\(\int \frac {(c+d x^3)^{3/2}}{x^7 (8 c-d x^3)^2} \, dx\) [590]

Optimal result

Integrand size = 27, antiderivative size = 161 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}+\frac {15 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2048 c^{5/2}}-\frac {17 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{5/2}} \] Output:

7/512*d^2*(d*x^3+c)^(1/2)/c^2/(-d*x^3+8*c)-1/48*(d*x^3+c)^(1/2)/x^6/(-d*x^ 
3+8*c)-23/384*d*(d*x^3+c)^(1/2)/c/x^3/(-d*x^3+8*c)+15/2048*d^2*arctanh(1/3 
*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-17/2048*d^2*arctanh((d*x^3+c)^(1/2)/c^(1 
/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {4 \sqrt {c} \sqrt {c+d x^3} \left (32 c^2+92 c d x^3-21 d^2 x^6\right )}{-8 c x^6+d x^9}+45 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-51 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{6144 c^{5/2}} \] Input:

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

Output:

((4*Sqrt[c]*Sqrt[c + d*x^3]*(32*c^2 + 92*c*d*x^3 - 21*d^2*x^6))/(-8*c*x^6 
+ d*x^9) + 45*d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 51*d^2*ArcTanh[Sq 
rt[c + d*x^3]/Sqrt[c]])/(6144*c^(5/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {948, 109, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{3} \left (-\frac {\int -\frac {c d \left (37 d x^3+46 c\right )}{2 x^6 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{16 c}-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \int \frac {37 d x^3+46 c}{x^6 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (-\frac {\int -\frac {3 c d \left (23 d x^3+68 c\right )}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c^2}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \int \frac {23 d x^3+68 c}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \left (\frac {7 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}-\frac {\int -\frac {18 c d \left (7 d x^3+34 c\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{72 c^2 d}\right )}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \left (\frac {\int \frac {7 d x^3+34 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}+\frac {7 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \left (\frac {\frac {17}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {45}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}+\frac {7 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \left (\frac {\frac {45}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {17 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}}{4 c}+\frac {7 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \left (\frac {\frac {17 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {15 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}+\frac {7 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {1}{32} d \left (\frac {3 d \left (\frac {\frac {15 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {17 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}+\frac {7 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )}{8 c}-\frac {23 \sqrt {c+d x^3}}{4 c x^3 \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{16 x^6 \left (8 c-d x^3\right )}\right )\)

Input:

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

Output:

(-1/16*Sqrt[c + d*x^3]/(x^6*(8*c - d*x^3)) + (d*((-23*Sqrt[c + d*x^3])/(4* 
c*x^3*(8*c - d*x^3)) + (3*d*((7*Sqrt[c + d*x^3])/(2*c*(8*c - d*x^3)) + ((1 
5*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2*Sqrt[c]) - (17*ArcTanh[Sqrt[c + 
 d*x^3]/Sqrt[c]])/(2*Sqrt[c]))/(4*c)))/(8*c)))/32)/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 = 1.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(\frac {d^{2} \left (-\frac {\sqrt {d \,x^{3}+c}\, \left (3 d \,x^{3}+c \right )}{d^{2} x^{6}}-\frac {51 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{16 \sqrt {c}}+\frac {9 \sqrt {d \,x^{3}+c}}{4 \left (-d \,x^{3}+8 c \right )}+\frac {45 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{16 \sqrt {c}}\right )}{384 c^{2}}\)	\(97\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}\, \left (3 d \,x^{3}+c \right )}{384 c^{2} x^{6}}-\frac {3 d^{2} \left (\frac {17 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{24 \sqrt {c}}-\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{24 \sqrt {c}}-\frac {c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{2}\right )}{256 c^{2}}\)	\(124\)
default	\(\frac {-\frac {c \sqrt {d \,x^{3}+c}}{6 x^{6}}-\frac {5 d \sqrt {d \,x^{3}+c}}{12 x^{3}}-\frac {d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{4 \sqrt {c}}}{64 c^{2}}+\frac {d \left (-\frac {c \sqrt {d \,x^{3}+c}}{3 x^{3}}+\frac {2 d \sqrt {d \,x^{3}+c}}{3}-\sqrt {c}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )\right )}{256 c^{3}}+\frac {3 d^{2} \left (\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{4096 c^{4}}+\frac {d^{2} \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )\right )}{512 c^{3}}+\frac {d^{2} \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{6144 c^{4}}\)	\(284\)
elliptic	\(\text {Expression too large to display}\)	\(1580\)

Input:

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

Output:

1/384*d^2/c^2*(-(d*x^3+c)^(1/2)*(3*d*x^3+c)/d^2/x^6-51/16*arctanh((d*x^3+c 
)^(1/2)/c^(1/2))/c^(1/2)+9/4*(d*x^3+c)^(1/2)/(-d*x^3+8*c)+45/16*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 = 304, normalized size of antiderivative = 1.89 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\left [\frac {45 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 51 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 8 \, {\left (21 \, c d^{2} x^{6} - 92 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt {d x^{3} + c}}{12288 \, {\left (c^{3} d x^{9} - 8 \, c^{4} x^{6}\right )}}, -\frac {45 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - 51 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 4 \, {\left (21 \, c d^{2} x^{6} - 92 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt {d x^{3} + c}}{6144 \, {\left (c^{3} d x^{9} - 8 \, c^{4} x^{6}\right )}}\right ] \] Input:

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

Output:

[1/12288*(45*(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)) + 51*(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) - 8*(21*c*d^2*x^6 - 92*c^2* 
d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^3*d*x^9 - 8*c^4*x^6), -1/6144*(45*(d^3 
*x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) - 51*(d^3* 
x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^3 + c)) + 4*(21*c*d^2 
*x^6 - 92*c^2*d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^3*d*x^9 - 8*c^4*x^6)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {17 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{2048 \, \sqrt {-c} c^{2}} - \frac {15 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{2048 \, \sqrt {-c} c^{2}} - \frac {3 \, \sqrt {d x^{3} + c} d^{2}}{512 \, {\left (d x^{3} - 8 \, c\right )} c^{2}} - \frac {3 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} - 2 \, \sqrt {d x^{3} + c} c d^{2}}{384 \, c^{2} d^{2} x^{6}} \] Input:

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

Output:

17/2048*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 15/2048*d^2* 
arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 3/512*sqrt(d*x^3 + c 
)*d^2/((d*x^3 - 8*c)*c^2) - 1/384*(3*(d*x^3 + c)^(3/2)*d^2 - 2*sqrt(d*x^3 
+ c)*c*d^2)/(c^2*d^2*x^6)
 

Mupad [B] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {81\,d^2\,\sqrt {d\,x^3+c}}{512}-\frac {67\,d^2\,{\left (d\,x^3+c\right )}^{3/2}}{256\,c}+\frac {21\,d^2\,{\left (d\,x^3+c\right )}^{5/2}}{512\,c^2}}{33\,c\,{\left (d\,x^3+c\right )}^2-57\,c^2\,\left (d\,x^3+c\right )-3\,{\left (d\,x^3+c\right )}^3+27\,c^3}+\frac {d^2\,\left (\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )\,15{}\mathrm {i}}{17}\right )\,17{}\mathrm {i}}{2048\,\sqrt {c^5}} \] Input:

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

Output:

((81*d^2*(c + d*x^3)^(1/2))/512 - (67*d^2*(c + d*x^3)^(3/2))/(256*c) + (21 
*d^2*(c + d*x^3)^(5/2))/(512*c^2))/(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^2*(c + d*x^3)^(1/2))/(c^5)^ 
(1/2))*1i - (atanh((c^2*(c + d*x^3)^(1/2))/(3*(c^5)^(1/2)))*15i)/17)*17i)/ 
(2048*(c^5)^(1/2))
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}+1104 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{13}-15 c \,d^{2} x^{10}+48 c^{2} d \,x^{7}+64 c^{3} x^{4}}d x \right ) c^{2} d \,x^{6}-138 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{13}-15 c \,d^{2} x^{10}+48 c^{2} d \,x^{7}+64 c^{3} x^{4}}d x \right ) c \,d^{2} x^{9}+888 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{10}-15 c \,d^{2} x^{7}+48 c^{2} d \,x^{4}+64 c^{3} x}d x \right ) c \,d^{2} x^{6}-111 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{10}-15 c \,d^{2} x^{7}+48 c^{2} d \,x^{4}+64 c^{3} x}d x \right ) d^{3} x^{9}}{96 x^{6} \left (-d \,x^{3}+8 c \right )} \] Input:

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

Output:

( - 2*sqrt(c + d*x**3) + 1104*int(sqrt(c + d*x**3)/(64*c**3*x**4 + 48*c**2 
*d*x**7 - 15*c*d**2*x**10 + d**3*x**13),x)*c**2*d*x**6 - 138*int(sqrt(c + 
d*x**3)/(64*c**3*x**4 + 48*c**2*d*x**7 - 15*c*d**2*x**10 + d**3*x**13),x)* 
c*d**2*x**9 + 888*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*d**2*x**6 - 111*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)*d**3*x**9)/(96*x**6* 
(8*c - d*x**3))