\(\int \frac {1}{x (a+b x^3)^2 \sqrt {c+d x^3}} \, dx\) [656]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {b \sqrt {c+d x^3}}{3 a (b c-a d) \left (a+b x^3\right )}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}+\frac {\sqrt {b} (2 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}} \] Output:

1/3*b*(d*x^3+c)^(1/2)/a/(-a*d+b*c)/(b*x^3+a)-2/3*arctanh((d*x^3+c)^(1/2)/c 
^(1/2))/a^2/c^(1/2)+1/3*b^(1/2)*(-3*a*d+2*b*c)*arctanh(b^(1/2)*(d*x^3+c)^( 
1/2)/(-a*d+b*c)^(1/2))/a^2/(-a*d+b*c)^(3/2)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {-\frac {a b \sqrt {c+d x^3}}{(-b c+a d) \left (a+b x^3\right )}+\frac {\sqrt {b} (2 b c-3 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}}{3 a^2} \] Input:

Integrate[1/(x*(a + b*x^3)^2*Sqrt[c + d*x^3]),x]
 

Output:

(-((a*b*Sqrt[c + d*x^3])/((-(b*c) + a*d)*(a + b*x^3))) + (Sqrt[b]*(2*b*c - 
 3*a*d)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a* 
d)^(3/2) - (2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/Sqrt[c])/(3*a^2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {948, 114, 27, 174, 73, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {\frac {4 (b c-a d) \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{a d}-\frac {2 b (2 b c-3 a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{a d}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^3}}{a \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {\frac {2 \sqrt {b} (2 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {4 (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{a \sqrt {c}}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^3}}{a \left (a+b x^3\right ) (b c-a d)}\right )\)

Input:

Int[1/(x*(a + b*x^3)^2*Sqrt[c + d*x^3]),x]
 

Output:

((b*Sqrt[c + d*x^3])/(a*(b*c - a*d)*(a + b*x^3)) + ((-4*(b*c - a*d)*ArcTan 
h[Sqrt[c + d*x^3]/Sqrt[c]])/(a*Sqrt[c]) + (2*Sqrt[b]*(2*b*c - 3*a*d)*ArcTa 
nh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(2*a*( 
b*c - a*d)))/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 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 = 1.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.06

method	result	size
default	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a^{2} \sqrt {c}}-\frac {2 b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {b \left (\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a \left (a d -b c \right )}\)	\(140\)
pseudoelliptic	\(\frac {\frac {2 \sqrt {c}\, \left (b \,x^{3}+a \right ) \left (b c -\frac {3 a d}{2}\right ) b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3}-\frac {\left (2 \left (a d -b c \right ) \left (b \,x^{3}+a \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )+\sqrt {c}\, \sqrt {d \,x^{3}+c}\, a b \right ) \sqrt {\left (a d -b c \right ) b}}{3}}{a^{2} \sqrt {c}\, \left (a d -b c \right ) \left (b \,x^{3}+a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(146\)
elliptic	\(\text {Expression too large to display}\)	\(1677\)

Input:

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

Output:

-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))/a^2/c^(1/2)-2/3*b/a^2/((a*d-b*c)*b)^ 
(1/2)*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2))-1/3*b/a/(a*d-b*c)*((d* 
x^3+c)^(1/2)/(b*x^3+a)+d/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x^3+c)^(1/2)/((a* 
d-b*c)*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 816, normalized size of antiderivative = 6.18 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/6*(2*sqrt(d*x^3 + c)*a*b*c + (2*a*b*c^2 - 3*a^2*c*d + (2*b^2*c^2 - 3*a* 
b*c*d)*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 
+ c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) + 2*((b^2*c - a*b*d)*x^ 
3 + a*b*c - a^2*d)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x 
^3))/(a^3*b*c^2 - a^4*c*d + (a^2*b^2*c^2 - a^3*b*c*d)*x^3), 1/3*(sqrt(d*x^ 
3 + c)*a*b*c - (2*a*b*c^2 - 3*a^2*c*d + (2*b^2*c^2 - 3*a*b*c*d)*x^3)*sqrt( 
-b/(b*c - a*d))*arctan(sqrt(d*x^3 + c)*sqrt(-b/(b*c - a*d))) + ((b^2*c - a 
*b*d)*x^3 + a*b*c - a^2*d)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) 
+ 2*c)/x^3))/(a^3*b*c^2 - a^4*c*d + (a^2*b^2*c^2 - a^3*b*c*d)*x^3), 1/6*(2 
*sqrt(d*x^3 + c)*a*b*c + 4*((b^2*c - a*b*d)*x^3 + a*b*c - a^2*d)*sqrt(-c)* 
arctan(sqrt(-c)/sqrt(d*x^3 + c)) + (2*a*b*c^2 - 3*a^2*c*d + (2*b^2*c^2 - 3 
*a*b*c*d)*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x 
^3 + c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)))/(a^3*b*c^2 - a^4*c* 
d + (a^2*b^2*c^2 - a^3*b*c*d)*x^3), 1/3*(sqrt(d*x^3 + c)*a*b*c - (2*a*b*c^ 
2 - 3*a^2*c*d + (2*b^2*c^2 - 3*a*b*c*d)*x^3)*sqrt(-b/(b*c - a*d))*arctan(s 
qrt(d*x^3 + c)*sqrt(-b/(b*c - a*d))) + 2*((b^2*c - a*b*d)*x^3 + a*b*c - a^ 
2*d)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^3 + c)))/(a^3*b*c^2 - a^4*c*d + (a^ 
2*b^2*c^2 - a^3*b*c*d)*x^3)]
 

Sympy [F]

\[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{x \left (a + b x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:

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

Output:

Integral(1/(x*(a + b*x**3)**2*sqrt(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c} x} \,d x } \] Input:

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

Output:

integrate(1/((b*x^3 + a)^2*sqrt(d*x^3 + c)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\sqrt {d x^{3} + c} b d}{3 \, {\left (a b c - a^{2} d\right )} {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )}} - \frac {{\left (2 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c}} \] Input:

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

Output:

1/3*sqrt(d*x^3 + c)*b*d/((a*b*c - a^2*d)*((d*x^3 + c)*b - b*c + a*d)) - 1/ 
3*(2*b^2*c - 3*a*b*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2 
*b*c - a^3*d)*sqrt(-b^2*c + a*b*d)) + 2/3*arctan(sqrt(d*x^3 + c)/sqrt(-c)) 
/(a^2*sqrt(-c))
 

Mupad [B] (verification not implemented)

Time = 8.75 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )}{3\,a^2\,\sqrt {c}}+\frac {b^2\,\sqrt {d\,x^3+c}}{3\,a\,\left (b\,x^3+a\right )\,\left (b^2\,c-a\,b\,d\right )}+\frac {\sqrt {b}\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (3\,a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,a^2\,{\left (a\,d-b\,c\right )}^{3/2}} \] Input:

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

Output:

log((((c + d*x^3)^(1/2) - c^(1/2))^3*((c + d*x^3)^(1/2) + c^(1/2)))/x^6)/( 
3*a^2*c^(1/2)) + (b^2*(c + d*x^3)^(1/2))/(3*a*(a + b*x^3)*(b^2*c - a*b*d)) 
 + (b^(1/2)*log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2) 
*2i - b*d*x^3)/(a + b*x^3))*(3*a*d - 2*b*c)*1i)/(6*a^2*(a*d - b*c)^(3/2))
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {1}{x \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d \,x^{10}+2 a b d \,x^{7}+b^{2} c \,x^{7}+a^{2} d \,x^{4}+2 a b c \,x^{4}+a^{2} c x}d x \] Input:

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

Output:

int(sqrt(c + d*x**3)/(a**2*c*x + a**2*d*x**4 + 2*a*b*c*x**4 + 2*a*b*d*x**7 
 + b**2*c*x**7 + b**2*d*x**10),x)