\(\int \frac {1}{x \sqrt [3]{a+b x^3} (c+d x^3)} \, dx\) [734]

Optimal result

Integrand size = 24, antiderivative size = 244 \[ \int \frac {1}{x \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} c}+\frac {\sqrt [3]{d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c \sqrt [3]{b c-a d}}-\frac {\log (x)}{2 \sqrt [3]{a} c}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 c \sqrt [3]{b c-a d}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a} c}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c \sqrt [3]{b c-a d}} \] Output:

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {2 \sqrt {3} \sqrt [3]{d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt [3]{b c-a d}}+\frac {2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{a}}+\frac {2 \sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b c-a d}}-\frac {\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{a}}-\frac {\sqrt [3]{d} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b c-a d}}}{6 c} \] Input:

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

Output:

((2*Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]])/a^(1/3) + 
 (2*Sqrt[3]*d^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^ 
(1/3))/Sqrt[3]])/(b*c - a*d)^(1/3) + (2*Log[-a^(1/3) + (a + b*x^3)^(1/3)]) 
/a^(1/3) + (2*d^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/ 
(b*c - a*d)^(1/3) - Log[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^ 
(2/3)]/a^(1/3) - (d^(1/3)*Log[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3 
)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(b*c - a*d)^(1/3))/(6*c)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 97, 67, 16, 68, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 97

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 68

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{3} \left (\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^3\right )}{2 \sqrt [3]{a}}}{c}-\frac {d \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{c}\right )\)

Input:

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

Output:

(((Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]])/a^(1/3) - 
Log[x^3]/(2*a^(1/3)) + (3*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(2*a^(1/3)))/c 
 - (d*(-((Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1 
/3))/Sqrt[3]])/(d^(2/3)*(b*c - a*d)^(1/3))) + Log[c + d*x^3]/(2*d^(2/3)*(b 
*c - a*d)^(1/3)) - (3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/ 
(2*d^(2/3)*(b*c - a*d)^(1/3))))/c)/3
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 68

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[p]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

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]]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {\frac {\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right ) a^{\frac {1}{3}}}{2}-\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) a^{\frac {1}{3}}-\frac {\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}{2}-\ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right ) a^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (\arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 a^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} a^{\frac {1}{3}} c}\)	\(256\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 628, normalized size of antiderivative = 2.57 \[ \int \frac {1}{x \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c^{2} - a c d\right )}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c^{2} d - \sqrt {3} a c d^{2}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c^{2} d - a c d^{2}\right )}} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {1}{3}} c} - \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {1}{3}} c} + \frac {\log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {1}{3}} c} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 702, normalized size of antiderivative = 2.88 \[ \int \frac {1}{x \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

log(b^5*d^4*(a + b*x^3)^(1/3) - (d*(27*b^4*c^2*d^3*(a + b*x^3)^(1/3)*(2*a^ 
2*d^2 + b^2*c^2 - 2*a*b*c*d) - 243*a*b^4*c^4*d^3*(d/(27*b*c^4 - 27*a*c^3*d 
))^(2/3)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)))/(27*b*c^4 - 27*a*c^3*d))*(d/( 
27*b*c^4 - 27*a*c^3*d))^(1/3) + log((a + b*x^3)^(1/3) - a*c^2*(1/(a*c^3))^ 
(2/3))*(1/(27*a*c^3))^(1/3) + (log(b^5*d^4*(a + b*x^3)^(1/3) - (d*(3^(1/2) 
*1i - 1)^3*(27*b^4*c^2*d^3*(a + b*x^3)^(1/3)*(2*a^2*d^2 + b^2*c^2 - 2*a*b* 
c*d) - (243*a*b^4*c^4*d^3*(3^(1/2)*1i - 1)^2*(d/(27*b*c^4 - 27*a*c^3*d))^( 
2/3)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d))/4))/(8*(27*b*c^4 - 27*a*c^3*d)))*( 
3^(1/2)*1i - 1)*(d/(27*b*c^4 - 27*a*c^3*d))^(1/3))/2 - (log(b^5*d^4*(a + b 
*x^3)^(1/3) + (d*(3^(1/2)*1i + 1)^3*(27*b^4*c^2*d^3*(a + b*x^3)^(1/3)*(2*a 
^2*d^2 + b^2*c^2 - 2*a*b*c*d) - (243*a*b^4*c^4*d^3*(3^(1/2)*1i + 1)^2*(d/( 
27*b*c^4 - 27*a*c^3*d))^(2/3)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d))/4))/(8*(2 
7*b*c^4 - 27*a*c^3*d)))*(3^(1/2)*1i + 1)*(d/(27*b*c^4 - 27*a*c^3*d))^(1/3) 
)/2 - log((a + b*x^3)^(1/3)*2i + a*c^2*(1/(a*c^3))^(2/3)*1i + 3^(1/2)*a*c^ 
2*(1/(a*c^3))^(2/3))*((3^(1/2)*1i)/2 + 1/2)*(1/(27*a*c^3))^(1/3) + log((a 
+ b*x^3)^(1/3)*2i + a*c^2*(1/(a*c^3))^(2/3)*1i - 3^(1/2)*a*c^2*(1/(a*c^3)) 
^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(1/(27*a*c^3))^(1/3)
 

Reduce [F]

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

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

Output:

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