\(\int \frac {\sqrt {c+d x^3}}{x (8 c-d x^3)^2} \, dx\) [575]

Optimal result

Integrand size = 27, antiderivative size = 88 \[ \int \frac {\sqrt {c+d x^3}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {\sqrt {c+d x^3}}{24 c \left (8 c-d x^3\right )}+\frac {5 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{288 c^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{3/2}} \] Output:

1/24*(d*x^3+c)^(1/2)/c/(-d*x^3+8*c)+5/288*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1 
/2))/c^(3/2)-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {c+d x^3}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {12 \sqrt {c} \sqrt {c+d x^3}}{8 c-d x^3}+5 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{288 c^{3/2}} \] Input:

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

Output:

((12*Sqrt[c]*Sqrt[c + d*x^3])/(8*c - d*x^3) + 5*ArcTanh[Sqrt[c + d*x^3]/(3 
*Sqrt[c])] - 3*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(288*c^(3/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {948, 110, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(Sqrt[c + d*x^3]/(8*c*(8*c - d*x^3)) + ((5*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt 
[c])])/(6*Sqrt[c]) - ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(2*Sqrt[c]))/(16*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 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 = 1.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{96 c^{\frac {3}{2}}}+\frac {\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \left (d \,x^{3}-8 c \right )}{\sqrt {c}}-12 \sqrt {d \,x^{3}+c}}{288 \left (d \,x^{3}-8 c \right ) c}\)	\(78\)
default	\(\frac {\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}}{64 c^{2}}+\frac {\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 \sqrt {c}}}{24 c}+\frac {-2 \sqrt {d \,x^{3}+c}+6 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{192 c^{2}}\)	\(123\)
elliptic	\(\text {Expression too large to display}\)	\(1534\)

Input:

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

Output:

-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)+1/288*(5*arctanh(1/3*(d*x^3 
+c)^(1/2)/c^(1/2))/c^(1/2)*(d*x^3-8*c)-12*(d*x^3+c)^(1/2))/(d*x^3-8*c)/c
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.50 \[ \int \frac {\sqrt {c+d x^3}}{x \left (8 c-d x^3\right )^2} \, dx=\left [\frac {5 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 3 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, \sqrt {d x^{3} + c} c}{576 \, {\left (c^{2} d x^{3} - 8 \, c^{3}\right )}}, -\frac {5 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - 3 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 12 \, \sqrt {d x^{3} + c} c}{288 \, {\left (c^{2} d x^{3} - 8 \, c^{3}\right )}}\right ] \] Input:

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

Output:

[1/576*(5*(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)) + 3*(d*x^3 - 8*c)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + 
c)*sqrt(c) + 2*c)/x^3) - 24*sqrt(d*x^3 + c)*c)/(c^2*d*x^3 - 8*c^3), -1/288 
*(5*(d*x^3 - 8*c)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) - 3*(d*x^3 - 
 8*c)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^3 + c)) + 12*sqrt(d*x^3 + c)*c)/(c 
^2*d*x^3 - 8*c^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {c+d x^3}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c} c} - \frac {5 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{288 \, \sqrt {-c} c} - \frac {\sqrt {d x^{3} + c}}{24 \, {\left (d x^{3} - 8 \, c\right )} c} \] Input:

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

Output:

1/96*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 5/288*arctan(1/3*sqrt 
(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 1/24*sqrt(d*x^3 + c)/((d*x^3 - 8*c)*c 
)
 

Mupad [B] (verification not implemented)

Time = 2.42 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {c+d x^3}}{x \left (8 c-d x^3\right )^2} \, dx=\frac {5\,\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^3}}\right )}{288\,\sqrt {c^3}}-\frac {\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{\sqrt {c^3}}\right )}{96\,\sqrt {c^3}}+\frac {\sqrt {d\,x^3+c}}{8\,c\,\left (24\,c-3\,d\,x^3\right )} \] Input:

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

Output:

(5*atanh((c*(c + d*x^3)^(1/2))/(3*(c^3)^(1/2))))/(288*(c^3)^(1/2)) - atanh 
((c*(c + d*x^3)^(1/2))/(c^3)^(1/2))/(96*(c^3)^(1/2)) + (c + d*x^3)^(1/2)/( 
8*c*(24*c - 3*d*x^3))
 

Reduce [F]

\[ \int \frac {\sqrt {c+d x^3}}{x \left (8 c-d x^3\right )^2} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d^{2} x^{7}-16 c d \,x^{4}+64 c^{2} x}d x \] Input:

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

Output:

int(sqrt(c + d*x**3)/(64*c**2*x - 16*c*d*x**4 + d**2*x**7),x)