\(\int \frac {(c+d x^3)^{3/2}}{x^4 (8 c-d x^3)} \, dx\) [475]

Optimal result

Integrand size = 27, antiderivative size = 78 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{24 x^3}+\frac {9 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{32 \sqrt {c}}-\frac {13 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 \sqrt {c}} \] Output:

-1/24*(d*x^3+c)^(1/2)/x^3+9/32*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(1 
/2)-13/96*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{24 x^3}+\frac {9 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{32 \sqrt {c}}-\frac {13 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 \sqrt {c}} \] Input:

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

Output:

-1/24*Sqrt[c + d*x^3]/x^3 + (9*d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(32 
*Sqrt[c]) - (13*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(96*Sqrt[c])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{3} \left (-\frac {\int -\frac {c d \left (17 d x^3+26 c\right )}{2 x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{8 c}-\frac {\sqrt {c+d x^3}}{8 x^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \int \frac {17 d x^3+26 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {\sqrt {c+d x^3}}{8 x^3}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {13}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {81}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\right )-\frac {\sqrt {c+d x^3}}{8 x^3}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {81}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {13 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}\right )-\frac {\sqrt {c+d x^3}}{8 x^3}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {13 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {27 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}\right )-\frac {\sqrt {c+d x^3}}{8 x^3}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {27 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {13 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )-\frac {\sqrt {c+d x^3}}{8 x^3}\right )\)

Input:

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

Output:

(-1/8*Sqrt[c + d*x^3]/x^3 + (d*((27*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/ 
(2*Sqrt[c]) - (13*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 = 1.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{24 x^{3}}+\frac {d \left (-\frac {13 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{6 \sqrt {c}}+\frac {9 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{16}\)	\(59\)
pseudoelliptic	\(\frac {-13 d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) x^{3}+27 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) d \,x^{3}-4 \sqrt {d \,x^{3}+c}\, \sqrt {c}}{96 x^{3} \sqrt {c}}\)	\(65\)
default	\(\frac {-\frac {c \sqrt {d \,x^{3}+c}}{3 x^{3}}+\frac {2 d \sqrt {d \,x^{3}+c}}{3}-\sqrt {c}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{8 c}+\frac {d \left (\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{64 c^{2}}+\frac {d \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{288 c^{2}}\)	\(155\)
elliptic	\(\text {Expression too large to display}\)	\(1514\)

Input:

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

Output:

-1/24*(d*x^3+c)^(1/2)/x^3+1/16*d*(-13/6*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c 
^(1/2)+9/2*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.31 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=\left [\frac {27 \, \sqrt {c} d x^{3} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 13 \, \sqrt {c} d x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 8 \, \sqrt {d x^{3} + c} c}{192 \, c x^{3}}, -\frac {27 \, \sqrt {-c} d x^{3} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - 13 \, \sqrt {-c} d x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 4 \, \sqrt {d x^{3} + c} c}{96 \, c x^{3}}\right ] \] Input:

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

Output:

[1/192*(27*sqrt(c)*d*x^3*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d 
*x^3 - 8*c)) + 13*sqrt(c)*d*x^3*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2 
*c)/x^3) - 8*sqrt(d*x^3 + c)*c)/(c*x^3), -1/96*(27*sqrt(-c)*d*x^3*arctan(3 
*sqrt(-c)/sqrt(d*x^3 + c)) - 13*sqrt(-c)*d*x^3*arctan(sqrt(-c)/sqrt(d*x^3 
+ c)) + 4*sqrt(d*x^3 + c)*c)/(c*x^3)]
 

Sympy [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=- \int \frac {c \sqrt {c + d x^{3}}}{- 8 c x^{4} + d x^{7}}\, dx - \int \frac {d x^{3} \sqrt {c + d x^{3}}}{- 8 c x^{4} + d x^{7}}\, dx \] Input:

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

Output:

-Integral(c*sqrt(c + d*x**3)/(-8*c*x**4 + d*x**7), x) - Integral(d*x**3*sq 
rt(c + d*x**3)/(-8*c*x**4 + d*x**7), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=\frac {13 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c}} - \frac {9 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{32 \, \sqrt {-c}} - \frac {\sqrt {d x^{3} + c}}{24 \, x^{3}} \] Input:

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

Output:

13/96*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c) - 9/32*d*arctan(1/3*sqrt 
(d*x^3 + c)/sqrt(-c))/sqrt(-c) - 1/24*sqrt(d*x^3 + c)/x^3
 

Mupad [B] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=\frac {9\,d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^3+c}}{3\,\sqrt {c}}\right )}{32\,\sqrt {c}}-\frac {13\,d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^3+c}}{\sqrt {c}}\right )}{96\,\sqrt {c}}-\frac {\sqrt {d\,x^3+c}}{24\,x^3} \] Input:

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

Output:

(9*d*atanh((c + d*x^3)^(1/2)/(3*c^(1/2))))/(32*c^(1/2)) - (13*d*atanh((c + 
 d*x^3)^(1/2)/c^(1/2)))/(96*c^(1/2)) - (c + d*x^3)^(1/2)/(24*x^3)
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )} \, dx=\frac {-256 \sqrt {d \,x^{3}+c}\, c +162 \sqrt {d \,x^{3}+c}\, d \,x^{3}+416 \sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{3}+c}-\sqrt {c}\right ) d \,x^{3}-416 \sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{3}+c}+\sqrt {c}\right ) d \,x^{3}+243 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \right ) d^{3} x^{3}+5832 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \right ) c \,d^{2} x^{3}}{6144 c \,x^{3}} \] Input:

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

Output:

( - 256*sqrt(c + d*x**3)*c + 162*sqrt(c + d*x**3)*d*x**3 + 416*sqrt(c)*log 
(sqrt(c + d*x**3) - sqrt(c))*d*x**3 - 416*sqrt(c)*log(sqrt(c + d*x**3) + s 
qrt(c))*d*x**3 + 243*int((sqrt(c + d*x**3)*x**5)/(8*c**2 + 7*c*d*x**3 - d* 
*2*x**6),x)*d**3*x**3 + 5832*int((sqrt(c + d*x**3)*x**2)/(8*c**2 + 7*c*d*x 
**3 - d**2*x**6),x)*c*d**2*x**3)/(6144*c*x**3)