\(\int \frac {x^8 (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\) [585]

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {416 c^2 \sqrt {c+d x^3}}{3 d^3}+\frac {192 c^3 \sqrt {c+d x^3}}{d^3 \left (8 c-d x^3\right )}+\frac {32 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {480 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3} \] Output:

416/3*c^2*(d*x^3+c)^(1/2)/d^3+192*c^3*(d*x^3+c)^(1/2)/d^3/(-d*x^3+8*c)+32/ 
9*c*(d*x^3+c)^(3/2)/d^3+2/15*(d*x^3+c)^(5/2)/d^3-480*c^(5/2)*arctanh(1/3*( 
d*x^3+c)^(1/2)/c^(1/2))/d^3
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.77 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \sqrt {c+d x^3} \left (-29944 c^3+2515 c^2 d x^3+62 c d^2 x^6+3 d^3 x^9\right )}{45 d^3 \left (-8 c+d x^3\right )}-\frac {480 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3} \] Input:

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

Output:

(2*Sqrt[c + d*x^3]*(-29944*c^3 + 2515*c^2*d*x^3 + 62*c*d^2*x^6 + 3*d^3*x^9 
))/(45*d^3*(-8*c + d*x^3)) - (480*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[ 
c])])/d^3
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {948, 100, 27, 90, 60, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {\int \frac {3 c d \left (d x^3+c\right )^{3/2} \left (3 d x^3+56 c\right )}{8 c-d x^3}dx^3}{9 c d^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {\int \frac {\left (d x^3+c\right )^{3/2} \left (3 d x^3+56 c\right )}{8 c-d x^3}dx^3}{3 d^2}\right )\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {80 c \int \frac {\left (d x^3+c\right )^{3/2}}{8 c-d x^3}dx^3-\frac {6 \left (c+d x^3\right )^{5/2}}{5 d}}{3 d^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {80 c \left (9 c \int \frac {\sqrt {d x^3+c}}{8 c-d x^3}dx^3-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )-\frac {6 \left (c+d x^3\right )^{5/2}}{5 d}}{3 d^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {80 c \left (9 c \left (9 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )-\frac {6 \left (c+d x^3\right )^{5/2}}{5 d}}{3 d^2}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {80 c \left (9 c \left (\frac {18 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )-\frac {6 \left (c+d x^3\right )^{5/2}}{5 d}}{3 d^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{5/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {80 c \left (9 c \left (\frac {6 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )-\frac {6 \left (c+d x^3\right )^{5/2}}{5 d}}{3 d^2}\right )\)

Input:

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

Output:

((64*c*(c + d*x^3)^(5/2))/(9*d^3*(8*c - d*x^3)) - ((-6*(c + d*x^3)^(5/2))/ 
(5*d) + 80*c*((-2*(c + d*x^3)^(3/2))/(3*d) + 9*c*((-2*Sqrt[c + d*x^3])/d + 
 (6*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d)))/(3*d^2))/3
 

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 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 90

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

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 = 1.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(-\frac {3840 \left (c^{3} \left (-\frac {d \,x^{3}}{8}+c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )+\frac {\sqrt {d \,x^{3}+c}\, \left (\sqrt {c}\, d^{3} x^{9}+\frac {62 c^{\frac {3}{2}} d^{2} x^{6}}{3}+\frac {2515 c^{\frac {5}{2}} d \,x^{3}}{3}-\frac {29944 c^{\frac {7}{2}}}{3}\right )}{28800}\right )}{\sqrt {c}\, \left (-d^{4} x^{3}+8 c \,d^{3}\right )}\)	\(98\)
risch	\(\frac {2 \left (3 d^{2} x^{6}+86 c d \,x^{3}+3203 c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d^{3}}+\frac {144 c^{3} \left (-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 \sqrt {c}\, d}+\frac {4 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{3 d}\right )}{d^{2}}\)	\(121\)
default	\(\frac {2 \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 d^{3}}-\frac {32 c \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{9 d^{3}}+\frac {64 c^{2} \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )\right )}{d^{3}}\)	\(129\)
elliptic	\(\frac {192 c^{3} \sqrt {d \,x^{3}+c}}{d^{3} \left (-d \,x^{3}+8 c \right )}+\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15 d}+\frac {172 c \,x^{3} \sqrt {d \,x^{3}+c}}{45 d^{2}}+\frac {6406 c^{2} \sqrt {d \,x^{3}+c}}{45 d^{3}}+\frac {80 i c^{2} \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^{5}}\)	\(495\)

Input:

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

Output:

-3840/c^(1/2)*(c^3*(-1/8*d*x^3+c)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))+1/2 
8800*(d*x^3+c)^(1/2)*(c^(1/2)*d^3*x^9+62/3*c^(3/2)*d^2*x^6+2515/3*c^(5/2)* 
d*x^3-29944/3*c^(7/2)))/(-d^4*x^3+8*c*d^3)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.79 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\left [\frac {2 \, {\left (5400 \, {\left (c^{2} d x^{3} - 8 \, c^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + {\left (3 \, d^{3} x^{9} + 62 \, c d^{2} x^{6} + 2515 \, c^{2} d x^{3} - 29944 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, {\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}, \frac {2 \, {\left (10800 \, {\left (c^{2} d x^{3} - 8 \, c^{3}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (3 \, d^{3} x^{9} + 62 \, c d^{2} x^{6} + 2515 \, c^{2} d x^{3} - 29944 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, {\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}\right ] \] Input:

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

Output:

[2/45*(5400*(c^2*d*x^3 - 8*c^3)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqr 
t(c) + 10*c)/(d*x^3 - 8*c)) + (3*d^3*x^9 + 62*c*d^2*x^6 + 2515*c^2*d*x^3 - 
 29944*c^3)*sqrt(d*x^3 + c))/(d^4*x^3 - 8*c*d^3), 2/45*(10800*(c^2*d*x^3 - 
 8*c^3)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) + (3*d^3*x^9 + 62*c*d^ 
2*x^6 + 2515*c^2*d*x^3 - 29944*c^3)*sqrt(d*x^3 + c))/(d^4*x^3 - 8*c*d^3)]
 

Sympy [F(-1)]

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

integrate(x**8*(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 = 107, normalized size of antiderivative = 0.88 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \, {\left (5400 \, c^{\frac {5}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} + 80 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c + 3120 \, \sqrt {d x^{3} + c} c^{2} - \frac {4320 \, \sqrt {d x^{3} + c} c^{3}}{d x^{3} - 8 \, c}\right )}}{45 \, d^{3}} \] Input:

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

Output:

2/45*(5400*c^(5/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3* 
sqrt(c))) + 3*(d*x^3 + c)^(5/2) + 80*(d*x^3 + c)^(3/2)*c + 3120*sqrt(d*x^3 
 + c)*c^2 - 4320*sqrt(d*x^3 + c)*c^3/(d*x^3 - 8*c))/d^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {480 \, c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{3}} - \frac {192 \, \sqrt {d x^{3} + c} c^{3}}{{\left (d x^{3} - 8 \, c\right )} d^{3}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} d^{12} + 80 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c d^{12} + 3120 \, \sqrt {d x^{3} + c} c^{2} d^{12}\right )}}{45 \, d^{15}} \] Input:

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

Output:

480*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 192*sqrt(d*x 
^3 + c)*c^3/((d*x^3 - 8*c)*d^3) + 2/45*(3*(d*x^3 + c)^(5/2)*d^12 + 80*(d*x 
^3 + c)^(3/2)*c*d^12 + 3120*sqrt(d*x^3 + c)*c^2*d^12)/d^15
 

Mupad [B] (verification not implemented)

Time = 2.61 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {240\,c^{5/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^3}+\frac {6406\,c^2\,\sqrt {d\,x^3+c}}{45\,d^3}+\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15\,d}+\frac {172\,c\,x^3\,\sqrt {d\,x^3+c}}{45\,d^2}+\frac {192\,c^3\,\sqrt {d\,x^3+c}}{d^3\,\left (8\,c-d\,x^3\right )} \] Input:

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

Output:

(240*c^(5/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3 
)))/d^3 + (6406*c^2*(c + d*x^3)^(1/2))/(45*d^3) + (2*x^6*(c + d*x^3)^(1/2) 
)/(15*d) + (172*c*x^3*(c + d*x^3)^(1/2))/(45*d^2) + (192*c^3*(c + d*x^3)^( 
1/2))/(d^3*(8*c - d*x^3))
 

Reduce [F]

\[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {\frac {8048 \sqrt {d \,x^{3}+c}\, c^{3}}{45}-\frac {1006 \sqrt {d \,x^{3}+c}\, c^{2} d \,x^{3}}{9}-\frac {124 \sqrt {d \,x^{3}+c}\, c \,d^{2} x^{6}}{45}-\frac {2 \sqrt {d \,x^{3}+c}\, d^{3} x^{9}}{15}+31104 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{4} d^{2}-3888 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{3} d^{3} x^{3}}{d^{3} \left (-d \,x^{3}+8 c \right )} \] Input:

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

Output:

(2*(4024*sqrt(c + d*x**3)*c**3 - 2515*sqrt(c + d*x**3)*c**2*d*x**3 - 62*sq 
rt(c + d*x**3)*c*d**2*x**6 - 3*sqrt(c + d*x**3)*d**3*x**9 + 699840*int((sq 
rt(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**2 - 87480*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**3*d**3*x**3))/(45*d**3*(8*c - d*x* 
*3))