\(\int \frac {\sqrt {c+d x^3}}{x^7 (8 c-d x^3)^2} \, dx\) [577]

Optimal result

Integrand size = 27, antiderivative size = 164 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {5 d^2 \sqrt {c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}-\frac {7 d \sqrt {c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}+\frac {23 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{18432 c^{7/2}}-\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{7/2}} \] Output:

5/1536*d^2*(d*x^3+c)^(1/2)/c^3/(-d*x^3+8*c)-1/48*(d*x^3+c)^(1/2)/c/x^6/(-d 
*x^3+8*c)-7/384*d*(d*x^3+c)^(1/2)/c^2/x^3/(-d*x^3+8*c)+23/18432*d^2*arctan 
h(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(7/2)-1/2048*d^2*arctanh((d*x^3+c)^(1/2)/ 
c^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {12 \sqrt {c} \sqrt {c+d x^3} \left (32 c^2+28 c d x^3-5 d^2 x^6\right )}{-8 c x^6+d x^9}+23 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-9 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{18432 c^{7/2}} \] Input:

Integrate[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)^2),x]
 

Output:

((12*Sqrt[c]*Sqrt[c + d*x^3]*(32*c^2 + 28*c*d*x^3 - 5*d^2*x^6))/(-8*c*x^6 
+ d*x^9) + 23*d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 9*d^2*ArcTanh[Sqr 
t[c + d*x^3]/Sqrt[c]])/(18432*c^(7/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {948, 110, 27, 168, 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.

\(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 110

\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {d \left (5 d x^3+14 c\right )}{2 x^6 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{16 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {d \int \frac {5 d x^3+14 c}{x^6 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (\frac {d \left (-\frac {\int -\frac {3 c d \left (7 d x^3+4 c\right )}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c^2}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \int \frac {7 d x^3+4 c}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \left (\frac {5 \sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}-\frac {\int -\frac {6 c d \left (5 d x^3+6 c\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{72 c^2 d}\right )}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \left (\frac {\int \frac {5 d x^3+6 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{12 c}+\frac {5 \sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \left (\frac {\frac {3}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {23}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{12 c}+\frac {5 \sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \left (\frac {\frac {23}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {3 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}}{12 c}+\frac {5 \sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \left (\frac {\frac {3 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {23 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}}{12 c}+\frac {5 \sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 d \left (\frac {\frac {23 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{12 c}+\frac {5 \sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {7 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6 \left (8 c-d x^3\right )}\right )\)

Input:

Int[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)^2),x]
 

Output:

(-1/16*Sqrt[c + d*x^3]/(c*x^6*(8*c - d*x^3)) + (d*((-7*Sqrt[c + d*x^3])/(4 
*c*x^3*(8*c - d*x^3)) + (3*d*((5*Sqrt[c + d*x^3])/(6*c*(8*c - d*x^3)) + (( 
23*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(6*Sqrt[c]) - (3*ArcTanh[Sqrt[c + 
 d*x^3]/Sqrt[c]])/(2*Sqrt[c]))/(12*c)))/(8*c)))/(32*c))/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 110

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 + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (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 = 1.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.54

method	result	size
pseudoelliptic	\(\frac {d^{2} \left (-\frac {\left (d \,x^{3}+c \right )^{\frac {3}{2}}}{d^{2} x^{6}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{16 \sqrt {c}}+\frac {\sqrt {d \,x^{3}+c}}{-4 d \,x^{3}+32 c}+\frac {23 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{48 \sqrt {c}}\right )}{384 c^{3}}\)	\(89\)
risch	\(-\frac {\left (d \,x^{3}+c \right )^{\frac {3}{2}}}{384 c^{3} x^{6}}-\frac {d^{2} \left (\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{8 \sqrt {c}}-\frac {19 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{72 \sqrt {c}}-\frac {c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{6}\right )}{256 c^{3}}\)	\(116\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{6 x^{6}}-\frac {d \sqrt {d \,x^{3}+c}}{12 c \,x^{3}}+\frac {d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{12 c^{\frac {3}{2}}}}{64 c^{2}}+\frac {d \left (-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}\right )}{256 c^{3}}+\frac {3 d^{2} \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{4096 c^{4}}+\frac {d^{2} \left (\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 \sqrt {c}}\right )}{1536 c^{3}}+\frac {d^{2} \left (-2 \sqrt {d \,x^{3}+c}+6 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )\right )}{4096 c^{4}}\)	\(233\)
elliptic	\(\text {Expression too large to display}\)	\(1580\)

Input:

int((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/384*d^2/c^3*(-(d*x^3+c)^(3/2)/d^2/x^6-3/16*arctanh((d*x^3+c)^(1/2)/c^(1/ 
2))/c^(1/2)+1/4*(d*x^3+c)^(1/2)/(-d*x^3+8*c)+23/48*arctanh(1/3*(d*x^3+c)^( 
1/2)/c^(1/2))/c^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\left [\frac {23 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 9 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, {\left (5 \, c d^{2} x^{6} - 28 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt {d x^{3} + c}}{36864 \, {\left (c^{4} d x^{9} - 8 \, c^{5} x^{6}\right )}}, -\frac {23 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - 9 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 12 \, {\left (5 \, c d^{2} x^{6} - 28 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt {d x^{3} + c}}{18432 \, {\left (c^{4} d x^{9} - 8 \, c^{5} x^{6}\right )}}\right ] \] Input:

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c)^2,x, algorithm="fricas")
 

Output:

[1/36864*(23*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c 
)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 9*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(c)*log(( 
d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*(5*c*d^2*x^6 - 28*c^2*d 
*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^4*d*x^9 - 8*c^5*x^6), -1/18432*(23*(d^3 
*x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) - 9*(d^3*x 
^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^3 + c)) + 12*(5*c*d^2* 
x^6 - 28*c^2*d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^4*d*x^9 - 8*c^5*x^6)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\text {Timed out} \] Input:

integrate((d*x**3+c)**(1/2)/x**7/(-d*x**3+8*c)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{7}} \,d x } \] Input:

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c)^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)^2*x^7), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{2048 \, \sqrt {-c} c^{3}} - \frac {23 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{18432 \, \sqrt {-c} c^{3}} - \frac {\sqrt {d x^{3} + c} d^{2}}{1536 \, {\left (d x^{3} - 8 \, c\right )} c^{3}} - \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{384 \, c^{3} x^{6}} \] Input:

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c)^2,x, algorithm="giac")
 

Output:

1/2048*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 23/18432*d^2* 
arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 1/1536*sqrt(d*x^3 + 
c)*d^2/((d*x^3 - 8*c)*c^3) - 1/384*(d*x^3 + c)^(3/2)/(c^3*x^6)
 

Mupad [B] (verification not implemented)

Time = 2.96 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {d^2\,\sqrt {d\,x^3+c}}{512\,c}-\frac {19\,d^2\,{\left (d\,x^3+c\right )}^{3/2}}{256\,c^2}+\frac {5\,d^2\,{\left (d\,x^3+c\right )}^{5/2}}{512\,c^3}}{33\,c\,{\left (d\,x^3+c\right )}^2-57\,c^2\,\left (d\,x^3+c\right )-3\,{\left (d\,x^3+c\right )}^3+27\,c^3}+\frac {d^2\,\left (\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{\sqrt {c^7}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^7}}\right )\,23{}\mathrm {i}}{9}\right )\,1{}\mathrm {i}}{2048\,\sqrt {c^7}} \] Input:

int((c + d*x^3)^(1/2)/(x^7*(8*c - d*x^3)^2),x)
 

Output:

((d^2*(c + d*x^3)^(1/2))/(512*c) - (19*d^2*(c + d*x^3)^(3/2))/(256*c^2) + 
(5*d^2*(c + d*x^3)^(5/2))/(512*c^3))/(33*c*(c + d*x^3)^2 - 57*c^2*(c + d*x 
^3) - 3*(c + d*x^3)^3 + 27*c^3) + (d^2*(atanh((c^3*(c + d*x^3)^(1/2))/(c^7 
)^(1/2))*1i - (atanh((c^3*(c + d*x^3)^(1/2))/(3*(c^7)^(1/2)))*23i)/9)*1i)/ 
(2048*(c^7)^(1/2))
 

Reduce [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d^{2} x^{13}-16 c d \,x^{10}+64 c^{2} x^{7}}d x \] Input:

int((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c)^2,x)
 

Output:

int(sqrt(c + d*x**3)/(64*c**2*x**7 - 16*c*d*x**10 + d**2*x**13),x)