\(\int \frac {1}{x^4 (a+b x^3) (c+d x^3)^{3/2}} \, dx\) [565]

Optimal result

Integrand size = 24, antiderivative size = 158 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}+\frac {(2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{5/2}}-\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}} \] Output:

-1/3*d*(-3*a*d+b*c)/a/c^2/(-a*d+b*c)/(d*x^3+c)^(1/2)-1/3/a/c/x^3/(d*x^3+c) 
^(1/2)+1/3*(3*a*d+2*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1/2))/a^2/c^(5/2)-2/3* 
b^(5/2)*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 = 0.53 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {\frac {a \left (-b c \left (c+d x^3\right )+a d \left (c+3 d x^3\right )\right )}{c^2 (b c-a d) x^3 \sqrt {c+d x^3}}-\frac {2 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{3/2}}+\frac {(2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{c^{5/2}}}{3 a^2} \] Input:

Integrate[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.19, 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.

\(\displaystyle \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
 

Output:

(-(1/(a*c*x^3*Sqrt[c + d*x^3])) - ((2*d*(b*c - 3*a*d))/(c*(b*c - a*d)*Sqrt 
[c + d*x^3]) + ((-2*(b*c - a*d)*(2*b*c + 3*a*d)*ArcTanh[Sqrt[c + d*x^3]/Sq 
rt[c]])/(a*Sqrt[c]) + (4*b^(5/2)*c^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqr 
t[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(c*(b*c - a*d)))/(2*a*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 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 = 1.46 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(\frac {d^{2} \left (\frac {-\frac {a \sqrt {d \,x^{3}+c}}{x^{3}}+\frac {\left (3 a d +2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{\sqrt {c}}}{a^{2} c^{2} d^{2}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {2}{c^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{3}\)	\(141\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} a \,x^{3}}-\frac {-\frac {2 \left (3 a d +2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a \sqrt {c}}+\frac {4 a \,d^{2}}{3 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}+\frac {4 b^{3} c^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \left (a d -b c \right ) a \sqrt {\left (a d -b c \right ) b}}}{2 a \,c^{2}}\)	\(144\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} x^{3}}-\frac {2 d}{3 c^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}}{a}+\frac {2 b^{2} \left (-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {1}{\sqrt {d \,x^{3}+c}}\right )}{3 a^{2} \left (a d -b c \right )}-\frac {b \left (\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{a^{2}}\)	\(175\)
elliptic	\(\text {Expression too large to display}\)	\(1682\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1073, normalized size of antiderivative = 6.79 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(1/(x**4*(a + b*x**3)*(c + d*x**3)**(3/2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \, b^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {{\left (d x^{3} + c\right )} b c d - 3 \, {\left (d x^{3} + c\right )} a d^{2} + 2 \, a c d^{2}}{3 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - \sqrt {d x^{3} + c} c\right )}} - \frac {{\left (2 \, b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c} c^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.78 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )\,\left (3\,a\,d+2\,b\,c\right )}{6\,a^2\,c^{5/2}}-\frac {\sqrt {d\,x^3+c}}{3\,a\,c^2\,x^3}-\frac {\frac {c\,\left (\frac {c\,\left (\frac {c\,\left (\frac {3\,a^2\,d^4+24\,a\,b\,c\,d^3+15\,b^2\,c^2\,d^2}{8\,a^3\,c^5}+\frac {c\,\left (\frac {c\,\left (\frac {3\,b^2\,d^4}{8\,a^3\,c^5}+\frac {b^2\,d^4\,\left (5\,a\,d-3\,b\,c\right )}{8\,a^3\,c^4\,\left (b\,c^2-a\,c\,d\right )}-\frac {b\,d^4\,\left (a\,d+2\,b\,c\right )\,\left (5\,a\,d-3\,b\,c\right )}{4\,a^3\,c^5\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {3\,b\,d^3\,\left (a\,d+2\,b\,c\right )}{4\,a^3\,c^5}+\frac {d\,\left (5\,a\,d-3\,b\,c\right )\,\left (3\,a^2\,d^4+24\,a\,b\,c\,d^3+15\,b^2\,c^2\,d^2\right )}{24\,a^3\,c^5\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {d^2\,\left (5\,a\,d-3\,b\,c\right )\,\left (6\,a^2\,d^2+14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{12\,a^3\,c^4\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {d\,\left (6\,a^2\,d^2+14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{4\,a^3\,c^4}+\frac {d^2\,\left (5\,a\,d-3\,b\,c\right )\,\left (13\,a\,d+18\,b\,c\right )}{24\,a^2\,c^3\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}+\frac {d\,\left (13\,a\,d+18\,b\,c\right )}{8\,a^2\,c^3}-\frac {d\,\left (3\,a\,d+2\,b\,c\right )\,\left (5\,a\,d-3\,b\,c\right )}{6\,a^2\,c^2\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {3\,a\,d+2\,b\,c}{2\,a^2\,c^2}}{\sqrt {d\,x^3+c}}+\frac {b^{5/2}\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,a^2\,{\left (a\,d-b\,c\right )}^{3/2}} \] Input:

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

Output:

(log((((c + d*x^3)^(1/2) - c^(1/2))*((c + d*x^3)^(1/2) + c^(1/2))^3)/x^6)* 
(3*a*d + 2*b*c))/(6*a^2*c^(5/2)) - (c + d*x^3)^(1/2)/(3*a*c^2*x^3) - ((c*( 
(c*((c*((3*a^2*d^4 + 15*b^2*c^2*d^2 + 24*a*b*c*d^3)/(8*a^3*c^5) + (c*((c*( 
(3*b^2*d^4)/(8*a^3*c^5) + (b^2*d^4*(5*a*d - 3*b*c))/(8*a^3*c^4*(b*c^2 - a* 
c*d)) - (b*d^4*(a*d + 2*b*c)*(5*a*d - 3*b*c))/(4*a^3*c^5*(b*c^2 - a*c*d))) 
)/d - (3*b*d^3*(a*d + 2*b*c))/(4*a^3*c^5) + (d*(5*a*d - 3*b*c)*(3*a^2*d^4 
+ 15*b^2*c^2*d^2 + 24*a*b*c*d^3))/(24*a^3*c^5*(b*c^2 - a*c*d))))/d - (d^2* 
(5*a*d - 3*b*c)*(6*a^2*d^2 + 3*b^2*c^2 + 14*a*b*c*d))/(12*a^3*c^4*(b*c^2 - 
 a*c*d))))/d - (d*(6*a^2*d^2 + 3*b^2*c^2 + 14*a*b*c*d))/(4*a^3*c^4) + (d^2 
*(5*a*d - 3*b*c)*(13*a*d + 18*b*c))/(24*a^2*c^3*(b*c^2 - a*c*d))))/d + (d* 
(13*a*d + 18*b*c))/(8*a^2*c^3) - (d*(3*a*d + 2*b*c)*(5*a*d - 3*b*c))/(6*a^ 
2*c^2*(b*c^2 - a*c*d))))/d - (3*a*d + 2*b*c)/(2*a^2*c^2))/(c + d*x^3)^(1/2 
) + (b^(5/2)*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))*1i)/(3*a^2*(a*d - b*c)^(3/2))
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-9 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{10}+a \,d^{2} x^{7}+2 b c d \,x^{7}+2 a c d \,x^{4}+b \,c^{2} x^{4}+a \,c^{2} x}d x \right ) a c d \,x^{3}-9 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{10}+a \,d^{2} x^{7}+2 b c d \,x^{7}+2 a c d \,x^{4}+b \,c^{2} x^{4}+a \,c^{2} x}d x \right ) a \,d^{2} x^{6}-6 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{10}+a \,d^{2} x^{7}+2 b c d \,x^{7}+2 a c d \,x^{4}+b \,c^{2} x^{4}+a \,c^{2} x}d x \right ) b \,c^{2} x^{3}-6 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{10}+a \,d^{2} x^{7}+2 b c d \,x^{7}+2 a c d \,x^{4}+b \,c^{2} x^{4}+a \,c^{2} x}d x \right ) b c d \,x^{6}-9 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b c d \,x^{3}-9 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b \,d^{2} x^{6}}{6 a c \,x^{3} \left (d \,x^{3}+c \right )} \] Input:

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

Output:

( - 2*sqrt(c + d*x**3) - 9*int(sqrt(c + d*x**3)/(a*c**2*x + 2*a*c*d*x**4 + 
 a*d**2*x**7 + b*c**2*x**4 + 2*b*c*d*x**7 + b*d**2*x**10),x)*a*c*d*x**3 - 
9*int(sqrt(c + d*x**3)/(a*c**2*x + 2*a*c*d*x**4 + a*d**2*x**7 + b*c**2*x** 
4 + 2*b*c*d*x**7 + b*d**2*x**10),x)*a*d**2*x**6 - 6*int(sqrt(c + d*x**3)/( 
a*c**2*x + 2*a*c*d*x**4 + a*d**2*x**7 + b*c**2*x**4 + 2*b*c*d*x**7 + b*d** 
2*x**10),x)*b*c**2*x**3 - 6*int(sqrt(c + d*x**3)/(a*c**2*x + 2*a*c*d*x**4 
+ a*d**2*x**7 + b*c**2*x**4 + 2*b*c*d*x**7 + b*d**2*x**10),x)*b*c*d*x**6 - 
 9*int((sqrt(c + d*x**3)*x**2)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c* 
*2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*b*c*d*x**3 - 9*int((sqrt(c + d*x* 
*3)*x**2)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x** 
6 + b*d**2*x**9),x)*b*d**2*x**6)/(6*a*c*x**3*(c + d*x**3))