\(\int \frac {(c+d x^3)^{3/2}}{x (8 c-d x^3)^2} \, dx\) [588]

Optimal result

Integrand size = 27, antiderivative size = 85 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {3 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{32 \sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 \sqrt {c}} \] Output:

3*(d*x^3+c)^(1/2)/(-8*d*x^3+64*c)-3/32*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2) 
)/c^(1/2)-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {3 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{32 \sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 \sqrt {c}} \] Input:

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

Output:

(3*Sqrt[c + d*x^3])/(8*(8*c - d*x^3)) - (3*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt 
[c])])/(32*Sqrt[c]) - ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(96*Sqrt[c])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {948, 109, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((9*Sqrt[c + d*x^3])/(8*(8*c - d*x^3)) + ((-9*ArcTanh[Sqrt[c + d*x^3]/(3*S 
qrt[c])])/(2*Sqrt[c]) - ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(2*Sqrt[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 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 = 1.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {3 \sqrt {d \,x^{3}+c}}{-8 d \,x^{3}+64 c}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{32 \sqrt {c}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{96 \sqrt {c}}\)	\(64\)
default	\(\frac {\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}}{64 c^{2}}+\frac {\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )}{8 c}+\frac {81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}}{288 c^{2}}\)	\(163\)
elliptic	\(\text {Expression too large to display}\)	\(1525\)

Input:

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

Output:

3*(d*x^3+c)^(1/2)/(-8*d*x^3+64*c)-3/32*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2) 
)/c^(1/2)-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.52 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx=\left [\frac {9 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 72 \, \sqrt {d x^{3} + c} c}{192 \, {\left (c d x^{3} - 8 \, c^{2}\right )}}, \frac {9 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - 36 \, \sqrt {d x^{3} + c} c}{96 \, {\left (c d x^{3} - 8 \, c^{2}\right )}}\right ] \] Input:

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

Output:

[1/192*(9*(d*x^3 - 8*c)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 1 
0*c)/(d*x^3 - 8*c)) + (d*x^3 - 8*c)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c) 
*sqrt(c) + 2*c)/x^3) - 72*sqrt(d*x^3 + c)*c)/(c*d*x^3 - 8*c^2), 1/96*(9*(d 
*x^3 - 8*c)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) + (d*x^3 - 8*c)*sq 
rt(-c)*arctan(sqrt(-c)/sqrt(d*x^3 + c)) - 36*sqrt(d*x^3 + c)*c)/(c*d*x^3 - 
 8*c^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c}} + \frac {3 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{32 \, \sqrt {-c}} - \frac {3 \, \sqrt {d x^{3} + c}}{8 \, {\left (d x^{3} - 8 \, c\right )}} \] Input:

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

Output:

1/96*arctan(sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c) + 3/32*arctan(1/3*sqrt(d*x^ 
3 + c)/sqrt(-c))/sqrt(-c) - 3/8*sqrt(d*x^3 + c)/(d*x^3 - 8*c)
 

Mupad [B] (verification not implemented)

Time = 3.44 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.19 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {3\,\sqrt {d\,x^3+c}}{8\,\left (8\,c-d\,x^3\right )}+\frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )\,{\left (10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}\right )}^9}{x^6\,{\left (8\,c-d\,x^3\right )}^9}\right )}{192\,\sqrt {c}} \] Input:

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

Output:

(3*(c + d*x^3)^(1/2))/(8*(8*c - d*x^3)) + log((((c + d*x^3)^(1/2) - c^(1/2 
))^3*((c + d*x^3)^(1/2) + c^(1/2))*(10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^( 
1/2))^9)/(x^6*(8*c - d*x^3)^9))/(192*c^(1/2))
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {2 \sqrt {d \,x^{3}+c}+120 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{10}-15 c \,d^{2} x^{7}+48 c^{2} d \,x^{4}+64 c^{3} x}d x \right ) c^{3}-15 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{10}-15 c \,d^{2} x^{7}+48 c^{2} d \,x^{4}+64 c^{3} x}d x \right ) c^{2} d \,x^{3}+96 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{2}-12 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) d^{3} x^{3}}{-15 d \,x^{3}+120 c} \] Input:

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

Output:

(2*sqrt(c + d*x**3) + 120*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 - 15*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*x**3 + 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**2 - 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**3*x**3)/(15*(8*c - d*x**3))