\(\int \frac {x^3}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\) [737]

Optimal result

Integrand size = 24, antiderivative size = 233 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b} d}-\frac {\sqrt [3]{c} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d \sqrt [3]{b c-a d}}-\frac {\sqrt [3]{c} \log \left (c+d x^3\right )}{6 d \sqrt [3]{b c-a d}}+\frac {\sqrt [3]{c} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d \sqrt [3]{b c-a d}}-\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b} d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.22 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.82 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\frac {4 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b}}+\frac {2 \sqrt {-6+6 i \sqrt {3}} \sqrt [3]{c} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b c-a d}}-\frac {4 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b}}-\frac {2 i \left (-i+\sqrt {3}\right ) \sqrt [3]{c} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b c-a d}}+\frac {2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b}}+\frac {\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b c-a d}}}{12 d} \] Input:

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

Output:

((4*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))]) 
/b^(1/3) + (2*Sqrt[-6 + (6*I)*Sqrt[3]]*c^(1/3)*ArcTan[(3*(b*c - a*d)^(1/3) 
*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/ 
3))])/(b*c - a*d)^(1/3) - (4*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/b^(1/3 
) - ((2*I)*(-I + Sqrt[3])*c^(1/3)*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)^(1/3) + (2*Log[b^(2/3)*x^2 + b 
^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/b^(1/3) + ((1 + I*Sqrt[3] 
)*c^(1/3)*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)^(1/3))/(12*d)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {983, 769, 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.

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

\(\Big \downarrow \) 983

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, 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 983

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^( 
n_)), x_Symbol] :> Simp[e^n/b   Int[(e*x)^(m - n)*(c + d*x^n)^q, x], x] - S 
imp[a*(e^n/b)   Int[(e*x)^(m - n)*((c + d*x^n)^q/(a + b*x^n)), x], x] /; Fr 
eeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, 
m, 2*n - 1] && IntBinomialQ[a, b, c, d, e, m, n, -1, q, x]
 

Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.27

method	result	size
pseudoelliptic	\(-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) b^{\frac {1}{3}}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{2}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) b^{\frac {1}{3}}-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) b^{\frac {1}{3}}}{2}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} b^{\frac {1}{3}} d}\)	\(297\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 761, normalized size of antiderivative = 3.27 \[ \int \frac {x^3}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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