\(\int \frac {1}{x (a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\) [666]

Optimal result

Integrand size = 24, antiderivative size = 172 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {d (b c+2 a d)}{3 a c (b c-a d)^2 \sqrt {c+d x^3}}+\frac {b}{3 a (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{3/2}}+\frac {b^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{5/2}} \] Output:

1/3*d*(2*a*d+b*c)/a/c/(-a*d+b*c)^2/(d*x^3+c)^(1/2)+1/3*b/a/(-a*d+b*c)/(b*x 
^3+a)/(d*x^3+c)^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))/a^2/c^(3/2)+1/3 
*b^(3/2)*(-5*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)^(5/2)
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {\frac {a \left (2 a^2 d^2+2 a b d^2 x^3+b^2 c \left (c+d x^3\right )\right )}{c (b c-a d)^2 \left (a+b x^3\right ) \sqrt {c+d x^3}}-\frac {b^{3/2} (2 b c-5 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{c^{3/2}}}{3 a^2} \] Input:

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

Output:

((a*(2*a^2*d^2 + 2*a*b*d^2*x^3 + b^2*c*(c + d*x^3)))/(c*(b*c - a*d)^2*(a + 
 b*x^3)*Sqrt[c + d*x^3]) - (b^(3/2)*(2*b*c - 5*a*d)*ArcTan[(Sqrt[b]*Sqrt[c 
 + d*x^3])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(5/2) - (2*ArcTanh[Sqrt[c + 
 d*x^3]/Sqrt[c]])/c^(3/2))/(3*a^2)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 114, 27, 169, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {2 (b c-a d)^2 \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3}{a}-\frac {b^2 c (2 b c-5 a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{a}}{c (b c-a d)}+\frac {2 d (2 a d+b c)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {4 (b c-a d)^2 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{a d}-\frac {2 b^2 c (2 b c-5 a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{a d}}{c (b c-a d)}+\frac {2 d (2 a d+b c)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {2 b^{3/2} c (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {4 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{a \sqrt {c}}}{c (b c-a d)}+\frac {2 d (2 a d+b c)}{c \sqrt {c+d x^3} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{a \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}\right )\)

Input:

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

Output:

(b/(a*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*x^3]) + ((2*d*(b*c + 2*a*d))/(c*( 
b*c - a*d)*Sqrt[c + d*x^3]) + ((-4*(b*c - a*d)^2*ArcTanh[Sqrt[c + d*x^3]/S 
qrt[c]])/(a*Sqrt[c]) + (2*b^(3/2)*c*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[ 
c + d*x^3])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(c*(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 169

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

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 = 1.35 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.16

method	result	size
pseudoelliptic	\(\frac {-2 \sqrt {d \,x^{3}+c}\, c^{\frac {5}{2}} \left (b \,x^{3}+a \right ) \left (-\frac {5 a d}{2}+b c \right ) b^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {\left (a d -b c \right ) b}\, \left (-2 c \sqrt {d \,x^{3}+c}\, \left (b \,x^{3}+a \right ) \left (a d -b c \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )+a \,c^{\frac {3}{2}} \left (c \left (d \,x^{3}+c \right ) b^{2}+2 x^{3} a b \,d^{2}+2 a^{2} d^{2}\right )\right )}{3 \sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{3}+c}\, c^{\frac {5}{2}} a^{2} \left (b \,x^{3}+a \right ) \left (a d -b c \right )^{2}}\)	\(199\)
default	\(\frac {\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{a^{2}}-\frac {2 b \left (-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {1}{\sqrt {d \,x^{3}+c}}\right )}{3 a^{2} \left (a d -b c \right )}-\frac {b d \left (-\frac {b \sqrt {d \,x^{3}+c}}{d \left (b \,x^{3}+a \right )}-\frac {3 b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {2}{\sqrt {d \,x^{3}+c}}\right )}{3 a \left (a d -b c \right )^{2}}\)	\(203\)
elliptic	\(\text {Expression too large to display}\)	\(1720\)

Input:

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

Output:

1/3/((a*d-b*c)*b)^(1/2)*(-2*(d*x^3+c)^(1/2)*c^(5/2)*(b*x^3+a)*(-5/2*a*d+b* 
c)*b^2*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2))+((a*d-b*c)*b)^(1/2)*( 
-2*c*(d*x^3+c)^(1/2)*(b*x^3+a)*(a*d-b*c)^2*arctanh((d*x^3+c)^(1/2)/c^(1/2) 
)+a*c^(3/2)*(c*(d*x^3+c)*b^2+2*x^3*a*b*d^2+2*a^2*d^2)))/(d*x^3+c)^(1/2)/c^ 
(5/2)/a^2/(b*x^3+a)/(a*d-b*c)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (144) = 288\).

Time = 0.39 (sec) , antiderivative size = 1774, normalized size of antiderivative = 10.31 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/6*((2*a*b^2*c^4 - 5*a^2*b*c^3*d + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^6 
+ (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*d^2)*x^3)*sqrt(b/(b*c - a*d))*l 
og((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^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^6 + a*b^2 
*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + 
a^3*d^3)*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 
 2*(a*b^2*c^3 + 2*a^3*c*d^2 + (a*b^2*c^2*d + 2*a^2*b*c*d^2)*x^3)*sqrt(d*x^ 
3 + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a^ 
3*b^2*c^3*d^2 + a^4*b*c^2*d^3)*x^6 + (a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b* 
c^3*d^2 + a^5*c^2*d^3)*x^3), -1/3*((2*a*b^2*c^4 - 5*a^2*b*c^3*d + (2*b^3*c 
^3*d - 5*a*b^2*c^2*d^2)*x^6 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*d^2 
)*x^3)*sqrt(-b/(b*c - a*d))*arctan(sqrt(d*x^3 + c)*sqrt(-b/(b*c - a*d))) - 
 ((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^6 + a*b^2*c^3 - 2*a^2*b*c^2*d 
+ a^3*c*d^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^3)*sqrt(c) 
*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - (a*b^2*c^3 + 2*a^3*c 
*d^2 + (a*b^2*c^2*d + 2*a^2*b*c*d^2)*x^3)*sqrt(d*x^3 + c))/(a^3*b^2*c^5 - 
2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 + a^4*b*c 
^2*d^3)*x^6 + (a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)* 
x^3), 1/6*(4*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^6 + a*b^2*c^3 - 2* 
a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.31 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {{\left (2 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {{\left (d x^{3} + c\right )} b^{2} c d + 2 \, {\left (d x^{3} + c\right )} a b d^{2} - 2 \, a b c d^{2} + 2 \, a^{2} d^{3}}{3 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{3} + c} b c + \sqrt {d x^{3} + c} a d\right )}} + \frac {2 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c} c} \] Input:

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

Output:

-1/3*(2*b^3*c - 5*a*b^2*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/ 
((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*sqrt(-b^2*c + a*b*d)) + 1/3*((d*x^3 
 + c)*b^2*c*d + 2*(d*x^3 + c)*a*b*d^2 - 2*a*b*c*d^2 + 2*a^2*d^3)/((a*b^2*c 
^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*((d*x^3 + c)^(3/2)*b - sqrt(d*x^3 + c)*b*c 
 + sqrt(d*x^3 + c)*a*d)) + 2/3*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^2*sqrt( 
-c)*c)
 

Mupad [B] (verification not implemented)

Time = 11.21 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )}{3\,a^2\,c^{3/2}}+\frac {\left (\frac {{\left (2\,a\,d+b\,c\right )}^4+{\left (2\,a\,d+b\,c\right )}^2\,\left (\left (a\,d+2\,b\,c\right )\,\left (2\,a\,d+b\,c\right )-9\,a\,b\,c\,d\right )}{9\,a\,c\,{\left (2\,a\,d+b\,c\right )}^2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {b\,d\,x^3\,\left (2\,a\,d+b\,c\right )}{3\,a\,c\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,\sqrt {d\,x^3+c}}{b\,d\,x^6+\left (a\,d+b\,c\right )\,x^3+a\,c}+\frac {b^{3/2}\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (5\,a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,a^2\,{\left (a\,d-b\,c\right )}^{5/2}} \] Input:

int(1/(x*(a + b*x^3)^2*(c + d*x^3)^(3/2)),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^(3/2)) + ((((2*a*d + b*c)^4 + (2*a*d + b*c)^2*((a*d + 2*b*c)*(2*a* 
d + b*c) - 9*a*b*c*d))/(9*a*c*(2*a*d + b*c)^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c 
*d)) + (b*d*x^3*(2*a*d + b*c))/(3*a*c*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))*(c 
 + d*x^3)^(1/2))/(a*c + x^3*(a*d + b*c) + b*d*x^6) + (b^(3/2)*log((2*b*c - 
 a*d + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i + b*d*x^3)/(a + b*x^ 
3))*(5*a*d - 2*b*c)*1i)/(6*a^2*(a*d - b*c)^(5/2))
 

Reduce [F]

\[ \int \frac {1}{x \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d^{2} x^{13}+2 a b \,d^{2} x^{10}+2 b^{2} c d \,x^{10}+a^{2} d^{2} x^{7}+4 a b c d \,x^{7}+b^{2} c^{2} x^{7}+2 a^{2} c d \,x^{4}+2 a b \,c^{2} x^{4}+a^{2} c^{2} x}d x \] Input:

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

Output:

int(sqrt(c + d*x**3)/(a**2*c**2*x + 2*a**2*c*d*x**4 + a**2*d**2*x**7 + 2*a 
*b*c**2*x**4 + 4*a*b*c*d*x**7 + 2*a*b*d**2*x**10 + b**2*c**2*x**7 + 2*b**2 
*c*d*x**10 + b**2*d**2*x**13),x)