^{2/3} c^{4/3} d}-\frac {3 \int \frac {1}{2^{2/3} c^{4/3}+\sqrt [3]{2} \sqrt [3]{c^2-c d x} c^{2/3}+\left (c^2-c d x\right )^{2/3}}d\sqrt [3]{c^2-c d x}}{2 \sqrt [3]{2} c^{2/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )-\frac {\sqrt [3]{c^2-c d x}}{2 d (c+d x)^2}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{6} c \left (\frac {-\frac {3 \int \frac {1}{2^{2/3} c^{4/3}+\sqrt [3]{2} \sqrt [3]{c^2-c d x} c^{2/3}+\left (c^2-c d x\right )^{2/3}}d\sqrt [3]{c^2-c d x}}{2 \sqrt [3]{2} c^{2/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}\right )}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )-\frac {\sqrt [3]{c^2-c d x}}{2 d (c+d x)^2}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{6} c \left (\frac {\frac {3 \int \frac {1}{-\left (c^2-c d x\right )^{2/3}-3}d\left (\frac {2^{2/3} \sqrt [3]{c^2-c d x}}{c^{2/3}}+1\right )}{2^{2/3} c^{4/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}\right )}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )-\frac {\sqrt [3]{c^2-c d x}}{2 d (c+d x)^2}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{6} c \left (\frac {-\frac {\sqrt {3} \arctan \left (\frac {\frac {2^{2/3} \sqrt [3]{c^2-c d x}}{c^{2/3}}+1}{\sqrt {3}}\right )}{2^{2/3} c^{4/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}\right )}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )-\frac {\sqrt [3]{c^2-c d x}}{2 d (c+d x)^2}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

Input:

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

Output:

((c^2 - d^2*x^2)^(4/3)*(-1/2*(c^2 - c*d*x)^(1/3)/(d*(c + d*x)^2) - (c*(-1/ 
2*(c^2 - c*d*x)^(1/3)/(c^2*d*(c + d*x)) + (-((Sqrt[3]*ArcTan[(1 + (2^(2/3) 
*(c^2 - c*d*x)^(1/3))/c^(2/3))/Sqrt[3]])/(2^(2/3)*c^(4/3)*d)) - Log[c + d* 
x]/(2*2^(2/3)*c^(4/3)*d) + (3*Log[2^(1/3)*c^(2/3) - (c^2 - c*d*x)^(1/3)])/ 
(2*2^(2/3)*c^(4/3)*d))/(3*c)))/6))/((c + d*x)^(1/3)*(1 + (d*x)/c)*(c^2 - c 
*d*x)^(4/3))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 474

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n])   Int[(1 + d 
*(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + 
 a*d^2, 0] &&  !(IntegerQ[n] || GtQ[c, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [F]

\[\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {10}{3}}}d x\]

Input:

int((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(10/3),x)
 

Output:

int((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(10/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\frac {4^{\frac {2}{3}} {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \left (-c^{2}\right )^{\frac {2}{3}} \log \left (\frac {4^{\frac {2}{3}} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} c - 2 \cdot 4^{\frac {1}{3}} {\left (c d x + c^{2}\right )} \left (-c^{2}\right )^{\frac {1}{3}}}{d x + c}\right ) - 2 \cdot 4^{\frac {2}{3}} {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \left (-c^{2}\right )^{\frac {2}{3}} \log \left (-\frac {4^{\frac {2}{3}} \left (-c^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )} - 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} c}{d x + c}\right ) + 12 \, \sqrt {\frac {1}{3}} {\left (c d^{3} x^{3} + 3 \, c^{2} d^{2} x^{2} + 3 \, c^{3} d x + c^{4}\right )} \sqrt {-4^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {1}{3}}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (4^{\frac {2}{3}} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} - 4^{\frac {1}{3}} {\left (c d x + c^{2}\right )} \left (-c^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-4^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {1}{3}}}}{2 \, {\left (c^{2} d x + c^{3}\right )}}\right ) + 12 \, {\left (c^{2} d x - 5 \, c^{3}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{144 \, {\left (c^{3} d^{4} x^{3} + 3 \, c^{4} d^{3} x^{2} + 3 \, c^{5} d^{2} x + c^{6} d\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(10/3),x, algorithm="fricas")
 

Output:

1/144*(4^(2/3)*(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(-c^2)^(2/3)*log( 
(4^(2/3)*(-d^2*x^2 + c^2)^(1/3)*(-c^2)^(2/3)*(d*x + c)^(2/3) + 2*(-d^2*x^2 
 + c^2)^(2/3)*(d*x + c)^(1/3)*c - 2*4^(1/3)*(c*d*x + c^2)*(-c^2)^(1/3))/(d 
*x + c)) - 2*4^(2/3)*(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(-c^2)^(2/3 
)*log(-(4^(2/3)*(-c^2)^(2/3)*(d*x + c) - 2*(-d^2*x^2 + c^2)^(1/3)*(d*x + c 
)^(2/3)*c)/(d*x + c)) + 12*sqrt(1/3)*(c*d^3*x^3 + 3*c^2*d^2*x^2 + 3*c^3*d* 
x + c^4)*sqrt(-4^(1/3)*(-c^2)^(1/3))*arctan(1/2*sqrt(1/3)*(4^(2/3)*(-d^2*x 
^2 + c^2)^(1/3)*(-c^2)^(2/3)*(d*x + c)^(2/3) - 4^(1/3)*(c*d*x + c^2)*(-c^2 
)^(1/3))*sqrt(-4^(1/3)*(-c^2)^(1/3))/(c^2*d*x + c^3)) + 12*(c^2*d*x - 5*c^ 
3)*(-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(2/3))/(c^3*d^4*x^3 + 3*c^4*d^3*x^2 + 
3*c^5*d^2*x + c^6*d)
 

Sympy [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\int \frac {\sqrt [3]{- \left (- c + d x\right ) \left (c + d x\right )}}{\left (c + d x\right )^{\frac {10}{3}}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(1/3)/(d*x+c)**(10/3),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(1/3)/(c + d*x)**(10/3), x)
 

Maxima [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(10/3),x, algorithm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(1/3)/(d*x + c)^(10/3), x)
 

Giac [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(10/3),x, algorithm="giac")
 

Output:

integrate((-d^2*x^2 + c^2)^(1/3)/(d*x + c)^(10/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{1/3}}{{\left (c+d\,x\right )}^{10/3}} \,d x \] Input:

int((c^2 - d^2*x^2)^(1/3)/(c + d*x)^(10/3),x)
 

Output:

int((c^2 - d^2*x^2)^(1/3)/(c + d*x)^(10/3), x)
 

Reduce [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}} \, dx=\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}} c^{3}+3 \left (d x +c \right )^{\frac {1}{3}} c^{2} d x +3 \left (d x +c \right )^{\frac {1}{3}} c \,d^{2} x^{2}+\left (d x +c \right )^{\frac {1}{3}} d^{3} x^{3}}d x \] Input:

int((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(10/3),x)
 

Output:

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