\(\int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx\) [279]

Optimal result

Integrand size = 26, antiderivative size = 329 \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=-\frac {\sqrt [3]{c^2-d^2 x^2}}{3 d (c+d x)^{10/3}}+\frac {\sqrt [3]{c^2-d^2 x^2}}{36 c d (c+d x)^{7/3}}+\frac {5 \sqrt [3]{c^2-d^2 x^2}}{216 c^2 d (c+d x)^{4/3}}+\frac {5 \sqrt [3]{c^2-d^2 x^2} \arctan \left (\frac {\sqrt [3]{c}+2^{2/3} \sqrt [3]{c-d x}}{\sqrt {3} \sqrt [3]{c}}\right )}{108\ 2^{2/3} \sqrt {3} c^{8/3} d \sqrt [3]{c-d x} \sqrt [3]{c+d x}}+\frac {5 \sqrt [3]{c^2-d^2 x^2} \log (c+d x)}{648\ 2^{2/3} c^{8/3} d \sqrt [3]{c-d x} \sqrt [3]{c+d x}}-\frac {5 \sqrt [3]{c^2-d^2 x^2} \log \left (\sqrt [3]{2} \sqrt [3]{c}-\sqrt [3]{c-d x}\right )}{216\ 2^{2/3} c^{8/3} d \sqrt [3]{c-d x} \sqrt [3]{c+d x}} \] Output:

-1/3*(-d^2*x^2+c^2)^(1/3)/d/(d*x+c)^(10/3)+1/36*(-d^2*x^2+c^2)^(1/3)/c/d/( 
d*x+c)^(7/3)+5/216*(-d^2*x^2+c^2)^(1/3)/c^2/d/(d*x+c)^(4/3)+5/648*(-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^(8/3)/d/(-d*x+c)^(1/3)/(d*x+c)^(1/3)+5/1296*(-d^2*x^2+c^ 
2)^(1/3)*ln(d*x+c)*2^(1/3)/c^(8/3)/d/(-d*x+c)^(1/3)/(d*x+c)^(1/3)-5/432*(- 
d^2*x^2+c^2)^(1/3)*ln(2^(1/3)*c^(1/3)-(-d*x+c)^(1/3))*2^(1/3)/c^(8/3)/d/(- 
d*x+c)^(1/3)/(d*x+c)^(1/3)
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\frac {-\frac {6 c^{2/3} \left (61 c^2-16 c d x-5 d^2 x^2\right ) \sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{10/3}}-10 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{c+d x}}{\sqrt [3]{c} \sqrt [3]{c+d x}+2^{2/3} \sqrt [3]{c^2-d^2 x^2}}\right )-10 \sqrt [3]{2} \log \left (-2 \sqrt [3]{c} \sqrt [3]{c+d x}+2^{2/3} \sqrt [3]{c^2-d^2 x^2}\right )+5 \sqrt [3]{2} \log \left (2 c^{2/3} (c+d x)^{2/3}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c+d x} \sqrt [3]{c^2-d^2 x^2}+\sqrt [3]{2} \left (c^2-d^2 x^2\right )^{2/3}\right )}{1296 c^{8/3} d} \] Input:

Integrate[(c^2 - d^2*x^2)^(1/3)/(c + d*x)^(13/3),x]
 

Output:

((-6*c^(2/3)*(61*c^2 - 16*c*d*x - 5*d^2*x^2)*(c^2 - d^2*x^2)^(1/3))/(c + d 
*x)^(10/3) - 10*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))] - 10*2^(1/3)*Log 
[-2*c^(1/3)*(c + d*x)^(1/3) + 2^(2/3)*(c^2 - d^2*x^2)^(1/3)] + 5*2^(1/3)*L 
og[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)])/(1296*c^(8/3)*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.85, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {474, 473, 27, 51, 52, 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.

\(\displaystyle \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx\)

\(\Big \downarrow \) 474

\(\displaystyle \frac {\sqrt [3]{\frac {d x}{c}+1} \int \frac {\sqrt [3]{c^2-d^2 x^2}}{\left (\frac {d x}{c}+1\right )^{13/3}}dx}{c^4 \sqrt [3]{c+d x}}\)

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 51

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \int \frac {1}{(c+d x)^3 \left (c^2-c d x\right )^{2/3}}dx-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \left (\frac {5 \int \frac {1}{(c+d x)^2 \left (c^2-c d x\right )^{2/3}}dx}{12 c}-\frac {\sqrt [3]{c^2-c d x}}{4 c^2 d (c+d x)^2}\right )-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \left (\frac {5 \left (\frac {\int \frac {1}{(c+d x) \left (c^2-c d x\right )^{2/3}}dx}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )}{12 c}-\frac {\sqrt [3]{c^2-c d x}}{4 c^2 d (c+d x)^2}\right )-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 69

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \left (\frac {5 \left (\frac {-\frac {3 \int \frac {1}{\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}}d\sqrt [3]{c^2-c d x}}{2\ 2^{2/3} c^{4/3} d}-\frac {3 \int \frac {1}{2^{2/3} c^{4/3}+\sqrt [3]{2} \sqrt [3]{c^2-c d x} c^{2/3}+\left (c^2-c d x\right )^{2/3}}d\sqrt [3]{c^2-c d x}}{2 \sqrt [3]{2} c^{2/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )}{12 c}-\frac {\sqrt [3]{c^2-c d x}}{4 c^2 d (c+d x)^2}\right )-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \left (\frac {5 \left (\frac {-\frac {3 \int \frac {1}{2^{2/3} c^{4/3}+\sqrt [3]{2} \sqrt [3]{c^2-c d x} c^{2/3}+\left (c^2-c d x\right )^{2/3}}d\sqrt [3]{c^2-c d x}}{2 \sqrt [3]{2} c^{2/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}\right )}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )}{12 c}-\frac {\sqrt [3]{c^2-c d x}}{4 c^2 d (c+d x)^2}\right )-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \left (\frac {5 \left (\frac {\frac {3 \int \frac {1}{-\left (c^2-c d x\right )^{2/3}-3}d\left (\frac {2^{2/3} \sqrt [3]{c^2-c d x}}{c^{2/3}}+1\right )}{2^{2/3} c^{4/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}\right )}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )}{12 c}-\frac {\sqrt [3]{c^2-c d x}}{4 c^2 d (c+d x)^2}\right )-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{4/3} \left (-\frac {1}{9} c \left (\frac {5 \left (\frac {-\frac {\sqrt {3} \arctan \left (\frac {\frac {2^{2/3} \sqrt [3]{c^2-c d x}}{c^{2/3}}+1}{\sqrt {3}}\right )}{2^{2/3} c^{4/3} d}-\frac {\log (c+d x)}{2\ 2^{2/3} c^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} c^{2/3}-\sqrt [3]{c^2-c d x}\right )}{2\ 2^{2/3} c^{4/3} d}}{3 c}-\frac {\sqrt [3]{c^2-c d x}}{2 c^2 d (c+d x)}\right )}{12 c}-\frac {\sqrt [3]{c^2-c d x}}{4 c^2 d (c+d x)^2}\right )-\frac {\sqrt [3]{c^2-c d x}}{3 d (c+d x)^3}\right )}{\sqrt [3]{c+d x} \left (\frac {d x}{c}+1\right ) \left (c^2-c d x\right )^{4/3}}\)

Input:

Int[(c^2 - d^2*x^2)^(1/3)/(c + d*x)^(13/3),x]
 

Output:

((c^2 - d^2*x^2)^(4/3)*(-1/3*(c^2 - c*d*x)^(1/3)/(d*(c + d*x)^3) - (c*(-1/ 
4*(c^2 - c*d*x)^(1/3)/(c^2*d*(c + d*x)^2) + (5*(-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)))/(12*c)))/9))/((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]

\[\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {13}{3}}}d x\]

Input:

int((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(13/3),x)
 

Output:

int((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(13/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\frac {5 \cdot 4^{\frac {2}{3}} {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\right )} \left (-c^{2}\right )^{\frac {2}{3}} \log \left (\frac {4^{\frac {2}{3}} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} c - 2 \cdot 4^{\frac {1}{3}} {\left (c d x + c^{2}\right )} \left (-c^{2}\right )^{\frac {1}{3}}}{d x + c}\right ) - 10 \cdot 4^{\frac {2}{3}} {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\right )} \left (-c^{2}\right )^{\frac {2}{3}} \log \left (-\frac {4^{\frac {2}{3}} \left (-c^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )} - 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} c}{d x + c}\right ) + 60 \, \sqrt {\frac {1}{3}} {\left (c d^{4} x^{4} + 4 \, c^{2} d^{3} x^{3} + 6 \, c^{3} d^{2} x^{2} + 4 \, c^{4} d x + c^{5}\right )} \sqrt {-4^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {1}{3}}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (4^{\frac {2}{3}} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} - 4^{\frac {1}{3}} {\left (c d x + c^{2}\right )} \left (-c^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-4^{\frac {1}{3}} \left (-c^{2}\right )^{\frac {1}{3}}}}{2 \, {\left (c^{2} d x + c^{3}\right )}}\right ) + 12 \, {\left (5 \, c^{2} d^{2} x^{2} + 16 \, c^{3} d x - 61 \, c^{4}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{2592 \, {\left (c^{4} d^{5} x^{4} + 4 \, c^{5} d^{4} x^{3} + 6 \, c^{6} d^{3} x^{2} + 4 \, c^{7} d^{2} x + c^{8} d\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(13/3),x, algorithm="fricas")
 

Output:

1/2592*(5*4^(2/3)*(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4 
)*(-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)) - 10*4^(2/3)*(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^ 
2*x^2 + 4*c^3*d*x + c^4)*(-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)) + 60*sqrt(1/3)*( 
c*d^4*x^4 + 4*c^2*d^3*x^3 + 6*c^3*d^2*x^2 + 4*c^4*d*x + c^5)*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*(5*c^2*d^2*x^2 + 16*c^3*d*x - 61*c^4) 
*(-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(2/3))/(c^4*d^5*x^4 + 4*c^5*d^4*x^3 + 6* 
c^6*d^3*x^2 + 4*c^7*d^2*x + c^8*d)
 

Sympy [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\int \frac {\sqrt [3]{- \left (- c + d x\right ) \left (c + d x\right )}}{\left (c + d x\right )^{\frac {13}{3}}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(1/3)/(d*x+c)**(13/3),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(1/3)/(c + d*x)**(13/3), x)
 

Maxima [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {13}{3}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(13/3),x, algorithm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(1/3)/(d*x + c)^(13/3), x)
 

Giac [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {13}{3}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(13/3),x, algorithm="giac")
 

Output:

integrate((-d^2*x^2 + c^2)^(1/3)/(d*x + c)^(13/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{1/3}}{{\left (c+d\,x\right )}^{13/3}} \,d x \] Input:

int((c^2 - d^2*x^2)^(1/3)/(c + d*x)^(13/3),x)
 

Output:

int((c^2 - d^2*x^2)^(1/3)/(c + d*x)^(13/3), x)
 

Reduce [F]

\[ \int \frac {\sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{13/3}} \, dx=\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}} c^{4}+4 \left (d x +c \right )^{\frac {1}{3}} c^{3} d x +6 \left (d x +c \right )^{\frac {1}{3}} c^{2} d^{2} x^{2}+4 \left (d x +c \right )^{\frac {1}{3}} c \,d^{3} x^{3}+\left (d x +c \right )^{\frac {1}{3}} d^{4} x^{4}}d x \] Input:

int((-d^2*x^2+c^2)^(1/3)/(d*x+c)^(13/3),x)
 

Output:

int((c**2 - d**2*x**2)**(1/3)/((c + d*x)**(1/3)*c**4 + 4*(c + d*x)**(1/3)* 
c**3*d*x + 6*(c + d*x)**(1/3)*c**2*d**2*x**2 + 4*(c + d*x)**(1/3)*c*d**3*x 
**3 + (c + d*x)**(1/3)*d**4*x**4),x)