\(\int \frac {1}{x^4 (a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\) [667]

Optimal result

Integrand size = 24, antiderivative size = 241 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {d \left (2 b^2 c^2-2 a b c d+3 a^2 d^2\right )}{3 a^2 c^2 (b c-a d)^2 \sqrt {c+d x^3}}-\frac {b (2 b c-a d)}{3 a^2 c (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}}+\frac {(4 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 c^{5/2}}-\frac {b^{5/2} (4 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 (b c-a d)^{5/2}} \] Output:

-1/3*d*(3*a^2*d^2-2*a*b*c*d+2*b^2*c^2)/a^2/c^2/(-a*d+b*c)^2/(d*x^3+c)^(1/2 
)-1/3*b*(-a*d+2*b*c)/a^2/c/(-a*d+b*c)/(b*x^3+a)/(d*x^3+c)^(1/2)-1/3/a/c/x^ 
3/(b*x^3+a)/(d*x^3+c)^(1/2)+1/3*(3*a*d+4*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1 
/2))/a^3/c^(5/2)-1/3*b^(5/2)*(-7*a*d+4*b*c)*arctanh(b^(1/2)*(d*x^3+c)^(1/2 
)/(-a*d+b*c)^(1/2))/a^3/(-a*d+b*c)^(5/2)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {-\frac {a \left (2 b^3 c^2 x^3 \left (c+d x^3\right )+a^3 d^2 \left (c+3 d x^3\right )+a b^2 c \left (c^2-c d x^3-2 d^2 x^6\right )+a^2 b d \left (-2 c^2-c d x^3+3 d^2 x^6\right )\right )}{c^2 (b c-a d)^2 x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}}+\frac {b^{5/2} (4 b c-7 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}+\frac {(4 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{c^{5/2}}}{3 a^3} \] Input:

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

Output:

(-((a*(2*b^3*c^2*x^3*(c + d*x^3) + a^3*d^2*(c + 3*d*x^3) + a*b^2*c*(c^2 - 
c*d*x^3 - 2*d^2*x^6) + a^2*b*d*(-2*c^2 - c*d*x^3 + 3*d^2*x^6)))/(c^2*(b*c 
- a*d)^2*x^3*(a + b*x^3)*Sqrt[c + d*x^3])) + (b^(5/2)*(4*b*c - 7*a*d)*ArcT 
an[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(5/2) + ( 
(4*b*c + 3*a*d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/c^(5/2))/(3*a^3)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {948, 114, 27, 168, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\frac {\frac {(b c-a d)^2 (3 a d+4 b c) \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3}{a}-\frac {b^3 c^2 (4 b c-7 a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{a}}{c (b c-a d)}+\frac {2 d \left (3 a^2 d^2-2 a b c d+2 b^2 c^2\right )}{c \sqrt {c+d x^3} (b c-a d)}}{a (b c-a d)}+\frac {2 b (2 b c-a d)}{a \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\frac {\frac {2 (b c-a d)^2 (3 a d+4 b c) \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{a d}-\frac {2 b^3 c^2 (4 b c-7 a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{a d}}{c (b c-a d)}+\frac {2 d \left (3 a^2 d^2-2 a b c d+2 b^2 c^2\right )}{c \sqrt {c+d x^3} (b c-a d)}}{a (b c-a d)}+\frac {2 b (2 b c-a d)}{a \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (-\frac {\frac {\frac {2 d \left (3 a^2 d^2-2 a b c d+2 b^2 c^2\right )}{c \sqrt {c+d x^3} (b c-a d)}+\frac {\frac {2 b^{5/2} c^2 (4 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {2 (b c-a d)^2 (3 a d+4 b c) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{a \sqrt {c}}}{c (b c-a d)}}{a (b c-a d)}+\frac {2 b (2 b c-a d)}{a \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}}{2 a c}-\frac {1}{a c x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}}\right )\)

Input:

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

Output:

(-(1/(a*c*x^3*(a + b*x^3)*Sqrt[c + d*x^3])) - ((2*b*(2*b*c - a*d))/(a*(b*c 
 - a*d)*(a + b*x^3)*Sqrt[c + d*x^3]) + ((2*d*(2*b^2*c^2 - 2*a*b*c*d + 3*a^ 
2*d^2))/(c*(b*c - a*d)*Sqrt[c + d*x^3]) + ((-2*(b*c - a*d)^2*(4*b*c + 3*a* 
d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(a*Sqrt[c]) + (2*b^(5/2)*c^2*(4*b*c - 
 7*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)))/(a*(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 168

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] && ILtQ[m, -1]
 

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 = 1.51 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} a^{2} x^{3}}-\frac {-\frac {2 \left (3 a d +4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a \sqrt {c}}+\frac {4 a^{2} d^{3}}{3 \left (a d -b c \right )^{2} \sqrt {d \,x^{3}+c}}+\frac {2 b^{3} c^{2} \left (\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \left (a d -b c \right )^{2}}+\frac {4 b^{3} c^{2} \left (3 a d -2 b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a \left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}}{2 a^{2} c^{2}}\)	\(231\)
pseudoelliptic	\(\frac {d^{3} \left (\frac {-\frac {a \sqrt {d \,x^{3}+c}}{x^{3}}+\frac {\left (3 a d +4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{\sqrt {c}}}{a^{3} c^{2} d^{3}}-\frac {\sqrt {d \,x^{3}+c}\, b^{3}}{a^{2} d^{3} \left (b \,x^{3}+a \right ) \left (a d -b c \right )^{2}}-\frac {7 \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) b^{3}}{\sqrt {\left (a d -b c \right ) b}\, a^{2} d^{2} \left (a d -b c \right )^{2}}+\frac {4 \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) b^{4} c}{\sqrt {\left (a d -b c \right ) b}\, a^{3} d^{3} \left (a d -b c \right )^{2}}-\frac {2}{c^{2} \left (a d -b c \right )^{2} \sqrt {d \,x^{3}+c}}\right )}{3}\)	\(238\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} x^{3}}-\frac {2 d}{3 c^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}}{a^{2}}+\frac {b^{2} d \left (-\frac {b \sqrt {d \,x^{3}+c}}{d \left (b \,x^{3}+a \right )}-\frac {3 b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {2}{\sqrt {d \,x^{3}+c}}\right )}{3 a^{2} \left (a d -b c \right )^{2}}-\frac {2 b \left (\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{a^{3}}+\frac {4 b^{2} \left (-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {1}{\sqrt {d \,x^{3}+c}}\right )}{3 a^{3} \left (a d -b c \right )}\)	\(269\)
elliptic	\(\text {Expression too large to display}\)	\(1763\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 563 vs. \(2 (209) = 418\).

Time = 0.67 (sec) , antiderivative size = 2337, normalized size of antiderivative = 9.70 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/6*(((4*b^4*c^4*d - 7*a*b^3*c^3*d^2)*x^9 + (4*b^4*c^5 - 3*a*b^3*c^4*d - 
 7*a^2*b^2*c^3*d^2)*x^6 + (4*a*b^3*c^5 - 7*a^2*b^2*c^4*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)) - ((4*b^4*c^3*d - 5*a*b^3*c^2*d^2 - 2*a^2*b^2* 
c*d^3 + 3*a^3*b*d^4)*x^9 + (4*b^4*c^4 - a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 + 
a^3*b*c*d^3 + 3*a^4*d^4)*x^6 + (4*a*b^3*c^4 - 5*a^2*b^2*c^3*d - 2*a^3*b*c^ 
2*d^2 + 3*a^4*c*d^3)*x^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 
 2*c)/x^3) + 2*(a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 + (2*a*b^3*c^3*d 
 - 2*a^2*b^2*c^2*d^2 + 3*a^3*b*c*d^3)*x^6 + (2*a*b^3*c^4 - a^2*b^2*c^3*d - 
 a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^3)*sqrt(d*x^3 + c))/((a^3*b^3*c^5*d - 2*a^ 
4*b^2*c^4*d^2 + a^5*b*c^3*d^3)*x^9 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b* 
c^4*d^2 + a^6*c^3*d^3)*x^6 + (a^4*b^2*c^6 - 2*a^5*b*c^5*d + a^6*c^4*d^2)*x 
^3), 1/6*(2*((4*b^4*c^4*d - 7*a*b^3*c^3*d^2)*x^9 + (4*b^4*c^5 - 3*a*b^3*c^ 
4*d - 7*a^2*b^2*c^3*d^2)*x^6 + (4*a*b^3*c^5 - 7*a^2*b^2*c^4*d)*x^3)*sqrt(- 
b/(b*c - a*d))*arctan(sqrt(d*x^3 + c)*sqrt(-b/(b*c - a*d))) + ((4*b^4*c^3* 
d - 5*a*b^3*c^2*d^2 - 2*a^2*b^2*c*d^3 + 3*a^3*b*d^4)*x^9 + (4*b^4*c^4 - a* 
b^3*c^3*d - 7*a^2*b^2*c^2*d^2 + a^3*b*c*d^3 + 3*a^4*d^4)*x^6 + (4*a*b^3*c^ 
4 - 5*a^2*b^2*c^3*d - 2*a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^3)*sqrt(c)*log((d*x 
^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 2*(a^2*b^2*c^4 - 2*a^3*b*c^3* 
d + a^4*c^2*d^2 + (2*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 3*a^3*b*c*d^3)*x...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/x^4/(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^4), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.52 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {{\left (4 \, b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (d x^{3} + c\right )}^{2} b^{3} c^{2} d - 2 \, {\left (d x^{3} + c\right )} b^{3} c^{3} d - 2 \, {\left (d x^{3} + c\right )}^{2} a b^{2} c d^{2} + 3 \, {\left (d x^{3} + c\right )} a b^{2} c^{2} d^{2} + 3 \, {\left (d x^{3} + c\right )}^{2} a^{2} b d^{3} - 7 \, {\left (d x^{3} + c\right )} a^{2} b c d^{3} + 2 \, a^{2} b c^{2} d^{3} + 3 \, {\left (d x^{3} + c\right )} a^{3} d^{4} - 2 \, a^{3} c d^{4}}{3 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {5}{2}} b - 2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} b c + \sqrt {d x^{3} + c} b c^{2} + {\left (d x^{3} + c\right )}^{\frac {3}{2}} a d - \sqrt {d x^{3} + c} a c d\right )}} - \frac {{\left (4 \, b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{3} \sqrt {-c} c^{2}} \] Input:

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

Output:

1/3*(4*b^4*c - 7*a*b^3*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/( 
(a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)*sqrt(-b^2*c + a*b*d)) - 1/3*(2*(d*x^ 
3 + c)^2*b^3*c^2*d - 2*(d*x^3 + c)*b^3*c^3*d - 2*(d*x^3 + c)^2*a*b^2*c*d^2 
 + 3*(d*x^3 + c)*a*b^2*c^2*d^2 + 3*(d*x^3 + c)^2*a^2*b*d^3 - 7*(d*x^3 + c) 
*a^2*b*c*d^3 + 2*a^2*b*c^2*d^3 + 3*(d*x^3 + c)*a^3*d^4 - 2*a^3*c*d^4)/((a^ 
2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2)*((d*x^3 + c)^(5/2)*b - 2*(d*x^3 + 
 c)^(3/2)*b*c + sqrt(d*x^3 + c)*b*c^2 + (d*x^3 + c)^(3/2)*a*d - sqrt(d*x^3 
 + c)*a*c*d)) - 1/3*(4*b*c + 3*a*d)*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^3* 
sqrt(-c)*c^2)
 

Mupad [B] (verification not implemented)

Time = 18.79 (sec) , antiderivative size = 18847, normalized size of antiderivative = 78.20 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*b*log(1/x^6))/(3*a^3*c^(3/2)) - (c + d*x^3)^(1/2)/(3*a^2*c^2*x^3) + (d* 
log(1/x^6))/(2*a^2*c^(5/2)) + (2*b*log(c^(3/2)*(c + d*x^3)^(1/2) - c^(1/2) 
*(c + d*x^3)^(3/2) + d^2*x^6 + 2*c*d*x^3 + 3*c^(1/2)*d*x^3*(c + d*x^3)^(1/ 
2)))/(3*a^3*c^(3/2)) + (d*log(c^(3/2)*(c + d*x^3)^(1/2) - c^(1/2)*(c + d*x 
^3)^(3/2) + d^2*x^6 + 2*c*d*x^3 + 3*c^(1/2)*d*x^3*(c + d*x^3)^(1/2)))/(2*a 
^2*c^(5/2)) - (b^7*c^9*x^4*(c + d*x^3)^(1/2))/(2*(2*a^9*c^6*d^5*x + 2*a^9* 
c^5*d^6*x^4 + a^5*b^4*c^9*d^2*x^4 + a^6*b^3*c^8*d^3*x^4 - 3*a^7*b^2*c^7*d^ 
4*x^4 + a^5*b^4*c^8*d^3*x^7 - 3*a^7*b^2*c^6*d^5*x^7 - 3*a^8*b*c^7*d^4*x + 
a^6*b^3*c^9*d^2*x - a^8*b*c^6*d^5*x^4 + 2*a^8*b*c^5*d^6*x^7)) - (5*a^9*d^7 
*x^4*(c + d*x^3)^(1/2))/(4*(a^6*b^5*c^9*x + a^5*b^6*c^9*x^4 - 3*a^7*b^4*c^ 
7*d^2*x^4 - a^8*b^3*c^6*d^3*x^4 + 2*a^9*b^2*c^5*d^4*x^4 - 3*a^7*b^4*c^6*d^ 
3*x^7 + 2*a^8*b^3*c^5*d^4*x^7 - 3*a^8*b^3*c^7*d^2*x + 2*a^9*b^2*c^6*d^3*x 
+ a^6*b^5*c^8*d*x^4 + a^5*b^6*c^8*d*x^7)) + (3*a^2*d^2*x*(c + d*x^3)^(1/2) 
)/(a^2*b^2*c^4*x^4 + 2*a^4*c^2*d^2*x^4 + a^3*b*c^4*x + 2*a^4*c^3*d*x + 3*a 
^3*b*c^3*d*x^4 + a^2*b^2*c^3*d*x^7 + 2*a^3*b*c^2*d^2*x^7) + (2*b^2*c^2*x*( 
c + d*x^3)^(1/2))/(a^2*b^2*c^4*x^4 + 2*a^4*c^2*d^2*x^4 + a^3*b*c^4*x + 2*a 
^4*c^3*d*x + 3*a^3*b*c^3*d*x^4 + a^2*b^2*c^3*d*x^7 + 2*a^3*b*c^2*d^2*x^7) 
- (b^(7/2)*c*log((a^6*b^(15/2)*c^10*36i)/(a*(a*d - b*c)^(1/2) + b*x^3*(a*d 
 - b*c)^(1/2)) - (a^7*b^(13/2)*c^9*d*198i)/(a*(a*d - b*c)^(1/2) + b*x^3*(a 
*d - b*c)^(1/2)) + (a^12*b^(3/2)*c^4*d^6*18i)/(a*(a*d - b*c)^(1/2) + b*...
 

Reduce [F]

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

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

Output:

( - 2*sqrt(c + d*x**3) - 9*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**1 
0 + b**2*c**2*x**7 + 2*b**2*c*d*x**10 + b**2*d**2*x**13),x)*a**2*c*d*x**3 
- 9*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)*a**2*d**2*x**6 - 12*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)*a*b*c**2*x**3 - 21*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)*a*b*c* 
d*x**6 - 9*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)*a*b*d**2*x**9 - 12*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)*b**2*c**2*x**6 - 12*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) 
*b**2*c*d*x**9 - 15*int((sqrt(c + d*x**3)*x**2)/(a**2*c**2 + 2*a**2*c*d...