\(\int \frac {\sqrt {c+d x^3}}{x^4 (4 c+d x^3)} \, dx\) [438]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{12 c x^3}-\frac {d \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{8 \sqrt {3} c^{3/2}}-\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{24 c^{3/2}} \] Output:

-1/12*(d*x^3+c)^(1/2)/c/x^3-1/24*d*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1 
/2))*3^(1/2)/c^(3/2)-1/24*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{12 c x^3}-\frac {d \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{8 \sqrt {3} c^{3/2}}-\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{24 c^{3/2}} \] Input:

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

Output:

-1/12*Sqrt[c + d*x^3]/(c*x^3) - (d*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c] 
)])/(8*Sqrt[3]*c^(3/2)) - (d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(24*c^(3/2) 
)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {948, 110, 27, 174, 73, 216, 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.

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

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {\sqrt {d x^3+c}}{x^6 \left (d x^3+4 c\right )}dx^3\)

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {\int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{d}-3 \int \frac {1}{x^6+3 c}d\sqrt {d x^3+c}\right )}{8 c}-\frac {\sqrt {c+d x^3}}{4 c x^3}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {\int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{d}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {c}}\right )}{8 c}-\frac {\sqrt {c+d x^3}}{4 c x^3}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {d \left (-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{8 c}-\frac {\sqrt {c+d x^3}}{4 c x^3}\right )\)

Input:

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

Output:

(-1/4*Sqrt[c + d*x^3]/(c*x^3) + (d*(-((Sqrt[3]*ArcTan[Sqrt[c + d*x^3]/(Sqr 
t[3]*Sqrt[c])])/Sqrt[c]) - ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/Sqrt[c]))/(8*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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[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 = 1.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{12 c \,x^{3}}-\frac {d \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{24 c^{\frac {3}{2}}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{24 c^{\frac {3}{2}}}\)	\(66\)
pseudoelliptic	\(-\frac {\sqrt {d \,x^{3}+c}}{12 c \,x^{3}}-\frac {d \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{24 c^{\frac {3}{2}}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{24 c^{\frac {3}{2}}}\)	\(66\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}}{4 c}-\frac {d \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{16 c^{2}}+\frac {d \left (2 \sqrt {d \,x^{3}+c}-2 \sqrt {c}\, \sqrt {3}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right )\right )}{48 c^{2}}\)	\(123\)
elliptic	\(\text {Expression too large to display}\)	\(1523\)

Input:

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

Output:

-1/12*(d*x^3+c)^(1/2)/c/x^3-1/24*d*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1 
/2))*3^(1/2)/c^(3/2)-1/24*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (4 c+d x^3\right )} \, dx=\left [\frac {2 \, \sqrt {3} \sqrt {c} d x^{3} \arctan \left (\frac {\sqrt {3} \sqrt {c}}{\sqrt {d x^{3} + c}}\right ) + \sqrt {c} d x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 4 \, \sqrt {d x^{3} + c} c}{48 \, c^{2} x^{3}}, -\frac {\sqrt {3} \sqrt {-c} d x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) - 2 \, \sqrt {-c} d x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 4 \, \sqrt {d x^{3} + c} c}{48 \, c^{2} x^{3}}\right ] \] Input:

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

Output:

[1/48*(2*sqrt(3)*sqrt(c)*d*x^3*arctan(sqrt(3)*sqrt(c)/sqrt(d*x^3 + c)) + s 
qrt(c)*d*x^3*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 4*sqrt(d 
*x^3 + c)*c)/(c^2*x^3), -1/48*(sqrt(3)*sqrt(-c)*d*x^3*log((d*x^3 + 2*sqrt( 
3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) - 2*sqrt(-c)*d*x^3*arcta 
n(sqrt(-c)/sqrt(d*x^3 + c)) + 4*sqrt(d*x^3 + c)*c)/(c^2*x^3)]
 

Sympy [F]

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

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

Output:

Integral(sqrt(c + d*x**3)/(x**4*(4*c + d*x**3)), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {3} d \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{24 \, c^{\frac {3}{2}}} + \frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{24 \, \sqrt {-c} c} - \frac {\sqrt {d x^{3} + c}}{12 \, c x^{3}} \] Input:

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

Output:

-1/24*sqrt(3)*d*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/c^(3/2) + 1/24 
*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 1/12*sqrt(d*x^3 + c)/(c 
*x^3)
 

Mupad [B] (verification not implemented)

Time = 3.55 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (4 c+d x^3\right )} \, dx=\frac {d\,\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )}{48\,c^{3/2}}-\frac {\sqrt {d\,x^3+c}}{12\,c\,x^3}+\frac {\sqrt {3}\,d\,\ln \left (\frac {2\,\sqrt {3}\,c-\sqrt {3}\,d\,x^3+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,1{}\mathrm {i}}{48\,c^{3/2}} \] Input:

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

Output:

(d*log((((c + d*x^3)^(1/2) - c^(1/2))^3*((c + d*x^3)^(1/2) + c^(1/2)))/x^6 
))/(48*c^(3/2)) - (c + d*x^3)^(1/2)/(12*c*x^3) + (3^(1/2)*d*log((2*3^(1/2) 
*c + c^(1/2)*(c + d*x^3)^(1/2)*6i - 3^(1/2)*d*x^3)/(4*c + d*x^3))*1i)/(48* 
c^(3/2))
 

Reduce [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (4 c+d x^3\right )} \, dx=\frac {-32 \sqrt {d \,x^{3}+c}\, c +6 \sqrt {d \,x^{3}+c}\, d \,x^{3}+8 \sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{3}+c}-\sqrt {c}\right ) d \,x^{3}-8 \sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{3}+c}+\sqrt {c}\right ) d \,x^{3}-9 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) d^{3} x^{3}-108 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) c \,d^{2} x^{3}}{384 c^{2} x^{3}} \] Input:

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

Output:

( - 32*sqrt(c + d*x**3)*c + 6*sqrt(c + d*x**3)*d*x**3 + 8*sqrt(c)*log(sqrt 
(c + d*x**3) - sqrt(c))*d*x**3 - 8*sqrt(c)*log(sqrt(c + d*x**3) + sqrt(c)) 
*d*x**3 - 9*int((sqrt(c + d*x**3)*x**5)/(4*c**2 + 5*c*d*x**3 + d**2*x**6), 
x)*d**3*x**3 - 108*int((sqrt(c + d*x**3)*x**2)/(4*c**2 + 5*c*d*x**3 + d**2 
*x**6),x)*c*d**2*x**3)/(384*c**2*x**3)