\(\int \frac {1}{\sqrt [3]{a+b x^3} (c+d x^3)^3} \, dx\) [166]

Optimal result

Integrand size = 21, antiderivative size = 307 \[ \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )^3} \, dx=-\frac {d x \left (a+b x^3\right )^{2/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}-\frac {d (9 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d)^2 \left (c+d x^3\right )}+\frac {\left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} (b c-a d)^{7/3}}+\frac {\left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{7/3}}-\frac {\left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{7/3}} \] Output:

-1/6*d*x*(b*x^3+a)^(2/3)/c/(-a*d+b*c)/(d*x^3+c)^2-1/18*d*(-5*a*d+9*b*c)*x* 
(b*x^3+a)^(2/3)/c^2/(-a*d+b*c)^2/(d*x^3+c)+1/27*(5*a^2*d^2-12*a*b*c*d+9*b^ 
2*c^2)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2) 
)*3^(1/2)/c^(8/3)/(-a*d+b*c)^(7/3)+1/54*(5*a^2*d^2-12*a*b*c*d+9*b^2*c^2)*l 
n(d*x^3+c)/c^(8/3)/(-a*d+b*c)^(7/3)-1/18*(5*a^2*d^2-12*a*b*c*d+9*b^2*c^2)* 
ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(8/3)/(-a*d+b*c)^(7/3)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.67 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )^3} \, dx=\frac {\frac {6 c^{2/3} d x \left (a+b x^3\right )^{2/3} \left (-3 b c \left (4 c+3 d x^3\right )+a d \left (8 c+5 d x^3\right )\right )}{(b c-a d)^2 \left (c+d x^3\right )^2}+\frac {2 \left (3-i \sqrt {3}\right ) \left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {i+\frac {\left (-i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d} x}}{\sqrt {3}}\right )}{(b c-a d)^{7/3}}+\frac {2 \left (1+i \sqrt {3}\right ) \left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{7/3}}-\frac {i \left (-i+\sqrt {3}\right ) \left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{7/3}}}{108 c^{8/3}} \] Input:

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

Output:

((6*c^(2/3)*d*x*(a + b*x^3)^(2/3)*(-3*b*c*(4*c + 3*d*x^3) + a*d*(8*c + 5*d 
*x^3)))/((b*c - a*d)^2*(c + d*x^3)^2) + (2*(3 - I*Sqrt[3])*(9*b^2*c^2 - 12 
*a*b*c*d + 5*a^2*d^2)*ArcTanh[(I + ((-I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/ 
3))/((b*c - a*d)^(1/3)*x))/Sqrt[3]])/(b*c - a*d)^(7/3) + (2*(1 + I*Sqrt[3] 
)*(9*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I* 
Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(7/3) - (I*(-I + Sqrt[3]) 
*(9*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Log[2*(b*c - a*d)^(2/3)*x^2 + (-1 - 
I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqrt[3]) 
*c^(2/3)*(a + b*x^3)^(2/3)])/(b*c - a*d)^(7/3))/(108*c^(8/3))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {931, 1024, 27, 901}

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.

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

\(\Big \downarrow \) 931

\(\displaystyle \frac {\int \frac {-3 b d x^3+6 b c-5 a d}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )^2}dx}{6 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {\int \frac {2 \left (9 b^2 c^2-12 a b d c+5 a^2 d^2\right )}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3} (9 b c-5 a d)}{3 c \left (c+d x^3\right ) (b c-a d)}}{6 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3} (9 b c-5 a d)}{3 c \left (c+d x^3\right ) (b c-a d)}}{6 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {\frac {2 \left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{3 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3} (9 b c-5 a d)}{3 c \left (c+d x^3\right ) (b c-a d)}}{6 c (b c-a d)}-\frac {d x \left (a+b x^3\right )^{2/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)}\)

Input:

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

Output:

-1/6*(d*x*(a + b*x^3)^(2/3))/(c*(b*c - a*d)*(c + d*x^3)^2) + (-1/3*(d*(9*b 
*c - 5*a*d)*x*(a + b*x^3)^(2/3))/(c*(b*c - a*d)*(c + d*x^3)) + (2*(9*b^2*c 
^2 - 12*a*b*c*d + 5*a^2*d^2)*(ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3) 
*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(1/3)) + Log[c 
+ d*x^3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d)^(1/3)*x)/c^(1/3) 
 - (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/(3*c*(b*c - a*d)))/( 
6*c*(b*c - a*d))
 

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 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] 
 + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0]
 

rule 931

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - 
 a*d))), x] + Simp[1/(a*n*(p + 1)*(b*c - a*d))   Int[(a + b*x^n)^(p + 1)*(c 
 + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, 
 x], x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, 
-1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, 
 c, d, n, p, q, x]
 

rule 1024

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f 
_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c 
+ d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*( 
p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b 
*c - a*d)*(p + 1) + d*(b*e - a*f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 1.95 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {\frac {5 \left (d \,x^{3}+c \right )^{2} \left (a^{2} d^{2}-\frac {12}{5} a b c d +\frac {9}{5} b^{2} c^{2}\right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )}{27}+\frac {4 c d \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-\frac {3 b \,c^{2}}{2}+d \left (-\frac {9 b \,x^{3}}{8}+a \right ) c +\frac {5 a \,d^{2} x^{3}}{8}\right ) x \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{9}+\frac {5 \left (d \,x^{3}+c \right )^{2} \left (a^{2} d^{2}-\frac {12}{5} a b c d +\frac {9}{5} b^{2} c^{2}\right ) \left (\arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )}{27}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (a d -b c \right )^{2} c^{3} \left (d \,x^{3}+c \right )^{2}}\)	\(292\)

Input:

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

Output:

5/27/((a*d-b*c)/c)^(1/3)*((d*x^3+c)^2*(a^2*d^2-12/5*a*b*c*d+9/5*b^2*c^2)*l 
n((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)+12/5*c*d*(b*x^3+a)^(2/3)*(-3/ 
2*b*c^2+d*(-9/8*b*x^3+a)*c+5/8*a*d^2*x^3)*x*((a*d-b*c)/c)^(1/3)+(d*x^3+c)^ 
2*(a^2*d^2-12/5*a*b*c*d+9/5*b^2*c^2)*(arctan(1/3*3^(1/2)*(-2/((a*d-b*c)/c) 
^(1/3)*(b*x^3+a)^(1/3)+x)/x)*3^(1/2)-1/2*ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d 
-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)))/(a*d-b*c)^2/c^3/( 
d*x^3+c)^2
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(1/((a + b*x**3)**(1/3)*c**3 + 3*(a + b*x**3)**(1/3)*c**2*d*x**3 + 3*(a 
 + b*x**3)**(1/3)*c*d**2*x**6 + (a + b*x**3)**(1/3)*d**3*x**9),x)