\(\int \frac {1}{x^4 (8 c-d x^3) \sqrt {c+d x^3}} \, dx\) [488]

Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {c+d x^3}}{24 c^2 x^3}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{288 c^{5/2}}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{32 c^{5/2}} \] Output:

-1/24*(d*x^3+c)^(1/2)/c^2/x^3+1/288*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2)) 
/c^(5/2)+1/32*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {c+d x^3}}{24 c^2 x^3}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{288 c^{5/2}}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{32 c^{5/2}} \] Input:

Integrate[1/(x^4*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]
 

Output:

-1/24*Sqrt[c + d*x^3]/(c^2*x^3) + (d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]) 
/(288*c^(5/2)) + (d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(32*c^(5/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 92, 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, 114, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(x^4*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]
 

Output:

(-1/8*Sqrt[c + d*x^3]/(c^2*x^3) - (d*(-1/6*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt 
[c])]/Sqrt[c] - (3*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c])))/(16*c^2 
))/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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(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] + Simp[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*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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 = 1.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {9 d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) x^{3}+\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) d \,x^{3}-12 \sqrt {d \,x^{3}+c}\, \sqrt {c}}{288 c^{\frac {5}{2}} x^{3}}\)	\(64\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{24 c^{2} x^{3}}-\frac {d \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{2 \sqrt {c}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{18 \sqrt {c}}\right )}{16 c^{2}}\)	\(65\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c \,x^{3}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{8 c}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{96 c^{\frac {5}{2}}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{288 c^{\frac {5}{2}}}\)	\(86\)
elliptic	\(\text {Expression too large to display}\)	\(1523\)

Input:

int(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/288/c^(5/2)*(9*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))*x^3+arctanh(1/3*(d*x^3 
+c)^(1/2)/c^(1/2))*d*x^3-12*(d*x^3+c)^(1/2)*c^(1/2))/x^3
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.20 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {\sqrt {c} d x^{3} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 9 \, \sqrt {c} d x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, \sqrt {d x^{3} + c} c}{576 \, c^{3} x^{3}}, -\frac {\sqrt {-c} d x^{3} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 9 \, \sqrt {-c} d x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 12 \, \sqrt {d x^{3} + c} c}{288 \, c^{3} x^{3}}\right ] \] Input:

integrate(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/576*(sqrt(c)*d*x^3*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^ 
3 - 8*c)) + 9*sqrt(c)*d*x^3*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/ 
x^3) - 24*sqrt(d*x^3 + c)*c)/(c^3*x^3), -1/288*(sqrt(-c)*d*x^3*arctan(3*sq 
rt(-c)/sqrt(d*x^3 + c)) + 9*sqrt(-c)*d*x^3*arctan(sqrt(-c)/sqrt(d*x^3 + c) 
) + 12*sqrt(d*x^3 + c)*c)/(c^3*x^3)]
 

Sympy [F]

\[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=- \int \frac {1}{- 8 c x^{4} \sqrt {c + d x^{3}} + d x^{7} \sqrt {c + d x^{3}}}\, dx \] Input:

integrate(1/x**4/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)
 

Output:

-Integral(1/(-8*c*x**4*sqrt(c + d*x**3) + d*x**7*sqrt(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int { -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{4}} \,d x } \] Input:

integrate(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")
 

Output:

-integrate(1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)*x^4), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{32 \, \sqrt {-c} c^{2}} - \frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{288 \, \sqrt {-c} c^{2}} - \frac {\sqrt {d x^{3} + c}}{24 \, c^{2} x^{3}} \] Input:

integrate(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="giac")
 

Output:

-1/32*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/288*d*arctan(1 
/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/24*sqrt(d*x^3 + c)/(c^2*x^ 
3)
 

Mupad [B] (verification not implemented)

Time = 1.84 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {d\,\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )}{32\,\sqrt {c^5}}+\frac {d\,\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )}{288\,\sqrt {c^5}}-\frac {\sqrt {d\,x^3+c}}{24\,c^2\,x^3} \] Input:

int(1/(x^4*(c + d*x^3)^(1/2)*(8*c - d*x^3)),x)
                                                                                    
                                                                                    
 

Output:

(d*atanh((c^2*(c + d*x^3)^(1/2))/(c^5)^(1/2)))/(32*(c^5)^(1/2)) + (d*atanh 
((c^2*(c + d*x^3)^(1/2))/(3*(c^5)^(1/2))))/(288*(c^5)^(1/2)) - (c + d*x^3) 
^(1/2)/(24*c^2*x^3)
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-64 \sqrt {d \,x^{3}+c}\, c +2 \sqrt {d \,x^{3}+c}\, d \,x^{3}-24 \sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{3}+c}-\sqrt {c}\right ) d \,x^{3}+24 \sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{3}+c}+\sqrt {c}\right ) d \,x^{3}+3 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \right ) d^{3} x^{3}}{1536 c^{3} x^{3}} \] Input:

int(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)
 

Output:

( - 64*sqrt(c + d*x**3)*c + 2*sqrt(c + d*x**3)*d*x**3 - 24*sqrt(c)*log(sqr 
t(c + d*x**3) - sqrt(c))*d*x**3 + 24*sqrt(c)*log(sqrt(c + d*x**3) + sqrt(c 
))*d*x**3 + 3*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)/(1536*c**3*x**3)