\(\int \frac {(c+d x^3)^{3/2}}{x^4 (8 c-d x^3)^2} \, dx\) [589]

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {5 d \sqrt {c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}+\frac {3 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{128 c^{3/2}}-\frac {7 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{3/2}} \] Output:

5/96*d*(d*x^3+c)^(1/2)/c/(-d*x^3+8*c)-1/24*(d*x^3+c)^(1/2)/x^3/(-d*x^3+8*c 
)+3/128*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)-7/384*d*arctanh((d* 
x^3+c)^(1/2)/c^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {4 \sqrt {c} \left (4 c-5 d x^3\right ) \sqrt {c+d x^3}}{-8 c x^3+d x^6}+9 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-7 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{3/2}} \] Input:

Integrate[(c + d*x^3)^(3/2)/(x^4*(8*c - d*x^3)^2),x]
 

Output:

((4*Sqrt[c]*(4*c - 5*d*x^3)*Sqrt[c + d*x^3])/(-8*c*x^3 + d*x^6) + 9*d*ArcT 
anh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 7*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/( 
384*c^(3/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 109, 27, 168, 27, 174, 73, 219, 221}

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.

\(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{3} \left (-\frac {\int -\frac {c d \left (19 d x^3+28 c\right )}{2 x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c}-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \int \frac {19 d x^3+28 c}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}-\frac {\int -\frac {18 c d \left (5 d x^3+14 c\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{72 c^2 d}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\int \frac {5 d x^3+14 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {7}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {27}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {27}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {7 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {7 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {9 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {9 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\)

Input:

Int[(c + d*x^3)^(3/2)/(x^4*(8*c - d*x^3)^2),x]
 

Output:

(-1/8*Sqrt[c + d*x^3]/(x^3*(8*c - d*x^3)) + (d*((5*Sqrt[c + d*x^3])/(2*c*( 
8*c - d*x^3)) + ((9*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2*Sqrt[c]) - (7 
*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c]))/(4*c)))/16)/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 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 109

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

rule 168

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

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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 221

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

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 = 1.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {d \left (-\frac {\sqrt {d \,x^{3}+c}}{d \,x^{3}}-\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{2 \sqrt {c}}+\frac {9 \sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {9 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{192 c}\)	\(87\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{192 c \,x^{3}}-\frac {d \left (\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}-6 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )\right )}{128 c}\)	\(114\)
default	\(\frac {-\frac {c \sqrt {d \,x^{3}+c}}{3 x^{3}}+\frac {2 d \sqrt {d \,x^{3}+c}}{3}-\sqrt {c}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{64 c^{2}}+\frac {d \left (\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{256 c^{3}}+\frac {d \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )\right )}{64 c^{2}}+\frac {d \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{1152 c^{3}}\)	\(219\)
elliptic	\(\text {Expression too large to display}\)	\(1550\)

Input:

int((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/192*d/c*(-(d*x^3+c)^(1/2)/d/x^3-7/2*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^( 
1/2)+9*(d*x^3+c)^(1/2)/(-d*x^3+8*c)+9/2*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2 
))/c^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.26 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\left [\frac {9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 8 \, {\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{768 \, {\left (c^{2} d x^{6} - 8 \, c^{3} x^{3}\right )}}, -\frac {9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - 7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 4 \, {\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{384 \, {\left (c^{2} d x^{6} - 8 \, c^{3} x^{3}\right )}}\right ] \] Input:

integrate((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x, algorithm="fricas")
 

Output:

[1/768*(9*(d^2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqr 
t(c) + 10*c)/(d*x^3 - 8*c)) + 7*(d^2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 - 
 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 8*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 
+ c))/(c^2*d*x^6 - 8*c^3*x^3), -1/384*(9*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*ar 
ctan(3*sqrt(-c)/sqrt(d*x^3 + c)) - 7*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan 
(sqrt(-c)/sqrt(d*x^3 + c)) + 4*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/(c^2*d 
*x^6 - 8*c^3*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\text {Timed out} \] Input:

integrate((d*x**3+c)**(3/2)/x**4/(-d*x**3+8*c)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{4}} \,d x } \] Input:

integrate((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x, algorithm="maxima")
 

Output:

integrate((d*x^3 + c)^(3/2)/((d*x^3 - 8*c)^2*x^4), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {7 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{384 \, \sqrt {-c} c} - \frac {3 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{128 \, \sqrt {-c} c} - \frac {5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d - 9 \, \sqrt {d x^{3} + c} c d}{96 \, {\left ({\left (d x^{3} + c\right )}^{2} - 10 \, {\left (d x^{3} + c\right )} c + 9 \, c^{2}\right )} c} \] Input:

integrate((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x, algorithm="giac")
 

Output:

7/384*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 3/128*d*arctan(1/3 
*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 1/96*(5*(d*x^3 + c)^(3/2)*d - 9* 
sqrt(d*x^3 + c)*c*d)/(((d*x^3 + c)^2 - 10*(d*x^3 + c)*c + 9*c^2)*c)
 

Mupad [B] (verification not implemented)

Time = 2.84 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {9\,d\,\sqrt {d\,x^3+c}}{32}-\frac {5\,d\,{\left (d\,x^3+c\right )}^{3/2}}{32\,c}}{3\,{\left (d\,x^3+c\right )}^2-30\,c\,\left (d\,x^3+c\right )+27\,c^2}+\frac {d\,\left (\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{\sqrt {c^3}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^3}}\right )\,9{}\mathrm {i}}{7}\right )\,7{}\mathrm {i}}{384\,\sqrt {c^3}} \] Input:

int((c + d*x^3)^(3/2)/(x^4*(8*c - d*x^3)^2),x)
 

Output:

((9*d*(c + d*x^3)^(1/2))/32 - (5*d*(c + d*x^3)^(3/2))/(32*c))/(3*(c + d*x^ 
3)^2 - 30*c*(c + d*x^3) + 27*c^2) + (d*(atanh((c*(c + d*x^3)^(1/2))/(c^3)^ 
(1/2))*1i - (atanh((c*(c + d*x^3)^(1/2))/(3*(c^3)^(1/2)))*9i)/7)*7i)/(384* 
(c^3)^(1/2))
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}\, c +8 \sqrt {d \,x^{3}+c}\, d \,x^{3}+672 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{10}-15 c \,d^{2} x^{7}+48 c^{2} d \,x^{4}+64 c^{3} x}d x \right ) c^{3} d \,x^{3}-84 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{10}-15 c \,d^{2} x^{7}+48 c^{2} d \,x^{4}+64 c^{3} x}d x \right ) c^{2} d^{2} x^{6}-96 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{3} x^{3}+12 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) d^{4} x^{6}-504 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{2} d^{2} x^{3}+63 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{3} x^{6}}{48 c \,x^{3} \left (-d \,x^{3}+8 c \right )} \] Input:

int((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x)
 

Output:

( - 2*sqrt(c + d*x**3)*c + 8*sqrt(c + d*x**3)*d*x**3 + 672*int(sqrt(c + d* 
x**3)/(64*c**3*x + 48*c**2*d*x**4 - 15*c*d**2*x**7 + d**3*x**10),x)*c**3*d 
*x**3 - 84*int(sqrt(c + d*x**3)/(64*c**3*x + 48*c**2*d*x**4 - 15*c*d**2*x* 
*7 + d**3*x**10),x)*c**2*d**2*x**6 - 96*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*d**3*x**3 + 12*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)*d**4*x**6 - 504*int((sqrt(c + d*x**3)*x**2)/(64*c**3 + 48*c**2*d* 
x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c**2*d**2*x**3 + 63*int((sqrt(c + d* 
x**3)*x**2)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c*d 
**3*x**6)/(48*c*x**3*(8*c - d*x**3))