\(\int \frac {x^{11} (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\) [584]

Optimal result

Integrand size = 27, antiderivative size = 140 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {4480 c^3 \sqrt {c+d x^3}}{3 d^4}+\frac {1536 c^4 \sqrt {c+d x^3}}{d^4 \left (8 c-d x^3\right )}+\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}+\frac {2 c \left (c+d x^3\right )^{5/2}}{d^4}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4}-\frac {4992 c^{7/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4} \] Output:

4480/3*c^3*(d*x^3+c)^(1/2)/d^4+1536*c^4*(d*x^3+c)^(1/2)/d^4/(-d*x^3+8*c)+1 
28/3*c^2*(d*x^3+c)^(3/2)/d^4+2*c*(d*x^3+c)^(5/2)/d^4+2/21*(d*x^3+c)^(7/2)/ 
d^4-4992*c^(7/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.74 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \sqrt {c+d x^3} \left (-145328 c^4+12206 c^3 d x^3+301 c^2 d^2 x^6+16 c d^3 x^9+d^4 x^{12}\right )}{21 d^4 \left (-8 c+d x^3\right )}-\frac {4992 c^{7/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4} \] Input:

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

Output:

(2*Sqrt[c + d*x^3]*(-145328*c^4 + 12206*c^3*d*x^3 + 301*c^2*d^2*x^6 + 16*c 
*d^3*x^9 + d^4*x^12))/(21*d^4*(-8*c + d*x^3)) - (4992*c^(7/2)*ArcTanh[Sqrt 
[c + d*x^3]/(3*Sqrt[c])])/d^4
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {948, 108, 27, 170, 25, 27, 164, 60, 73, 219}

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.

\(\displaystyle \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^9 \left (d x^3+c\right )^{3/2}}{\left (8 c-d x^3\right )^2}dx^3\)

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {\int \frac {3 x^6 \sqrt {d x^3+c} \left (3 d x^3+2 c\right )}{2 \left (8 c-d x^3\right )}dx^3}{d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \int \frac {x^6 \sqrt {d x^3+c} \left (3 d x^3+2 c\right )}{8 c-d x^3}dx^3}{2 d}\right )\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (-\frac {2 \int -\frac {c d x^3 \sqrt {d x^3+c} \left (85 d x^3+48 c\right )}{8 c-d x^3}dx^3}{7 d^2}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (\frac {2 \int \frac {c d x^3 \sqrt {d x^3+c} \left (85 d x^3+48 c\right )}{8 c-d x^3}dx^3}{7 d^2}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (\frac {2 c \int \frac {x^3 \sqrt {d x^3+c} \left (85 d x^3+48 c\right )}{8 c-d x^3}dx^3}{7 d}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (\frac {2 c \left (\frac {5824 c^2 \int \frac {\sqrt {d x^3+c}}{8 c-d x^3}dx^3}{d}-\frac {2 \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{3 d^2}\right )}{7 d}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (\frac {2 c \left (\frac {5824 c^2 \left (9 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {2 \sqrt {c+d x^3}}{d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{3 d^2}\right )}{7 d}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (\frac {2 c \left (\frac {5824 c^2 \left (\frac {18 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{3 d^2}\right )}{7 d}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {x^9 \left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3 \left (\frac {2 c \left (\frac {5824 c^2 \left (\frac {6 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{3 d^2}\right )}{7 d}-\frac {6 x^6 \left (c+d x^3\right )^{3/2}}{7 d}\right )}{2 d}\right )\)

Input:

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

Output:

((x^9*(c + d*x^3)^(3/2))/(d*(8*c - d*x^3)) - (3*((-6*x^6*(c + d*x^3)^(3/2) 
)/(7*d) + (2*c*((-2*(c + d*x^3)^(3/2)*(694*c + 51*d*x^3))/(3*d^2) + (5824* 
c^2*((-2*Sqrt[c + d*x^3])/d + (6*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c 
])])/d))/d))/(7*d)))/(2*d))/3
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 60

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

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 108

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 
*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
 

rule 170

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegerQ[m]
 

rule 219

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

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(-\frac {39936 \left (c^{4} \left (-\frac {d \,x^{3}}{8}+c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )+\frac {\left (\sqrt {c}\, d^{4} x^{12}+16 c^{\frac {3}{2}} d^{3} x^{9}+301 c^{\frac {5}{2}} d^{2} x^{6}+12206 c^{\frac {7}{2}} d \,x^{3}-145328 c^{\frac {9}{2}}\right ) \sqrt {d \,x^{3}+c}}{419328}\right )}{\sqrt {c}\, \left (-d^{5} x^{3}+8 c \,d^{4}\right )}\)	\(109\)
risch	\(\frac {2 \left (d^{3} x^{9}+24 c \,d^{2} x^{6}+493 c^{2} d \,x^{3}+16150 c^{3}\right ) \sqrt {d \,x^{3}+c}}{21 d^{4}}+\frac {576 c^{4} \left (-\frac {86 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 \sqrt {c}\, d}+\frac {8 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{3 d}\right )}{d^{3}}\)	\(131\)
default	\(\frac {d \left (\frac {2 d \,x^{9} \sqrt {d \,x^{3}+c}}{21}+\frac {16 c \,x^{6} \sqrt {d \,x^{3}+c}}{105}+\frac {2 c^{2} x^{3} \sqrt {d \,x^{3}+c}}{105 d}-\frac {4 c^{3} \sqrt {d \,x^{3}+c}}{105 d^{2}}\right )+\frac {32 c \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 d}}{d^{3}}-\frac {128 c^{2} \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{3 d^{4}}+\frac {512 c^{3} \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )\right )}{d^{4}}\)	\(207\)
elliptic	\(\frac {1536 c^{4} \sqrt {d \,x^{3}+c}}{d^{4} \left (-d \,x^{3}+8 c \right )}+\frac {2 x^{9} \sqrt {d \,x^{3}+c}}{21 d}+\frac {16 c \,x^{6} \sqrt {d \,x^{3}+c}}{7 d^{2}}+\frac {986 c^{2} x^{3} \sqrt {d \,x^{3}+c}}{21 d^{3}}+\frac {32300 c^{3} \sqrt {d \,x^{3}+c}}{21 d^{4}}+\frac {832 i c^{3} \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{d^{6}}\)	\(515\)

Input:

int(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
 

Output:

-39936*(c^4*(-1/8*d*x^3+c)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))+1/419328*( 
c^(1/2)*d^4*x^12+16*c^(3/2)*d^3*x^9+301*c^(5/2)*d^2*x^6+12206*c^(7/2)*d*x^ 
3-145328*c^(9/2))*(d*x^3+c)^(1/2))/c^(1/2)/(-d^5*x^3+8*c*d^4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.69 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\left [\frac {2 \, {\left (26208 \, {\left (c^{3} d x^{3} - 8 \, c^{4}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + {\left (d^{4} x^{12} + 16 \, c d^{3} x^{9} + 301 \, c^{2} d^{2} x^{6} + 12206 \, c^{3} d x^{3} - 145328 \, c^{4}\right )} \sqrt {d x^{3} + c}\right )}}{21 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}, \frac {2 \, {\left (52416 \, {\left (c^{3} d x^{3} - 8 \, c^{4}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (d^{4} x^{12} + 16 \, c d^{3} x^{9} + 301 \, c^{2} d^{2} x^{6} + 12206 \, c^{3} d x^{3} - 145328 \, c^{4}\right )} \sqrt {d x^{3} + c}\right )}}{21 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}\right ] \] Input:

integrate(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")
 

Output:

[2/21*(26208*(c^3*d*x^3 - 8*c^4)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sq 
rt(c) + 10*c)/(d*x^3 - 8*c)) + (d^4*x^12 + 16*c*d^3*x^9 + 301*c^2*d^2*x^6 
+ 12206*c^3*d*x^3 - 145328*c^4)*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4), 2/21 
*(52416*(c^3*d*x^3 - 8*c^4)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) + 
(d^4*x^12 + 16*c*d^3*x^9 + 301*c^2*d^2*x^6 + 12206*c^3*d*x^3 - 145328*c^4) 
*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**11*(d*x**3+c)**(3/2)/(-d*x**3+8*c)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.85 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \, {\left (26208 \, c^{\frac {7}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {7}{2}} + 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c + 448 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} + 15680 \, \sqrt {d x^{3} + c} c^{3} - \frac {16128 \, \sqrt {d x^{3} + c} c^{4}}{d x^{3} - 8 \, c}\right )}}{21 \, d^{4}} \] Input:

integrate(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
 

Output:

2/21*(26208*c^(7/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3 
*sqrt(c))) + (d*x^3 + c)^(7/2) + 21*(d*x^3 + c)^(5/2)*c + 448*(d*x^3 + c)^ 
(3/2)*c^2 + 15680*sqrt(d*x^3 + c)*c^3 - 16128*sqrt(d*x^3 + c)*c^4/(d*x^3 - 
 8*c))/d^4
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {4992 \, c^{4} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{4}} - \frac {1536 \, \sqrt {d x^{3} + c} c^{4}}{{\left (d x^{3} - 8 \, c\right )} d^{4}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {7}{2}} d^{24} + 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c d^{24} + 448 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} d^{24} + 15680 \, \sqrt {d x^{3} + c} c^{3} d^{24}\right )}}{21 \, d^{28}} \] Input:

integrate(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")
 

Output:

4992*c^4*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 1536*sqrt(d 
*x^3 + c)*c^4/((d*x^3 - 8*c)*d^4) + 2/21*((d*x^3 + c)^(7/2)*d^24 + 21*(d*x 
^3 + c)^(5/2)*c*d^24 + 448*(d*x^3 + c)^(3/2)*c^2*d^24 + 15680*sqrt(d*x^3 + 
 c)*c^3*d^24)/d^28
 

Mupad [B] (verification not implemented)

Time = 2.68 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2496\,c^{7/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^4}+\frac {32300\,c^3\,\sqrt {d\,x^3+c}}{21\,d^4}+\frac {2\,x^9\,\sqrt {d\,x^3+c}}{21\,d}+\frac {16\,c\,x^6\,\sqrt {d\,x^3+c}}{7\,d^2}+\frac {986\,c^2\,x^3\,\sqrt {d\,x^3+c}}{21\,d^3}+\frac {1536\,c^4\,\sqrt {d\,x^3+c}}{d^4\,\left (8\,c-d\,x^3\right )} \] Input:

int((x^11*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x)
 

Output:

(2496*c^(7/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^ 
3)))/d^4 + (32300*c^3*(c + d*x^3)^(1/2))/(21*d^4) + (2*x^9*(c + d*x^3)^(1/ 
2))/(21*d) + (16*c*x^6*(c + d*x^3)^(1/2))/(7*d^2) + (986*c^2*x^3*(c + d*x^ 
3)^(1/2))/(21*d^3) + (1536*c^4*(c + d*x^3)^(1/2))/(d^4*(8*c - d*x^3))
 

Reduce [F]

\[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {\frac {195296 \sqrt {d \,x^{3}+c}\, c^{4}}{105}-\frac {24412 \sqrt {d \,x^{3}+c}\, c^{3} d \,x^{3}}{21}-\frac {86 \sqrt {d \,x^{3}+c}\, c^{2} d^{2} x^{6}}{3}-\frac {32 \sqrt {d \,x^{3}+c}\, c \,d^{3} x^{9}}{21}-\frac {2 \sqrt {d \,x^{3}+c}\, d^{4} x^{12}}{21}+\frac {1617408 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{5} d^{2}}{5}-\frac {202176 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{4} d^{3} x^{3}}{5}}{d^{4} \left (-d \,x^{3}+8 c \right )} \] Input:

int(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x)
 

Output:

(2*(97648*sqrt(c + d*x**3)*c**4 - 61030*sqrt(c + d*x**3)*c**3*d*x**3 - 150 
5*sqrt(c + d*x**3)*c**2*d**2*x**6 - 80*sqrt(c + d*x**3)*c*d**3*x**9 - 5*sq 
rt(c + d*x**3)*d**4*x**12 + 16982784*int((sqrt(c + d*x**3)*x**5)/(64*c**3 
+ 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c**5*d**2 - 2122848*int( 
(sqrt(c + d*x**3)*x**5)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3* 
x**9),x)*c**4*d**3*x**3))/(105*d**4*(8*c - d*x**3))